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stringlengths 30
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stringlengths 0
61.9k
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|---|---|---|---|---|---|
http://rosettacode.org/wiki/Roots_of_unity
|
Roots of unity
|
The purpose of this task is to explore working with complex numbers.
Task
Given n, find the nth roots of unity.
|
#Nim
|
Nim
|
import complex, math, sequtils, strformat, strutils
proc roots(n: Positive): seq[Complex64] =
for k in 0..<n:
result.add rect(1.0, 2 * k.float * Pi / n.float)
proc toString(z: Complex64): string =
&"{z.re:.3f} + {z.im:.3f}i"
for nr in 2..10:
let result = roots(nr).map(toString).join(", ")
echo &"{nr:2}: {result}"
|
http://rosettacode.org/wiki/Roots_of_unity
|
Roots of unity
|
The purpose of this task is to explore working with complex numbers.
Task
Given n, find the nth roots of unity.
|
#OCaml
|
OCaml
|
open Complex
let pi = 4. *. atan 1.
let () =
for n = 1 to 10 do
Printf.printf "%2d " n;
for k = 1 to n do
let ret = polar 1. (2. *. pi *. float_of_int k /. float_of_int n) in
Printf.printf "(%f + %f i)" ret.re ret.im
done;
print_newline ()
done
|
http://rosettacode.org/wiki/Roots_of_a_quadratic_function
|
Roots of a quadratic function
|
This task has been clarified. Its programming examples are in need of review to ensure that they still fit the requirements of the task.
Write a program to find the roots of a quadratic equation, i.e., solve the equation
a
x
2
+
b
x
+
c
=
0
{\displaystyle ax^{2}+bx+c=0}
.
Your program must correctly handle non-real roots, but it need not check that
a
≠
0
{\displaystyle a\neq 0}
.
The problem of solving a quadratic equation is a good example of how dangerous it can be to ignore the peculiarities of floating-point arithmetic.
The obvious way to implement the quadratic formula suffers catastrophic loss of accuracy when one of the roots to be found is much closer to 0 than the other.
In their classic textbook on numeric methods Computer Methods for Mathematical Computations, George Forsythe, Michael Malcolm, and Cleve Moler suggest trying the naive algorithm with
a
=
1
{\displaystyle a=1}
,
b
=
−
10
5
{\displaystyle b=-10^{5}}
, and
c
=
1
{\displaystyle c=1}
.
(For double-precision floats, set
b
=
−
10
9
{\displaystyle b=-10^{9}}
.)
Consider the following implementation in Ada:
with Ada.Text_IO; use Ada.Text_IO;
with Ada.Numerics.Elementary_Functions; use Ada.Numerics.Elementary_Functions;
procedure Quadratic_Equation is
type Roots is array (1..2) of Float;
function Solve (A, B, C : Float) return Roots is
SD : constant Float := sqrt (B**2 - 4.0 * A * C);
AA : constant Float := 2.0 * A;
begin
return ((- B + SD) / AA, (- B - SD) / AA);
end Solve;
R : constant Roots := Solve (1.0, -10.0E5, 1.0);
begin
Put_Line ("X1 =" & Float'Image (R (1)) & " X2 =" & Float'Image (R (2)));
end Quadratic_Equation;
Output:
X1 = 1.00000E+06 X2 = 0.00000E+00
As we can see, the second root has lost all significant figures. The right answer is that X2 is about
10
−
6
{\displaystyle 10^{-6}}
. The naive method is numerically unstable.
Suggested by Middlebrook (D-OA), a better numerical method: to define two parameters
q
=
a
c
/
b
{\displaystyle q={\sqrt {ac}}/b}
and
f
=
1
/
2
+
1
−
4
q
2
/
2
{\displaystyle f=1/2+{\sqrt {1-4q^{2}}}/2}
and the two roots of the quardratic are:
−
b
a
f
{\displaystyle {\frac {-b}{a}}f}
and
−
c
b
f
{\displaystyle {\frac {-c}{bf}}}
Task: do it better. This means that given
a
=
1
{\displaystyle a=1}
,
b
=
−
10
9
{\displaystyle b=-10^{9}}
, and
c
=
1
{\displaystyle c=1}
, both of the roots your program returns should be greater than
10
−
11
{\displaystyle 10^{-11}}
. Or, if your language can't do floating-point arithmetic any more precisely than single precision, your program should be able to handle
b
=
−
10
6
{\displaystyle b=-10^{6}}
. Either way, show what your program gives as the roots of the quadratic in question. See page 9 of
"What Every Scientist Should Know About Floating-Point Arithmetic" for a possible algorithm.
|
#Nim
|
Nim
|
import math, complex, strformat
const Epsilon = 1e-15
type
SolKind = enum solDouble, solFloat, solComplex
Roots = object
case kind: SolKind
of solDouble:
fvalue: float
of solFloat:
fvalues: (float, float)
of solComplex:
cvalues: (Complex64, Complex64)
func quadRoots(a, b, c: float): Roots =
if a == 0:
raise newException(ValueError, "first coefficient cannot be null.")
let den = a * 2
let Δ = b * b - a * c * 4
if abs(Δ) < Epsilon:
result = Roots(kind: solDouble, fvalue: -b / den)
elif Δ < 0:
let r = -b / den
let i = sqrt(-Δ) / den
result = Roots(kind: solComplex, cvalues: (complex64(r, i), complex64(r, -i)))
else:
let r = (if b < 0: -b + sqrt(Δ) else: -b - sqrt(Δ)) / den
result = Roots(kind: solFloat, fvalues: (r, c / (a * r)))
func `$`(r: Roots): string =
case r.kind
of solDouble:
result = $r.fvalue
of solFloat:
result = &"{r.fvalues[0]}, {r.fvalues[1]}"
of solComplex:
result = &"{r.cvalues[0].re} + {r.cvalues[0].im}i, {r.cvalues[1].re} + {r.cvalues[1].im}i"
when isMainModule:
const Equations = [(1.0, -2.0, 1.0),
(10.0, 1.0, 1.0),
(1.0, -10.0, 1.0),
(1.0, -1000.0, 1.0),
(1.0, -1e9, 1.0)]
for (a, b, c) in Equations:
echo &"Equation: {a=}, {b=}, {c=}"
let roots = quadRoots(a, b, c)
let plural = if roots.kind == solDouble: "" else: "s"
echo &" root{plural}: {roots}"
|
http://rosettacode.org/wiki/Roots_of_a_quadratic_function
|
Roots of a quadratic function
|
This task has been clarified. Its programming examples are in need of review to ensure that they still fit the requirements of the task.
Write a program to find the roots of a quadratic equation, i.e., solve the equation
a
x
2
+
b
x
+
c
=
0
{\displaystyle ax^{2}+bx+c=0}
.
Your program must correctly handle non-real roots, but it need not check that
a
≠
0
{\displaystyle a\neq 0}
.
The problem of solving a quadratic equation is a good example of how dangerous it can be to ignore the peculiarities of floating-point arithmetic.
The obvious way to implement the quadratic formula suffers catastrophic loss of accuracy when one of the roots to be found is much closer to 0 than the other.
In their classic textbook on numeric methods Computer Methods for Mathematical Computations, George Forsythe, Michael Malcolm, and Cleve Moler suggest trying the naive algorithm with
a
=
1
{\displaystyle a=1}
,
b
=
−
10
5
{\displaystyle b=-10^{5}}
, and
c
=
1
{\displaystyle c=1}
.
(For double-precision floats, set
b
=
−
10
9
{\displaystyle b=-10^{9}}
.)
Consider the following implementation in Ada:
with Ada.Text_IO; use Ada.Text_IO;
with Ada.Numerics.Elementary_Functions; use Ada.Numerics.Elementary_Functions;
procedure Quadratic_Equation is
type Roots is array (1..2) of Float;
function Solve (A, B, C : Float) return Roots is
SD : constant Float := sqrt (B**2 - 4.0 * A * C);
AA : constant Float := 2.0 * A;
begin
return ((- B + SD) / AA, (- B - SD) / AA);
end Solve;
R : constant Roots := Solve (1.0, -10.0E5, 1.0);
begin
Put_Line ("X1 =" & Float'Image (R (1)) & " X2 =" & Float'Image (R (2)));
end Quadratic_Equation;
Output:
X1 = 1.00000E+06 X2 = 0.00000E+00
As we can see, the second root has lost all significant figures. The right answer is that X2 is about
10
−
6
{\displaystyle 10^{-6}}
. The naive method is numerically unstable.
Suggested by Middlebrook (D-OA), a better numerical method: to define two parameters
q
=
a
c
/
b
{\displaystyle q={\sqrt {ac}}/b}
and
f
=
1
/
2
+
1
−
4
q
2
/
2
{\displaystyle f=1/2+{\sqrt {1-4q^{2}}}/2}
and the two roots of the quardratic are:
−
b
a
f
{\displaystyle {\frac {-b}{a}}f}
and
−
c
b
f
{\displaystyle {\frac {-c}{bf}}}
Task: do it better. This means that given
a
=
1
{\displaystyle a=1}
,
b
=
−
10
9
{\displaystyle b=-10^{9}}
, and
c
=
1
{\displaystyle c=1}
, both of the roots your program returns should be greater than
10
−
11
{\displaystyle 10^{-11}}
. Or, if your language can't do floating-point arithmetic any more precisely than single precision, your program should be able to handle
b
=
−
10
6
{\displaystyle b=-10^{6}}
. Either way, show what your program gives as the roots of the quadratic in question. See page 9 of
"What Every Scientist Should Know About Floating-Point Arithmetic" for a possible algorithm.
|
#OCaml
|
OCaml
|
type quadroots =
| RealRoots of float * float
| ComplexRoots of Complex.t * Complex.t ;;
let quadsolve a b c =
let d = (b *. b) -. (4.0 *. a *. c) in
if d < 0.0
then
let r = -. b /. (2.0 *. a)
and i = sqrt(-. d) /. (2.0 *. a) in
ComplexRoots ({ Complex.re = r; Complex.im = i },
{ Complex.re = r; Complex.im = (-.i) })
else
let r =
if b < 0.0
then ((sqrt d) -. b) /. (2.0 *. a)
else ((sqrt d) +. b) /. (-2.0 *. a)
in
RealRoots (r, c /. (r *. a))
;;
|
http://rosettacode.org/wiki/Rot-13
|
Rot-13
|
Task
Implement a rot-13 function (or procedure, class, subroutine, or other "callable" object as appropriate to your programming environment).
Optionally wrap this function in a utility program (like tr, which acts like a common UNIX utility, performing a line-by-line rot-13 encoding of every line of input contained in each file listed on its command line, or (if no filenames are passed thereon) acting as a filter on its "standard input."
(A number of UNIX scripting languages and utilities, such as awk and sed either default to processing files in this way or have command line switches or modules to easily implement these wrapper semantics, e.g., Perl and Python).
The rot-13 encoding is commonly known from the early days of Usenet "Netnews" as a way of obfuscating text to prevent casual reading of spoiler or potentially offensive material.
Many news reader and mail user agent programs have built-in rot-13 encoder/decoders or have the ability to feed a message through any external utility script for performing this (or other) actions.
The definition of the rot-13 function is to simply replace every letter of the ASCII alphabet with the letter which is "rotated" 13 characters "around" the 26 letter alphabet from its normal cardinal position (wrapping around from z to a as necessary).
Thus the letters abc become nop and so on.
Technically rot-13 is a "mono-alphabetic substitution cipher" with a trivial "key".
A proper implementation should work on upper and lower case letters, preserve case, and pass all non-alphabetic characters
in the input stream through without alteration.
Related tasks
Caesar cipher
Substitution Cipher
Vigenère Cipher/Cryptanalysis
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
|
#D.C3.A9j.C3.A0_Vu
|
Déjà Vu
|
rot-13:
)
for ch in chars swap:
ord ch
if <= 65 dup:
if >= 90 dup:
+ 13 - swap 65
+ 65 % swap 26
if <= 97 dup:
if >= 122 dup:
+ 13 - swap 97
+ 97 % swap 26
chr
concat(
!print rot-13 "Snape kills Frodo with Rosebud."
|
http://rosettacode.org/wiki/Runge-Kutta_method
|
Runge-Kutta method
|
Given the example Differential equation:
y
′
(
t
)
=
t
×
y
(
t
)
{\displaystyle y'(t)=t\times {\sqrt {y(t)}}}
With initial condition:
t
0
=
0
{\displaystyle t_{0}=0}
and
y
0
=
y
(
t
0
)
=
y
(
0
)
=
1
{\displaystyle y_{0}=y(t_{0})=y(0)=1}
This equation has an exact solution:
y
(
t
)
=
1
16
(
t
2
+
4
)
2
{\displaystyle y(t)={\tfrac {1}{16}}(t^{2}+4)^{2}}
Task
Demonstrate the commonly used explicit fourth-order Runge–Kutta method to solve the above differential equation.
Solve the given differential equation over the range
t
=
0
…
10
{\displaystyle t=0\ldots 10}
with a step value of
δ
t
=
0.1
{\displaystyle \delta t=0.1}
(101 total points, the first being given)
Print the calculated values of
y
{\displaystyle y}
at whole numbered
t
{\displaystyle t}
's (
0.0
,
1.0
,
…
10.0
{\displaystyle 0.0,1.0,\ldots 10.0}
) along with error as compared to the exact solution.
Method summary
Starting with a given
y
n
{\displaystyle y_{n}}
and
t
n
{\displaystyle t_{n}}
calculate:
δ
y
1
=
δ
t
×
y
′
(
t
n
,
y
n
)
{\displaystyle \delta y_{1}=\delta t\times y'(t_{n},y_{n})\quad }
δ
y
2
=
δ
t
×
y
′
(
t
n
+
1
2
δ
t
,
y
n
+
1
2
δ
y
1
)
{\displaystyle \delta y_{2}=\delta t\times y'(t_{n}+{\tfrac {1}{2}}\delta t,y_{n}+{\tfrac {1}{2}}\delta y_{1})}
δ
y
3
=
δ
t
×
y
′
(
t
n
+
1
2
δ
t
,
y
n
+
1
2
δ
y
2
)
{\displaystyle \delta y_{3}=\delta t\times y'(t_{n}+{\tfrac {1}{2}}\delta t,y_{n}+{\tfrac {1}{2}}\delta y_{2})}
δ
y
4
=
δ
t
×
y
′
(
t
n
+
δ
t
,
y
n
+
δ
y
3
)
{\displaystyle \delta y_{4}=\delta t\times y'(t_{n}+\delta t,y_{n}+\delta y_{3})\quad }
then:
y
n
+
1
=
y
n
+
1
6
(
δ
y
1
+
2
δ
y
2
+
2
δ
y
3
+
δ
y
4
)
{\displaystyle y_{n+1}=y_{n}+{\tfrac {1}{6}}(\delta y_{1}+2\delta y_{2}+2\delta y_{3}+\delta y_{4})}
t
n
+
1
=
t
n
+
δ
t
{\displaystyle t_{n+1}=t_{n}+\delta t\quad }
|
#Raku
|
Raku
|
sub runge-kutta(&yp) {
return -> \t, \y, \δt {
my $a = δt * yp( t, y );
my $b = δt * yp( t + δt/2, y + $a/2 );
my $c = δt * yp( t + δt/2, y + $b/2 );
my $d = δt * yp( t + δt, y + $c );
($a + 2*($b + $c) + $d) / 6;
}
}
constant δt = .1;
my &δy = runge-kutta { $^t * sqrt($^y) };
loop (
my ($t, $y) = (0, 1);
$t <= 10;
($t, $y) »+=« (δt, δy($t, $y, δt))
) {
printf "y(%2d) = %12f ± %e\n", $t, $y, abs($y - ($t**2 + 4)**2 / 16)
if $t %% 1;
}
|
http://rosettacode.org/wiki/S-expressions
|
S-expressions
|
S-Expressions are one convenient way to parse and store data.
Task
Write a simple reader and writer for S-Expressions that handles quoted and unquoted strings, integers and floats.
The reader should read a single but nested S-Expression from a string and store it in a suitable datastructure (list, array, etc).
Newlines and other whitespace may be ignored unless contained within a quoted string.
“()” inside quoted strings are not interpreted, but treated as part of the string.
Handling escaped quotes inside a string is optional; thus “(foo"bar)” maybe treated as a string “foo"bar”, or as an error.
For this, the reader need not recognize “\” for escaping, but should, in addition, recognize numbers if the language has appropriate datatypes.
Languages that support it may treat unquoted strings as symbols.
Note that with the exception of “()"” (“\” if escaping is supported) and whitespace there are no special characters. Anything else is allowed without quotes.
The reader should be able to read the following input
((data "quoted data" 123 4.5)
(data (!@# (4.5) "(more" "data)")))
and turn it into a native datastructure. (see the Pike, Python and Ruby implementations for examples of native data structures.)
The writer should be able to take the produced list and turn it into a new S-Expression.
Strings that don't contain whitespace or parentheses () don't need to be quoted in the resulting S-Expression, but as a simplification, any string may be quoted.
Extra Credit
Let the writer produce pretty printed output with indenting and line-breaks.
|
#Sidef
|
Sidef
|
func sexpr(txt) {
txt.trim!
if (txt.match(/^\((.*)\)$/s)) {|m|
txt = m[0]
}
else {
die "Invalid: <<#{txt}>>"
}
var w
var ret = []
while (!txt.is_empty) {
given (txt.first) {
when('(') {
(w, txt) = txt.extract_bracketed('()');
w = sexpr(w)
}
when ('"') {
(w, txt) = txt.extract_delimited('"')
w.sub!(/^"(.*)"/, {|s1| s1 })
}
else {
txt.sub!(/^(\S+)/, {|s1| w = s1; '' })
}
}
ret << w
txt.trim_beg!
}
return ret
}
func sexpr2txt(String e) {
e ~~ /[\s"\(\)]/ ? do { e.gsub!('"', '\\"'); %Q("#{e}") } : e
}
func sexpr2txt(expr) {
'(' + expr.map {|e| sexpr2txt(e) }.join(' ') + ')'
}
var s = sexpr(%q{
((data "quoted data" 123 4.5)
(data (!@# (4.5) "(more" "data)")))
})
say s # dump structure
say sexpr2txt(s) # convert back
|
http://rosettacode.org/wiki/RPG_attributes_generator
|
RPG attributes generator
|
RPG = Role Playing Game.
You're running a tabletop RPG, and your players are creating characters.
Each character has six core attributes: strength, dexterity, constitution, intelligence, wisdom, and charisma.
One way of generating values for these attributes is to roll four, 6-sided dice (d6) and sum the three highest rolls, discarding the lowest roll.
Some players like to assign values to their attributes in the order they're rolled.
To ensure generated characters don't put players at a disadvantage, the following requirements must be satisfied:
The total of all character attributes must be at least 75.
At least two of the attributes must be at least 15.
However, this can require a lot of manual dice rolling. A programatic solution would be much faster.
Task
Write a program that:
Generates 4 random, whole values between 1 and 6.
Saves the sum of the 3 largest values.
Generates a total of 6 values this way.
Displays the total, and all 6 values once finished.
The order in which each value was generated must be preserved.
The total of all 6 values must be at least 75.
At least 2 of the values must be 15 or more.
|
#Ring
|
Ring
|
# Project : RPG Attributes Generator
load "stdlib.ring"
attributestotal = 0
count = 0
while attributestotal < 75 or count < 2
attributes = []
for attribute = 0 to 6
rolls = []
largest3 = []
for roll = 0 to 4
result = random(5)+1
add(rolls,result)
next
sortedrolls = sort(rolls)
sortedrolls = reverse(sortedrolls)
for n = 1 to 3
add(largest3,sortedrolls[n])
next
rollstotal = sum(largest3)
if rollstotal >= 15
count = count + 1
ok
add(attributes,rollstotal)
next
attributestotal = sum(attributes)
end
showline()
func sum(aList)
num = 0
for n = 1 to len(aList)
num = num + aList[n]
next
return num
func showline()
line = "(" + attributestotal + ", ["
for n = 1 to len(attributes)
line = line + attributes[n] + ", "
next
line = left(line,len(line)-2)
line = line + "])"
see line + nl
|
http://rosettacode.org/wiki/RPG_attributes_generator
|
RPG attributes generator
|
RPG = Role Playing Game.
You're running a tabletop RPG, and your players are creating characters.
Each character has six core attributes: strength, dexterity, constitution, intelligence, wisdom, and charisma.
One way of generating values for these attributes is to roll four, 6-sided dice (d6) and sum the three highest rolls, discarding the lowest roll.
Some players like to assign values to their attributes in the order they're rolled.
To ensure generated characters don't put players at a disadvantage, the following requirements must be satisfied:
The total of all character attributes must be at least 75.
At least two of the attributes must be at least 15.
However, this can require a lot of manual dice rolling. A programatic solution would be much faster.
Task
Write a program that:
Generates 4 random, whole values between 1 and 6.
Saves the sum of the 3 largest values.
Generates a total of 6 values this way.
Displays the total, and all 6 values once finished.
The order in which each value was generated must be preserved.
The total of all 6 values must be at least 75.
At least 2 of the values must be 15 or more.
|
#Ruby
|
Ruby
|
res = []
until res.sum >= 75 && res.count{|n| n >= 15} >= 2 do
res = Array.new(6) do
a = Array.new(4){rand(1..6)}
a.sum - a.min
end
end
p res
puts "sum: #{res.sum}"
|
http://rosettacode.org/wiki/Sieve_of_Eratosthenes
|
Sieve of Eratosthenes
|
This task has been clarified. Its programming examples are in need of review to ensure that they still fit the requirements of the task.
The Sieve of Eratosthenes is a simple algorithm that finds the prime numbers up to a given integer.
Task
Implement the Sieve of Eratosthenes algorithm, with the only allowed optimization that the outer loop can stop at the square root of the limit, and the inner loop may start at the square of the prime just found.
That means especially that you shouldn't optimize by using pre-computed wheels, i.e. don't assume you need only to cross out odd numbers (wheel based on 2), numbers equal to 1 or 5 modulo 6 (wheel based on 2 and 3), or similar wheels based on low primes.
If there's an easy way to add such a wheel based optimization, implement it as an alternative version.
Note
It is important that the sieve algorithm be the actual algorithm used to find prime numbers for the task.
Related tasks
Emirp primes
count in factors
prime decomposition
factors of an integer
extensible prime generator
primality by trial division
factors of a Mersenne number
trial factoring of a Mersenne number
partition an integer X into N primes
sequence of primes by Trial Division
|
#UNIX_Shell
|
UNIX Shell
|
function primes {
typeset -a a
typeset i j
a[1]=""
for (( i = 2; i <= $1; i++ )); do
a[$i]=$i
done
for (( i = 2; i * i <= $1; i++ )); do
if [[ ! -z ${a[$i]} ]]; then
for (( j = i * i; j <= $1; j += i )); do
a[$j]=""
done
fi
done
printf '%s' "${a[2]}"
printf ' %s' ${a[*]:3}
printf '\n'
}
primes 1000
|
http://rosettacode.org/wiki/Rosetta_Code/Count_examples
|
Rosetta Code/Count examples
|
task
Essentially, count the number of occurrences of =={{header| on each task page.
Output:
100 doors: 20 examples.
99 Bottles of Beer: 29 examples.
Abstract type: 10 examples.
Total: X examples.
For a full output, updated periodically, see Rosetta Code/Count examples/Full list.
You'll need to use the Media Wiki API, which you can find out about locally, here, or in Media Wiki's API documentation at, API:Query
|
#Sidef
|
Sidef
|
var lwp = require('LWP::UserAgent').new(agent => 'Mozilla/5.0');
var site = 'http://rosettacode.org';
var list_url = '/mw/api.php?action=query&list=categorymembers&'+
'cmtitle=Category:Programming_Tasks&cmlimit=500&format=xml';
var content = lwp.get(site + list_url).decoded_content;
while (var m = content.match(/cm.*?title="(.*?)"/g)) {
(var slug = m[0]).gsub!(' ', '_');
var count = lwp.get("#{site}/wiki/#{slug}").decoded_content.count(/toclevel-1/g);
say "#{m[0]}: #{count} examples";
}
|
http://rosettacode.org/wiki/Rosetta_Code/Count_examples
|
Rosetta Code/Count examples
|
task
Essentially, count the number of occurrences of =={{header| on each task page.
Output:
100 doors: 20 examples.
99 Bottles of Beer: 29 examples.
Abstract type: 10 examples.
Total: X examples.
For a full output, updated periodically, see Rosetta Code/Count examples/Full list.
You'll need to use the Media Wiki API, which you can find out about locally, here, or in Media Wiki's API documentation at, API:Query
|
#Tcl
|
Tcl
|
package require Tcl 8.5
package require http
package require json
fconfigure stdout -buffering none
proc get_tasks {category} {
set start [clock milliseconds]
puts -nonewline "getting $category members..."
set base_url http://www.rosettacode.org/w/api.php
set query {action query list categorymembers cmtitle Category:%s format json cmlimit 500}
set this_query [dict create {*}[split [format $query $category]]]
set tasks [list]
while {1} {
set url [join [list $base_url [http::formatQuery {*}$this_query]] ?]
set response [http::geturl $url]
if {[set s [http::status $response]] ne "ok" || [http::ncode $response] != 200} {
error "Oops: url=$url\nstatus=$s\nhttp code=[http::code $response]"
}
set data [json::json2dict [http::data $response]]
http::cleanup $response
# add tasks to list
foreach task [dict get $data query categorymembers] {
lappend tasks [dict get [dict create {*}$task] title]
}
if {[catch {dict get $data query-continue categorymembers cmcontinue} continue_task] != 0} {
# no more continuations, we're done
break
}
dict set this_query cmcontinue $continue_task
}
puts " found [llength $tasks] tasks in [expr {[clock milliseconds] - $start}] milliseconds"
return $tasks
}
# This proc can be replaced by a single regexp command:
# set count [regexp -all "***=$needle" $haystack]
# However this proc is more efficient -- we're dealing with plain strings only.
proc count_substrings {needle haystack} {
set count 0
set idx 0
while {[set idx [string first $needle $haystack $idx]] != -1} {
incr count
incr idx
}
return $count
}
set total 0
foreach task [get_tasks Programming_Tasks] {
set url [format "http://www.rosettacode.org/w/index.php?title=%s&action=raw" [string map {{ } _} $task]]
set response [http::geturl $url]
if {[set s [http::status $response]] ne "ok" || [http::ncode $response] != 200} {
error "Oops: url=$url\nstatus=$s\nhttp code=[http::code $response]"
}
set count [count_substrings "\{\{header|" [http::data $response]]
puts [format "%3d examples in %s" $count $task]
http::cleanup $response
incr total $count
}
puts "\nTotal: $total examples"
|
http://rosettacode.org/wiki/Search_a_list
|
Search a list
|
Task[edit]
Find the index of a string (needle) in an indexable, ordered collection of strings (haystack).
Raise an exception if the needle is missing.
If there is more than one occurrence then return the smallest index to the needle.
Extra credit
Return the largest index to a needle that has multiple occurrences in the haystack.
See also
Search a list of records
|
#Objeck
|
Objeck
|
use Collection;
class Test {
function : Main(args : String[]) ~ Nil {
haystack := ["Zig","Zag","Wally","Ronald","Bush","Krusty","Charlie","Bush","Bozo"];
values := CompareVector->New();
each(i : haystack) {
values->AddBack(haystack[i]->As(Compare));
};
needles := ["Washington", "Bush"];
each(i : needles) {
values->Has(needles[i]->As(Compare))->PrintLine();
};
}
}
|
http://rosettacode.org/wiki/Rosetta_Code/Rank_languages_by_popularity
|
Rosetta Code/Rank languages by popularity
|
Rosetta Code/Rank languages by popularity
You are encouraged to solve this task according to the task description, using any language you may know.
Task
Sort the most popular computer programming languages based in number of members in Rosetta Code categories.
Sample output on 01 juin 2022 at 14:13 +02
Rank: 1 (1,540 entries) Phix
Rank: 2 (1,531 entries) Wren
Rank: 3 (1,507 entries) Julia
Rank: 4 (1,494 entries) Go
Rank: 5 (1,488 entries) Raku
Rank: 6 (1,448 entries) Perl
Rank: 7 (1,402 entries) Nim
Rank: 8 (1,382 entries) Python
Rank: 9 (1,204 entries) C
Rank: 10 (1,152 entries) REXX
...
Notes
Each language typically demonstrates one or two methods of accessing the data:
with web scraping (via http://www.rosettacode.org/mw/index.php?title=Special:Categories&limit=5000)
with the API method (examples below for Awk, Perl, Ruby, Tcl, etc).
The scraping and API solutions can be separate subsections, see the Tcl example.
Filtering wrong results is optional. You can check against Special:MostLinkedCategories (if using web scraping)
If you use the API, and do elect to filter, you may check your results against this complete, accurate, sortable, wikitable listing of all 869 programming languages, updated periodically, typically weekly.
A complete ranked listing of all 813 languages (from the REXX example) is included here ──► output from the REXX program.
|
#Python
|
Python
|
import requests
import re
response = requests.get("http://rosettacode.org/wiki/Category:Programming_Languages").text
languages = re.findall('title="Category:(.*?)">',response)[:-3] # strip last 3
response = requests.get("http://rosettacode.org/mw/index.php?title=Special:Categories&limit=5000").text
response = re.sub('(\d+),(\d+)',r'\1'+r'\2',response) # strip ',' from popular languages above 999 members
members = re.findall('<li><a[^>]+>([^<]+)</a>[^(]*[(](\\d+) member[s]*[)]</li>',response) # find language and members
for cnt, (language, members) in enumerate(sorted(members, key=lambda x: -int(x[1]))[:15]): # show only top 15 languages
if language in languages:
print("{:4d} {:4d} - {}".format(cnt+1, int(members), language))
|
http://rosettacode.org/wiki/Roman_numerals/Encode
|
Roman numerals/Encode
|
Task
Create a function taking a positive integer as its parameter and returning a string containing the Roman numeral representation of that integer. Modern Roman numerals are written by expressing each digit separately, starting with the left most digit and skipping any digit with a value of zero.
In Roman numerals:
1990 is rendered: 1000=M, 900=CM, 90=XC; resulting in MCMXC
2008 is written as 2000=MM, 8=VIII; or MMVIII
1666 uses each Roman symbol in descending order: MDCLXVI
|
#Batch_File
|
Batch File
|
@echo off
setlocal enabledelayedexpansion
set cnt=0&for %%A in (1000,900,500,400,100,90,50,40,10,9,5,4,1) do (set arab!cnt!=%%A&set /a cnt+=1)
set cnt=0&for %%R in (M,CM,D,CD,C,XC,L,XL,X,IX,V,IV,I) do (set rom!cnt!=%%R&set /a cnt+=1)
::Testing
call :toRoman 2009
echo 2009 = !result!
call :toRoman 1666
echo 1666 = !result!
call :toRoman 3888
echo 3888 = !result!
pause>nul
exit/b 0
::The "function"...
:toRoman
set value=%1
set result=
for /l %%i in (0,1,12) do (
set a=%%i
call :add_val
)
goto :EOF
:add_val
if !value! lss !arab%a%! goto :EOF
set result=!result!!rom%a%!
set /a value-=!arab%a%!
goto add_val
|
http://rosettacode.org/wiki/Roman_numerals/Decode
|
Roman numerals/Decode
|
Task
Create a function that takes a Roman numeral as its argument and returns its value as a numeric decimal integer.
You don't need to validate the form of the Roman numeral.
Modern Roman numerals are written by expressing each decimal digit of the number to be encoded separately,
starting with the leftmost decimal digit and skipping any 0s (zeroes).
1990 is rendered as MCMXC (1000 = M, 900 = CM, 90 = XC) and
2008 is rendered as MMVIII (2000 = MM, 8 = VIII).
The Roman numeral for 1666, MDCLXVI, uses each letter in descending order.
|
#BQN
|
BQN
|
⟨ToArabic⇐A⟩ ← {
c ← "IVXLCDM" # Characters
v ← ⥊ (10⋆↕4) ×⌜ 1‿5 # Their values
A ⇐ +´∘(⊢ׯ1⋆<⟜«) v ⊏˜ c ⊐ ⊢
}
|
http://rosettacode.org/wiki/Roots_of_a_function
|
Roots of a function
|
Task
Create a program that finds and outputs the roots of a given function, range and (if applicable) step width.
The program should identify whether the root is exact or approximate.
For this task, use: ƒ(x) = x3 - 3x2 + 2x
|
#FreeBASIC
|
FreeBASIC
|
#Include "crt.bi"
const iterations=20000000
sub bisect( f1 as function(as double) as double,min as double,max as double,byref O as double,a() as double)
dim as double last,st=(max-min)/iterations,v
for n as double=min to max step st
v=f1(n)
if sgn(v)<>sgn(last) then
redim preserve a(1 to ubound(a)+1)
a(ubound(a))=n
O=n+st:exit sub
end if
last=v
next
end sub
function roots(f1 as function(as double) as double,min as double,max as double, a() as double) as long
redim a(0)
dim as double last,O,st=(max-min)/iterations,v
for n as double=min to max step st
v=f1(n)
if sgn(v)<>sgn(last) and n>min then bisect(f1,n-st,n,O,a()):n=O
last=v
next
return ubound(a)
end function
Function CRound(Byval x As Double,Byval precision As Integer=30) As String
If precision>30 Then precision=30
Dim As zstring * 40 z:Var s="%." &str(Abs(precision)) &"f"
sprintf(z,s,x)
If Val(z) Then Return Rtrim(Rtrim(z,"0"),".")Else Return "0"
End Function
function defn(x as double) as double
return x^3-3*x^2+2*x
end function
redim as double r()
print
if roots(@defn,-20,20,r()) then
print "in range -20 to 20"
print "All roots approximate"
print "number","root to 6 dec places","function value at root"
for n as long=1 to ubound(r)
print n,CRound(r(n),6),,defn(r(n))
next n
end if
sleep
|
http://rosettacode.org/wiki/Rock-paper-scissors
|
Rock-paper-scissors
|
Task
Implement the classic children's game Rock-paper-scissors, as well as a simple predictive AI (artificial intelligence) player.
Rock Paper Scissors is a two player game.
Each player chooses one of rock, paper or scissors, without knowing the other player's choice.
The winner is decided by a set of rules:
Rock beats scissors
Scissors beat paper
Paper beats rock
If both players choose the same thing, there is no winner for that round.
For this task, the computer will be one of the players.
The operator will select Rock, Paper or Scissors and the computer will keep a record of the choice frequency, and use that information to make a weighted random choice in an attempt to defeat its opponent.
Extra credit
Support additional choices additional weapons.
|
#Erlang
|
Erlang
|
-module(rps).
-compile(export_all).
play() ->
loop([1,1,1]).
loop([R,P,S]=Odds) ->
case io:fread("What is your move? (R,P,S,Q) ","~c") of
{ok,["Q"]} -> io:fwrite("Good bye!~n");
{ok,[[Human]]} when Human == $R; Human == $P; Human == $S ->
io:fwrite("Your move is ~s.~n",
[play_to_string(Human)]),
Computer = select_play(Odds),
io:fwrite("My move is ~s~n",
[play_to_string(Computer)]),
case {beats(Human,Computer),beats(Computer,Human)} of
{true,_} -> io:fwrite("You win!~n");
{_,true} -> io:fwrite("I win!~n");
_ -> io:fwrite("Draw~n")
end,
case Human of
$R -> loop([R+1,P,S]);
$P -> loop([R,P+1,S]);
$S -> loop([R,P,S+1])
end;
_ ->
io:fwrite("Invalid play~n"),
loop(Odds)
end.
beats($R,$S) -> true;
beats($P,$R) -> true;
beats($S,$P) -> true;
beats(_,_) -> false.
play_to_string($R) -> "Rock";
play_to_string($P) -> "Paper";
play_to_string($S) -> "Scissors".
select_play([R,P,S]) ->
N = random:uniform(R+P+S),
if
N =< R -> $P;
N =< R+P -> $S;
true -> $R
end.
|
http://rosettacode.org/wiki/Run-length_encoding
|
Run-length encoding
|
Run-length encoding
You are encouraged to solve this task according to the task description, using any language you may know.
Task
Given a string containing uppercase characters (A-Z), compress repeated 'runs' of the same character by storing the length of that run, and provide a function to reverse the compression.
The output can be anything, as long as you can recreate the input with it.
Example
Input: WWWWWWWWWWWWBWWWWWWWWWWWWBBBWWWWWWWWWWWWWWWWWWWWWWWWBWWWWWWWWWWWWWW
Output: 12W1B12W3B24W1B14W
Note: the encoding step in the above example is the same as a step of the Look-and-say sequence.
|
#Fan
|
Fan
|
**
** Generates a run-length encoding for a string
**
class RLE
{
Run[] encode(Str s)
{
runs := Run[,]
s.size.times |i|
{
ch := s[i]
if (runs.size==0 || runs.last.char != ch)
runs.add(Run(ch))
runs.last.inc
}
return runs
}
Str decode(Run[] runs)
{
buf := StrBuf()
runs.each |run|
{
run.count.times { buf.add(run.char.toChar) }
}
return buf.toStr
}
Void main()
{
echo(decode(encode(
"WWWWWWWWWWWWBWWWWWWWWWWWWBBBWWWWWWWWWWWWWWWWWWWWWWWWBWWWWWWWWWWWWWW"
)))
}
}
internal class Run
{
Int char
Int count := 0
new make(Int ch) { char = ch }
Void inc() { ++count }
override Str toStr() { return "${count}${char.toChar}" }
}
|
http://rosettacode.org/wiki/Roots_of_unity
|
Roots of unity
|
The purpose of this task is to explore working with complex numbers.
Task
Given n, find the nth roots of unity.
|
#Octave
|
Octave
|
for j = 2 : 10
printf("*** %d\n", j);
for n = 1 : j
disp(exp(2i*pi*n/j));
endfor
disp("");
endfor
|
http://rosettacode.org/wiki/Roots_of_unity
|
Roots of unity
|
The purpose of this task is to explore working with complex numbers.
Task
Given n, find the nth roots of unity.
|
#OoRexx
|
OoRexx
|
/*REXX program computes the K roots of unity (which include complex roots).*/
parse Version v
Say v
parse arg n frac . /*get optional arguments from the C.L. */
if n=='' then n=1 /*Not specified? Then use the default.*/
if frac='' then frac=5 /* " " " " " " */
start=abs(n) /*assume only one K is wanted. */
if n<0 then start=1 /*Negative? Then use a range of K's. */
/*display unity roots for a range, or */
do k=start to abs(n) /* just for one K. */
say right(k 'roots of unity',40,"-") /*display a pretty separator with title*/
do angle=0 by 360/k for k /*compute the angle for each root. */
rp=adjust(rxCalcCos(angle,,'D')) /*compute real part via COS function.*/
if left(rp,1)\=='-' then rp=" "rp /*not negative? Then pad with a blank.*/
ip=adjust(rxCalcSin(angle,,'D')) /*compute imaginary part via SIN funct.*/
if left(ip,1)\=='-' then ip="+"ip /*Not negative? Then pad with + char.*/
if ip=0 then say rp /*Only real part? Ignore imaginary part*/
else say left(rp,frac+4)ip'i' /*show the real & imaginary part*/
end /*angle*/
end /*k*/
exit /*stick a fork in it, we're all done. */
/*----------------------------------------------------------------------------*/
adjust: parse arg x; near0='1e-' || (digits()-digits()%10) /*compute small #*/
if abs(x)<near0 then x=0 /*if near zero, then assume zero.*/
return format(x,,frac)/1 /*fraction digits past dec point.*/
::requires rxMath library
|
http://rosettacode.org/wiki/Roots_of_a_quadratic_function
|
Roots of a quadratic function
|
This task has been clarified. Its programming examples are in need of review to ensure that they still fit the requirements of the task.
Write a program to find the roots of a quadratic equation, i.e., solve the equation
a
x
2
+
b
x
+
c
=
0
{\displaystyle ax^{2}+bx+c=0}
.
Your program must correctly handle non-real roots, but it need not check that
a
≠
0
{\displaystyle a\neq 0}
.
The problem of solving a quadratic equation is a good example of how dangerous it can be to ignore the peculiarities of floating-point arithmetic.
The obvious way to implement the quadratic formula suffers catastrophic loss of accuracy when one of the roots to be found is much closer to 0 than the other.
In their classic textbook on numeric methods Computer Methods for Mathematical Computations, George Forsythe, Michael Malcolm, and Cleve Moler suggest trying the naive algorithm with
a
=
1
{\displaystyle a=1}
,
b
=
−
10
5
{\displaystyle b=-10^{5}}
, and
c
=
1
{\displaystyle c=1}
.
(For double-precision floats, set
b
=
−
10
9
{\displaystyle b=-10^{9}}
.)
Consider the following implementation in Ada:
with Ada.Text_IO; use Ada.Text_IO;
with Ada.Numerics.Elementary_Functions; use Ada.Numerics.Elementary_Functions;
procedure Quadratic_Equation is
type Roots is array (1..2) of Float;
function Solve (A, B, C : Float) return Roots is
SD : constant Float := sqrt (B**2 - 4.0 * A * C);
AA : constant Float := 2.0 * A;
begin
return ((- B + SD) / AA, (- B - SD) / AA);
end Solve;
R : constant Roots := Solve (1.0, -10.0E5, 1.0);
begin
Put_Line ("X1 =" & Float'Image (R (1)) & " X2 =" & Float'Image (R (2)));
end Quadratic_Equation;
Output:
X1 = 1.00000E+06 X2 = 0.00000E+00
As we can see, the second root has lost all significant figures. The right answer is that X2 is about
10
−
6
{\displaystyle 10^{-6}}
. The naive method is numerically unstable.
Suggested by Middlebrook (D-OA), a better numerical method: to define two parameters
q
=
a
c
/
b
{\displaystyle q={\sqrt {ac}}/b}
and
f
=
1
/
2
+
1
−
4
q
2
/
2
{\displaystyle f=1/2+{\sqrt {1-4q^{2}}}/2}
and the two roots of the quardratic are:
−
b
a
f
{\displaystyle {\frac {-b}{a}}f}
and
−
c
b
f
{\displaystyle {\frac {-c}{bf}}}
Task: do it better. This means that given
a
=
1
{\displaystyle a=1}
,
b
=
−
10
9
{\displaystyle b=-10^{9}}
, and
c
=
1
{\displaystyle c=1}
, both of the roots your program returns should be greater than
10
−
11
{\displaystyle 10^{-11}}
. Or, if your language can't do floating-point arithmetic any more precisely than single precision, your program should be able to handle
b
=
−
10
6
{\displaystyle b=-10^{6}}
. Either way, show what your program gives as the roots of the quadratic in question. See page 9 of
"What Every Scientist Should Know About Floating-Point Arithmetic" for a possible algorithm.
|
#Octave
|
Octave
|
roots(a,b,c)=polrootsreal(Pol([a,b,c]))
|
http://rosettacode.org/wiki/Roots_of_a_quadratic_function
|
Roots of a quadratic function
|
This task has been clarified. Its programming examples are in need of review to ensure that they still fit the requirements of the task.
Write a program to find the roots of a quadratic equation, i.e., solve the equation
a
x
2
+
b
x
+
c
=
0
{\displaystyle ax^{2}+bx+c=0}
.
Your program must correctly handle non-real roots, but it need not check that
a
≠
0
{\displaystyle a\neq 0}
.
The problem of solving a quadratic equation is a good example of how dangerous it can be to ignore the peculiarities of floating-point arithmetic.
The obvious way to implement the quadratic formula suffers catastrophic loss of accuracy when one of the roots to be found is much closer to 0 than the other.
In their classic textbook on numeric methods Computer Methods for Mathematical Computations, George Forsythe, Michael Malcolm, and Cleve Moler suggest trying the naive algorithm with
a
=
1
{\displaystyle a=1}
,
b
=
−
10
5
{\displaystyle b=-10^{5}}
, and
c
=
1
{\displaystyle c=1}
.
(For double-precision floats, set
b
=
−
10
9
{\displaystyle b=-10^{9}}
.)
Consider the following implementation in Ada:
with Ada.Text_IO; use Ada.Text_IO;
with Ada.Numerics.Elementary_Functions; use Ada.Numerics.Elementary_Functions;
procedure Quadratic_Equation is
type Roots is array (1..2) of Float;
function Solve (A, B, C : Float) return Roots is
SD : constant Float := sqrt (B**2 - 4.0 * A * C);
AA : constant Float := 2.0 * A;
begin
return ((- B + SD) / AA, (- B - SD) / AA);
end Solve;
R : constant Roots := Solve (1.0, -10.0E5, 1.0);
begin
Put_Line ("X1 =" & Float'Image (R (1)) & " X2 =" & Float'Image (R (2)));
end Quadratic_Equation;
Output:
X1 = 1.00000E+06 X2 = 0.00000E+00
As we can see, the second root has lost all significant figures. The right answer is that X2 is about
10
−
6
{\displaystyle 10^{-6}}
. The naive method is numerically unstable.
Suggested by Middlebrook (D-OA), a better numerical method: to define two parameters
q
=
a
c
/
b
{\displaystyle q={\sqrt {ac}}/b}
and
f
=
1
/
2
+
1
−
4
q
2
/
2
{\displaystyle f=1/2+{\sqrt {1-4q^{2}}}/2}
and the two roots of the quardratic are:
−
b
a
f
{\displaystyle {\frac {-b}{a}}f}
and
−
c
b
f
{\displaystyle {\frac {-c}{bf}}}
Task: do it better. This means that given
a
=
1
{\displaystyle a=1}
,
b
=
−
10
9
{\displaystyle b=-10^{9}}
, and
c
=
1
{\displaystyle c=1}
, both of the roots your program returns should be greater than
10
−
11
{\displaystyle 10^{-11}}
. Or, if your language can't do floating-point arithmetic any more precisely than single precision, your program should be able to handle
b
=
−
10
6
{\displaystyle b=-10^{6}}
. Either way, show what your program gives as the roots of the quadratic in question. See page 9 of
"What Every Scientist Should Know About Floating-Point Arithmetic" for a possible algorithm.
|
#PARI.2FGP
|
PARI/GP
|
roots(a,b,c)=polrootsreal(Pol([a,b,c]))
|
http://rosettacode.org/wiki/Rot-13
|
Rot-13
|
Task
Implement a rot-13 function (or procedure, class, subroutine, or other "callable" object as appropriate to your programming environment).
Optionally wrap this function in a utility program (like tr, which acts like a common UNIX utility, performing a line-by-line rot-13 encoding of every line of input contained in each file listed on its command line, or (if no filenames are passed thereon) acting as a filter on its "standard input."
(A number of UNIX scripting languages and utilities, such as awk and sed either default to processing files in this way or have command line switches or modules to easily implement these wrapper semantics, e.g., Perl and Python).
The rot-13 encoding is commonly known from the early days of Usenet "Netnews" as a way of obfuscating text to prevent casual reading of spoiler or potentially offensive material.
Many news reader and mail user agent programs have built-in rot-13 encoder/decoders or have the ability to feed a message through any external utility script for performing this (or other) actions.
The definition of the rot-13 function is to simply replace every letter of the ASCII alphabet with the letter which is "rotated" 13 characters "around" the 26 letter alphabet from its normal cardinal position (wrapping around from z to a as necessary).
Thus the letters abc become nop and so on.
Technically rot-13 is a "mono-alphabetic substitution cipher" with a trivial "key".
A proper implementation should work on upper and lower case letters, preserve case, and pass all non-alphabetic characters
in the input stream through without alteration.
Related tasks
Caesar cipher
Substitution Cipher
Vigenère Cipher/Cryptanalysis
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
|
#E
|
E
|
pragma.enable("accumulator")
var rot13Map := [].asMap()
for a in ['a', 'A'] {
for i in 0..!26 {
rot13Map with= (a + i, E.toString(a + (i + 13) % 26))
}
}
def rot13(s :String) {
return accum "" for c in s { _ + rot13Map.fetch(c, fn{ c }) }
}
|
http://rosettacode.org/wiki/Runge-Kutta_method
|
Runge-Kutta method
|
Given the example Differential equation:
y
′
(
t
)
=
t
×
y
(
t
)
{\displaystyle y'(t)=t\times {\sqrt {y(t)}}}
With initial condition:
t
0
=
0
{\displaystyle t_{0}=0}
and
y
0
=
y
(
t
0
)
=
y
(
0
)
=
1
{\displaystyle y_{0}=y(t_{0})=y(0)=1}
This equation has an exact solution:
y
(
t
)
=
1
16
(
t
2
+
4
)
2
{\displaystyle y(t)={\tfrac {1}{16}}(t^{2}+4)^{2}}
Task
Demonstrate the commonly used explicit fourth-order Runge–Kutta method to solve the above differential equation.
Solve the given differential equation over the range
t
=
0
…
10
{\displaystyle t=0\ldots 10}
with a step value of
δ
t
=
0.1
{\displaystyle \delta t=0.1}
(101 total points, the first being given)
Print the calculated values of
y
{\displaystyle y}
at whole numbered
t
{\displaystyle t}
's (
0.0
,
1.0
,
…
10.0
{\displaystyle 0.0,1.0,\ldots 10.0}
) along with error as compared to the exact solution.
Method summary
Starting with a given
y
n
{\displaystyle y_{n}}
and
t
n
{\displaystyle t_{n}}
calculate:
δ
y
1
=
δ
t
×
y
′
(
t
n
,
y
n
)
{\displaystyle \delta y_{1}=\delta t\times y'(t_{n},y_{n})\quad }
δ
y
2
=
δ
t
×
y
′
(
t
n
+
1
2
δ
t
,
y
n
+
1
2
δ
y
1
)
{\displaystyle \delta y_{2}=\delta t\times y'(t_{n}+{\tfrac {1}{2}}\delta t,y_{n}+{\tfrac {1}{2}}\delta y_{1})}
δ
y
3
=
δ
t
×
y
′
(
t
n
+
1
2
δ
t
,
y
n
+
1
2
δ
y
2
)
{\displaystyle \delta y_{3}=\delta t\times y'(t_{n}+{\tfrac {1}{2}}\delta t,y_{n}+{\tfrac {1}{2}}\delta y_{2})}
δ
y
4
=
δ
t
×
y
′
(
t
n
+
δ
t
,
y
n
+
δ
y
3
)
{\displaystyle \delta y_{4}=\delta t\times y'(t_{n}+\delta t,y_{n}+\delta y_{3})\quad }
then:
y
n
+
1
=
y
n
+
1
6
(
δ
y
1
+
2
δ
y
2
+
2
δ
y
3
+
δ
y
4
)
{\displaystyle y_{n+1}=y_{n}+{\tfrac {1}{6}}(\delta y_{1}+2\delta y_{2}+2\delta y_{3}+\delta y_{4})}
t
n
+
1
=
t
n
+
δ
t
{\displaystyle t_{n+1}=t_{n}+\delta t\quad }
|
#REXX
|
REXX
|
The Runge─Kutta method is used to solve the following differential equation:
y'(t) = t2 √ y(t)
The exact solution: y(t) = (t2+4)2 ÷ 16
|
http://rosettacode.org/wiki/Runge-Kutta_method
|
Runge-Kutta method
|
Given the example Differential equation:
y
′
(
t
)
=
t
×
y
(
t
)
{\displaystyle y'(t)=t\times {\sqrt {y(t)}}}
With initial condition:
t
0
=
0
{\displaystyle t_{0}=0}
and
y
0
=
y
(
t
0
)
=
y
(
0
)
=
1
{\displaystyle y_{0}=y(t_{0})=y(0)=1}
This equation has an exact solution:
y
(
t
)
=
1
16
(
t
2
+
4
)
2
{\displaystyle y(t)={\tfrac {1}{16}}(t^{2}+4)^{2}}
Task
Demonstrate the commonly used explicit fourth-order Runge–Kutta method to solve the above differential equation.
Solve the given differential equation over the range
t
=
0
…
10
{\displaystyle t=0\ldots 10}
with a step value of
δ
t
=
0.1
{\displaystyle \delta t=0.1}
(101 total points, the first being given)
Print the calculated values of
y
{\displaystyle y}
at whole numbered
t
{\displaystyle t}
's (
0.0
,
1.0
,
…
10.0
{\displaystyle 0.0,1.0,\ldots 10.0}
) along with error as compared to the exact solution.
Method summary
Starting with a given
y
n
{\displaystyle y_{n}}
and
t
n
{\displaystyle t_{n}}
calculate:
δ
y
1
=
δ
t
×
y
′
(
t
n
,
y
n
)
{\displaystyle \delta y_{1}=\delta t\times y'(t_{n},y_{n})\quad }
δ
y
2
=
δ
t
×
y
′
(
t
n
+
1
2
δ
t
,
y
n
+
1
2
δ
y
1
)
{\displaystyle \delta y_{2}=\delta t\times y'(t_{n}+{\tfrac {1}{2}}\delta t,y_{n}+{\tfrac {1}{2}}\delta y_{1})}
δ
y
3
=
δ
t
×
y
′
(
t
n
+
1
2
δ
t
,
y
n
+
1
2
δ
y
2
)
{\displaystyle \delta y_{3}=\delta t\times y'(t_{n}+{\tfrac {1}{2}}\delta t,y_{n}+{\tfrac {1}{2}}\delta y_{2})}
δ
y
4
=
δ
t
×
y
′
(
t
n
+
δ
t
,
y
n
+
δ
y
3
)
{\displaystyle \delta y_{4}=\delta t\times y'(t_{n}+\delta t,y_{n}+\delta y_{3})\quad }
then:
y
n
+
1
=
y
n
+
1
6
(
δ
y
1
+
2
δ
y
2
+
2
δ
y
3
+
δ
y
4
)
{\displaystyle y_{n+1}=y_{n}+{\tfrac {1}{6}}(\delta y_{1}+2\delta y_{2}+2\delta y_{3}+\delta y_{4})}
t
n
+
1
=
t
n
+
δ
t
{\displaystyle t_{n+1}=t_{n}+\delta t\quad }
|
#Ring
|
Ring
|
decimals(8)
y = 1.0
for i = 0 to 100
t = i / 10
if t = floor(t)
actual = (pow((pow(t,2) + 4),2)) / 16
see "y(" + t + ") = " + y + " error = " + (actual - y) + nl ok
k1 = t * sqrt(y)
k2 = (t + 0.05) * sqrt(y + 0.05 * k1)
k3 = (t + 0.05) * sqrt(y + 0.05 * k2)
k4 = (t + 0.10) * sqrt(y + 0.10 * k3)
y += 0.1 * (k1 + 2 * (k2 + k3) + k4) / 6
next
|
http://rosettacode.org/wiki/S-expressions
|
S-expressions
|
S-Expressions are one convenient way to parse and store data.
Task
Write a simple reader and writer for S-Expressions that handles quoted and unquoted strings, integers and floats.
The reader should read a single but nested S-Expression from a string and store it in a suitable datastructure (list, array, etc).
Newlines and other whitespace may be ignored unless contained within a quoted string.
“()” inside quoted strings are not interpreted, but treated as part of the string.
Handling escaped quotes inside a string is optional; thus “(foo"bar)” maybe treated as a string “foo"bar”, or as an error.
For this, the reader need not recognize “\” for escaping, but should, in addition, recognize numbers if the language has appropriate datatypes.
Languages that support it may treat unquoted strings as symbols.
Note that with the exception of “()"” (“\” if escaping is supported) and whitespace there are no special characters. Anything else is allowed without quotes.
The reader should be able to read the following input
((data "quoted data" 123 4.5)
(data (!@# (4.5) "(more" "data)")))
and turn it into a native datastructure. (see the Pike, Python and Ruby implementations for examples of native data structures.)
The writer should be able to take the produced list and turn it into a new S-Expression.
Strings that don't contain whitespace or parentheses () don't need to be quoted in the resulting S-Expression, but as a simplification, any string may be quoted.
Extra Credit
Let the writer produce pretty printed output with indenting and line-breaks.
|
#Tcl
|
Tcl
|
package require Tcl 8.5
proc fromSexp {str} {
set tokenizer {[()]|"(?:[^\\""]|\\.)*"|(?:[^()""\s\\]|\\.)+|[""]}
set stack {}
set accum {}
foreach token [regexp -inline -all $tokenizer $str] {
if {$token eq "("} {
lappend stack $accum
set accum {}
} elseif {$token eq ")"} {
if {![llength $stack]} {error "unbalanced"}
set accum [list {*}[lindex $stack end] [list list {*}$accum]]
set stack [lrange $stack 0 end-1]
} elseif {$token eq "\""} {
error "bad quote"
} elseif {[string match {"*"} $token]} {
set token [string range $token 1 end-1]
lappend accum [list string [regsub -all {\\(.)} $token {\1}]]
} else {
if {[string is integer -strict $token]} {
set type int
} elseif {[string is double -strict $token]} {
set type real
} else {
set type atom
}
lappend accum [list $type [regsub -all {\\(.)} $token {\1}]]
}
}
if {[llength $stack]} {error "unbalanced"}
return [lindex $accum 0]
}
proc toSexp {tokList} {
set content [lassign $tokList type]
if {$type eq "list"} {
set s "("
set sep ""
foreach bit $content {
append s $sep [toSexp $bit]
set sep " "
}
return [append s ")"]
} elseif {$type eq "string"} {
return "\"[regsub -all {[\\""]} [lindex $content 0] {\\\0}]\""
} else {
return [lindex $content 0]
}
}
|
http://rosettacode.org/wiki/RPG_attributes_generator
|
RPG attributes generator
|
RPG = Role Playing Game.
You're running a tabletop RPG, and your players are creating characters.
Each character has six core attributes: strength, dexterity, constitution, intelligence, wisdom, and charisma.
One way of generating values for these attributes is to roll four, 6-sided dice (d6) and sum the three highest rolls, discarding the lowest roll.
Some players like to assign values to their attributes in the order they're rolled.
To ensure generated characters don't put players at a disadvantage, the following requirements must be satisfied:
The total of all character attributes must be at least 75.
At least two of the attributes must be at least 15.
However, this can require a lot of manual dice rolling. A programatic solution would be much faster.
Task
Write a program that:
Generates 4 random, whole values between 1 and 6.
Saves the sum of the 3 largest values.
Generates a total of 6 values this way.
Displays the total, and all 6 values once finished.
The order in which each value was generated must be preserved.
The total of all 6 values must be at least 75.
At least 2 of the values must be 15 or more.
|
#Run_BASIC
|
Run BASIC
|
dim statnames$(6)
data "STR", "CON", "DEX", "INT", "WIS", "CHA"
for i = 1 to 6
read statnames$(i)
next i
dim stat(6)
acceptable = false
while 1
sum = 0 : n15 = 0
for i = 1 to 6
stat(i) = rollstat()
sum = sum + stat(i)
if stat(i) >= 15 then n15 = n15 + 1
next i
if sum >= 75 and n15 >= 2 then exit while
wend
for i = 1 to 6
print statnames$(i); ": "; stat(i)
next i
print "-------"
print "TOT: "; sum
end
function d6()
d6 = 1 + int(rnd(1) * 6)
end function
function rollstat()
a = d6() : b = d6() : c = d6() : d = d6()
rollstat = a + b + c + d - min(min(a, b), min(c, d))
end function
|
http://rosettacode.org/wiki/RPG_attributes_generator
|
RPG attributes generator
|
RPG = Role Playing Game.
You're running a tabletop RPG, and your players are creating characters.
Each character has six core attributes: strength, dexterity, constitution, intelligence, wisdom, and charisma.
One way of generating values for these attributes is to roll four, 6-sided dice (d6) and sum the three highest rolls, discarding the lowest roll.
Some players like to assign values to their attributes in the order they're rolled.
To ensure generated characters don't put players at a disadvantage, the following requirements must be satisfied:
The total of all character attributes must be at least 75.
At least two of the attributes must be at least 15.
However, this can require a lot of manual dice rolling. A programatic solution would be much faster.
Task
Write a program that:
Generates 4 random, whole values between 1 and 6.
Saves the sum of the 3 largest values.
Generates a total of 6 values this way.
Displays the total, and all 6 values once finished.
The order in which each value was generated must be preserved.
The total of all 6 values must be at least 75.
At least 2 of the values must be 15 or more.
|
#Rust
|
Rust
|
use rand::distributions::Uniform;
use rand::prelude::{thread_rng, ThreadRng};
use rand::Rng;
fn main() {
for _ in 0..=10 {
attributes_engine();
}
}
#[derive(Copy, Clone, Debug)]
pub struct Dice {
amount: i32,
range: Uniform<i32>,
rng: ThreadRng,
}
impl Dice {
// Modeled after d20 polyhederal dice use and notation.
// roll_pool() - returns Vec<i32> with length of vector determined by dice amount.
// attribute_out() - returns i32, by sorting a dice pool of 4d6, dropping the lowest integer, and summing all elements.
pub fn new(amount: i32, size: i32) -> Self {
Self {
amount,
range: Uniform::new(1, size + 1),
rng: thread_rng(),
}
}
fn roll_pool(mut self) -> Vec<i32> {
(0..self.amount)
.map(|_| self.rng.sample(self.range))
.collect()
}
fn attribute_out(&self) -> i32 {
// Sort dice pool lowest to high and drain all results to exclude the lowest before summing.
let mut attribute_array: Vec<i32> = self.roll_pool();
attribute_array.sort();
attribute_array.drain(1..=3).sum()
}
}
fn attributes_finalizer() -> (Vec<i32>, i32, bool) {
let die: Dice = Dice::new(4, 6);
let mut attributes: Vec<i32> = Vec::new();
for _ in 0..6 {
attributes.push(die.attribute_out())
}
let attributes_total: i32 = attributes.iter().sum();
let numerical_condition: bool = attributes
.iter()
.filter(|attribute| **attribute >= 15)
.count()
>= 2;
(attributes, attributes_total, numerical_condition)
}
fn attributes_engine() {
loop {
let (attributes, attributes_total, numerical_condition) = attributes_finalizer();
if (attributes_total >= 75) && (numerical_condition) {
println!(
"{:?} | sum: {:?}",
attributes, attributes_total
);
break;
} else {
continue;
}
}
}
|
http://rosettacode.org/wiki/Sieve_of_Eratosthenes
|
Sieve of Eratosthenes
|
This task has been clarified. Its programming examples are in need of review to ensure that they still fit the requirements of the task.
The Sieve of Eratosthenes is a simple algorithm that finds the prime numbers up to a given integer.
Task
Implement the Sieve of Eratosthenes algorithm, with the only allowed optimization that the outer loop can stop at the square root of the limit, and the inner loop may start at the square of the prime just found.
That means especially that you shouldn't optimize by using pre-computed wheels, i.e. don't assume you need only to cross out odd numbers (wheel based on 2), numbers equal to 1 or 5 modulo 6 (wheel based on 2 and 3), or similar wheels based on low primes.
If there's an easy way to add such a wheel based optimization, implement it as an alternative version.
Note
It is important that the sieve algorithm be the actual algorithm used to find prime numbers for the task.
Related tasks
Emirp primes
count in factors
prime decomposition
factors of an integer
extensible prime generator
primality by trial division
factors of a Mersenne number
trial factoring of a Mersenne number
partition an integer X into N primes
sequence of primes by Trial Division
|
#Ursala
|
Ursala
|
#import nat
sieve = ~<{0,1}&& iota; @NttPX ~&r->lx ^/~&rhPlC remainder@rlX~|@r
|
http://rosettacode.org/wiki/Rosetta_Code/Count_examples
|
Rosetta Code/Count examples
|
task
Essentially, count the number of occurrences of =={{header| on each task page.
Output:
100 doors: 20 examples.
99 Bottles of Beer: 29 examples.
Abstract type: 10 examples.
Total: X examples.
For a full output, updated periodically, see Rosetta Code/Count examples/Full list.
You'll need to use the Media Wiki API, which you can find out about locally, here, or in Media Wiki's API documentation at, API:Query
|
#TUSCRIPT
|
TUSCRIPT
|
$$ MODE TUSCRIPT
url="http://www.rosettacode.org/w/api.php?action=query&list=categorymembers&cmtitle=Category:Programming_Tasks&cmlimit=500&format=xml"
data=REQUEST (url)
BUILD S_TABLE beg=*
DATA :title=":
BUILD S_TABLE end=*
DATA :":
titles=EXTRACT (data,beg|,end,1,0,"~~")
titles=SPLIT (titles,":~~:")
sz_titles=SIZE (titles)
BUILD R_TABLE header=":==\{\{header|:"
all=*
ERROR/STOP CREATE ("tasks",seq-e,-std-)
COMPILE
LOOP title=titles
ask=*
ask =SET_VALUE(ask,"title",title)
ask =SET_VALUE(ask,"action","raw")
ask =ENCODE (ask,cgi)
http ="http://www.rosettacode.org/mw/index.php"
url =CONCAT (http,"?",ask)
data =REQUEST (url)
header =FILTER_INDEX (data,header,-)
sz_header=SIZE(header)
line =CONCAT (title,"=",sz_header," members")
FILE "tasks" = line
all =APPEND(all,sz_header)
ENDLOOP
ENDCOMPILE
all =JOIN(all),sum=SUM(all),time=time()
line=CONCAT (time,": ", sz_titles, " Programing Tasks: ", sum, " solutions")
FILE "tasks" = line
|
http://rosettacode.org/wiki/Rosetta_Code/Count_examples
|
Rosetta Code/Count examples
|
task
Essentially, count the number of occurrences of =={{header| on each task page.
Output:
100 doors: 20 examples.
99 Bottles of Beer: 29 examples.
Abstract type: 10 examples.
Total: X examples.
For a full output, updated periodically, see Rosetta Code/Count examples/Full list.
You'll need to use the Media Wiki API, which you can find out about locally, here, or in Media Wiki's API documentation at, API:Query
|
#UNIX_Shell
|
UNIX Shell
|
#!/usr/bin/env bash
SITE=https://www.rosettacode.org
API=$SITE/mw/api.php
PAGES=$SITE/mw/index.php
query="$API?action=query"
query+=$(printf '&%s' \
list=categorymembers \
cmtitle=Category:Programming_Tasks \
cmlimit=500)
total=0
while read title; do
t=${title// /_}
tasks=$(curl -s "$PAGES?title=$t&action=raw" \
| grep -c '{{header|')
printf '%s: %d examples.\n' "$title" "$tasks"
let total+=tasks
done < <(curl -s "$query&format=json" \
| jq -r '.query.categorymembers[].title')
printf '\nTotal: %d examples.\n' "$total"
|
http://rosettacode.org/wiki/Search_a_list
|
Search a list
|
Task[edit]
Find the index of a string (needle) in an indexable, ordered collection of strings (haystack).
Raise an exception if the needle is missing.
If there is more than one occurrence then return the smallest index to the needle.
Extra credit
Return the largest index to a needle that has multiple occurrences in the haystack.
See also
Search a list of records
|
#Objective-C
|
Objective-C
|
NSArray *haystack = @[@"Zig",@"Zag",@"Wally",@"Ronald",@"Bush",@"Krusty",@"Charlie",@"Bush",@"Bozo"];
for (id needle in @[@"Washington",@"Bush"]) {
int index = [haystack indexOfObject:needle];
if (index == NSNotFound)
NSLog(@"%@ is not in haystack", needle);
else
NSLog(@"%i %@", index, needle);
}
|
http://rosettacode.org/wiki/Rosetta_Code/Rank_languages_by_popularity
|
Rosetta Code/Rank languages by popularity
|
Rosetta Code/Rank languages by popularity
You are encouraged to solve this task according to the task description, using any language you may know.
Task
Sort the most popular computer programming languages based in number of members in Rosetta Code categories.
Sample output on 01 juin 2022 at 14:13 +02
Rank: 1 (1,540 entries) Phix
Rank: 2 (1,531 entries) Wren
Rank: 3 (1,507 entries) Julia
Rank: 4 (1,494 entries) Go
Rank: 5 (1,488 entries) Raku
Rank: 6 (1,448 entries) Perl
Rank: 7 (1,402 entries) Nim
Rank: 8 (1,382 entries) Python
Rank: 9 (1,204 entries) C
Rank: 10 (1,152 entries) REXX
...
Notes
Each language typically demonstrates one or two methods of accessing the data:
with web scraping (via http://www.rosettacode.org/mw/index.php?title=Special:Categories&limit=5000)
with the API method (examples below for Awk, Perl, Ruby, Tcl, etc).
The scraping and API solutions can be separate subsections, see the Tcl example.
Filtering wrong results is optional. You can check against Special:MostLinkedCategories (if using web scraping)
If you use the API, and do elect to filter, you may check your results against this complete, accurate, sortable, wikitable listing of all 869 programming languages, updated periodically, typically weekly.
A complete ranked listing of all 813 languages (from the REXX example) is included here ──► output from the REXX program.
|
#R
|
R
|
library(rvest)
library(plyr)
library(dplyr)
options(stringsAsFactors=FALSE)
langUrl <- "http://rosettacode.org/mw/api.php?format=xml&action=query&generator=categorymembers&gcmtitle=Category:Programming%20Languages&prop=categoryinfo&gcmlimit=5000"
langs <- html(langUrl) %>%
html_nodes('page')
ff <- function(xml_node) {
language <- xml_node %>% html_attr("title")
language <- sub("^Category:", "", language)
npages <- xml_node %>% html_nodes('categoryinfo') %>%
html_attr("pages")
c(language, npages)
}
tbl <- ldply(sapply(langs, ff), rbind)
names(tbl) <- c("language", "n")
tbl %>%
mutate(n=as.integer(n)) %>%
arrange(desc(n)) %>%
head
|
http://rosettacode.org/wiki/Roman_numerals/Encode
|
Roman numerals/Encode
|
Task
Create a function taking a positive integer as its parameter and returning a string containing the Roman numeral representation of that integer. Modern Roman numerals are written by expressing each digit separately, starting with the left most digit and skipping any digit with a value of zero.
In Roman numerals:
1990 is rendered: 1000=M, 900=CM, 90=XC; resulting in MCMXC
2008 is written as 2000=MM, 8=VIII; or MMVIII
1666 uses each Roman symbol in descending order: MDCLXVI
|
#BBC_BASIC
|
BBC BASIC
|
PRINT ;1999, FNroman(1999)
PRINT ;2012, FNroman(2012)
PRINT ;1666, FNroman(1666)
PRINT ;3888, FNroman(3888)
END
DEF FNroman(n%)
LOCAL i%, r$, arabic%(), roman$()
DIM arabic%(12), roman$(12)
arabic%() = 1, 4, 5, 9, 10, 40, 50, 90, 100, 400, 500, 900,1000
roman$() = "I","IV", "V","IX", "X","XL", "L","XC", "C","CD", "D","CM", "M"
FOR i% = 12 TO 0 STEP -1
WHILE n% >= arabic%(i%)
r$ += roman$(i%)
n% -= arabic%(i%)
ENDWHILE
NEXT
= r$
|
http://rosettacode.org/wiki/Roman_numerals/Decode
|
Roman numerals/Decode
|
Task
Create a function that takes a Roman numeral as its argument and returns its value as a numeric decimal integer.
You don't need to validate the form of the Roman numeral.
Modern Roman numerals are written by expressing each decimal digit of the number to be encoded separately,
starting with the leftmost decimal digit and skipping any 0s (zeroes).
1990 is rendered as MCMXC (1000 = M, 900 = CM, 90 = XC) and
2008 is rendered as MMVIII (2000 = MM, 8 = VIII).
The Roman numeral for 1666, MDCLXVI, uses each letter in descending order.
|
#Bracmat
|
Bracmat
|
( unroman
= nbr,lastVal,val
. 0:?nbr:?lastVal
& @( low$!arg
: ?
%@?L
( ?
& (m.1000)
(d.500)
(c.100)
(l.50)
(x.10)
(v.5)
(i.1)
: ? (!L.?val) ?
& (!val:~>!lastVal|!val+-2*!lastVal)
+ !nbr
: ?nbr
& !val:?lastVal
& ~
)
)
| !nbr
)
& (M.1000)
(MCXI.1111)
(CMXI.911)
(MCM.1900)
(MCMXC.1990)
(MMVIII.2008)
(MMIX.2009)
(MCDXLIV.1444)
(MDCLXVI.1666)
(MMXII.2012)
: ?years
& (test=.out$(!arg unroman$!arg))
& ( !years
: ? (?L.?D) (?&test$!L&~)
| done
);
|
http://rosettacode.org/wiki/Roots_of_a_function
|
Roots of a function
|
Task
Create a program that finds and outputs the roots of a given function, range and (if applicable) step width.
The program should identify whether the root is exact or approximate.
For this task, use: ƒ(x) = x3 - 3x2 + 2x
|
#Go
|
Go
|
package main
import (
"fmt"
"math"
)
func main() {
example := func(x float64) float64 { return x*x*x - 3*x*x + 2*x }
findroots(example, -.5, 2.6, 1)
}
func findroots(f func(float64) float64, lower, upper, step float64) {
for x0, x1 := lower, lower+step; x0 < upper; x0, x1 = x1, x1+step {
x1 = math.Min(x1, upper)
r, status := secant(f, x0, x1)
if status != "" && r >= x0 && r < x1 {
fmt.Printf(" %6.3f %s\n", r, status)
}
}
}
func secant(f func(float64) float64, x0, x1 float64) (float64, string) {
var f0 float64
f1 := f(x0)
for i := 0; i < 100; i++ {
f0, f1 = f1, f(x1)
switch {
case f1 == 0:
return x1, "exact"
case math.Abs(x1-x0) < 1e-6:
return x1, "approximate"
}
x0, x1 = x1, x1-f1*(x1-x0)/(f1-f0)
}
return 0, ""
}
|
http://rosettacode.org/wiki/Rock-paper-scissors
|
Rock-paper-scissors
|
Task
Implement the classic children's game Rock-paper-scissors, as well as a simple predictive AI (artificial intelligence) player.
Rock Paper Scissors is a two player game.
Each player chooses one of rock, paper or scissors, without knowing the other player's choice.
The winner is decided by a set of rules:
Rock beats scissors
Scissors beat paper
Paper beats rock
If both players choose the same thing, there is no winner for that round.
For this task, the computer will be one of the players.
The operator will select Rock, Paper or Scissors and the computer will keep a record of the choice frequency, and use that information to make a weighted random choice in an attempt to defeat its opponent.
Extra credit
Support additional choices additional weapons.
|
#Euphoria
|
Euphoria
|
function weighted_rand(sequence table)
integer sum,r
sum = 0
for i = 1 to length(table) do
sum += table[i]
end for
r = rand(sum)
for i = 1 to length(table)-1 do
r -= table[i]
if r <= 0 then
return i
end if
end for
return length(table)
end function
constant names = { "Rock", "Paper", "Scissors" }
constant winner = { "We tied.", "Meself winned.", "You win." }
integer user_action, my_action, key, win
sequence user_rec, score
user_rec = {1,1,1}
score = {0,0}
while 1 do
my_action = remainder(weighted_rand(user_rec)+1,3)+1
puts(1,"Your choice [1-3]:\n")
puts(1," 1. Rock\n 2. Paper\n 3. Scissors\n> ")
key = -1
while (key < '1' or key > '3') and key != 'q' do
key = get_key()
end while
puts(1,key)
puts(1,'\n')
if key = 'q' then
exit
end if
user_action = key-'0'
win = remainder(my_action-user_action+3,3)
printf(1,"You chose %s; I chose %s. %s\n",
{ names[user_action],
names[my_action],
winner[win+1] })
if win then
score[win] += 1
end if
printf(1,"\nScore %d:%d\n",score)
user_rec[user_action] += 1
end while
|
http://rosettacode.org/wiki/Run-length_encoding
|
Run-length encoding
|
Run-length encoding
You are encouraged to solve this task according to the task description, using any language you may know.
Task
Given a string containing uppercase characters (A-Z), compress repeated 'runs' of the same character by storing the length of that run, and provide a function to reverse the compression.
The output can be anything, as long as you can recreate the input with it.
Example
Input: WWWWWWWWWWWWBWWWWWWWWWWWWBBBWWWWWWWWWWWWWWWWWWWWWWWWBWWWWWWWWWWWWWW
Output: 12W1B12W3B24W1B14W
Note: the encoding step in the above example is the same as a step of the Look-and-say sequence.
|
#Forth
|
Forth
|
variable a
: n>a (.) tuck a @ swap move a +! ;
: >a a @ c! 1 a +! ;
: encode ( c-addr +n a -- a n' )
dup a ! -rot over c@ 1 2swap 1 /string bounds ?do
over i c@ = if 1+
else n>a >a i c@ 1 then
loop n>a >a a @ over - ;
: digit? [char] 0 [ char 9 1+ literal ] within ;
: decode ( c-addr +n a -- a n' )
dup a ! 0 2swap bounds ?do
i c@ digit? if 10 * i c@ [char] 0 - + else
a @ over i c@ fill a +! 0 then
loop drop a @ over - ;
|
http://rosettacode.org/wiki/Roots_of_unity
|
Roots of unity
|
The purpose of this task is to explore working with complex numbers.
Task
Given n, find the nth roots of unity.
|
#PARI.2FGP
|
PARI/GP
|
vector(n,k,exp(2*Pi*I*k/n))
|
http://rosettacode.org/wiki/Roots_of_unity
|
Roots of unity
|
The purpose of this task is to explore working with complex numbers.
Task
Given n, find the nth roots of unity.
|
#Pascal
|
Pascal
|
Program Roots;
var
root: record // poor man's complex type.
r: real;
i: real;
end;
i, n: integer;
angle: real;
begin
for n := 2 to 7 do
begin
angle := 0.0;
write(n, ': ');
for i := 1 to n do
begin
root.r := cos(angle);
root.i := sin(angle);
write(root.r:8:5, root.i:8:5, 'i ');
angle := angle + (2.0 * pi / n);
end;
writeln;
end;
end.
|
http://rosettacode.org/wiki/Roots_of_a_quadratic_function
|
Roots of a quadratic function
|
This task has been clarified. Its programming examples are in need of review to ensure that they still fit the requirements of the task.
Write a program to find the roots of a quadratic equation, i.e., solve the equation
a
x
2
+
b
x
+
c
=
0
{\displaystyle ax^{2}+bx+c=0}
.
Your program must correctly handle non-real roots, but it need not check that
a
≠
0
{\displaystyle a\neq 0}
.
The problem of solving a quadratic equation is a good example of how dangerous it can be to ignore the peculiarities of floating-point arithmetic.
The obvious way to implement the quadratic formula suffers catastrophic loss of accuracy when one of the roots to be found is much closer to 0 than the other.
In their classic textbook on numeric methods Computer Methods for Mathematical Computations, George Forsythe, Michael Malcolm, and Cleve Moler suggest trying the naive algorithm with
a
=
1
{\displaystyle a=1}
,
b
=
−
10
5
{\displaystyle b=-10^{5}}
, and
c
=
1
{\displaystyle c=1}
.
(For double-precision floats, set
b
=
−
10
9
{\displaystyle b=-10^{9}}
.)
Consider the following implementation in Ada:
with Ada.Text_IO; use Ada.Text_IO;
with Ada.Numerics.Elementary_Functions; use Ada.Numerics.Elementary_Functions;
procedure Quadratic_Equation is
type Roots is array (1..2) of Float;
function Solve (A, B, C : Float) return Roots is
SD : constant Float := sqrt (B**2 - 4.0 * A * C);
AA : constant Float := 2.0 * A;
begin
return ((- B + SD) / AA, (- B - SD) / AA);
end Solve;
R : constant Roots := Solve (1.0, -10.0E5, 1.0);
begin
Put_Line ("X1 =" & Float'Image (R (1)) & " X2 =" & Float'Image (R (2)));
end Quadratic_Equation;
Output:
X1 = 1.00000E+06 X2 = 0.00000E+00
As we can see, the second root has lost all significant figures. The right answer is that X2 is about
10
−
6
{\displaystyle 10^{-6}}
. The naive method is numerically unstable.
Suggested by Middlebrook (D-OA), a better numerical method: to define two parameters
q
=
a
c
/
b
{\displaystyle q={\sqrt {ac}}/b}
and
f
=
1
/
2
+
1
−
4
q
2
/
2
{\displaystyle f=1/2+{\sqrt {1-4q^{2}}}/2}
and the two roots of the quardratic are:
−
b
a
f
{\displaystyle {\frac {-b}{a}}f}
and
−
c
b
f
{\displaystyle {\frac {-c}{bf}}}
Task: do it better. This means that given
a
=
1
{\displaystyle a=1}
,
b
=
−
10
9
{\displaystyle b=-10^{9}}
, and
c
=
1
{\displaystyle c=1}
, both of the roots your program returns should be greater than
10
−
11
{\displaystyle 10^{-11}}
. Or, if your language can't do floating-point arithmetic any more precisely than single precision, your program should be able to handle
b
=
−
10
6
{\displaystyle b=-10^{6}}
. Either way, show what your program gives as the roots of the quadratic in question. See page 9 of
"What Every Scientist Should Know About Floating-Point Arithmetic" for a possible algorithm.
|
#Pascal
|
Pascal
|
Program QuadraticRoots;
var
a, b, c, q, f: double;
begin
a := 1;
b := -10e9;
c := 1;
q := sqrt(a * c) / b;
f := (1 + sqrt(1 - 4 * q * q)) / 2;
writeln ('Version 1:');
writeln ('x1: ', (-b * f / a):16, ', x2: ', (-c / (b * f)):16);
writeln ('Version 2:');
q := sqrt(b * b - 4 * a * c);
if b < 0 then
begin
f := (-b + q) / 2 * a;
writeln ('x1: ', f:16, ', x2: ', (c / (a * f)):16);
end
else
begin
f := (-b - q) / 2 * a;
writeln ('x1: ', (c / (a * f)):16, ', x2: ', f:16);
end;
end.
|
http://rosettacode.org/wiki/Rot-13
|
Rot-13
|
Task
Implement a rot-13 function (or procedure, class, subroutine, or other "callable" object as appropriate to your programming environment).
Optionally wrap this function in a utility program (like tr, which acts like a common UNIX utility, performing a line-by-line rot-13 encoding of every line of input contained in each file listed on its command line, or (if no filenames are passed thereon) acting as a filter on its "standard input."
(A number of UNIX scripting languages and utilities, such as awk and sed either default to processing files in this way or have command line switches or modules to easily implement these wrapper semantics, e.g., Perl and Python).
The rot-13 encoding is commonly known from the early days of Usenet "Netnews" as a way of obfuscating text to prevent casual reading of spoiler or potentially offensive material.
Many news reader and mail user agent programs have built-in rot-13 encoder/decoders or have the ability to feed a message through any external utility script for performing this (or other) actions.
The definition of the rot-13 function is to simply replace every letter of the ASCII alphabet with the letter which is "rotated" 13 characters "around" the 26 letter alphabet from its normal cardinal position (wrapping around from z to a as necessary).
Thus the letters abc become nop and so on.
Technically rot-13 is a "mono-alphabetic substitution cipher" with a trivial "key".
A proper implementation should work on upper and lower case letters, preserve case, and pass all non-alphabetic characters
in the input stream through without alteration.
Related tasks
Caesar cipher
Substitution Cipher
Vigenère Cipher/Cryptanalysis
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
|
#Elena
|
Elena
|
import system'routines;
import extensions;
import extensions'text;
singleton rotConvertor
{
char convert(char ch)
{
if (($97 <= ch && ch <= $109) || ($65 <= ch && ch <= $77))
{
^ (ch.toInt() + 13).toChar()
};
if (($110 <= ch && ch <= $122) || ($78 <= ch && ch <= $90))
{
^ (ch.toInt() - 13).toChar()
};
^ ch
}
string convert(string s)
= s.selectBy:(ch => rotConvertor.convert(ch)).summarize(new StringWriter());
}
public program()
{
if (program_arguments.Length > 1)
{
console.printLine(rotConvertor.convert(program_arguments[1]))
}
}
|
http://rosettacode.org/wiki/Runge-Kutta_method
|
Runge-Kutta method
|
Given the example Differential equation:
y
′
(
t
)
=
t
×
y
(
t
)
{\displaystyle y'(t)=t\times {\sqrt {y(t)}}}
With initial condition:
t
0
=
0
{\displaystyle t_{0}=0}
and
y
0
=
y
(
t
0
)
=
y
(
0
)
=
1
{\displaystyle y_{0}=y(t_{0})=y(0)=1}
This equation has an exact solution:
y
(
t
)
=
1
16
(
t
2
+
4
)
2
{\displaystyle y(t)={\tfrac {1}{16}}(t^{2}+4)^{2}}
Task
Demonstrate the commonly used explicit fourth-order Runge–Kutta method to solve the above differential equation.
Solve the given differential equation over the range
t
=
0
…
10
{\displaystyle t=0\ldots 10}
with a step value of
δ
t
=
0.1
{\displaystyle \delta t=0.1}
(101 total points, the first being given)
Print the calculated values of
y
{\displaystyle y}
at whole numbered
t
{\displaystyle t}
's (
0.0
,
1.0
,
…
10.0
{\displaystyle 0.0,1.0,\ldots 10.0}
) along with error as compared to the exact solution.
Method summary
Starting with a given
y
n
{\displaystyle y_{n}}
and
t
n
{\displaystyle t_{n}}
calculate:
δ
y
1
=
δ
t
×
y
′
(
t
n
,
y
n
)
{\displaystyle \delta y_{1}=\delta t\times y'(t_{n},y_{n})\quad }
δ
y
2
=
δ
t
×
y
′
(
t
n
+
1
2
δ
t
,
y
n
+
1
2
δ
y
1
)
{\displaystyle \delta y_{2}=\delta t\times y'(t_{n}+{\tfrac {1}{2}}\delta t,y_{n}+{\tfrac {1}{2}}\delta y_{1})}
δ
y
3
=
δ
t
×
y
′
(
t
n
+
1
2
δ
t
,
y
n
+
1
2
δ
y
2
)
{\displaystyle \delta y_{3}=\delta t\times y'(t_{n}+{\tfrac {1}{2}}\delta t,y_{n}+{\tfrac {1}{2}}\delta y_{2})}
δ
y
4
=
δ
t
×
y
′
(
t
n
+
δ
t
,
y
n
+
δ
y
3
)
{\displaystyle \delta y_{4}=\delta t\times y'(t_{n}+\delta t,y_{n}+\delta y_{3})\quad }
then:
y
n
+
1
=
y
n
+
1
6
(
δ
y
1
+
2
δ
y
2
+
2
δ
y
3
+
δ
y
4
)
{\displaystyle y_{n+1}=y_{n}+{\tfrac {1}{6}}(\delta y_{1}+2\delta y_{2}+2\delta y_{3}+\delta y_{4})}
t
n
+
1
=
t
n
+
δ
t
{\displaystyle t_{n+1}=t_{n}+\delta t\quad }
|
#Ruby
|
Ruby
|
def calc_rk4(f)
return ->(t,y,dt){
->(dy1 ){
->(dy2 ){
->(dy3 ){
->(dy4 ){ ( dy1 + 2*dy2 + 2*dy3 + dy4 ) / 6 }.call(
dt * f.call( t + dt , y + dy3 ))}.call(
dt * f.call( t + dt/2, y + dy2/2 ))}.call(
dt * f.call( t + dt/2, y + dy1/2 ))}.call(
dt * f.call( t , y ))}
end
TIME_MAXIMUM, WHOLE_TOLERANCE = 10.0, 1.0e-5
T_START, Y_START, DT = 0.0, 1.0, 0.10
def my_diff_eqn(t,y) ; t * Math.sqrt(y) ; end
def my_solution(t ) ; (t**2 + 4)**2 / 16 ; end
def find_error(t,y) ; (y - my_solution(t)).abs ; end
def is_whole?(t ) ; (t.round - t).abs < WHOLE_TOLERANCE ; end
dy = calc_rk4( ->(t,y){my_diff_eqn(t,y)} )
t, y = T_START, Y_START
while t <= TIME_MAXIMUM
printf("y(%4.1f)\t= %12.6f \t error: %12.6e\n",t,y,find_error(t,y)) if is_whole?(t)
t, y = t + DT, y + dy.call(t,y,DT)
end
|
http://rosettacode.org/wiki/S-expressions
|
S-expressions
|
S-Expressions are one convenient way to parse and store data.
Task
Write a simple reader and writer for S-Expressions that handles quoted and unquoted strings, integers and floats.
The reader should read a single but nested S-Expression from a string and store it in a suitable datastructure (list, array, etc).
Newlines and other whitespace may be ignored unless contained within a quoted string.
“()” inside quoted strings are not interpreted, but treated as part of the string.
Handling escaped quotes inside a string is optional; thus “(foo"bar)” maybe treated as a string “foo"bar”, or as an error.
For this, the reader need not recognize “\” for escaping, but should, in addition, recognize numbers if the language has appropriate datatypes.
Languages that support it may treat unquoted strings as symbols.
Note that with the exception of “()"” (“\” if escaping is supported) and whitespace there are no special characters. Anything else is allowed without quotes.
The reader should be able to read the following input
((data "quoted data" 123 4.5)
(data (!@# (4.5) "(more" "data)")))
and turn it into a native datastructure. (see the Pike, Python and Ruby implementations for examples of native data structures.)
The writer should be able to take the produced list and turn it into a new S-Expression.
Strings that don't contain whitespace or parentheses () don't need to be quoted in the resulting S-Expression, but as a simplification, any string may be quoted.
Extra Credit
Let the writer produce pretty printed output with indenting and line-breaks.
|
#TXR
|
TXR
|
$ txr -p '(read)'
((data "quoted data" 123 4.5)
(data (!@# (4.5) "(more" "data)")))
[Ctrl-D][Enter]
((data "quoted data" 123 4.5) (data (! (sys:var #(4.5)) "(more" "data)")))
|
http://rosettacode.org/wiki/RPG_attributes_generator
|
RPG attributes generator
|
RPG = Role Playing Game.
You're running a tabletop RPG, and your players are creating characters.
Each character has six core attributes: strength, dexterity, constitution, intelligence, wisdom, and charisma.
One way of generating values for these attributes is to roll four, 6-sided dice (d6) and sum the three highest rolls, discarding the lowest roll.
Some players like to assign values to their attributes in the order they're rolled.
To ensure generated characters don't put players at a disadvantage, the following requirements must be satisfied:
The total of all character attributes must be at least 75.
At least two of the attributes must be at least 15.
However, this can require a lot of manual dice rolling. A programatic solution would be much faster.
Task
Write a program that:
Generates 4 random, whole values between 1 and 6.
Saves the sum of the 3 largest values.
Generates a total of 6 values this way.
Displays the total, and all 6 values once finished.
The order in which each value was generated must be preserved.
The total of all 6 values must be at least 75.
At least 2 of the values must be 15 or more.
|
#Scala
|
Scala
|
import scala.util.Random
Random.setSeed(1)
def rollDice():Int = {
val v4 = Stream.continually(Random.nextInt(6)+1).take(4)
v4.sum - v4.min
}
def getAttributes():Seq[Int] = Stream.continually(rollDice()).take(6)
def getCharacter():Seq[Int] = {
val attrs = getAttributes()
println("generated => " + attrs.mkString("[",",", "]"))
(attrs.sum, attrs.filter(_>15).size) match {
case (a, b) if (a < 75 || b < 2) => getCharacter
case _ => attrs
}
}
println("picked => " + getCharacter.mkString("[", ",", "]"))
|
http://rosettacode.org/wiki/Sieve_of_Eratosthenes
|
Sieve of Eratosthenes
|
This task has been clarified. Its programming examples are in need of review to ensure that they still fit the requirements of the task.
The Sieve of Eratosthenes is a simple algorithm that finds the prime numbers up to a given integer.
Task
Implement the Sieve of Eratosthenes algorithm, with the only allowed optimization that the outer loop can stop at the square root of the limit, and the inner loop may start at the square of the prime just found.
That means especially that you shouldn't optimize by using pre-computed wheels, i.e. don't assume you need only to cross out odd numbers (wheel based on 2), numbers equal to 1 or 5 modulo 6 (wheel based on 2 and 3), or similar wheels based on low primes.
If there's an easy way to add such a wheel based optimization, implement it as an alternative version.
Note
It is important that the sieve algorithm be the actual algorithm used to find prime numbers for the task.
Related tasks
Emirp primes
count in factors
prime decomposition
factors of an integer
extensible prime generator
primality by trial division
factors of a Mersenne number
trial factoring of a Mersenne number
partition an integer X into N primes
sequence of primes by Trial Division
|
#Vala
|
Vala
|
using Gee;
ArrayList<int> primes(int limit){
var sieve = new ArrayList<bool>();
var prime_list = new ArrayList<int>();
for(int i = 0; i <= limit; i++){
sieve.add(true);
}
sieve[0] = false;
sieve[1] = false;
for (int i = 2; i <= limit/2; i++){
if (sieve[i] != false){
for (int j = 2; i*j <= limit; j++){
sieve[i*j] = false;
}
}
}
for (int i = 0; i < sieve.size; i++){
if (sieve[i] != false){
prime_list.add(i);
}
}
return prime_list;
} // end primes
public static void main(){
var prime_list = primes(50);
foreach(var prime in prime_list)
stdout.printf("%s ", prime.to_string());
stdout.printf("\n");
}
|
http://rosettacode.org/wiki/Rosetta_Code/Count_examples
|
Rosetta Code/Count examples
|
task
Essentially, count the number of occurrences of =={{header| on each task page.
Output:
100 doors: 20 examples.
99 Bottles of Beer: 29 examples.
Abstract type: 10 examples.
Total: X examples.
For a full output, updated periodically, see Rosetta Code/Count examples/Full list.
You'll need to use the Media Wiki API, which you can find out about locally, here, or in Media Wiki's API documentation at, API:Query
|
#Wren
|
Wren
|
/* rc_count_examples.wren */
import "./pattern" for Pattern
var CURLOPT_URL = 10002
var CURLOPT_FOLLOWLOCATION = 52
var CURLOPT_WRITEFUNCTION = 20011
var CURLOPT_WRITEDATA = 10001
foreign class Buffer {
construct new() {} // C will allocate buffer of a suitable size
foreign value // returns buffer contents as a string
}
foreign class Curl {
construct easyInit() {}
foreign easySetOpt(opt, param)
foreign easyPerform()
foreign easyCleanup()
}
var curl = Curl.easyInit()
var getContent = Fn.new { |url|
var buffer = Buffer.new()
curl.easySetOpt(CURLOPT_URL, url)
curl.easySetOpt(CURLOPT_FOLLOWLOCATION, 1)
curl.easySetOpt(CURLOPT_WRITEFUNCTION, 0) // write function to be supplied by C
curl.easySetOpt(CURLOPT_WRITEDATA, buffer)
curl.easyPerform()
return buffer.value
}
var url = "https://www.rosettacode.org/mw/api.php?action=query&list=categorymembers&cmtitle=Category:Programming_Tasks&cmlimit=500&format=xml"
var content = getContent.call(url)
var p = Pattern.new("title/=\"[+1^\"]\"")
var matches = p.findAll(content)
for (m in matches) {
var title = m.capsText[0].replace("'", "'").replace(""", "\"")
var title2 = title.replace(" ", "_").replace("+", "\%2B")
var taskUrl = "https://www.rosettacode.org/mw/index.php?title=%(title2)&action=raw"
var taskContent = getContent.call(taskUrl)
var lines = taskContent.split("\n")
var count = lines.count { |line| line.trim().startsWith("=={{header|") }
System.print("%(title) : %(count) examples")
}
curl.easyCleanup()
|
http://rosettacode.org/wiki/Search_a_list
|
Search a list
|
Task[edit]
Find the index of a string (needle) in an indexable, ordered collection of strings (haystack).
Raise an exception if the needle is missing.
If there is more than one occurrence then return the smallest index to the needle.
Extra credit
Return the largest index to a needle that has multiple occurrences in the haystack.
See also
Search a list of records
|
#OCaml
|
OCaml
|
# let find_index pred lst =
let rec loop n = function
[] -> raise Not_found
| x::xs -> if pred x then n
else loop (n+1) xs
in
loop 0 lst;;
val find_index : ('a -> bool) -> 'a list -> int = <fun>
# let haystack =
["Zig";"Zag";"Wally";"Ronald";"Bush";"Krusty";"Charlie";"Bush";"Bozo"];;
val haystack : string list =
["Zig"; "Zag"; "Wally"; "Ronald"; "Bush"; "Krusty"; "Charlie"; "Bush";
"Bozo"]
# List.iter (fun needle ->
try
Printf.printf "%i %s\n" (find_index ((=) needle) haystack) needle
with Not_found ->
Printf.printf "%s is not in haystack\n" needle)
["Washington"; "Bush"];;
Washington is not in haystack
4 Bush
- : unit = ()
|
http://rosettacode.org/wiki/Rosetta_Code/Rank_languages_by_popularity
|
Rosetta Code/Rank languages by popularity
|
Rosetta Code/Rank languages by popularity
You are encouraged to solve this task according to the task description, using any language you may know.
Task
Sort the most popular computer programming languages based in number of members in Rosetta Code categories.
Sample output on 01 juin 2022 at 14:13 +02
Rank: 1 (1,540 entries) Phix
Rank: 2 (1,531 entries) Wren
Rank: 3 (1,507 entries) Julia
Rank: 4 (1,494 entries) Go
Rank: 5 (1,488 entries) Raku
Rank: 6 (1,448 entries) Perl
Rank: 7 (1,402 entries) Nim
Rank: 8 (1,382 entries) Python
Rank: 9 (1,204 entries) C
Rank: 10 (1,152 entries) REXX
...
Notes
Each language typically demonstrates one or two methods of accessing the data:
with web scraping (via http://www.rosettacode.org/mw/index.php?title=Special:Categories&limit=5000)
with the API method (examples below for Awk, Perl, Ruby, Tcl, etc).
The scraping and API solutions can be separate subsections, see the Tcl example.
Filtering wrong results is optional. You can check against Special:MostLinkedCategories (if using web scraping)
If you use the API, and do elect to filter, you may check your results against this complete, accurate, sortable, wikitable listing of all 869 programming languages, updated periodically, typically weekly.
A complete ranked listing of all 813 languages (from the REXX example) is included here ──► output from the REXX program.
|
#Racket
|
Racket
|
#lang racket
(require racket/hash
net/url
json)
(define limit 15)
(define (replacer cat) (regexp-replace #rx"^Category:(.*?)$" cat "\\1"))
(define category "Category:Programming Languages")
(define entries "entries")
(define api-url (string->url "http://rosettacode.org/mw/api.php"))
(define (make-complete-url gcmcontinue)
(struct-copy url api-url
[query `([format . "json"]
[action . "query"]
[generator . "categorymembers"]
[gcmtitle . ,category]
[gcmlimit . "200"]
[gcmcontinue . ,gcmcontinue]
[continue . ""]
[prop . "categoryinfo"])]))
(define @ hash-ref)
(define table (make-hash))
(let loop ([gcmcontinue ""])
(define resp (read-json (get-pure-port (make-complete-url gcmcontinue))))
(hash-union! table
(for/hash ([(k v) (in-hash (@ (@ resp 'query) 'pages))])
(values (@ v 'title #f) (@ (@ v 'categoryinfo (hash)) 'size 0))))
(cond [(@ resp 'continue #f) => (λ (c) (loop (@ c 'gcmcontinue)))]))
(for/fold ([prev #f] [rank #f] #:result (void))
([item (in-list (sort (hash->list table) > #:key cdr))] [i (in-range limit)])
(match-define (cons cat size) item)
(define this-rank (if (equal? prev size) rank (add1 i)))
(printf "Rank: ~a ~a ~a\n"
(~a this-rank #:align 'right #:min-width 2)
(~a (format "(~a ~a)" size entries) #:align 'right #:min-width 14)
(replacer cat))
(values size this-rank))
|
http://rosettacode.org/wiki/Roman_numerals/Encode
|
Roman numerals/Encode
|
Task
Create a function taking a positive integer as its parameter and returning a string containing the Roman numeral representation of that integer. Modern Roman numerals are written by expressing each digit separately, starting with the left most digit and skipping any digit with a value of zero.
In Roman numerals:
1990 is rendered: 1000=M, 900=CM, 90=XC; resulting in MCMXC
2008 is written as 2000=MM, 8=VIII; or MMVIII
1666 uses each Roman symbol in descending order: MDCLXVI
|
#BCPL
|
BCPL
|
get "libhdr"
let toroman(n, v) = valof
$( let extract(n, val, rmn, v) = valof
$( while n >= val
$( n := n - val;
for i=1 to rmn%0 do v%(v%0+i) := rmn%i
v%0 := v%0 + rmn%0
$)
resultis n
$)
v%0 := 0
n := extract(n, 1000, "M", v)
n := extract(n, 900, "CM", v)
n := extract(n, 500, "D", v)
n := extract(n, 400, "CD", v)
n := extract(n, 100, "C", v)
n := extract(n, 90, "XC", v)
n := extract(n, 50, "L", v)
n := extract(n, 40, "XL", v)
n := extract(n, 10, "X", v)
n := extract(n, 9, "IX", v)
n := extract(n, 5, "V", v)
n := extract(n, 4, "IV", v)
n := extract(n, 1, "I", v)
resultis v
$)
let show(n) be
$( let v = vec 50
writef("%I4 = %S*N", n, toroman(n, v))
$)
let start() be
$( show(1666)
show(2008)
show(1001)
show(1999)
show(3888)
show(2021)
$)
|
http://rosettacode.org/wiki/Roman_numerals/Decode
|
Roman numerals/Decode
|
Task
Create a function that takes a Roman numeral as its argument and returns its value as a numeric decimal integer.
You don't need to validate the form of the Roman numeral.
Modern Roman numerals are written by expressing each decimal digit of the number to be encoded separately,
starting with the leftmost decimal digit and skipping any 0s (zeroes).
1990 is rendered as MCMXC (1000 = M, 900 = CM, 90 = XC) and
2008 is rendered as MMVIII (2000 = MM, 8 = VIII).
The Roman numeral for 1666, MDCLXVI, uses each letter in descending order.
|
#C
|
C
|
#include <stdio.h>
int digits[26] = { 0, 0, 100, 500, 0, 0, 0, 0, 1, 1, 0, 50, 1000, 0, 0, 0, 0, 0, 0, 0, 5, 5, 0, 10, 0, 0 };
/* assuming ASCII, do upper case and get index in alphabet. could also be
inline int VALUE(char x) { return digits [ (~0x20 & x) - 'A' ]; }
if you think macros are evil */
#define VALUE(x) digits[(~0x20 & (x)) - 'A']
int decode(const char * roman)
{
const char *bigger;
int current;
int arabic = 0;
while (*roman != '\0') {
current = VALUE(*roman);
/* if (!current) return -1;
note: -1 can be used as error code; Romans didn't even have zero
*/
bigger = roman;
/* look for a larger digit, like IV or XM */
while (VALUE(*bigger) <= current && *++bigger != '\0');
if (*bigger == '\0')
arabic += current;
else {
arabic += VALUE(*bigger);
while (roman < bigger)
arabic -= VALUE(* (roman++) );
}
roman ++;
}
return arabic;
}
int main()
{
const char * romans[] = { "MCmxC", "MMVIII", "MDClXVI", "MCXLUJ" };
int i;
for (i = 0; i < 4; i++)
printf("%s\t%d\n", romans[i], decode(romans[i]));
return 0;
}
|
http://rosettacode.org/wiki/Roots_of_a_function
|
Roots of a function
|
Task
Create a program that finds and outputs the roots of a given function, range and (if applicable) step width.
The program should identify whether the root is exact or approximate.
For this task, use: ƒ(x) = x3 - 3x2 + 2x
|
#Haskell
|
Haskell
|
f x = x^3-3*x^2+2*x
findRoots start stop step eps =
[x | x <- [start, start+step .. stop], abs (f x) < eps]
|
http://rosettacode.org/wiki/Rock-paper-scissors
|
Rock-paper-scissors
|
Task
Implement the classic children's game Rock-paper-scissors, as well as a simple predictive AI (artificial intelligence) player.
Rock Paper Scissors is a two player game.
Each player chooses one of rock, paper or scissors, without knowing the other player's choice.
The winner is decided by a set of rules:
Rock beats scissors
Scissors beat paper
Paper beats rock
If both players choose the same thing, there is no winner for that round.
For this task, the computer will be one of the players.
The operator will select Rock, Paper or Scissors and the computer will keep a record of the choice frequency, and use that information to make a weighted random choice in an attempt to defeat its opponent.
Extra credit
Support additional choices additional weapons.
|
#F.23
|
F#
|
open System
let random = Random ()
let rand = random.NextDouble () //Gets a random number in the range (0.0, 1.0)
/// Union of possible choices for a round of rock-paper-scissors
type Choice =
| Rock
| Paper
| Scissors
/// Gets the string representation of a Choice
let getString = function
| Rock -> "Rock"
| Paper -> "Paper"
| Scissors -> "Scissors"
/// Defines rules for winning and losing
let beats (a : Choice, b : Choice) =
match a, b with
| Rock, Scissors -> true // Rock beats Scissors
| Paper, Rock -> true // Paper beats Rock
| Scissors, Paper -> true // Scissors beat Paper
| _, _ -> false
/// Generates the next move for the computer based on probability derived from previous player moves.
let genMove r p s =
let tot = r + p + s
let n = rand
if n <= s / tot then Rock
elif n <= (s + r) / tot then Paper
else Scissors
/// Gets the move chosen by the player
let rec getMove () =
printf "[R]ock, [P]aper, or [S]cissors?: "
let choice = Console.ReadLine ()
match choice with
| "r" | "R" -> Rock
| "p" | "P" -> Paper
| "s" | "S" -> Scissors
| _ ->
printf "Invalid choice.\n\n"
getMove ()
/// Place where all the game logic takes place.
let rec game (r : float, p : float, s : float) =
let comp = genMove r p s
let player = getMove ()
Console.WriteLine ("Player: {0} vs Computer: {1}", getString player, getString comp)
Console.WriteLine (
if beats(player, comp) then "Player Wins!\n"
elif beats(comp, player) then "Computer Wins!\n"
else "Draw!\n"
)
let nextR = if player = Rock then r + 1.0 else r
let nextP = if player = Paper then p + 1.0 else p
let nextS = if player = Scissors then s + 1.0 else s
game (nextR, nextP, nextS)
game(1.0, 1.0, 1.0)
|
http://rosettacode.org/wiki/Run-length_encoding
|
Run-length encoding
|
Run-length encoding
You are encouraged to solve this task according to the task description, using any language you may know.
Task
Given a string containing uppercase characters (A-Z), compress repeated 'runs' of the same character by storing the length of that run, and provide a function to reverse the compression.
The output can be anything, as long as you can recreate the input with it.
Example
Input: WWWWWWWWWWWWBWWWWWWWWWWWWBBBWWWWWWWWWWWWWWWWWWWWWWWWBWWWWWWWWWWWWWW
Output: 12W1B12W3B24W1B14W
Note: the encoding step in the above example is the same as a step of the Look-and-say sequence.
|
#Fortran
|
Fortran
|
program RLE
implicit none
integer, parameter :: bufsize = 100 ! Sets maximum size of coded and decoded strings, adjust as necessary
character(bufsize) :: teststr = "WWWWWWWWWWWWBWWWWWWWWWWWWBBBWWWWWWWWWWWWWWWWWWWWWWWWBWWWWWWWWWWWWWW"
character(bufsize) :: codedstr = "", decodedstr = ""
call Encode(teststr, codedstr)
write(*,"(a)") trim(codedstr)
call Decode(codedstr, decodedstr)
write(*,"(a)") trim(decodedstr)
contains
subroutine Encode(instr, outstr)
character(*), intent(in) :: instr
character(*), intent(out) :: outstr
character(8) :: tempstr = ""
character(26) :: validchars = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
integer :: a, b, c, i
if(verify(trim(instr), validchars) /= 0) then
outstr = "Invalid input"
return
end if
outstr = ""
c = 1
a = iachar(instr(1:1))
do i = 2, len(trim(instr))
b = iachar(instr(i:i))
if(a == b) then
c = c + 1
else
write(tempstr, "(i0)") c
outstr = trim(outstr) // trim(tempstr) // achar(a)
a = b
c = 1
end if
end do
write(tempstr, "(i0)") c
outstr = trim(outstr) // trim(tempstr) // achar(b)
end subroutine
subroutine Decode(instr, outstr)
character(*), intent(in) :: instr
character(*), intent(out) :: outstr
character(26) :: validchars = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
integer :: startn, endn, n
outstr = ""
startn = 1
do while(startn < len(trim(instr)))
endn = scan(instr(startn:), validchars) + startn - 1
read(instr(startn:endn-1), "(i8)") n
outstr = trim(outstr) // repeat(instr(endn:endn), n)
startn = endn + 1
end do
end subroutine
end program
|
http://rosettacode.org/wiki/Roots_of_unity
|
Roots of unity
|
The purpose of this task is to explore working with complex numbers.
Task
Given n, find the nth roots of unity.
|
#Perl
|
Perl
|
use Math::Complex;
foreach my $n (2 .. 10) {
printf "%2d", $n;
my @roots = root(1,$n);
foreach my $root (@roots) {
$root->display_format(style => 'cartesian', format => '%.3f');
print " $root";
}
print "\n";
}
|
http://rosettacode.org/wiki/Roots_of_unity
|
Roots of unity
|
The purpose of this task is to explore working with complex numbers.
Task
Given n, find the nth roots of unity.
|
#Phix
|
Phix
|
with javascript_semantics
for n=2 to 10 do
printf(1,"%2d:",n)
for root=0 to n-1 do
atom real = cos(2*PI*root/n)
atom imag = sin(2*PI*root/n)
printf(1,"%s %6.3f %6.3fi",{iff(root?",":""),real,imag})
end for
printf(1,"\n")
end for
|
http://rosettacode.org/wiki/Roots_of_a_quadratic_function
|
Roots of a quadratic function
|
This task has been clarified. Its programming examples are in need of review to ensure that they still fit the requirements of the task.
Write a program to find the roots of a quadratic equation, i.e., solve the equation
a
x
2
+
b
x
+
c
=
0
{\displaystyle ax^{2}+bx+c=0}
.
Your program must correctly handle non-real roots, but it need not check that
a
≠
0
{\displaystyle a\neq 0}
.
The problem of solving a quadratic equation is a good example of how dangerous it can be to ignore the peculiarities of floating-point arithmetic.
The obvious way to implement the quadratic formula suffers catastrophic loss of accuracy when one of the roots to be found is much closer to 0 than the other.
In their classic textbook on numeric methods Computer Methods for Mathematical Computations, George Forsythe, Michael Malcolm, and Cleve Moler suggest trying the naive algorithm with
a
=
1
{\displaystyle a=1}
,
b
=
−
10
5
{\displaystyle b=-10^{5}}
, and
c
=
1
{\displaystyle c=1}
.
(For double-precision floats, set
b
=
−
10
9
{\displaystyle b=-10^{9}}
.)
Consider the following implementation in Ada:
with Ada.Text_IO; use Ada.Text_IO;
with Ada.Numerics.Elementary_Functions; use Ada.Numerics.Elementary_Functions;
procedure Quadratic_Equation is
type Roots is array (1..2) of Float;
function Solve (A, B, C : Float) return Roots is
SD : constant Float := sqrt (B**2 - 4.0 * A * C);
AA : constant Float := 2.0 * A;
begin
return ((- B + SD) / AA, (- B - SD) / AA);
end Solve;
R : constant Roots := Solve (1.0, -10.0E5, 1.0);
begin
Put_Line ("X1 =" & Float'Image (R (1)) & " X2 =" & Float'Image (R (2)));
end Quadratic_Equation;
Output:
X1 = 1.00000E+06 X2 = 0.00000E+00
As we can see, the second root has lost all significant figures. The right answer is that X2 is about
10
−
6
{\displaystyle 10^{-6}}
. The naive method is numerically unstable.
Suggested by Middlebrook (D-OA), a better numerical method: to define two parameters
q
=
a
c
/
b
{\displaystyle q={\sqrt {ac}}/b}
and
f
=
1
/
2
+
1
−
4
q
2
/
2
{\displaystyle f=1/2+{\sqrt {1-4q^{2}}}/2}
and the two roots of the quardratic are:
−
b
a
f
{\displaystyle {\frac {-b}{a}}f}
and
−
c
b
f
{\displaystyle {\frac {-c}{bf}}}
Task: do it better. This means that given
a
=
1
{\displaystyle a=1}
,
b
=
−
10
9
{\displaystyle b=-10^{9}}
, and
c
=
1
{\displaystyle c=1}
, both of the roots your program returns should be greater than
10
−
11
{\displaystyle 10^{-11}}
. Or, if your language can't do floating-point arithmetic any more precisely than single precision, your program should be able to handle
b
=
−
10
6
{\displaystyle b=-10^{6}}
. Either way, show what your program gives as the roots of the quadratic in question. See page 9 of
"What Every Scientist Should Know About Floating-Point Arithmetic" for a possible algorithm.
|
#Perl
|
Perl
|
use Math::Complex;
($x1,$x2) = solveQuad(1,2,3);
print "x1 = $x1, x2 = $x2\n";
sub solveQuad
{
my ($a,$b,$c) = @_;
my $root = sqrt($b**2 - 4*$a*$c);
return ( -$b + $root )/(2*$a), ( -$b - $root )/(2*$a);
}
|
http://rosettacode.org/wiki/Roots_of_a_quadratic_function
|
Roots of a quadratic function
|
This task has been clarified. Its programming examples are in need of review to ensure that they still fit the requirements of the task.
Write a program to find the roots of a quadratic equation, i.e., solve the equation
a
x
2
+
b
x
+
c
=
0
{\displaystyle ax^{2}+bx+c=0}
.
Your program must correctly handle non-real roots, but it need not check that
a
≠
0
{\displaystyle a\neq 0}
.
The problem of solving a quadratic equation is a good example of how dangerous it can be to ignore the peculiarities of floating-point arithmetic.
The obvious way to implement the quadratic formula suffers catastrophic loss of accuracy when one of the roots to be found is much closer to 0 than the other.
In their classic textbook on numeric methods Computer Methods for Mathematical Computations, George Forsythe, Michael Malcolm, and Cleve Moler suggest trying the naive algorithm with
a
=
1
{\displaystyle a=1}
,
b
=
−
10
5
{\displaystyle b=-10^{5}}
, and
c
=
1
{\displaystyle c=1}
.
(For double-precision floats, set
b
=
−
10
9
{\displaystyle b=-10^{9}}
.)
Consider the following implementation in Ada:
with Ada.Text_IO; use Ada.Text_IO;
with Ada.Numerics.Elementary_Functions; use Ada.Numerics.Elementary_Functions;
procedure Quadratic_Equation is
type Roots is array (1..2) of Float;
function Solve (A, B, C : Float) return Roots is
SD : constant Float := sqrt (B**2 - 4.0 * A * C);
AA : constant Float := 2.0 * A;
begin
return ((- B + SD) / AA, (- B - SD) / AA);
end Solve;
R : constant Roots := Solve (1.0, -10.0E5, 1.0);
begin
Put_Line ("X1 =" & Float'Image (R (1)) & " X2 =" & Float'Image (R (2)));
end Quadratic_Equation;
Output:
X1 = 1.00000E+06 X2 = 0.00000E+00
As we can see, the second root has lost all significant figures. The right answer is that X2 is about
10
−
6
{\displaystyle 10^{-6}}
. The naive method is numerically unstable.
Suggested by Middlebrook (D-OA), a better numerical method: to define two parameters
q
=
a
c
/
b
{\displaystyle q={\sqrt {ac}}/b}
and
f
=
1
/
2
+
1
−
4
q
2
/
2
{\displaystyle f=1/2+{\sqrt {1-4q^{2}}}/2}
and the two roots of the quardratic are:
−
b
a
f
{\displaystyle {\frac {-b}{a}}f}
and
−
c
b
f
{\displaystyle {\frac {-c}{bf}}}
Task: do it better. This means that given
a
=
1
{\displaystyle a=1}
,
b
=
−
10
9
{\displaystyle b=-10^{9}}
, and
c
=
1
{\displaystyle c=1}
, both of the roots your program returns should be greater than
10
−
11
{\displaystyle 10^{-11}}
. Or, if your language can't do floating-point arithmetic any more precisely than single precision, your program should be able to handle
b
=
−
10
6
{\displaystyle b=-10^{6}}
. Either way, show what your program gives as the roots of the quadratic in question. See page 9 of
"What Every Scientist Should Know About Floating-Point Arithmetic" for a possible algorithm.
|
#Phix
|
Phix
|
procedure solve_quadratic(sequence t3)
atom {a,b,c} = t3,
d = b*b-4*a*c, f
string s = sprintf("for a=%g,b=%g,c=%g",t3), t
sequence u
if abs(d)<1e-6 then d=0 end if
switch sign(d) do
case 0: t = "single root is %g"
u = {-b/2/a}
case 1: t = "real roots are %g and %g"
f = (1+sqrt(1-4*a*c/(b*b)))/2
u = {-f*b/a,-c/b/f}
case-1: t = "complex roots are %g +/- %g*i"
u = {-b/2/a,sqrt(-d)/2/a}
end switch
printf(1,"%-25s the %s\n",{s,sprintf(t,u)})
end procedure
constant tests = {{1,-1E9,1},
{1,0,1},
{2,-1,-6},
{1,2,-2},
{0.5,1.4142135,1},
{1,3,2},
{3,4,5}}
papply(tests,solve_quadratic)
|
http://rosettacode.org/wiki/Rot-13
|
Rot-13
|
Task
Implement a rot-13 function (or procedure, class, subroutine, or other "callable" object as appropriate to your programming environment).
Optionally wrap this function in a utility program (like tr, which acts like a common UNIX utility, performing a line-by-line rot-13 encoding of every line of input contained in each file listed on its command line, or (if no filenames are passed thereon) acting as a filter on its "standard input."
(A number of UNIX scripting languages and utilities, such as awk and sed either default to processing files in this way or have command line switches or modules to easily implement these wrapper semantics, e.g., Perl and Python).
The rot-13 encoding is commonly known from the early days of Usenet "Netnews" as a way of obfuscating text to prevent casual reading of spoiler or potentially offensive material.
Many news reader and mail user agent programs have built-in rot-13 encoder/decoders or have the ability to feed a message through any external utility script for performing this (or other) actions.
The definition of the rot-13 function is to simply replace every letter of the ASCII alphabet with the letter which is "rotated" 13 characters "around" the 26 letter alphabet from its normal cardinal position (wrapping around from z to a as necessary).
Thus the letters abc become nop and so on.
Technically rot-13 is a "mono-alphabetic substitution cipher" with a trivial "key".
A proper implementation should work on upper and lower case letters, preserve case, and pass all non-alphabetic characters
in the input stream through without alteration.
Related tasks
Caesar cipher
Substitution Cipher
Vigenère Cipher/Cryptanalysis
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
|
#Elixir
|
Elixir
|
defmodule RC do
def rot13(clist) do
f = fn(c) when (?A <= c and c <= ?M) or (?a <= c and c <= ?m) -> c + 13
(c) when (?N <= c and c <= ?Z) or (?n <= c and c <= ?z) -> c - 13
(c) -> c
end
Enum.map(clist, f)
end
end
IO.inspect encode = RC.rot13('Rosetta Code')
IO.inspect RC.rot13(encode)
|
http://rosettacode.org/wiki/Runge-Kutta_method
|
Runge-Kutta method
|
Given the example Differential equation:
y
′
(
t
)
=
t
×
y
(
t
)
{\displaystyle y'(t)=t\times {\sqrt {y(t)}}}
With initial condition:
t
0
=
0
{\displaystyle t_{0}=0}
and
y
0
=
y
(
t
0
)
=
y
(
0
)
=
1
{\displaystyle y_{0}=y(t_{0})=y(0)=1}
This equation has an exact solution:
y
(
t
)
=
1
16
(
t
2
+
4
)
2
{\displaystyle y(t)={\tfrac {1}{16}}(t^{2}+4)^{2}}
Task
Demonstrate the commonly used explicit fourth-order Runge–Kutta method to solve the above differential equation.
Solve the given differential equation over the range
t
=
0
…
10
{\displaystyle t=0\ldots 10}
with a step value of
δ
t
=
0.1
{\displaystyle \delta t=0.1}
(101 total points, the first being given)
Print the calculated values of
y
{\displaystyle y}
at whole numbered
t
{\displaystyle t}
's (
0.0
,
1.0
,
…
10.0
{\displaystyle 0.0,1.0,\ldots 10.0}
) along with error as compared to the exact solution.
Method summary
Starting with a given
y
n
{\displaystyle y_{n}}
and
t
n
{\displaystyle t_{n}}
calculate:
δ
y
1
=
δ
t
×
y
′
(
t
n
,
y
n
)
{\displaystyle \delta y_{1}=\delta t\times y'(t_{n},y_{n})\quad }
δ
y
2
=
δ
t
×
y
′
(
t
n
+
1
2
δ
t
,
y
n
+
1
2
δ
y
1
)
{\displaystyle \delta y_{2}=\delta t\times y'(t_{n}+{\tfrac {1}{2}}\delta t,y_{n}+{\tfrac {1}{2}}\delta y_{1})}
δ
y
3
=
δ
t
×
y
′
(
t
n
+
1
2
δ
t
,
y
n
+
1
2
δ
y
2
)
{\displaystyle \delta y_{3}=\delta t\times y'(t_{n}+{\tfrac {1}{2}}\delta t,y_{n}+{\tfrac {1}{2}}\delta y_{2})}
δ
y
4
=
δ
t
×
y
′
(
t
n
+
δ
t
,
y
n
+
δ
y
3
)
{\displaystyle \delta y_{4}=\delta t\times y'(t_{n}+\delta t,y_{n}+\delta y_{3})\quad }
then:
y
n
+
1
=
y
n
+
1
6
(
δ
y
1
+
2
δ
y
2
+
2
δ
y
3
+
δ
y
4
)
{\displaystyle y_{n+1}=y_{n}+{\tfrac {1}{6}}(\delta y_{1}+2\delta y_{2}+2\delta y_{3}+\delta y_{4})}
t
n
+
1
=
t
n
+
δ
t
{\displaystyle t_{n+1}=t_{n}+\delta t\quad }
|
#Run_BASIC
|
Run BASIC
|
y = 1
while t <= 10
k1 = t * sqr(y)
k2 = (t + .05) * sqr(y + .05 * k1)
k3 = (t + .05) * sqr(y + .05 * k2)
k4 = (t + .1) * sqr(y + .1 * k3)
if right$(using("##.#",t),1) = "0" then print "y(";using("##",t);") ="; using("####.#######", y);chr$(9);"Error ="; (((t^2 + 4)^2) /16) -y
y = y + .1 *(k1 + 2 * (k2 + k3) + k4) / 6
t = t + .1
wend
end
|
http://rosettacode.org/wiki/Runge-Kutta_method
|
Runge-Kutta method
|
Given the example Differential equation:
y
′
(
t
)
=
t
×
y
(
t
)
{\displaystyle y'(t)=t\times {\sqrt {y(t)}}}
With initial condition:
t
0
=
0
{\displaystyle t_{0}=0}
and
y
0
=
y
(
t
0
)
=
y
(
0
)
=
1
{\displaystyle y_{0}=y(t_{0})=y(0)=1}
This equation has an exact solution:
y
(
t
)
=
1
16
(
t
2
+
4
)
2
{\displaystyle y(t)={\tfrac {1}{16}}(t^{2}+4)^{2}}
Task
Demonstrate the commonly used explicit fourth-order Runge–Kutta method to solve the above differential equation.
Solve the given differential equation over the range
t
=
0
…
10
{\displaystyle t=0\ldots 10}
with a step value of
δ
t
=
0.1
{\displaystyle \delta t=0.1}
(101 total points, the first being given)
Print the calculated values of
y
{\displaystyle y}
at whole numbered
t
{\displaystyle t}
's (
0.0
,
1.0
,
…
10.0
{\displaystyle 0.0,1.0,\ldots 10.0}
) along with error as compared to the exact solution.
Method summary
Starting with a given
y
n
{\displaystyle y_{n}}
and
t
n
{\displaystyle t_{n}}
calculate:
δ
y
1
=
δ
t
×
y
′
(
t
n
,
y
n
)
{\displaystyle \delta y_{1}=\delta t\times y'(t_{n},y_{n})\quad }
δ
y
2
=
δ
t
×
y
′
(
t
n
+
1
2
δ
t
,
y
n
+
1
2
δ
y
1
)
{\displaystyle \delta y_{2}=\delta t\times y'(t_{n}+{\tfrac {1}{2}}\delta t,y_{n}+{\tfrac {1}{2}}\delta y_{1})}
δ
y
3
=
δ
t
×
y
′
(
t
n
+
1
2
δ
t
,
y
n
+
1
2
δ
y
2
)
{\displaystyle \delta y_{3}=\delta t\times y'(t_{n}+{\tfrac {1}{2}}\delta t,y_{n}+{\tfrac {1}{2}}\delta y_{2})}
δ
y
4
=
δ
t
×
y
′
(
t
n
+
δ
t
,
y
n
+
δ
y
3
)
{\displaystyle \delta y_{4}=\delta t\times y'(t_{n}+\delta t,y_{n}+\delta y_{3})\quad }
then:
y
n
+
1
=
y
n
+
1
6
(
δ
y
1
+
2
δ
y
2
+
2
δ
y
3
+
δ
y
4
)
{\displaystyle y_{n+1}=y_{n}+{\tfrac {1}{6}}(\delta y_{1}+2\delta y_{2}+2\delta y_{3}+\delta y_{4})}
t
n
+
1
=
t
n
+
δ
t
{\displaystyle t_{n+1}=t_{n}+\delta t\quad }
|
#Rust
|
Rust
|
fn runge_kutta4(fx: &dyn Fn(f64, f64) -> f64, x: f64, y: f64, dx: f64) -> f64 {
let k1 = dx * fx(x, y);
let k2 = dx * fx(x + dx / 2.0, y + k1 / 2.0);
let k3 = dx * fx(x + dx / 2.0, y + k2 / 2.0);
let k4 = dx * fx(x + dx, y + k3);
y + (k1 + 2.0 * k2 + 2.0 * k3 + k4) / 6.0
}
fn f(x: f64, y: f64) -> f64 {
x * y.sqrt()
}
fn actual(x: f64) -> f64 {
(1.0 / 16.0) * (x * x + 4.0).powi(2)
}
fn main() {
let mut y = 1.0;
let mut x = 0.0;
let step = 0.1;
let max_steps = 101;
let sample_every_n = 10;
for steps in 0..max_steps {
if steps % sample_every_n == 0 {
println!("y({}):\t{:.10}\t\t {:E}", x, y, actual(x) - y)
}
y = runge_kutta4(&f, x, y, step);
x = ((x * 10.0) + (step * 10.0)) / 10.0;
}
}
|
http://rosettacode.org/wiki/S-expressions
|
S-expressions
|
S-Expressions are one convenient way to parse and store data.
Task
Write a simple reader and writer for S-Expressions that handles quoted and unquoted strings, integers and floats.
The reader should read a single but nested S-Expression from a string and store it in a suitable datastructure (list, array, etc).
Newlines and other whitespace may be ignored unless contained within a quoted string.
“()” inside quoted strings are not interpreted, but treated as part of the string.
Handling escaped quotes inside a string is optional; thus “(foo"bar)” maybe treated as a string “foo"bar”, or as an error.
For this, the reader need not recognize “\” for escaping, but should, in addition, recognize numbers if the language has appropriate datatypes.
Languages that support it may treat unquoted strings as symbols.
Note that with the exception of “()"” (“\” if escaping is supported) and whitespace there are no special characters. Anything else is allowed without quotes.
The reader should be able to read the following input
((data "quoted data" 123 4.5)
(data (!@# (4.5) "(more" "data)")))
and turn it into a native datastructure. (see the Pike, Python and Ruby implementations for examples of native data structures.)
The writer should be able to take the produced list and turn it into a new S-Expression.
Strings that don't contain whitespace or parentheses () don't need to be quoted in the resulting S-Expression, but as a simplification, any string may be quoted.
Extra Credit
Let the writer produce pretty printed output with indenting and line-breaks.
|
#Wren
|
Wren
|
import "/pattern" for Pattern
import "/fmt" for Fmt
var INDENT = 2
var parseSExpr = Fn.new { |str|
var ipat = " \t\n\f\v\r()\""
var p = Pattern.new("""+0/s["+0^""|(|)|"|+1/I]""", Pattern.within, ipat)
var t = p.findAll(str).map { |m| m.text }.toList
if (t.count == 0) return null
var o = false
var c = 0
for (i in t.count-1..0) {
var ti = t[i].trim()
var nd = Num.fromString(ti)
if (ti == "\"") return null
if (ti == "(") {
t[i] = "["
c = c + 1
} else if (ti == ")") {
t[i] = "]"
c = c - 1
} else if (nd) {
var ni = Num.fromString(ti)
t[i] = ni ? ni.toString : nd.toString
} else if (ti.startsWith("\"")) { // escape embedded double quotes
var temp = ti[1...-1]
t[i] = "\"" + temp.replace("\"", "\\\"") + "\""
}
if (i > 0 && t[i] != "]" && t[i - 1].trim() != "(") t.insert(i, ", ")
if (c == 0) {
if (!o) o = true else return null
}
}
return (c != 0) ? null : t
}
var toSExpr = Fn.new { |tokens|
for (i in 0...tokens.count) {
if (tokens[i] == "[") {
tokens[i] = "("
} else if (tokens[i] == "]") {
tokens[i] = ")"
} else if (tokens[i] == ", ") {
tokens[i] = " "
} else if (tokens[i].startsWith("\"")) { // unescape embedded quotes
var temp = tokens[i][1...-1]
tokens[i] = "\"" + temp.replace("\\\"", "\"") + "\""
}
}
return tokens.join()
}
var prettyPrint = Fn.new { |tokens|
var level = 0
for (t in tokens) {
var n
if (t == ", " || t == " ") {
continue
} else if (t == "[" || t == "(") {
n = level * INDENT + 1
level = level + 1
} else if (t == "]" || t == ")") {
level = level - 1
n = level * INDENT + 1
} else {
n = level * INDENT + t.count
}
Fmt.print("$*s", n, t)
}
}
var str = """((data "quoted data" 123 4.5)""" + "\n" +
""" (data (!@# (4.5) "(more" "data)")))"""
var tokens = parseSExpr.call(str)
if (!tokens) {
System.print("Invalid s-expr!")
} else {
System.print("Native data structure:")
System.print(tokens.join())
System.print("\nNative data structure (pretty print):")
prettyPrint.call(tokens)
System.print("\nRecovered S-Expression:")
System.print(toSExpr.call(tokens))
System.print("\nRecovered S-Expression (pretty print):")
prettyPrint.call(tokens)
}
|
http://rosettacode.org/wiki/RPG_attributes_generator
|
RPG attributes generator
|
RPG = Role Playing Game.
You're running a tabletop RPG, and your players are creating characters.
Each character has six core attributes: strength, dexterity, constitution, intelligence, wisdom, and charisma.
One way of generating values for these attributes is to roll four, 6-sided dice (d6) and sum the three highest rolls, discarding the lowest roll.
Some players like to assign values to their attributes in the order they're rolled.
To ensure generated characters don't put players at a disadvantage, the following requirements must be satisfied:
The total of all character attributes must be at least 75.
At least two of the attributes must be at least 15.
However, this can require a lot of manual dice rolling. A programatic solution would be much faster.
Task
Write a program that:
Generates 4 random, whole values between 1 and 6.
Saves the sum of the 3 largest values.
Generates a total of 6 values this way.
Displays the total, and all 6 values once finished.
The order in which each value was generated must be preserved.
The total of all 6 values must be at least 75.
At least 2 of the values must be 15 or more.
|
#Seed7
|
Seed7
|
$ include "seed7_05.s7i";
const proc: main is func
local
var integer: count is 0;
var integer: total is 0;
var integer: attribIdx is 0;
var integer: diceroll is 0;
var integer: sumOfRolls is 0;
var array integer: attribute is 6 times 0;
var array integer: dice is 4 times 0;
begin
repeat
count := 0;
total := 0;
for key attribIdx range attribute do
for key diceroll range dice do
dice[diceroll] := rand(1, 6);
end for;
dice := sort(dice);
sumOfRolls := 0;
for diceroll range 2 to maxIdx(dice) do # Discard the lowest roll
sumOfRolls +:= dice[diceroll];
end for;
attribute[attribIdx] := sumOfRolls;
total +:= sumOfRolls;
if sumOfRolls >= 15 then
incr(count);
end if;
end for;
until total >= 75 and count >= 2;
writeln("Attributes:");
for key attribIdx range attribute do
writeln(attribIdx <& " ..... " <& attribute[attribIdx] lpad 2);
end for;
writeln(" ----");
writeln("Total " <& total lpad 3);
end func;
|
http://rosettacode.org/wiki/RPG_attributes_generator
|
RPG attributes generator
|
RPG = Role Playing Game.
You're running a tabletop RPG, and your players are creating characters.
Each character has six core attributes: strength, dexterity, constitution, intelligence, wisdom, and charisma.
One way of generating values for these attributes is to roll four, 6-sided dice (d6) and sum the three highest rolls, discarding the lowest roll.
Some players like to assign values to their attributes in the order they're rolled.
To ensure generated characters don't put players at a disadvantage, the following requirements must be satisfied:
The total of all character attributes must be at least 75.
At least two of the attributes must be at least 15.
However, this can require a lot of manual dice rolling. A programatic solution would be much faster.
Task
Write a program that:
Generates 4 random, whole values between 1 and 6.
Saves the sum of the 3 largest values.
Generates a total of 6 values this way.
Displays the total, and all 6 values once finished.
The order in which each value was generated must be preserved.
The total of all 6 values must be at least 75.
At least 2 of the values must be 15 or more.
|
#True_BASIC
|
True BASIC
|
FUNCTION min(a, b)
IF a < b THEN LET min = a ELSE LET min = b
END FUNCTION
FUNCTION d6
LET d6 = 1 + INT(RND * 6)
END FUNCTION
FUNCTION rollstat
LET a = d6
LET b = d6
LET c = d6
LET d = d6
LET rollstat = a + b + c + d - min(min(a, b), min(c, d))
END FUNCTION
DIM statnames$(6)
DATA "STR", "CON", "DEX", "INT", "WIS", "CHA"
FOR i = 1 TO 6
READ statnames$(i)
NEXT i
DIM stat(6)
LET acceptable = 0
RANDOMIZE ! RANDOMIZE TIMER en QBasic
DO
LET sum = 0
LET n15 = 0
FOR i = 1 TO 6
LET stat(i) = rollstat
LET sum = sum + stat(i)
IF stat(i) >= 15 THEN LET n15 = n15 + 1
NEXT i
IF sum >= 75 AND n15 >= 2 THEN LET acceptable = 1
LOOP UNTIL acceptable = 1
FOR i = 1 TO 6
PRINT statnames$(i); ": "; stat(i)
NEXT i
PRINT "--------"
PRINT "TOT: "; sum
END
|
http://rosettacode.org/wiki/Sieve_of_Eratosthenes
|
Sieve of Eratosthenes
|
This task has been clarified. Its programming examples are in need of review to ensure that they still fit the requirements of the task.
The Sieve of Eratosthenes is a simple algorithm that finds the prime numbers up to a given integer.
Task
Implement the Sieve of Eratosthenes algorithm, with the only allowed optimization that the outer loop can stop at the square root of the limit, and the inner loop may start at the square of the prime just found.
That means especially that you shouldn't optimize by using pre-computed wheels, i.e. don't assume you need only to cross out odd numbers (wheel based on 2), numbers equal to 1 or 5 modulo 6 (wheel based on 2 and 3), or similar wheels based on low primes.
If there's an easy way to add such a wheel based optimization, implement it as an alternative version.
Note
It is important that the sieve algorithm be the actual algorithm used to find prime numbers for the task.
Related tasks
Emirp primes
count in factors
prime decomposition
factors of an integer
extensible prime generator
primality by trial division
factors of a Mersenne number
trial factoring of a Mersenne number
partition an integer X into N primes
sequence of primes by Trial Division
|
#VAX_Assembly
|
VAX Assembly
|
000F4240 0000 1 n=1000*1000
0000 0000 2 .entry main,0
7E 7CFD 0002 3 clro -(sp) ;result buffer
5E DD 0005 4 pushl sp ;pointer to buffer
10 DD 0007 5 pushl #16 ;descriptor -> len of buffer
0009 6
02 DD 0009 7 pushl #2 ;1st candidate
000B 8 test:
09 46'AF 6E E1 000B 9 bbc (sp), b^bits, found ;bc - bit clear
0010 10 next:
F3 6E 000F4240 8F F2 0010 11 aoblss #n, (sp), test ;+1: limit,index
04 0018 12 ret
0019 13 found:
04 AE 7F 0019 14 pushaq 4(sp) ;-> descriptor by ref
04 AE DF 001C 15 pushal 4(sp) ;-> prime on stack by ref
00000000'GF 02 FB 001F 16 calls #2, g^ots$cvt_l_ti ;convert integer to string
04 AE 7F 0026 17 pushaq 4(sp) ;
00000000'GF 01 FB 0029 18 calls #1, g^lib$put_output ;show result
0030 19
53 6E D0 0030 20 movl (sp), r3
0033 21 mult:
0002 53 6E 000F4240 8F F1 0033 22 acbl #n, (sp), r3, set_mult ;limit,add,index
D1 11 003D 23 brb next
003F 24 set_mult: ;set bits for multiples
EF 46'AF 53 E2 003F 25 bbss r3, b^bits, mult ;branch on bit set & set
ED 11 0044 26 brb mult
0046 27
0001E892 0046 28 bits: .blkl <n+2+31>/32
E892 29 .end main
|
http://rosettacode.org/wiki/Rosetta_Code/Count_examples
|
Rosetta Code/Count examples
|
task
Essentially, count the number of occurrences of =={{header| on each task page.
Output:
100 doors: 20 examples.
99 Bottles of Beer: 29 examples.
Abstract type: 10 examples.
Total: X examples.
For a full output, updated periodically, see Rosetta Code/Count examples/Full list.
You'll need to use the Media Wiki API, which you can find out about locally, here, or in Media Wiki's API documentation at, API:Query
|
#zkl
|
zkl
|
var [const] YAJL=Import("zklYAJL")[0], CURL=Import("zklCurl");
fcn getTasks(language){
continueValue,tasks:="",Data(0,String); // "nm\0nm\0...."
do{
page:=CURL().get(("http://rosettacode.org/mw/api.php?"
"action=query&cmlimit=500"
"&format=json"
"&list=categorymembers"
"&cmtitle=Category:%s"
"&cmcontinue=%s").fmt(language,continueValue));
page=page[0].del(0,page[1]); // get rid of HTML header
json:=YAJL().write(page).close();
json["query"]["categorymembers"].pump(tasks,T("get","title"));
continueValue=json.find("continue") //continue:-||,cmcontinue:page|954|19)
.toList().apply("concat","=").concat("&");
}while(continueValue);
tasks
}
re:=RegExp(0'!\s+==\s*{{\s*header\s*\|!); // == {{ header | zkl
foreach task in (getTasks("Programming_Tasks")){
page:=CURL().get(
"http://www.rosettacode.org/mw/index.php?title=%s&action=raw"
.fmt(CURL.urlEncode(task)));
page=page[0].del(0,page[1]); // get rid of HTML header
cnt,n:=0,0; while(re.search(page,True,n)){ cnt+=1; n=re.matched[0].sum(0); }
"%4d: %s".fmt(cnt,task).println();
}
|
http://rosettacode.org/wiki/Search_a_list
|
Search a list
|
Task[edit]
Find the index of a string (needle) in an indexable, ordered collection of strings (haystack).
Raise an exception if the needle is missing.
If there is more than one occurrence then return the smallest index to the needle.
Extra credit
Return the largest index to a needle that has multiple occurrences in the haystack.
See also
Search a list of records
|
#Oforth
|
Oforth
|
: needleIndex(needle, haystack)
haystack indexOf(needle) dup ifNull: [ drop ExRuntime throw("Not found", needle) ] ;
[ "Zig", "Zag", "Wally", "Ronald", "Bush", "Krusty", "Charlie", "Bush", "Boz" ] const: Haystack
needleIndex("Bush", Haystack) println
Haystack lastIndexOf("Bush") println
needleIndex("Washington", Haystack) println
|
http://rosettacode.org/wiki/Rosetta_Code/Rank_languages_by_popularity
|
Rosetta Code/Rank languages by popularity
|
Rosetta Code/Rank languages by popularity
You are encouraged to solve this task according to the task description, using any language you may know.
Task
Sort the most popular computer programming languages based in number of members in Rosetta Code categories.
Sample output on 01 juin 2022 at 14:13 +02
Rank: 1 (1,540 entries) Phix
Rank: 2 (1,531 entries) Wren
Rank: 3 (1,507 entries) Julia
Rank: 4 (1,494 entries) Go
Rank: 5 (1,488 entries) Raku
Rank: 6 (1,448 entries) Perl
Rank: 7 (1,402 entries) Nim
Rank: 8 (1,382 entries) Python
Rank: 9 (1,204 entries) C
Rank: 10 (1,152 entries) REXX
...
Notes
Each language typically demonstrates one or two methods of accessing the data:
with web scraping (via http://www.rosettacode.org/mw/index.php?title=Special:Categories&limit=5000)
with the API method (examples below for Awk, Perl, Ruby, Tcl, etc).
The scraping and API solutions can be separate subsections, see the Tcl example.
Filtering wrong results is optional. You can check against Special:MostLinkedCategories (if using web scraping)
If you use the API, and do elect to filter, you may check your results against this complete, accurate, sortable, wikitable listing of all 869 programming languages, updated periodically, typically weekly.
A complete ranked listing of all 813 languages (from the REXX example) is included here ──► output from the REXX program.
|
#Raku
|
Raku
|
use HTTP::UserAgent;
use URI::Escape;
use JSON::Fast;
use Sort::Naturally;
my $client = HTTP::UserAgent.new;
my $url = 'http://rosettacode.org/mw';
my $tablefile = './RC_Popularity.txt';
my %cat = (
'Programming_Tasks' => 'Task',
'Draft_Programming_Tasks' => 'Draft'
);
my %tasks;
for %cat.keys.sort -> $cat {
mediawiki-query(
$url, 'pages',
:generator<categorymembers>,
:gcmtitle("Category:$cat"),
:gcmlimit<350>,
:rawcontinue(),
:prop<title>
).map({ %tasks{%cat{$cat}}++ });
}
my %counts =
mediawiki-query(
$url, 'pages',
:generator<categorymembers>,
:gcmtitle<Category:Programming Languages>,
:gcmlimit<350>,
:rawcontinue(),
:prop<categoryinfo>
)
.map({
my $title = .<title>.subst(/^'Category:'/, '');
my $tasks = (.<categoryinfo><pages> || 0);
my $categories = (.<categoryinfo><subcats> || 0);
my $total = (.<categoryinfo><size> || 0);
$title => [$tasks ,$categories, $total]
});
my $out = open($tablefile, :w) or die "$!\n";
# Add table boilerplate and header
$out.say:
"\{|class=\"wikitable sortable\"\n",
"|+ As of { Date.today } :: {+%counts} Languages\n",
'!Rank!!Language!!Task<br>Entries!!Tasks<br>done %!!Non-task<br>Subcate-<br>gories!!Total<br>Categories'
;
my @bg = <#fff; #ccc;>;
my $ff = 0;
my $rank = 1;
my $ties = 0;
# Get sorted unique task counts
for %counts.values»[0].unique.sort: -* -> $count {
$ff++;
# Get list of tasks with this count
my @these = %counts.grep( *.value[0] == $count )».keys.sort: *.&naturally;
for @these {
$ties++;
$out.say:
"|- style=\"background-color: { @bg[$ff % 2] }\"\n"~
"|$rank\n"~
"|[[:Category:$_|]]\n"~
"|$count\n"~
"|{(100 * $count/%tasks<Draft Task>.sum).round(.01)} %\n"~
"|{%counts{$_}[1]}\n"~
"|{%counts{$_}[2]}"
}
$rank += $ties;
$ties = 0;
}
$out.say( "|}" );
$out.say('=' x 5, " query, download & processing: {(now - INIT now).round(.01)} seconds ", '=' x 5);
$out.close;
sub mediawiki-query ($site, $type, *%query) {
my $url = "$site/api.php?" ~ uri-query-string(
:action<query>, :format<json>, :formatversion<2>, |%query);
my $continue = '';
gather loop {
my $response = $client.get("$url&$continue");
my $data = from-json($response.content);
take $_ for $data.<query>.{$type}.values;
$continue = uri-query-string |($data.<query-continue>{*}».hash.hash or last);
}
}
sub uri-query-string (*%fields) {
join '&', %fields.map: { "{.key}={uri-escape .value}" }
}
|
http://rosettacode.org/wiki/Roman_numerals/Encode
|
Roman numerals/Encode
|
Task
Create a function taking a positive integer as its parameter and returning a string containing the Roman numeral representation of that integer. Modern Roman numerals are written by expressing each digit separately, starting with the left most digit and skipping any digit with a value of zero.
In Roman numerals:
1990 is rendered: 1000=M, 900=CM, 90=XC; resulting in MCMXC
2008 is written as 2000=MM, 8=VIII; or MMVIII
1666 uses each Roman symbol in descending order: MDCLXVI
|
#Befunge
|
Befunge
|
&>0\0>00p:#v_$ >:#,_ $ @
4-v >5+#:/#<\55+%:5/\5%:
vv_$9+00g+5g\00g8+>5g\00
g>\20p>:10p00g \#v _20gv
> 2+ v^-1g01\g5+8<^ +9 _
IVXLCDM
|
http://rosettacode.org/wiki/Roman_numerals/Decode
|
Roman numerals/Decode
|
Task
Create a function that takes a Roman numeral as its argument and returns its value as a numeric decimal integer.
You don't need to validate the form of the Roman numeral.
Modern Roman numerals are written by expressing each decimal digit of the number to be encoded separately,
starting with the leftmost decimal digit and skipping any 0s (zeroes).
1990 is rendered as MCMXC (1000 = M, 900 = CM, 90 = XC) and
2008 is rendered as MMVIII (2000 = MM, 8 = VIII).
The Roman numeral for 1666, MDCLXVI, uses each letter in descending order.
|
#C.23
|
C#
|
using System;
using System.Collections.Generic;
namespace Roman
{
internal class Program
{
private static void Main(string[] args)
{
// Decode and print the numerals.
Console.WriteLine("{0}: {1}", "MCMXC", Decode("MCMXC"));
Console.WriteLine("{0}: {1}", "MMVIII", Decode("MMVIII"));
Console.WriteLine("{0}: {1}", "MDCLXVI", Decode("MDCLXVI"));
}
// Dictionary to hold our numerals and their values.
private static readonly Dictionary<char, int> RomanDictionary = new Dictionary<char, int>
{
{'I', 1},
{'V', 5},
{'X', 10},
{'L', 50},
{'C', 100},
{'D', 500},
{'M', 1000}
};
private static int Decode(string roman)
{
/* Make the input string upper-case,
* because the dictionary doesn't support lower-case characters. */
roman = roman.ToUpper();
/* total = the current total value that will be returned.
* minus = value to subtract from next numeral. */
int total = 0, minus = 0;
for (int i = 0; i < roman.Length; i++) // Iterate through characters.
{
// Get the value for the current numeral. Takes subtraction into account.
int thisNumeral = RomanDictionary[roman[i]] - minus;
/* Checks if this is the last character in the string, or if the current numeral
* is greater than or equal to the next numeral. If so, we will reset our minus
* variable and add the current numeral to the total value. Otherwise, we will
* subtract the current numeral from the next numeral, and continue. */
if (i >= roman.Length - 1 ||
thisNumeral + minus >= RomanDictionary[roman[i + 1]])
{
total += thisNumeral;
minus = 0;
}
else
{
minus = thisNumeral;
}
}
return total; // Return the total.
}
}
}
|
http://rosettacode.org/wiki/Roots_of_a_function
|
Roots of a function
|
Task
Create a program that finds and outputs the roots of a given function, range and (if applicable) step width.
The program should identify whether the root is exact or approximate.
For this task, use: ƒ(x) = x3 - 3x2 + 2x
|
#HicEst
|
HicEst
|
OPEN(FIle='test.txt')
1 DLG(NameEdit=x0, DNum=3)
x = x0
chi2 = SOLVE(NUL=x^3 - 3*x^2 + 2*x, Unknown=x, I=iterations, NumDiff=1E-15)
EDIT(Text='approximate exact ', Word=(chi2 == 0), Parse=solution)
WRITE(FIle='test.txt', LENgth=6, Name) x0, x, solution, chi2, iterations
GOTO 1
|
http://rosettacode.org/wiki/Rock-paper-scissors
|
Rock-paper-scissors
|
Task
Implement the classic children's game Rock-paper-scissors, as well as a simple predictive AI (artificial intelligence) player.
Rock Paper Scissors is a two player game.
Each player chooses one of rock, paper or scissors, without knowing the other player's choice.
The winner is decided by a set of rules:
Rock beats scissors
Scissors beat paper
Paper beats rock
If both players choose the same thing, there is no winner for that round.
For this task, the computer will be one of the players.
The operator will select Rock, Paper or Scissors and the computer will keep a record of the choice frequency, and use that information to make a weighted random choice in an attempt to defeat its opponent.
Extra credit
Support additional choices additional weapons.
|
#Factor
|
Factor
|
USING: combinators formatting io kernel math math.ranges qw
random sequences ;
IN: rosetta-code.rock-paper-scissors
CONSTANT: thing qw{ rock paper scissors }
CONSTANT: msg { "I win." "Tie!" "You win." }
: ai-choice ( r p s -- n )
3dup + + nip [1,b] random {
{ [ 3dup nip >= ] [ 3drop 1 ] }
{ [ 3dup [ + ] dip >= ] [ 3drop 2 ] }
[ 3drop 0 ]
} cond ;
: player-choice ( -- n )
"r/p/s/q? " write readln qw{ r p s q } index dup
[ drop player-choice ] unless ;
! return:
! -1 for n1 loses to n2.
! 0 for n1 ties n2.
! 1 for n1 beats n2.
: outcome ( n1 n2 -- n3 ) - dup abs 1 > [ neg ] when sgn ;
: status. ( seq -- )
"My wins: %d Ties: %d Your wins: %d\n" vprintf ;
: choices. ( n1 n2 -- )
[ thing nth ] bi@ "You chose: %s\nI chose: %s\n" printf ;
: tally ( seq n -- seq' ) over [ 1 + ] change-nth ;
: game ( seq -- seq' )
dup status. player-choice dup 3 = [ drop ] [
[ 3 + tally ] keep over 3 tail* first3 ai-choice 2dup
choices. outcome 1 + dup [ tally ] dip msg nth print nl
game
] if ;
! The game state is a sequence where the members are:
! losses, ties, wins, #rock, #paper, #scissors
: main ( -- ) { 0 0 0 1 1 1 } clone game drop ;
MAIN: main
|
http://rosettacode.org/wiki/Run-length_encoding
|
Run-length encoding
|
Run-length encoding
You are encouraged to solve this task according to the task description, using any language you may know.
Task
Given a string containing uppercase characters (A-Z), compress repeated 'runs' of the same character by storing the length of that run, and provide a function to reverse the compression.
The output can be anything, as long as you can recreate the input with it.
Example
Input: WWWWWWWWWWWWBWWWWWWWWWWWWBBBWWWWWWWWWWWWWWWWWWWWWWWWBWWWWWWWWWWWWWW
Output: 12W1B12W3B24W1B14W
Note: the encoding step in the above example is the same as a step of the Look-and-say sequence.
|
#FreeBASIC
|
FreeBASIC
|
Dim As String initial, encoded, decoded
Function RLDecode(i As String) As String
Dim As Long Loop0
dim as string rCount, outP, m
For Loop0 = 1 To Len(i)
m = Mid(i, Loop0, 1)
Select Case m
Case "0" To "9"
rCount += m
Case Else
If Len(rCount) Then
outP += String(Val(rCount), m)
rCount = ""
Else
outP += m
End If
End Select
Next
RLDecode = outP
End Function
Function RLEncode(i As String) As String
Dim As String tmp1, tmp2, outP
Dim As Long Loop0, rCount
tmp1 = Mid(i, 1, 1)
tmp2 = tmp1
rCount = 1
For Loop0 = 2 To Len(i)
tmp1 = Mid(i, Loop0, 1)
If tmp1 <> tmp2 Then
outP += Ltrim(Rtrim(Str(rCount))) + tmp2
tmp2 = tmp1
rCount = 1
Else
rCount += 1
End If
Next
outP += Ltrim(Rtrim(Str(rCount)))
outP += tmp2
RLEncode = outP
End Function
Input "Type something: ", initial
encoded = RLEncode(initial)
decoded = RLDecode(encoded)
Print initial
Print encoded
Print decoded
End
|
http://rosettacode.org/wiki/Roots_of_unity
|
Roots of unity
|
The purpose of this task is to explore working with complex numbers.
Task
Given n, find the nth roots of unity.
|
#PicoLisp
|
PicoLisp
|
(load "@lib/math.l")
(for N (range 2 10)
(let Angle 0.0
(prin N ": ")
(for I N
(let Ipart (sin Angle)
(prin
(round (cos Angle) 4)
(if (lt0 Ipart) "-" "+")
"j"
(round (abs Ipart) 4)
" " ) )
(inc 'Angle (*/ 2 pi N)) )
(prinl) ) )
|
http://rosettacode.org/wiki/Roots_of_unity
|
Roots of unity
|
The purpose of this task is to explore working with complex numbers.
Task
Given n, find the nth roots of unity.
|
#PL.2FI
|
PL/I
|
complex_roots:
procedure (N);
declare N fixed binary nonassignable;
declare x float, c fixed decimal (10,8) complex;
declare twopi float initial ((4*asin(1.0)));
do x = 0 to twopi by twopi/N;
c = complex(cos(x), sin(x));
put skip list (c);
end;
end complex_roots;
1.00000000+0.00000000I
0.80901700+0.58778524I
0.30901697+0.95105654I
-0.30901703+0.95105648I
-0.80901706+0.58778518I
-1.00000000-0.00000008I
-0.80901694-0.58778536I
-0.30901709-0.95105648I
0.30901712-0.95105648I
0.80901724-0.58778494I
|
http://rosettacode.org/wiki/Roots_of_unity
|
Roots of unity
|
The purpose of this task is to explore working with complex numbers.
Task
Given n, find the nth roots of unity.
|
#Prolog
|
Prolog
|
roots(N, Rs) :-
succ(Pn, N), numlist(0, Pn, Ks),
maplist(root(N), Ks, Rs).
root(N, M, R2) :-
R0 is (2*M) rdiv N, % multiple of PI
(R0 > 1 -> R1 is R0 - 2; R1 = R0), % adjust for principal values
cis(R1, R2).
cis(0, 1) :- !.
cis(1, -1) :- !.
cis(1 rdiv 2, i) :- !.
cis(-1 rdiv 2, -i) :- !.
cis(-1 rdiv Q, exp(-i*pi/Q)) :- !.
cis(1 rdiv Q, exp(i*pi/Q)) :- !.
cis(P rdiv Q, exp(P*i*pi/Q)).
|
http://rosettacode.org/wiki/Roots_of_a_quadratic_function
|
Roots of a quadratic function
|
This task has been clarified. Its programming examples are in need of review to ensure that they still fit the requirements of the task.
Write a program to find the roots of a quadratic equation, i.e., solve the equation
a
x
2
+
b
x
+
c
=
0
{\displaystyle ax^{2}+bx+c=0}
.
Your program must correctly handle non-real roots, but it need not check that
a
≠
0
{\displaystyle a\neq 0}
.
The problem of solving a quadratic equation is a good example of how dangerous it can be to ignore the peculiarities of floating-point arithmetic.
The obvious way to implement the quadratic formula suffers catastrophic loss of accuracy when one of the roots to be found is much closer to 0 than the other.
In their classic textbook on numeric methods Computer Methods for Mathematical Computations, George Forsythe, Michael Malcolm, and Cleve Moler suggest trying the naive algorithm with
a
=
1
{\displaystyle a=1}
,
b
=
−
10
5
{\displaystyle b=-10^{5}}
, and
c
=
1
{\displaystyle c=1}
.
(For double-precision floats, set
b
=
−
10
9
{\displaystyle b=-10^{9}}
.)
Consider the following implementation in Ada:
with Ada.Text_IO; use Ada.Text_IO;
with Ada.Numerics.Elementary_Functions; use Ada.Numerics.Elementary_Functions;
procedure Quadratic_Equation is
type Roots is array (1..2) of Float;
function Solve (A, B, C : Float) return Roots is
SD : constant Float := sqrt (B**2 - 4.0 * A * C);
AA : constant Float := 2.0 * A;
begin
return ((- B + SD) / AA, (- B - SD) / AA);
end Solve;
R : constant Roots := Solve (1.0, -10.0E5, 1.0);
begin
Put_Line ("X1 =" & Float'Image (R (1)) & " X2 =" & Float'Image (R (2)));
end Quadratic_Equation;
Output:
X1 = 1.00000E+06 X2 = 0.00000E+00
As we can see, the second root has lost all significant figures. The right answer is that X2 is about
10
−
6
{\displaystyle 10^{-6}}
. The naive method is numerically unstable.
Suggested by Middlebrook (D-OA), a better numerical method: to define two parameters
q
=
a
c
/
b
{\displaystyle q={\sqrt {ac}}/b}
and
f
=
1
/
2
+
1
−
4
q
2
/
2
{\displaystyle f=1/2+{\sqrt {1-4q^{2}}}/2}
and the two roots of the quardratic are:
−
b
a
f
{\displaystyle {\frac {-b}{a}}f}
and
−
c
b
f
{\displaystyle {\frac {-c}{bf}}}
Task: do it better. This means that given
a
=
1
{\displaystyle a=1}
,
b
=
−
10
9
{\displaystyle b=-10^{9}}
, and
c
=
1
{\displaystyle c=1}
, both of the roots your program returns should be greater than
10
−
11
{\displaystyle 10^{-11}}
. Or, if your language can't do floating-point arithmetic any more precisely than single precision, your program should be able to handle
b
=
−
10
6
{\displaystyle b=-10^{6}}
. Either way, show what your program gives as the roots of the quadratic in question. See page 9 of
"What Every Scientist Should Know About Floating-Point Arithmetic" for a possible algorithm.
|
#PicoLisp
|
PicoLisp
|
(scl 40)
(de solveQuad (A B C)
(let SD (sqrt (- (* B B) (* 4 A C)))
(if (lt0 B)
(list
(*/ (- SD B) A 2.0)
(*/ C 2.0 (*/ A A (- SD B) `(* 1.0 1.0))) )
(list
(*/ C 2.0 (*/ A A (- 0 B SD) `(* 1.0 1.0)))
(*/ (- 0 B SD) A 2.0) ) ) ) )
(mapcar round
(solveQuad 1.0 -1000000.0 1.0)
(6 .) )
|
http://rosettacode.org/wiki/Roots_of_a_quadratic_function
|
Roots of a quadratic function
|
This task has been clarified. Its programming examples are in need of review to ensure that they still fit the requirements of the task.
Write a program to find the roots of a quadratic equation, i.e., solve the equation
a
x
2
+
b
x
+
c
=
0
{\displaystyle ax^{2}+bx+c=0}
.
Your program must correctly handle non-real roots, but it need not check that
a
≠
0
{\displaystyle a\neq 0}
.
The problem of solving a quadratic equation is a good example of how dangerous it can be to ignore the peculiarities of floating-point arithmetic.
The obvious way to implement the quadratic formula suffers catastrophic loss of accuracy when one of the roots to be found is much closer to 0 than the other.
In their classic textbook on numeric methods Computer Methods for Mathematical Computations, George Forsythe, Michael Malcolm, and Cleve Moler suggest trying the naive algorithm with
a
=
1
{\displaystyle a=1}
,
b
=
−
10
5
{\displaystyle b=-10^{5}}
, and
c
=
1
{\displaystyle c=1}
.
(For double-precision floats, set
b
=
−
10
9
{\displaystyle b=-10^{9}}
.)
Consider the following implementation in Ada:
with Ada.Text_IO; use Ada.Text_IO;
with Ada.Numerics.Elementary_Functions; use Ada.Numerics.Elementary_Functions;
procedure Quadratic_Equation is
type Roots is array (1..2) of Float;
function Solve (A, B, C : Float) return Roots is
SD : constant Float := sqrt (B**2 - 4.0 * A * C);
AA : constant Float := 2.0 * A;
begin
return ((- B + SD) / AA, (- B - SD) / AA);
end Solve;
R : constant Roots := Solve (1.0, -10.0E5, 1.0);
begin
Put_Line ("X1 =" & Float'Image (R (1)) & " X2 =" & Float'Image (R (2)));
end Quadratic_Equation;
Output:
X1 = 1.00000E+06 X2 = 0.00000E+00
As we can see, the second root has lost all significant figures. The right answer is that X2 is about
10
−
6
{\displaystyle 10^{-6}}
. The naive method is numerically unstable.
Suggested by Middlebrook (D-OA), a better numerical method: to define two parameters
q
=
a
c
/
b
{\displaystyle q={\sqrt {ac}}/b}
and
f
=
1
/
2
+
1
−
4
q
2
/
2
{\displaystyle f=1/2+{\sqrt {1-4q^{2}}}/2}
and the two roots of the quardratic are:
−
b
a
f
{\displaystyle {\frac {-b}{a}}f}
and
−
c
b
f
{\displaystyle {\frac {-c}{bf}}}
Task: do it better. This means that given
a
=
1
{\displaystyle a=1}
,
b
=
−
10
9
{\displaystyle b=-10^{9}}
, and
c
=
1
{\displaystyle c=1}
, both of the roots your program returns should be greater than
10
−
11
{\displaystyle 10^{-11}}
. Or, if your language can't do floating-point arithmetic any more precisely than single precision, your program should be able to handle
b
=
−
10
6
{\displaystyle b=-10^{6}}
. Either way, show what your program gives as the roots of the quadratic in question. See page 9 of
"What Every Scientist Should Know About Floating-Point Arithmetic" for a possible algorithm.
|
#PL.2FI
|
PL/I
|
declare (c1, c2) float complex,
(a, b, c, x1, x2) float;
get list (a, b, c);
if b**2 < 4*a*c then
do;
c1 = (-b + sqrt(b**2 - 4+0i*a*c)) / (2*a);
c2 = (-b - sqrt(b**2 - 4+0i*a*c)) / (2*a);
put data (c1, c2);
end;
else
do;
x1 = (-b + sqrt(b**2 - 4*a*c)) / (2*a);
x2 = (-b - sqrt(b**2 - 4*a*c)) / (2*a);
put data (x1, x2);
end;
|
http://rosettacode.org/wiki/Rot-13
|
Rot-13
|
Task
Implement a rot-13 function (or procedure, class, subroutine, or other "callable" object as appropriate to your programming environment).
Optionally wrap this function in a utility program (like tr, which acts like a common UNIX utility, performing a line-by-line rot-13 encoding of every line of input contained in each file listed on its command line, or (if no filenames are passed thereon) acting as a filter on its "standard input."
(A number of UNIX scripting languages and utilities, such as awk and sed either default to processing files in this way or have command line switches or modules to easily implement these wrapper semantics, e.g., Perl and Python).
The rot-13 encoding is commonly known from the early days of Usenet "Netnews" as a way of obfuscating text to prevent casual reading of spoiler or potentially offensive material.
Many news reader and mail user agent programs have built-in rot-13 encoder/decoders or have the ability to feed a message through any external utility script for performing this (or other) actions.
The definition of the rot-13 function is to simply replace every letter of the ASCII alphabet with the letter which is "rotated" 13 characters "around" the 26 letter alphabet from its normal cardinal position (wrapping around from z to a as necessary).
Thus the letters abc become nop and so on.
Technically rot-13 is a "mono-alphabetic substitution cipher" with a trivial "key".
A proper implementation should work on upper and lower case letters, preserve case, and pass all non-alphabetic characters
in the input stream through without alteration.
Related tasks
Caesar cipher
Substitution Cipher
Vigenère Cipher/Cryptanalysis
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
|
#Emacs_Lisp
|
Emacs Lisp
|
(rot13-string "abc") ;=> "nop"
(with-temp-buffer
(insert "abc")
(rot13-region (point-min) (point-max))
(buffer-string)) ;=> "nop"
|
http://rosettacode.org/wiki/Runge-Kutta_method
|
Runge-Kutta method
|
Given the example Differential equation:
y
′
(
t
)
=
t
×
y
(
t
)
{\displaystyle y'(t)=t\times {\sqrt {y(t)}}}
With initial condition:
t
0
=
0
{\displaystyle t_{0}=0}
and
y
0
=
y
(
t
0
)
=
y
(
0
)
=
1
{\displaystyle y_{0}=y(t_{0})=y(0)=1}
This equation has an exact solution:
y
(
t
)
=
1
16
(
t
2
+
4
)
2
{\displaystyle y(t)={\tfrac {1}{16}}(t^{2}+4)^{2}}
Task
Demonstrate the commonly used explicit fourth-order Runge–Kutta method to solve the above differential equation.
Solve the given differential equation over the range
t
=
0
…
10
{\displaystyle t=0\ldots 10}
with a step value of
δ
t
=
0.1
{\displaystyle \delta t=0.1}
(101 total points, the first being given)
Print the calculated values of
y
{\displaystyle y}
at whole numbered
t
{\displaystyle t}
's (
0.0
,
1.0
,
…
10.0
{\displaystyle 0.0,1.0,\ldots 10.0}
) along with error as compared to the exact solution.
Method summary
Starting with a given
y
n
{\displaystyle y_{n}}
and
t
n
{\displaystyle t_{n}}
calculate:
δ
y
1
=
δ
t
×
y
′
(
t
n
,
y
n
)
{\displaystyle \delta y_{1}=\delta t\times y'(t_{n},y_{n})\quad }
δ
y
2
=
δ
t
×
y
′
(
t
n
+
1
2
δ
t
,
y
n
+
1
2
δ
y
1
)
{\displaystyle \delta y_{2}=\delta t\times y'(t_{n}+{\tfrac {1}{2}}\delta t,y_{n}+{\tfrac {1}{2}}\delta y_{1})}
δ
y
3
=
δ
t
×
y
′
(
t
n
+
1
2
δ
t
,
y
n
+
1
2
δ
y
2
)
{\displaystyle \delta y_{3}=\delta t\times y'(t_{n}+{\tfrac {1}{2}}\delta t,y_{n}+{\tfrac {1}{2}}\delta y_{2})}
δ
y
4
=
δ
t
×
y
′
(
t
n
+
δ
t
,
y
n
+
δ
y
3
)
{\displaystyle \delta y_{4}=\delta t\times y'(t_{n}+\delta t,y_{n}+\delta y_{3})\quad }
then:
y
n
+
1
=
y
n
+
1
6
(
δ
y
1
+
2
δ
y
2
+
2
δ
y
3
+
δ
y
4
)
{\displaystyle y_{n+1}=y_{n}+{\tfrac {1}{6}}(\delta y_{1}+2\delta y_{2}+2\delta y_{3}+\delta y_{4})}
t
n
+
1
=
t
n
+
δ
t
{\displaystyle t_{n+1}=t_{n}+\delta t\quad }
|
#Scala
|
Scala
|
object Main extends App {
val f = (t: Double, y: Double) => t * Math.sqrt(y) // Runge-Kutta solution
val g = (t: Double) => Math.pow(t * t + 4, 2) / 16 // Exact solution
new Calculator(f, Some(g)).compute(100, 0, .1, 1)
}
class Calculator(f: (Double, Double) => Double, g: Option[Double => Double] = None) {
def compute(counter: Int, tn: Double, dt: Double, yn: Double): Unit = {
if (counter % 10 == 0) {
val c = (x: Double => Double) => (t: Double) => {
val err = Math.abs(x(t) - yn)
f" Error: $err%7.5e"
}
val s = g.map(c(_)).getOrElse((x: Double) => "") // If we don't have exact solution, just print nothing
println(f"y($tn%4.1f) = $yn%12.8f${s(tn)}") // Else, print Error estimation here
}
if (counter > 0) {
val dy1 = dt * f(tn, yn)
val dy2 = dt * f(tn + dt / 2, yn + dy1 / 2)
val dy3 = dt * f(tn + dt / 2, yn + dy2 / 2)
val dy4 = dt * f(tn + dt, yn + dy3)
val y = yn + (dy1 + 2 * dy2 + 2 * dy3 + dy4) / 6
val t = tn + dt
compute(counter - 1, t, dt, y)
}
}
}
|
http://rosettacode.org/wiki/RPG_attributes_generator
|
RPG attributes generator
|
RPG = Role Playing Game.
You're running a tabletop RPG, and your players are creating characters.
Each character has six core attributes: strength, dexterity, constitution, intelligence, wisdom, and charisma.
One way of generating values for these attributes is to roll four, 6-sided dice (d6) and sum the three highest rolls, discarding the lowest roll.
Some players like to assign values to their attributes in the order they're rolled.
To ensure generated characters don't put players at a disadvantage, the following requirements must be satisfied:
The total of all character attributes must be at least 75.
At least two of the attributes must be at least 15.
However, this can require a lot of manual dice rolling. A programatic solution would be much faster.
Task
Write a program that:
Generates 4 random, whole values between 1 and 6.
Saves the sum of the 3 largest values.
Generates a total of 6 values this way.
Displays the total, and all 6 values once finished.
The order in which each value was generated must be preserved.
The total of all 6 values must be at least 75.
At least 2 of the values must be 15 or more.
|
#UNIX_Shell
|
UNIX Shell
|
function main {
typeset attrs=(str dex con int wis cha)
typeset -A values
typeset attr
typeset -i value total fifteens
while true; do
fifteens=0
total=0
for attr in "${attrs[@]}"; do
# "random" values repeat in zsh if run in a subshell
r4d6drop >/tmp/$$
read value </tmp/$$
values[$attr]=$value
(( total += value ))
if (( value >= 15 )); then
(( fifteens += 1 ))
fi
done
if (( total >= 75 && fifteens >= 2 )); then
break
fi
done
rm -f /tmp/$$
for attr in "${attrs[@]}"; do
printf '%s: %d\n' "$attr" "${values[$attr]}"
done
}
function r4d6drop {
typeset -i d1=RANDOM%6+1 d2=RANDOM%6+1 d3=RANDOM%6+1 d4=RANDOM%6+1
typeset e=$(printf '%s\n' $d1 $d2 $d3 $d4 |
sort -n | tail -n +2 | tr $'\n' +)
printf '%d\n' $(( ${e%+} ))
}
main "$@"
|
http://rosettacode.org/wiki/RPG_attributes_generator
|
RPG attributes generator
|
RPG = Role Playing Game.
You're running a tabletop RPG, and your players are creating characters.
Each character has six core attributes: strength, dexterity, constitution, intelligence, wisdom, and charisma.
One way of generating values for these attributes is to roll four, 6-sided dice (d6) and sum the three highest rolls, discarding the lowest roll.
Some players like to assign values to their attributes in the order they're rolled.
To ensure generated characters don't put players at a disadvantage, the following requirements must be satisfied:
The total of all character attributes must be at least 75.
At least two of the attributes must be at least 15.
However, this can require a lot of manual dice rolling. A programatic solution would be much faster.
Task
Write a program that:
Generates 4 random, whole values between 1 and 6.
Saves the sum of the 3 largest values.
Generates a total of 6 values this way.
Displays the total, and all 6 values once finished.
The order in which each value was generated must be preserved.
The total of all 6 values must be at least 75.
At least 2 of the values must be 15 or more.
|
#Visual_Basic_.NET
|
Visual Basic .NET
|
Module Module1
Dim r As New Random
Function getThree(n As Integer) As List(Of Integer)
getThree = New List(Of Integer)
For i As Integer = 1 To 4 : getThree.Add(r.Next(n) + 1) : Next
getThree.Sort() : getThree.RemoveAt(0)
End Function
Function getSix() As List(Of Integer)
getSix = New List(Of Integer)
For i As Integer = 1 To 6 : getSix.Add(getThree(6).Sum) : Next
End Function
Sub Main(args As String())
Dim good As Boolean = False : Do
Dim gs As List(Of Integer) = getSix(), gss As Integer = gs.Sum,
hvc As Integer = gs.FindAll(Function(x) x > 14).Count
Console.Write("attribs: {0}, sum={1}, ({2} sum, high vals={3})",
String.Join(", ", gs), gss, If(gss >= 75, "good", "low"), hvc)
good = gs.Sum >= 75 AndAlso hvc > 1
Console.WriteLine(" - {0}", If(good, "success", "failure"))
Loop Until good
End Sub
End Module
|
http://rosettacode.org/wiki/Sieve_of_Eratosthenes
|
Sieve of Eratosthenes
|
This task has been clarified. Its programming examples are in need of review to ensure that they still fit the requirements of the task.
The Sieve of Eratosthenes is a simple algorithm that finds the prime numbers up to a given integer.
Task
Implement the Sieve of Eratosthenes algorithm, with the only allowed optimization that the outer loop can stop at the square root of the limit, and the inner loop may start at the square of the prime just found.
That means especially that you shouldn't optimize by using pre-computed wheels, i.e. don't assume you need only to cross out odd numbers (wheel based on 2), numbers equal to 1 or 5 modulo 6 (wheel based on 2 and 3), or similar wheels based on low primes.
If there's an easy way to add such a wheel based optimization, implement it as an alternative version.
Note
It is important that the sieve algorithm be the actual algorithm used to find prime numbers for the task.
Related tasks
Emirp primes
count in factors
prime decomposition
factors of an integer
extensible prime generator
primality by trial division
factors of a Mersenne number
trial factoring of a Mersenne number
partition an integer X into N primes
sequence of primes by Trial Division
|
#VBA
|
VBA
|
Sub primes()
'BRRJPA
'Prime calculation for VBA_Excel
'p is the superior limit of the range calculation
'This example calculates from 2 to 100000 and print it
'at the collum A
p = 100000
Dim nprimes(1 To 100000) As Integer
b = Sqr(p)
For n = 2 To b
For k = n * n To p Step n
nprimes(k) = 1
Next k
Next n
For a = 2 To p
If nprimes(a) = 0 Then
c = c + 1
Range("A" & c).Value = a
End If
Next a
End Sub
|
http://rosettacode.org/wiki/Search_a_list
|
Search a list
|
Task[edit]
Find the index of a string (needle) in an indexable, ordered collection of strings (haystack).
Raise an exception if the needle is missing.
If there is more than one occurrence then return the smallest index to the needle.
Extra credit
Return the largest index to a needle that has multiple occurrences in the haystack.
See also
Search a list of records
|
#ooRexx
|
ooRexx
|
-- ordered collections always return the first hit
a = .array~of(1,2,3,4,4,5)
say a~index(4)
a2 = .array~new(5,5) -- multidimensional
a2[3,3] = 4
-- the returned index is an array of values
say a2~index(4)~makestring('line', ',')
-- Note, list indexes are assigned when an item is added and
-- are not tied to relative position
l = .list~of(1,2,3,4,4,5)
say l~index(4)
q = .queue~of(1,2,3,4,4,5)
say q~index(4)
-- directories are unordered, so it is
-- undertermined which one is returned
d = .directory~new
d["foo"] = 4
d["bar"] = 4
say d~index(4)
|
http://rosettacode.org/wiki/Rosetta_Code/Rank_languages_by_popularity
|
Rosetta Code/Rank languages by popularity
|
Rosetta Code/Rank languages by popularity
You are encouraged to solve this task according to the task description, using any language you may know.
Task
Sort the most popular computer programming languages based in number of members in Rosetta Code categories.
Sample output on 01 juin 2022 at 14:13 +02
Rank: 1 (1,540 entries) Phix
Rank: 2 (1,531 entries) Wren
Rank: 3 (1,507 entries) Julia
Rank: 4 (1,494 entries) Go
Rank: 5 (1,488 entries) Raku
Rank: 6 (1,448 entries) Perl
Rank: 7 (1,402 entries) Nim
Rank: 8 (1,382 entries) Python
Rank: 9 (1,204 entries) C
Rank: 10 (1,152 entries) REXX
...
Notes
Each language typically demonstrates one or two methods of accessing the data:
with web scraping (via http://www.rosettacode.org/mw/index.php?title=Special:Categories&limit=5000)
with the API method (examples below for Awk, Perl, Ruby, Tcl, etc).
The scraping and API solutions can be separate subsections, see the Tcl example.
Filtering wrong results is optional. You can check against Special:MostLinkedCategories (if using web scraping)
If you use the API, and do elect to filter, you may check your results against this complete, accurate, sortable, wikitable listing of all 869 programming languages, updated periodically, typically weekly.
A complete ranked listing of all 813 languages (from the REXX example) is included here ──► output from the REXX program.
|
#Red
|
Red
|
Red []
data: read http://www.rosettacode.org/mw/index.php?title=Special:Categories&limit=5000
lb: make block! 500
;;data: read %data.html ;; for testing save html and use flat file
arr: split data newline
k: "Category:"
;; exclude list:
exrule: [thru ["programming"
| "users"
| "Implementations"
| "Solutions by "
| "Members"
| "WikipediaSourced"
| "Typing/Strong"
| "Impl needed"
]
to end
]
foreach line arr [
unless find line k [continue]
parse line [ thru k thru ">" copy lang to "<" to end ] ;; extract/parse language
if 20 < length? lang [continue]
if parse lang [exrule] [continue] ;; exclude invalid
cnt: 0
;; parse number of entries
parse line [thru "</a>" thru "(" copy cnt to " member" (cnt: to-integer cnt ) to end]
if cnt > 25 [ append lb reduce [to-string lang cnt] ] ;; only process lang with > 25 entries
]
lb: sort/skip/compare lb 2 2 ;; sort series by entries
print reduce [ "Rank Entries Language" ] ;; header
last: 0
rank: 0
lb: tail lb ;; process the series backwards
until [
lb: skip lb -2
cnt: second lb
if cnt <> last [
rank: rank + 1
]
print rejoin [ pad/left rank 4 "." pad/left cnt 5 " - " first lb ]
last: cnt
head? lb ;; until head reached
]
|
http://rosettacode.org/wiki/Roman_numerals/Encode
|
Roman numerals/Encode
|
Task
Create a function taking a positive integer as its parameter and returning a string containing the Roman numeral representation of that integer. Modern Roman numerals are written by expressing each digit separately, starting with the left most digit and skipping any digit with a value of zero.
In Roman numerals:
1990 is rendered: 1000=M, 900=CM, 90=XC; resulting in MCMXC
2008 is written as 2000=MM, 8=VIII; or MMVIII
1666 uses each Roman symbol in descending order: MDCLXVI
|
#BQN
|
BQN
|
⟨ToRoman⇐R⟩ ← {
ds ← 1↓¨(¯1+`⊏⊸=)⊸⊔" I IV V IX X XL L XC C CD D CM M"
vs ← 1e3∾˜ ⥊1‿4‿5‿9×⌜˜10⋆↕3
R ⇐ {
𝕨𝕊0: "";
(⊑⟜ds∾·𝕊𝕩-⊑⟜vs) 1-˜⊑vs⍋𝕩
}
}
|
http://rosettacode.org/wiki/Roman_numerals/Decode
|
Roman numerals/Decode
|
Task
Create a function that takes a Roman numeral as its argument and returns its value as a numeric decimal integer.
You don't need to validate the form of the Roman numeral.
Modern Roman numerals are written by expressing each decimal digit of the number to be encoded separately,
starting with the leftmost decimal digit and skipping any 0s (zeroes).
1990 is rendered as MCMXC (1000 = M, 900 = CM, 90 = XC) and
2008 is rendered as MMVIII (2000 = MM, 8 = VIII).
The Roman numeral for 1666, MDCLXVI, uses each letter in descending order.
|
#C.2B.2B
|
C++
|
#include <exception>
#include <string>
#include <iostream>
using namespace std;
namespace Roman
{
int ToInt(char c)
{
switch (c)
{
case 'I': return 1;
case 'V': return 5;
case 'X': return 10;
case 'L': return 50;
case 'C': return 100;
case 'D': return 500;
case 'M': return 1000;
}
throw exception("Invalid character");
}
int ToInt(const string& s)
{
int retval = 0, pvs = 0;
for (auto pc = s.rbegin(); pc != s.rend(); ++pc)
{
const int inc = ToInt(*pc);
retval += inc < pvs ? -inc : inc;
pvs = inc;
}
return retval;
}
}
int main(int argc, char* argv[])
{
try
{
cout << "MCMXC = " << Roman::ToInt("MCMXC") << "\n";
cout << "MMVIII = " << Roman::ToInt("MMVIII") << "\n";
cout << "MDCLXVI = " << Roman::ToInt("MDCLXVI") << "\n";
}
catch (exception& e)
{
cerr << e.what();
return -1;
}
return 0;
}
|
http://rosettacode.org/wiki/Roots_of_a_function
|
Roots of a function
|
Task
Create a program that finds and outputs the roots of a given function, range and (if applicable) step width.
The program should identify whether the root is exact or approximate.
For this task, use: ƒ(x) = x3 - 3x2 + 2x
|
#Icon_and_Unicon
|
Icon and Unicon
|
procedure main()
showRoots(f, -1.0, 4, 0.002)
end
procedure f(x)
return x^3 - 3*x^2 + 2*x
end
procedure showRoots(f, lb, ub, step)
ox := x := lb
oy := f(x)
os := sign(oy)
while x <= ub do {
if (s := sign(y := f(x))) = 0 then write(x)
else if s ~= os then {
dx := x-ox
dy := y-oy
cx := x-dx*(y/dy)
write("~",cx)
}
(ox := x, oy := y, os := s)
x +:= step
}
end
procedure sign(x)
return (x<0, -1) | (x>0, 1) | 0
end
|
http://rosettacode.org/wiki/Rock-paper-scissors
|
Rock-paper-scissors
|
Task
Implement the classic children's game Rock-paper-scissors, as well as a simple predictive AI (artificial intelligence) player.
Rock Paper Scissors is a two player game.
Each player chooses one of rock, paper or scissors, without knowing the other player's choice.
The winner is decided by a set of rules:
Rock beats scissors
Scissors beat paper
Paper beats rock
If both players choose the same thing, there is no winner for that round.
For this task, the computer will be one of the players.
The operator will select Rock, Paper or Scissors and the computer will keep a record of the choice frequency, and use that information to make a weighted random choice in an attempt to defeat its opponent.
Extra credit
Support additional choices additional weapons.
|
#Forth
|
Forth
|
include random.fs
0 value aiwins
0 value plwins
10 constant inlim
0 constant rock
1 constant paper
2 constant scissors
create rpshistory 0 , 0 , 0 ,
create inversemove 1 , 2 , 0 ,
3 constant historylen
: @a ( n array -- )
swap cells + @ ;
: sum-history ( -- n )
0 historylen 0 ?do
i rpshistory @a 1+ + loop ;
: probable-choice ( -- n ) \ Simple linear search
sum-history random
historylen 0 ?do
i rpshistory @a -
dup 0< if drop i leave then
loop inversemove @a ;
: rps-print ( addr u -- )
cr type ;
: rpslog. ( -- )
s" ROCK PAPER SCISSORS AI/W PL/W" rps-print cr
3 0 do i cells rpshistory + @ 9 u.r loop aiwins 7 u.r plwins 6 u.r cr ;
create rpswords ' rock , ' paper , ' scissors , ' quit ,
: update-history! ( n -- )
cells rpshistory + 1 swap +! ;
: thrown. ( n -- addr u )
cells rpswords + @ name>string ;
: throws. ( n n -- )
thrown. s" AI threw: " 2swap s+ rps-print
thrown. s" You threw: " 2swap s+ rps-print ;
: print-throws ( n n -- )
rpslog. throws. ;
: tie. ( n n -- )
s" Tie. " rps-print ;
: plwin ( n n -- )
1 +to plwins s" You win. " rps-print ;
: aiwin ( n n -- )
1 +to aiwins s" AI wins. " rps-print ;
create rpsstates ' tie. , ' plwin , ' aiwin ,
: determine-winner ( n n -- )
>r abs r> abs - 3 + 3 mod
cells rpsstates + @ execute ;
: rps-validate ( name -- ) ( Rude way of checking for only valid commands )
4 0 do i cells rpswords + @ over = swap loop drop or or or ;
: rps-prompt. ( -- )
s" Enter choice (rock, paper, scissors or quit): " rps-print ;
: player-choice ( -- n ) recursive
pad inlim accept pad swap find-name
dup rps-validate if execute
else drop rps-prompt. player-choice then ;
: update-log ( n n -- )
update-history! update-history! ;
: rps ( -- ) recursive
rps-prompt. player-choice probable-choice
2dup update-log 2dup print-throws
determine-winner rps ;
|
http://rosettacode.org/wiki/Run-length_encoding
|
Run-length encoding
|
Run-length encoding
You are encouraged to solve this task according to the task description, using any language you may know.
Task
Given a string containing uppercase characters (A-Z), compress repeated 'runs' of the same character by storing the length of that run, and provide a function to reverse the compression.
The output can be anything, as long as you can recreate the input with it.
Example
Input: WWWWWWWWWWWWBWWWWWWWWWWWWBBBWWWWWWWWWWWWWWWWWWWWWWWWBWWWWWWWWWWWWWW
Output: 12W1B12W3B24W1B14W
Note: the encoding step in the above example is the same as a step of the Look-and-say sequence.
|
#Gambas
|
Gambas
|
Public Sub Main()
Dim sString As String = "WWWWWWWWWWWWBWWWWWWWWWWWWBBBWWWWWWWWWWWWWWWWWWWWWWWWBWWWWWWWWWWWWWW"
Dim siCount As Short = 1
Dim siStart As Short = 1
Dim sHold As New String[]
Dim sTemp As String
sString &= " "
Repeat
sTemp = Mid(sString, siCount, 1)
Do
Inc siCount
If Mid(sString, siCount, 1) <> sTemp Then Break
If siCount = Len(sString) Then Break
Loop
sHold.add(Str(siCount - siStart) & sTemp)
siStart = siCount
Until siCount = Len(sString)
Print sString & gb.NewLine & sHold.Join(", ")
End
|
http://rosettacode.org/wiki/Roots_of_unity
|
Roots of unity
|
The purpose of this task is to explore working with complex numbers.
Task
Given n, find the nth roots of unity.
|
#PureBasic
|
PureBasic
|
OpenConsole()
For n = 2 To 10
angle = 0
PrintN(Str(n))
For i = 1 To n
x.f = Cos(Radian(angle))
y.f = Sin(Radian(angle))
PrintN( Str(i) + ": " + StrF(x, 6) + " / " + StrF(y, 6))
angle = angle + (360 / n)
Next
Next
Input()
|
http://rosettacode.org/wiki/Roots_of_unity
|
Roots of unity
|
The purpose of this task is to explore working with complex numbers.
Task
Given n, find the nth roots of unity.
|
#Python
|
Python
|
import cmath
class Complex(complex):
def __repr__(self):
rp = '%7.5f' % self.real if not self.pureImag() else ''
ip = '%7.5fj' % self.imag if not self.pureReal() else ''
conj = '' if (
self.pureImag() or self.pureReal() or self.imag < 0.0
) else '+'
return '0.0' if (
self.pureImag() and self.pureReal()
) else rp + conj + ip
def pureImag(self):
return abs(self.real) < 0.000005
def pureReal(self):
return abs(self.imag) < 0.000005
def croots(n):
if n <= 0:
return None
return (Complex(cmath.rect(1, 2 * k * cmath.pi / n)) for k in range(n))
# in pre-Python 2.6:
# return (Complex(cmath.exp(2j*k*cmath.pi/n)) for k in range(n))
for nr in range(2, 11):
print(nr, list(croots(nr)))
|
http://rosettacode.org/wiki/Roots_of_a_quadratic_function
|
Roots of a quadratic function
|
This task has been clarified. Its programming examples are in need of review to ensure that they still fit the requirements of the task.
Write a program to find the roots of a quadratic equation, i.e., solve the equation
a
x
2
+
b
x
+
c
=
0
{\displaystyle ax^{2}+bx+c=0}
.
Your program must correctly handle non-real roots, but it need not check that
a
≠
0
{\displaystyle a\neq 0}
.
The problem of solving a quadratic equation is a good example of how dangerous it can be to ignore the peculiarities of floating-point arithmetic.
The obvious way to implement the quadratic formula suffers catastrophic loss of accuracy when one of the roots to be found is much closer to 0 than the other.
In their classic textbook on numeric methods Computer Methods for Mathematical Computations, George Forsythe, Michael Malcolm, and Cleve Moler suggest trying the naive algorithm with
a
=
1
{\displaystyle a=1}
,
b
=
−
10
5
{\displaystyle b=-10^{5}}
, and
c
=
1
{\displaystyle c=1}
.
(For double-precision floats, set
b
=
−
10
9
{\displaystyle b=-10^{9}}
.)
Consider the following implementation in Ada:
with Ada.Text_IO; use Ada.Text_IO;
with Ada.Numerics.Elementary_Functions; use Ada.Numerics.Elementary_Functions;
procedure Quadratic_Equation is
type Roots is array (1..2) of Float;
function Solve (A, B, C : Float) return Roots is
SD : constant Float := sqrt (B**2 - 4.0 * A * C);
AA : constant Float := 2.0 * A;
begin
return ((- B + SD) / AA, (- B - SD) / AA);
end Solve;
R : constant Roots := Solve (1.0, -10.0E5, 1.0);
begin
Put_Line ("X1 =" & Float'Image (R (1)) & " X2 =" & Float'Image (R (2)));
end Quadratic_Equation;
Output:
X1 = 1.00000E+06 X2 = 0.00000E+00
As we can see, the second root has lost all significant figures. The right answer is that X2 is about
10
−
6
{\displaystyle 10^{-6}}
. The naive method is numerically unstable.
Suggested by Middlebrook (D-OA), a better numerical method: to define two parameters
q
=
a
c
/
b
{\displaystyle q={\sqrt {ac}}/b}
and
f
=
1
/
2
+
1
−
4
q
2
/
2
{\displaystyle f=1/2+{\sqrt {1-4q^{2}}}/2}
and the two roots of the quardratic are:
−
b
a
f
{\displaystyle {\frac {-b}{a}}f}
and
−
c
b
f
{\displaystyle {\frac {-c}{bf}}}
Task: do it better. This means that given
a
=
1
{\displaystyle a=1}
,
b
=
−
10
9
{\displaystyle b=-10^{9}}
, and
c
=
1
{\displaystyle c=1}
, both of the roots your program returns should be greater than
10
−
11
{\displaystyle 10^{-11}}
. Or, if your language can't do floating-point arithmetic any more precisely than single precision, your program should be able to handle
b
=
−
10
6
{\displaystyle b=-10^{6}}
. Either way, show what your program gives as the roots of the quadratic in question. See page 9 of
"What Every Scientist Should Know About Floating-Point Arithmetic" for a possible algorithm.
|
#Python
|
Python
|
#!/usr/bin/env python3
import math
import cmath
import numpy
def quad_discriminating_roots(a,b,c, entier = 1e-5):
"""For reference, the naive algorithm which shows complete loss of
precision on the quadratic in question. (This function also returns a
characterization of the roots.)"""
discriminant = b*b - 4*a*c
a,b,c,d =complex(a), complex(b), complex(c), complex(discriminant)
root1 = (-b + cmath.sqrt(d))/2./a
root2 = (-b - cmath.sqrt(d))/2./a
if abs(discriminant) < entier:
return "real and equal", abs(root1), abs(root1)
if discriminant > 0:
return "real", root1.real, root2.real
return "complex", root1, root2
def middlebrook(a, b, c):
try:
q = math.sqrt(a*c)/b
f = .5+ math.sqrt(1-4*q*q)/2
except ValueError:
q = cmath.sqrt(a*c)/b
f = .5+ cmath.sqrt(1-4*q*q)/2
return (-b/a)*f, -c/(b*f)
def whatevery(a, b, c):
try:
d = math.sqrt(b*b-4*a*c)
except ValueError:
d = cmath.sqrt(b*b-4*a*c)
if b > 0:
return div(2*c, (-b-d)), div((-b-d), 2*a)
else:
return div((-b+d), 2*a), div(2*c, (-b+d))
def div(n, d):
"""Divide, with a useful interpretation of division by zero."""
try:
return n/d
except ZeroDivisionError:
if n:
return n*float('inf')
return float('nan')
testcases = [
(3, 4, 4/3), # real, equal
(3, 2, -1), # real, unequal
(3, 2, 1), # complex
(1, -1e9, 1), # ill-conditioned "quadratic in question" required by task.
(1, -1e100, 1),
(1, -1e200, 1),
(1, -1e300, 1),
]
print('Naive:')
for c in testcases:
print("{} {:.5} {:.5}".format(*quad_discriminating_roots(*c)))
print('\nMiddlebrook:')
for c in testcases:
print(("{:.5} "*2).format(*middlebrook(*c)))
print('\nWhat Every...')
for c in testcases:
print(("{:.5} "*2).format(*whatevery(*c)))
print('\nNumpy:')
for c in testcases:
print(("{:.5} "*2).format(*numpy.roots(c)))
|
http://rosettacode.org/wiki/Rot-13
|
Rot-13
|
Task
Implement a rot-13 function (or procedure, class, subroutine, or other "callable" object as appropriate to your programming environment).
Optionally wrap this function in a utility program (like tr, which acts like a common UNIX utility, performing a line-by-line rot-13 encoding of every line of input contained in each file listed on its command line, or (if no filenames are passed thereon) acting as a filter on its "standard input."
(A number of UNIX scripting languages and utilities, such as awk and sed either default to processing files in this way or have command line switches or modules to easily implement these wrapper semantics, e.g., Perl and Python).
The rot-13 encoding is commonly known from the early days of Usenet "Netnews" as a way of obfuscating text to prevent casual reading of spoiler or potentially offensive material.
Many news reader and mail user agent programs have built-in rot-13 encoder/decoders or have the ability to feed a message through any external utility script for performing this (or other) actions.
The definition of the rot-13 function is to simply replace every letter of the ASCII alphabet with the letter which is "rotated" 13 characters "around" the 26 letter alphabet from its normal cardinal position (wrapping around from z to a as necessary).
Thus the letters abc become nop and so on.
Technically rot-13 is a "mono-alphabetic substitution cipher" with a trivial "key".
A proper implementation should work on upper and lower case letters, preserve case, and pass all non-alphabetic characters
in the input stream through without alteration.
Related tasks
Caesar cipher
Substitution Cipher
Vigenère Cipher/Cryptanalysis
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
|
#Erlang
|
Erlang
|
rot13(Str) ->
F = fun(C) when (C >= $A andalso C =< $M); (C >= $a andalso C =< $m) -> C + 13;
(C) when (C >= $N andalso C =< $Z); (C >= $n andalso C =< $z) -> C - 13;
(C) -> C
end,
lists:map(F, Str).
|
http://rosettacode.org/wiki/Runge-Kutta_method
|
Runge-Kutta method
|
Given the example Differential equation:
y
′
(
t
)
=
t
×
y
(
t
)
{\displaystyle y'(t)=t\times {\sqrt {y(t)}}}
With initial condition:
t
0
=
0
{\displaystyle t_{0}=0}
and
y
0
=
y
(
t
0
)
=
y
(
0
)
=
1
{\displaystyle y_{0}=y(t_{0})=y(0)=1}
This equation has an exact solution:
y
(
t
)
=
1
16
(
t
2
+
4
)
2
{\displaystyle y(t)={\tfrac {1}{16}}(t^{2}+4)^{2}}
Task
Demonstrate the commonly used explicit fourth-order Runge–Kutta method to solve the above differential equation.
Solve the given differential equation over the range
t
=
0
…
10
{\displaystyle t=0\ldots 10}
with a step value of
δ
t
=
0.1
{\displaystyle \delta t=0.1}
(101 total points, the first being given)
Print the calculated values of
y
{\displaystyle y}
at whole numbered
t
{\displaystyle t}
's (
0.0
,
1.0
,
…
10.0
{\displaystyle 0.0,1.0,\ldots 10.0}
) along with error as compared to the exact solution.
Method summary
Starting with a given
y
n
{\displaystyle y_{n}}
and
t
n
{\displaystyle t_{n}}
calculate:
δ
y
1
=
δ
t
×
y
′
(
t
n
,
y
n
)
{\displaystyle \delta y_{1}=\delta t\times y'(t_{n},y_{n})\quad }
δ
y
2
=
δ
t
×
y
′
(
t
n
+
1
2
δ
t
,
y
n
+
1
2
δ
y
1
)
{\displaystyle \delta y_{2}=\delta t\times y'(t_{n}+{\tfrac {1}{2}}\delta t,y_{n}+{\tfrac {1}{2}}\delta y_{1})}
δ
y
3
=
δ
t
×
y
′
(
t
n
+
1
2
δ
t
,
y
n
+
1
2
δ
y
2
)
{\displaystyle \delta y_{3}=\delta t\times y'(t_{n}+{\tfrac {1}{2}}\delta t,y_{n}+{\tfrac {1}{2}}\delta y_{2})}
δ
y
4
=
δ
t
×
y
′
(
t
n
+
δ
t
,
y
n
+
δ
y
3
)
{\displaystyle \delta y_{4}=\delta t\times y'(t_{n}+\delta t,y_{n}+\delta y_{3})\quad }
then:
y
n
+
1
=
y
n
+
1
6
(
δ
y
1
+
2
δ
y
2
+
2
δ
y
3
+
δ
y
4
)
{\displaystyle y_{n+1}=y_{n}+{\tfrac {1}{6}}(\delta y_{1}+2\delta y_{2}+2\delta y_{3}+\delta y_{4})}
t
n
+
1
=
t
n
+
δ
t
{\displaystyle t_{n+1}=t_{n}+\delta t\quad }
|
#Sidef
|
Sidef
|
func runge_kutta(yp) {
func (t, y, δt) {
var a = (δt * yp(t, y));
var b = (δt * yp(t + δt/2, y + a/2));
var c = (δt * yp(t + δt/2, y + b/2));
var d = (δt * yp(t + δt, y + c));
(a + 2*(b + c) + d) / 6;
}
}
define δt = 0.1;
var δy = runge_kutta(func(t, y) { t * y.sqrt });
var(t, y) = (0, 1);
loop {
t.is_int &&
printf("y(%2d) = %12f ± %e\n", t, y, abs(y - ((t**2 + 4)**2 / 16)));
t <= 10 || break;
y += δy(t, y, δt);
t += δt;
}
|
http://rosettacode.org/wiki/RPG_attributes_generator
|
RPG attributes generator
|
RPG = Role Playing Game.
You're running a tabletop RPG, and your players are creating characters.
Each character has six core attributes: strength, dexterity, constitution, intelligence, wisdom, and charisma.
One way of generating values for these attributes is to roll four, 6-sided dice (d6) and sum the three highest rolls, discarding the lowest roll.
Some players like to assign values to their attributes in the order they're rolled.
To ensure generated characters don't put players at a disadvantage, the following requirements must be satisfied:
The total of all character attributes must be at least 75.
At least two of the attributes must be at least 15.
However, this can require a lot of manual dice rolling. A programatic solution would be much faster.
Task
Write a program that:
Generates 4 random, whole values between 1 and 6.
Saves the sum of the 3 largest values.
Generates a total of 6 values this way.
Displays the total, and all 6 values once finished.
The order in which each value was generated must be preserved.
The total of all 6 values must be at least 75.
At least 2 of the values must be 15 or more.
|
#Wren
|
Wren
|
import "random" for Random
import "/sort" for Sort
var rand = Random.new()
var vals = List.filled(6, 0)
while (true) {
for (i in 0..5) {
var rns = List.filled(4, 0)
for (j in 0..3) rns[j] = rand.int(1, 7)
var sum = rns.reduce { |acc, n| acc + n }
Sort.insertion(rns)
vals[i] = sum - rns[0]
}
var total = vals.reduce { |acc, n| acc + n }
if (total >= 75) {
var fifteens = vals.count { |n| n >= 15 }
if (fifteens >= 2) {
System.print("The six values are: %(vals)")
System.print("Their total is: %(total)")
break
}
}
}
|
http://rosettacode.org/wiki/Sieve_of_Eratosthenes
|
Sieve of Eratosthenes
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This task has been clarified. Its programming examples are in need of review to ensure that they still fit the requirements of the task.
The Sieve of Eratosthenes is a simple algorithm that finds the prime numbers up to a given integer.
Task
Implement the Sieve of Eratosthenes algorithm, with the only allowed optimization that the outer loop can stop at the square root of the limit, and the inner loop may start at the square of the prime just found.
That means especially that you shouldn't optimize by using pre-computed wheels, i.e. don't assume you need only to cross out odd numbers (wheel based on 2), numbers equal to 1 or 5 modulo 6 (wheel based on 2 and 3), or similar wheels based on low primes.
If there's an easy way to add such a wheel based optimization, implement it as an alternative version.
Note
It is important that the sieve algorithm be the actual algorithm used to find prime numbers for the task.
Related tasks
Emirp primes
count in factors
prime decomposition
factors of an integer
extensible prime generator
primality by trial division
factors of a Mersenne number
trial factoring of a Mersenne number
partition an integer X into N primes
sequence of primes by Trial Division
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#VBScript
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VBScript
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Dim sieve()
If WScript.Arguments.Count>=1 Then
n = WScript.Arguments(0)
Else
n = 99
End If
ReDim sieve(n)
For i = 1 To n
sieve(i) = True
Next
For i = 2 To n
If sieve(i) Then
For j = i * 2 To n Step i
sieve(j) = False
Next
End If
Next
For i = 2 To n
If sieve(i) Then WScript.Echo i
Next
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http://rosettacode.org/wiki/Search_a_list
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Search a list
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Task[edit]
Find the index of a string (needle) in an indexable, ordered collection of strings (haystack).
Raise an exception if the needle is missing.
If there is more than one occurrence then return the smallest index to the needle.
Extra credit
Return the largest index to a needle that has multiple occurrences in the haystack.
See also
Search a list of records
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#Oz
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Oz
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declare
%% Lazy list of indices of Y in Xs.
fun {Indices Y Xs}
for
X in Xs
I in 1;I+1
yield:Yield
do
if Y == X then {Yield I} end
end
end
fun {Index Y Xs}
case {Indices Y Xs} of X|_ then X
else raise index(elementNotFound Y) end
end
end
Haystack = ["Zig" "Zag" "Wally" "Ronald" "Bush" "Krusty" "Charlie" "Bush" "Bozo"]
in
{Show {Index "Bush" Haystack}}
{Show {List.last {Indices "Bush" Haystack}}}
{Show {Index "Washington" Haystack}} %% throws
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http://rosettacode.org/wiki/Rosetta_Code/Rank_languages_by_popularity
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Rosetta Code/Rank languages by popularity
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Rosetta Code/Rank languages by popularity
You are encouraged to solve this task according to the task description, using any language you may know.
Task
Sort the most popular computer programming languages based in number of members in Rosetta Code categories.
Sample output on 01 juin 2022 at 14:13 +02
Rank: 1 (1,540 entries) Phix
Rank: 2 (1,531 entries) Wren
Rank: 3 (1,507 entries) Julia
Rank: 4 (1,494 entries) Go
Rank: 5 (1,488 entries) Raku
Rank: 6 (1,448 entries) Perl
Rank: 7 (1,402 entries) Nim
Rank: 8 (1,382 entries) Python
Rank: 9 (1,204 entries) C
Rank: 10 (1,152 entries) REXX
...
Notes
Each language typically demonstrates one or two methods of accessing the data:
with web scraping (via http://www.rosettacode.org/mw/index.php?title=Special:Categories&limit=5000)
with the API method (examples below for Awk, Perl, Ruby, Tcl, etc).
The scraping and API solutions can be separate subsections, see the Tcl example.
Filtering wrong results is optional. You can check against Special:MostLinkedCategories (if using web scraping)
If you use the API, and do elect to filter, you may check your results against this complete, accurate, sortable, wikitable listing of all 869 programming languages, updated periodically, typically weekly.
A complete ranked listing of all 813 languages (from the REXX example) is included here ──► output from the REXX program.
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#REXX
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REXX
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/*REXX program reads two files and displays a ranked list of Rosetta Code languages.*/
parse arg catFID lanFID outFID . /*obtain optional arguments from the CL*/
call init /*initialize some REXX variables. */
call get /*obtain data from two separate files. */
call eSort #,0 /*sort languages along with members. */
call tSort /* " " that are tied in rank.*/
call eSort #,1 /* " " along with members. */
call out /*create the RC_POP.OUT (output) file.*/
exit 0 /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
commas: parse arg _; do jc=length(_)-3 to 1 by -3; _= insert(",",_,jc); end; return _
s: if arg(1)==1 then return arg(3); return word(arg(2) 's',1) /*pluralizer.*/
/*──────────────────────────────────────────────────────────────────────────────────────*/
eSort: procedure expose #. @. !tr.; arg N,p2; h= N /*sort: number of members*/
do while h>1; h= h % 2 /*halve number of records*/
do i=1 for N-h; j= i; k= h + i /*sort this part of list.*/
if p2 then do while !tR.k==!tR.j & @.k>@.j /*this uses a hard swap ↓*/
@= @.j; #= !tR.j; @.j= @.k; !tR.j= !tR.k; @.k= @; !tR.k= #
if h>=j then leave; j= j - h; k= k - h
end /*while !tR.k==···*/
else do while #.k<#.j /*this uses a hard swap ↓*/
@= @.j; #= #.j; @.j= @.k; #.j= #.k; @.k= @; #.k= #
if h>=j then leave; j= j - h; k= k - h
end /*while #.k<···*/
end /*i*/ /*hard swaps needed for embedded blanks.*/
end /*while h>1*/; return
/*──────────────────────────────────────────────────────────────────────────────────────*/
get: langs= 0; call rdr 'languages' /*assign languages ───► L.ααα */
call rdr 'categories' /*append categories ───► catHeap */
#= 0
do j=1 until catHeap=='' /*process the heap of categories. */
parse var catHeap cat.j (sep) catHeap /*get a category from the catHeap. */
parse var cat.j cat.j '<' "(" mems . /*untangle the strange─looking string. */
cat.j= space(cat.j); ?=cat.j; upper ? /*remove any superfluous blanks. */
if ?=='' | \L.? then iterate /*it's blank or it's not a language. */
if pos(',', mems)\==0 then mems= space(translate(mems,,","), 0) /*¬commas.*/
if \datatype(mems, 'W') then iterate /*is the "members" number not numeric? */
#.0= #.0 + mems /*bump the number of members found. */
if u.?\==0 then do; do f=1 for # until [email protected]
end /*f*/
#.f= #.f + mems; iterate j /*languages in different cases.*/
end /* [↑] handle any possible duplicates.*/
u.?= u.? + 1; #= # + 1 /*bump a couple of counters. */
#.#= #.# + mems; @.#= cat.j; @u.#=? /*bump the counter; assign it (upper).*/
end /*j*/
!.=; @tno= '(total) number of' /*array holds indication of TIED langs.*/
call tell right(commas(#), 9) @tno 'languages detected in the category file'
call tell right(commas(langs),9) ' " " " " " " " language "
call tell right(commas(#.0), 9) @tno 'entries (solutions) detected', , 1; term= 0
return /*don't show any more msgs──►term. [↑] */
/*──────────────────────────────────────────────────────────────────────────────────────*/
init: sep= '█'; L.=0; #.=0; u.=#.; catHeap=; term=1; old.= /*assign some REXX vars*/
if catFID=='' then catFID= "RC_POP.CAT" /*Not specified? Then use the default.*/
if lanFID=='' then lanFID= "RC_POP.LAN" /* " " " " " " */
if outFID=='' then outFID= "RC_POP.OUT" /* " " " " " " */
call tell center('timestamp: ' date() time("Civil"),79,'═'), 2, 1; return
/*──────────────────────────────────────────────────────────────────────────────────────*/
out: w= length( commas(#) ); rank= 0 /* [↓] show by ascending rank of lang.*/
do t=# by -1 for #; rank= rank + 1 /*bump rank of a programming language. */
call tell right('rank:' right(commas(!tR.t), w), 20-1) right(!.t, 7),
right('('commas(#.t) left("entr"s(#.t, 'ies', "y")')', 9), 20) @.t
end /*#*/ /* [↑] S(···) pluralizes a word. */
call tell left('', 27) "☼ end─of─list. ☼", 1, 2; return /*bottom title.*/
/*──────────────────────────────────────────────────────────────────────────────────────*/
rdr: arg which 2; igAst= 1 /*ARG uppers WHICH, obtain the 1st char*/
if which=='L' then inFID= lanFID /*use this fileID for the languages. */
if which=='C' then inFID= catFID /* " " " " " categories. */
Uyir= 'உயிர்/Uyir' /*Unicode (in text) name for Uyir */
old.0= '╬£C++' ; new.0= "µC++" /*Unicode ╬£C++ ───► ASCII─8: µC++ */
old.1= 'UC++' ; new.1= "µC++" /*old UC++ ───► ASCII─8: µC++ */
old.2= '╨£╨Ü-' ; new.2= "MK-" /*Unicode ╨£╨Ü─ ───► ASCII-8: MK- */
old.3= 'D├⌐j├á' ; new.3= "Déjà" /*Unicode ├⌐j├á ───► ASCII─8: Déjà */
old.4= 'Cach├⌐' ; new.4= "Caché" /*Unicode Cach├⌐ ───► ASCII─8: Caché */
old.5= '??-61/52' ; new.5= "MK-61/52" /*somewhere past, a mis─translated: MK-*/
old.6= 'F┼ìrmul├ª' ; new.6= 'Fôrmulæ' /*Unicode ───► ASCII─8: Fôrmulæ */
old.7= '╨£iniZinc' ; new.7= 'MiniZinc' /*Unicode ───► ASCII─8: MiniZinc*/
old.8= Uyir ; new.8= 'Uyir' /*Unicode ───► ASCII─8: Uyir */
old.9= 'Perl 6' ; new.9= 'Raku' /* (old name) ───► (new name) */
do recs=0 while lines(inFID) \== 0 /*read a file, a single line at a time.*/
$= translate( linein(inFID), , '9'x) /*handle any stray TAB ('09'x) chars.*/
$$= space($); if $$=='' then iterate /*ignore all blank lines in the file(s)*/
do v=0 while old.v \== '' /*translate Unicode variations of langs*/
if pos(old.v, $$) \==0 then $$= changestr(old.v, $$, new.v)
end /*v*/ /* [↑] handle different lang spellings*/
if igAst then do; igAst= pos(' * ', $)==0; if igAst then iterate; end
if pos('RETRIEVED FROM', translate($$))\==0 then leave /*pseudo End─Of─Data?.*/
if which=='L' then do; if left($$, 1)\=="*" then iterate /*lang ¬legitimate?*/
parse upper var $$ '*' $$ "<"; $$= space($$)
if $$=='' then iterate; L.$$= 1
langs= langs + 1 /*bump number of languages found. */
iterate
end /* [↓] extract computer language name.*/
if left($$, 1)=='*' then $$= sep || space( substr($$, 2) )
catHeap= catHeap $$ /*append to the catHeap (CATegory) heap*/
end /*recs*/
call tell right( commas(recs), 9) 'records read from file: ' inFID
return
/*──────────────────────────────────────────────────────────────────────────────────────*/
tell: do '0'arg(2); call lineout outFID," " ; if term then say ; end
call lineout outFID,arg(1) ; if term then say arg(1)
do '0'arg(3); call lineout outFID," " ; if term then say ; end
return /*show BEFORE blank lines (if any), message, show AFTER blank lines.*/
/*──────────────────────────────────────────────────────────────────────────────────────*/
tSort: tied=; r= 0 /*add true rank (tR) ───► the entries. */
do j=# by -1 for #; r= r+1; tR= r; !tR.j= r; jp= j+1; jm= j-1
if tied=='' then pR= r; tied= /*handle when language rank is untied. */
if #.j==#.jp | #.j==#.jm then do; !.j= '[tied]'; tied= !.j; end
if #.j==#.jp then do; tR= pR; !tR.j= pR; end
else pR= r
end /*j*/; return
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