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http://rosettacode.org/wiki/Apply_a_callback_to_an_array | Apply a callback to an array | Task
Take a combined set of elements and apply a function to each element.
| #CoffeeScript | CoffeeScript |
map = (arr, f) -> (f(e) for e in arr)
arr = [1, 2, 3, 4, 5]
f = (x) -> x * x
console.log map arr, f # prints [1, 4, 9, 16, 25]
|
http://rosettacode.org/wiki/Averages/Mode | Averages/Mode | Task[edit]
Write a program to find the mode value of a collection.
The case where the collection is empty may be ignored. Care must be taken to handle the case where the mode is non-unique.
If it is not appropriate or possible to support a general collection, use a vector (array), if possible. If it is not appropriate or possible to support an unspecified value type, use integers.
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #PARI.2FGP | PARI/GP | mode(v)={
my(count=1,r=1,b=v[1]);
v=vecsort(v);
for(i=2,#v,
if(v[i]==v[i-1],
count++
,
if(count>r,
r=count;
b=v[i-1]
);
count=1
)
);
if(count>r,v[#v],b)
}; |
http://rosettacode.org/wiki/Associative_array/Iteration | Associative array/Iteration | Show how to iterate over the key-value pairs of an associative array, and print each pair out.
Also show how to iterate just over the keys, or the values, if there is a separate way to do that in your language.
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #Dyalect | Dyalect | var t = (x: 1, y: 2, z: 3)
for x in t.Keys() {
print("\(x)=\(t[x])")
} |
http://rosettacode.org/wiki/Associative_array/Iteration | Associative array/Iteration | Show how to iterate over the key-value pairs of an associative array, and print each pair out.
Also show how to iterate just over the keys, or the values, if there is a separate way to do that in your language.
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #E | E | def map := [
"a" => 1,
"b" => 2,
"c" => 3,
]
for key => value in map {
println(`$key $value`)
}
for value in map { # ignore keys
println(`. $value`)
}
for key => _ in map { # ignore values
println(`$key .`)
}
for key in map.domain() { # iterate over the set whose values are the keys
println(`$key .`)
} |
http://rosettacode.org/wiki/Apply_a_digital_filter_(direct_form_II_transposed) | Apply a digital filter (direct form II transposed) | Digital filters are used to apply a mathematical operation to a sampled signal. One of the common formulations is the "direct form II transposed" which can represent both infinite impulse response (IIR) and finite impulse response (FIR) filters, as well as being more numerically stable than other forms. [1]
Task
Filter a signal using an order 3 low-pass Butterworth filter. The coefficients for the filter are a=[1.00000000, -2.77555756e-16, 3.33333333e-01, -1.85037171e-17] and b = [0.16666667, 0.5, 0.5, 0.16666667]
The signal that needs filtering is the following vector: [-0.917843918645, 0.141984778794, 1.20536903482, 0.190286794412, -0.662370894973, -1.00700480494, -0.404707073677 ,0.800482325044, 0.743500089861, 1.01090520172, 0.741527555207, 0.277841675195, 0.400833448236, -0.2085993586, -0.172842103641, -0.134316096293, 0.0259303398477, 0.490105989562, 0.549391221511, 0.9047198589]
| #Phix | Phix | with javascript_semantics
function direct_form_II_transposed_filter(sequence a, b, signal)
sequence result = repeat(0,length(signal))
for i=1 to length(signal) do
atom tmp = 0
for j=1 to min(i,length(b)) do tmp += b[j]*signal[i-j+1] end for
for j=2 to min(i,length(a)) do tmp -= a[j]*result[i-j+1] end for
result[i] = tmp/a[1]
end for
return result
end function
constant acoef = {1.00000000, -2.77555756e-16, 3.33333333e-01, -1.85037171e-17},
bcoef = {0.16666667, 0.5, 0.5, 0.16666667},
signal = {-0.917843918645,0.141984778794,1.20536903482,0.190286794412,-0.662370894973,
-1.00700480494,-0.404707073677,0.800482325044,0.743500089861,1.01090520172,
0.741527555207,0.277841675195,0.400833448236,-0.2085993586,-0.172842103641,
-0.134316096293,0.0259303398477,0.490105989562,0.549391221511,0.9047198589}
pp(direct_form_II_transposed_filter(acoef, bcoef, signal),{pp_FltFmt,"%9.6f",pp_Maxlen,110})
|
http://rosettacode.org/wiki/Averages/Arithmetic_mean | Averages/Arithmetic mean | Task[edit]
Write a program to find the mean (arithmetic average) of a numeric vector.
In case of a zero-length input, since the mean of an empty set of numbers is ill-defined, the program may choose to behave in any way it deems appropriate, though if the programming language has an established convention for conveying math errors or undefined values, it's preferable to follow it.
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #Elixir | Elixir | defmodule Average do
def mean(list), do: Enum.sum(list) / length(list)
end |
http://rosettacode.org/wiki/Averages/Arithmetic_mean | Averages/Arithmetic mean | Task[edit]
Write a program to find the mean (arithmetic average) of a numeric vector.
In case of a zero-length input, since the mean of an empty set of numbers is ill-defined, the program may choose to behave in any way it deems appropriate, though if the programming language has an established convention for conveying math errors or undefined values, it's preferable to follow it.
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #Emacs_Lisp | Emacs Lisp | (defun mean (lst)
(/ (float (apply '+ lst)) (length lst)))
(mean '(1 2 3 4)) |
http://rosettacode.org/wiki/Average_loop_length | Average loop length | Let f be a uniformly-randomly chosen mapping from the numbers 1..N to the numbers 1..N (note: not necessarily a permutation of 1..N; the mapping could produce a number in more than one way or not at all). At some point, the sequence 1, f(1), f(f(1))... will contain a repetition, a number that occurring for the second time in the sequence.
Task
Write a program or a script that estimates, for each N, the average length until the first such repetition.
Also calculate this expected length using an analytical formula, and optionally compare the simulated result with the theoretical one.
This problem comes from the end of Donald Knuth's Christmas tree lecture 2011.
Example of expected output:
N average analytical (error)
=== ========= ============ =========
1 1.0000 1.0000 ( 0.00%)
2 1.4992 1.5000 ( 0.05%)
3 1.8784 1.8889 ( 0.56%)
4 2.2316 2.2188 ( 0.58%)
5 2.4982 2.5104 ( 0.49%)
6 2.7897 2.7747 ( 0.54%)
7 3.0153 3.0181 ( 0.09%)
8 3.2429 3.2450 ( 0.07%)
9 3.4536 3.4583 ( 0.14%)
10 3.6649 3.6602 ( 0.13%)
11 3.8091 3.8524 ( 1.12%)
12 3.9986 4.0361 ( 0.93%)
13 4.2074 4.2123 ( 0.12%)
14 4.3711 4.3820 ( 0.25%)
15 4.5275 4.5458 ( 0.40%)
16 4.6755 4.7043 ( 0.61%)
17 4.8877 4.8579 ( 0.61%)
18 4.9951 5.0071 ( 0.24%)
19 5.1312 5.1522 ( 0.41%)
20 5.2699 5.2936 ( 0.45%)
| #Seed7 | Seed7 | $ include "seed7_05.s7i";
include "float.s7i";
const integer: TESTS is 1000000;
const func float: factorial (in integer: number) is func
result
var float: factorial is 1.0;
local
var integer: i is 0;
begin
for i range 2 to number do
factorial *:= flt(i);
end for;
end func;
const func float: analytical (in integer: number) is func
result
var float: sum is 0.0;
local
var integer: i is 0;
begin
for i range 1 to number do
sum +:= factorial(number) / factorial(number - i) / flt(number)**i;
end for;
end func;
const func float: experimental (in integer: number) is func
result
var float: experimental is 0.0;
local
var integer: run is 0;
var set of integer: seen is EMPTY_SET;
var integer: current is 1;
var integer: count is 0;
begin
for run range 1 to TESTS do
current := 1;
seen := EMPTY_SET;
while current not in seen do
incr(count);
incl(seen, current);
current := rand(1, number);
end while;
end for;
experimental := flt(count) / flt(TESTS);
end func;
const proc: main is func
local
var integer: number is 0;
var float: analytical is 0.0;
var float: experimental is 0.0;
var float: err is 0.0;
begin
writeln(" N avg calc %diff");
for number range 1 to 20 do
analytical := analytical(number);
experimental := experimental(number);
err := abs(experimental - analytical) / analytical * 100.0;
writeln(number lpad 2 <& experimental digits 4 lpad 7 <&
analytical digits 4 lpad 7 <& err digits 3 lpad 7);
end for;
end func; |
http://rosettacode.org/wiki/Average_loop_length | Average loop length | Let f be a uniformly-randomly chosen mapping from the numbers 1..N to the numbers 1..N (note: not necessarily a permutation of 1..N; the mapping could produce a number in more than one way or not at all). At some point, the sequence 1, f(1), f(f(1))... will contain a repetition, a number that occurring for the second time in the sequence.
Task
Write a program or a script that estimates, for each N, the average length until the first such repetition.
Also calculate this expected length using an analytical formula, and optionally compare the simulated result with the theoretical one.
This problem comes from the end of Donald Knuth's Christmas tree lecture 2011.
Example of expected output:
N average analytical (error)
=== ========= ============ =========
1 1.0000 1.0000 ( 0.00%)
2 1.4992 1.5000 ( 0.05%)
3 1.8784 1.8889 ( 0.56%)
4 2.2316 2.2188 ( 0.58%)
5 2.4982 2.5104 ( 0.49%)
6 2.7897 2.7747 ( 0.54%)
7 3.0153 3.0181 ( 0.09%)
8 3.2429 3.2450 ( 0.07%)
9 3.4536 3.4583 ( 0.14%)
10 3.6649 3.6602 ( 0.13%)
11 3.8091 3.8524 ( 1.12%)
12 3.9986 4.0361 ( 0.93%)
13 4.2074 4.2123 ( 0.12%)
14 4.3711 4.3820 ( 0.25%)
15 4.5275 4.5458 ( 0.40%)
16 4.6755 4.7043 ( 0.61%)
17 4.8877 4.8579 ( 0.61%)
18 4.9951 5.0071 ( 0.24%)
19 5.1312 5.1522 ( 0.41%)
20 5.2699 5.2936 ( 0.45%)
| #Sidef | Sidef | func find_loop(n) {
var seen = Hash()
loop {
with (irand(1, n)) { |r|
seen.has(r) ? (return seen.len) : (seen{r} = true)
}
}
}
print " N empiric theoric (error)\n";
print "=== ========= ============ =========\n";
define MAX = 20
define TRIALS = 1000
for n in (1..MAX) {
var empiric = (1..TRIALS -> sum { find_loop(n) } / TRIALS)
var theoric = (1..n -> sum {|k| prod(n - k + 1 .. n) * k**2 / n**(k+1) })
printf("%3d %9.4f %12.4f (%5.2f%%)\n",
n, empiric, theoric, 100*(empiric-theoric)/theoric)
} |
http://rosettacode.org/wiki/Averages/Simple_moving_average | Averages/Simple moving average | Computing the simple moving average of a series of numbers.
Task[edit]
Create a stateful function/class/instance that takes a period and returns a routine that takes a number as argument and returns a simple moving average of its arguments so far.
Description
A simple moving average is a method for computing an average of a stream of numbers by only averaging the last P numbers from the stream, where P is known as the period.
It can be implemented by calling an initialing routine with P as its argument, I(P), which should then return a routine that when called with individual, successive members of a stream of numbers, computes the mean of (up to), the last P of them, lets call this SMA().
The word stateful in the task description refers to the need for SMA() to remember certain information between calls to it:
The period, P
An ordered container of at least the last P numbers from each of its individual calls.
Stateful also means that successive calls to I(), the initializer, should return separate routines that do not share saved state so they could be used on two independent streams of data.
Pseudo-code for an implementation of SMA is:
function SMA(number: N):
stateful integer: P
stateful list: stream
number: average
stream.append_last(N)
if stream.length() > P:
# Only average the last P elements of the stream
stream.delete_first()
if stream.length() == 0:
average = 0
else:
average = sum( stream.values() ) / stream.length()
return average
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #TI-83_BASIC | TI-83 BASIC | :1->C
:While 1
:Prompt I
:C->dim(L1)
:I->L1(C)
:Disp mean(L1)
:1+C->C
:End |
http://rosettacode.org/wiki/Attractive_numbers | Attractive numbers | A number is an attractive number if the number of its prime factors (whether distinct or not) is also prime.
Example
The number 20, whose prime decomposition is 2 × 2 × 5, is an attractive number because the number of its prime factors (3) is also prime.
Task
Show sequence items up to 120.
Reference
The OEIS entry: A063989: Numbers with a prime number of prime divisors.
| #Nanoquery | Nanoquery | MAX = 120
def is_prime(n)
d = 5
if (n < 2)
return false
end
if (n % 2) = 0
return n = 2
end
if (n % 3) = 0
return n = 3
end
while (d * d) <= n
if n % d = 0
return false
end
d += 2
if n % d = 0
return false
end
d += 4
end
return true
end
def count_prime_factors(n)
count = 0; f = 2
if n = 1
return 0
end
if is_prime(n)
return 1
end
while true
if (n % f) = 0
count += 1
n /= f
if n = 1
return count
end
if is_prime(n)
f = n
end
else if f >= 3
f += 2
else
f = 3
end
end
end
i = 0; n = 0; count = 0
println format("The attractive numbers up to and including %d are:\n", MAX)
for i in range(1, MAX)
n = count_prime_factors(i)
if is_prime(n)
print format("%4d", i)
count += 1
if (count % 20) = 0
println
end
end
end
println |
http://rosettacode.org/wiki/Attractive_numbers | Attractive numbers | A number is an attractive number if the number of its prime factors (whether distinct or not) is also prime.
Example
The number 20, whose prime decomposition is 2 × 2 × 5, is an attractive number because the number of its prime factors (3) is also prime.
Task
Show sequence items up to 120.
Reference
The OEIS entry: A063989: Numbers with a prime number of prime divisors.
| #NewLisp | NewLisp |
(define (prime? n)
(= (length (factor n)) 1))
(define (attractive? n)
(prime? (length (factor n))))
;
(filter attractive? (sequence 2 120))
|
http://rosettacode.org/wiki/Averages/Mean_time_of_day | Averages/Mean time of day | Task[edit]
A particular activity of bats occurs at these times of the day:
23:00:17, 23:40:20, 00:12:45, 00:17:19
Using the idea that there are twenty-four hours in a day,
which is analogous to there being 360 degrees in a circle,
map times of day to and from angles;
and using the ideas of Averages/Mean angle
compute and show the average time of the nocturnal activity
to an accuracy of one second of time.
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #Python | Python | from cmath import rect, phase
from math import radians, degrees
def mean_angle(deg):
return degrees(phase(sum(rect(1, radians(d)) for d in deg)/len(deg)))
def mean_time(times):
t = (time.split(':') for time in times)
seconds = ((float(s) + int(m) * 60 + int(h) * 3600)
for h, m, s in t)
day = 24 * 60 * 60
to_angles = [s * 360. / day for s in seconds]
mean_as_angle = mean_angle(to_angles)
mean_seconds = mean_as_angle * day / 360.
if mean_seconds < 0:
mean_seconds += day
h, m = divmod(mean_seconds, 3600)
m, s = divmod(m, 60)
return '%02i:%02i:%02i' % (h, m, s)
if __name__ == '__main__':
print( mean_time(["23:00:17", "23:40:20", "00:12:45", "00:17:19"]) ) |
http://rosettacode.org/wiki/AVL_tree | AVL tree |
This page uses content from Wikipedia. The original article was at AVL tree. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance)
In computer science, an AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; at no time do they differ by more than one because rebalancing is done ensure this is the case. Lookup, insertion, and deletion all take O(log n) time in both the average and worst cases, where n is the number of nodes in the tree prior to the operation. Insertions and deletions may require the tree to be rebalanced by one or more tree rotations. Note the tree of nodes comprise a set, so duplicate node keys are not allowed.
AVL trees are often compared with red-black trees because they support the same set of operations and because red-black trees also take O(log n) time for the basic operations. Because AVL trees are more rigidly balanced, they are faster than red-black trees for lookup-intensive applications. Similar to red-black trees, AVL trees are height-balanced, but in general not weight-balanced nor μ-balanced; that is, sibling nodes can have hugely differing numbers of descendants.
Task
Implement an AVL tree in the language of choice, and provide at least basic operations.
| #Tcl | Tcl | package require TclOO
namespace eval AVL {
# Class for the overall tree; manages real public API
oo::class create Tree {
variable root nil class
constructor {{nodeClass AVL::Node}} {
set class [oo::class create Node [list superclass $nodeClass]]
# Create a nil instance to act as a leaf sentinel
set nil [my NewNode ""]
set root [$nil ref]
# Make nil be special
oo::objdefine $nil {
method height {} {return 0}
method key {} {error "no key possible"}
method value {} {error "no value possible"}
method destroy {} {
# Do nothing (doesn't prohibit destruction entirely)
}
method print {indent increment} {
# Do nothing
}
}
}
# How to actually manufacture a new node
method NewNode {key} {
if {![info exists nil]} {set nil ""}
$class new $key $nil [list [namespace current]::my NewNode]
}
# Create a new node in the tree and return it
method insert {key} {
set node [my NewNode $key]
if {$root eq $nil} {
set root $node
} else {
$root insert $node
}
return $node
}
# Find the node for a particular key
method lookup {key} {
for {set node $root} {$node ne $nil} {} {
if {[$node key] == $key} {
return $node
} elseif {[$node key] > $key} {
set node [$node left]
} else {
set node [$node right]
}
}
error "no such node"
}
# Print a tree out, one node per line
method print {{indent 0} {increment 1}} {
$root print $indent $increment
return
}
}
# Class of an individual node; may be subclassed
oo::class create Node {
variable key value left right 0 refcount newNode
constructor {n nil instanceFactory} {
set newNode $instanceFactory
set 0 [expr {$nil eq "" ? [self] : $nil}]
set key $n
set value {}
set left [set right $0]
set refcount 0
}
method ref {} {
incr refcount
return [self]
}
method destroy {} {
if {[incr refcount -1] < 1} next
}
method New {key value} {
set n [{*}$newNode $key]
$n setValue $value
return $n
}
# Getters
method key {} {return $key}
method value {} {return $value}
method left {} {return $left}
method right {args} {return $right}
# Setters
method setValue {newValue} {
set value $newValue
}
method setLeft {node} {
# Non-trivial because of reference management
$node ref
$left destroy
set left $node
return
}
method setRight {node} {
# Non-trivial because of reference management
$node ref
$right destroy
set right $node
return
}
# Print a node and its descendents
method print {indent increment} {
puts [format "%s%s => %s" [string repeat " " $indent] $key $value]
incr indent $increment
$left print $indent $increment
$right print $indent $increment
}
method height {} {
return [expr {max([$left height], [$right height]) + 1}]
}
method balanceFactor {} {
expr {[$left height] - [$right height]}
}
method insert {node} {
# Simple insertion
if {$key > [$node key]} {
if {$left eq $0} {
my setLeft $node
} else {
$left insert $node
}
} else {
if {$right eq $0} {
my setRight $node
} else {
$right insert $node
}
}
# Rebalance this node
if {[my balanceFactor] > 1} {
if {[$left balanceFactor] < 0} {
$left rotateLeft
}
my rotateRight
} elseif {[my balanceFactor] < -1} {
if {[$right balanceFactor] > 0} {
$right rotateRight
}
my rotateLeft
}
}
# AVL Rotations
method rotateLeft {} {
set new [my New $key $value]
set key [$right key]
set value [$right value]
$new setLeft $left
$new setRight [$right left]
my setLeft $new
my setRight [$right right]
}
method rotateRight {} {
set new [my New $key $value]
set key [$left key]
set value [$left value]
$new setLeft [$left right]
$new setRight $right
my setLeft [$left left]
my setRight $new
}
}
} |
http://rosettacode.org/wiki/Averages/Mean_angle | Averages/Mean angle | When calculating the average or mean of an angle one has to take into account how angles wrap around so that any angle in degrees plus any integer multiple of 360 degrees is a measure of the same angle.
If one wanted an average direction of the wind over two readings where the first reading was of 350 degrees and the second was of 10 degrees then the average of the numbers is 180 degrees, whereas if you can note that 350 degrees is equivalent to -10 degrees and so you have two readings at 10 degrees either side of zero degrees leading to a more fitting mean angle of zero degrees.
To calculate the mean angle of several angles:
Assume all angles are on the unit circle and convert them to complex numbers expressed in real and imaginary form.
Compute the mean of the complex numbers.
Convert the complex mean to polar coordinates whereupon the phase of the complex mean is the required angular mean.
(Note that, since the mean is the sum divided by the number of numbers, and division by a positive real number does not affect the angle, you can also simply compute the sum for step 2.)
You can alternatively use this formula:
Given the angles
α
1
,
…
,
α
n
{\displaystyle \alpha _{1},\dots ,\alpha _{n}}
the mean is computed by
α
¯
=
atan2
(
1
n
⋅
∑
j
=
1
n
sin
α
j
,
1
n
⋅
∑
j
=
1
n
cos
α
j
)
{\displaystyle {\bar {\alpha }}=\operatorname {atan2} \left({\frac {1}{n}}\cdot \sum _{j=1}^{n}\sin \alpha _{j},{\frac {1}{n}}\cdot \sum _{j=1}^{n}\cos \alpha _{j}\right)}
Task[edit]
write a function/method/subroutine/... that given a list of angles in degrees returns their mean angle.
(You should use a built-in function if you have one that does this for degrees or radians).
Use the function to compute the means of these lists of angles (in degrees):
[350, 10]
[90, 180, 270, 360]
[10, 20, 30]
Show your output here.
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #Processing | Processing | void setup() {
println(meanAngle(350, 10));
println(meanAngle(90, 180, 270, 360));
println(meanAngle(10, 20, 30));
}
float meanAngle(float... angles) {
float sum1 = 0, sum2 = 0;
for (int i = 0; i < angles.length; i++) {
sum1 += sin(radians(angles[i])) / angles.length;
sum2 += cos(radians(angles[i])) / angles.length;
}
return degrees(atan2(sum1, sum2));
} |
http://rosettacode.org/wiki/Averages/Mean_angle | Averages/Mean angle | When calculating the average or mean of an angle one has to take into account how angles wrap around so that any angle in degrees plus any integer multiple of 360 degrees is a measure of the same angle.
If one wanted an average direction of the wind over two readings where the first reading was of 350 degrees and the second was of 10 degrees then the average of the numbers is 180 degrees, whereas if you can note that 350 degrees is equivalent to -10 degrees and so you have two readings at 10 degrees either side of zero degrees leading to a more fitting mean angle of zero degrees.
To calculate the mean angle of several angles:
Assume all angles are on the unit circle and convert them to complex numbers expressed in real and imaginary form.
Compute the mean of the complex numbers.
Convert the complex mean to polar coordinates whereupon the phase of the complex mean is the required angular mean.
(Note that, since the mean is the sum divided by the number of numbers, and division by a positive real number does not affect the angle, you can also simply compute the sum for step 2.)
You can alternatively use this formula:
Given the angles
α
1
,
…
,
α
n
{\displaystyle \alpha _{1},\dots ,\alpha _{n}}
the mean is computed by
α
¯
=
atan2
(
1
n
⋅
∑
j
=
1
n
sin
α
j
,
1
n
⋅
∑
j
=
1
n
cos
α
j
)
{\displaystyle {\bar {\alpha }}=\operatorname {atan2} \left({\frac {1}{n}}\cdot \sum _{j=1}^{n}\sin \alpha _{j},{\frac {1}{n}}\cdot \sum _{j=1}^{n}\cos \alpha _{j}\right)}
Task[edit]
write a function/method/subroutine/... that given a list of angles in degrees returns their mean angle.
(You should use a built-in function if you have one that does this for degrees or radians).
Use the function to compute the means of these lists of angles (in degrees):
[350, 10]
[90, 180, 270, 360]
[10, 20, 30]
Show your output here.
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #PureBasic | PureBasic | NewList angle.d()
Macro AE(x)
AddElement(angle()) : angle()=x
EndMacro
Procedure.d atan3(y.d,x.d)
If x<=0.0 : ProcedureReturn Sign(y)*#PI/2 : EndIf
If x>0.0 : ProcedureReturn ATan(y/x) : EndIf
If y>0.0 : ProcedureReturn ATan(y/x)+#PI : EndIf
ProcedureReturn ATan(y/x)-#PI
EndProcedure
Procedure.d mAngle(List angle.d())
Define.d sumS,sumC
ForEach angle()
sumS+Sin(Radian(angle())) : sumC+Cos(Radian(angle()))
Next
ProcedureReturn Degree(atan3(sumS,sumC))
EndProcedure
AE(350.0) : AE(10.0)
Debug StrD(mAngle(angle()),6) : ClearList(angle())
AE(90.0) : AE(180.0) : AE(270.0) : AE(360.0)
Debug StrD(mAngle(angle()),6) : ClearList(angle())
AE(10.0) : AE(20.0) : AE(30.0)
Debug StrD(mAngle(angle()),6) : ClearList(angle())
|
http://rosettacode.org/wiki/Averages/Median | Averages/Median | Task[edit]
Write a program to find the median value of a vector of floating-point numbers.
The program need not handle the case where the vector is empty, but must handle the case where there are an even number of elements. In that case, return the average of the two middle values.
There are several approaches to this. One is to sort the elements, and then pick the element(s) in the middle.
Sorting would take at least O(n logn). Another approach would be to build a priority queue from the elements, and then extract half of the elements to get to the middle element(s). This would also take O(n logn). The best solution is to use the selection algorithm to find the median in O(n) time.
See also
Quickselect_algorithm
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #Haskell | Haskell | import Data.List (partition)
nth :: Ord t => [t] -> Int -> t
nth (x:xs) n
| k == n = x
| k > n = nth ys n
| otherwise = nth zs $ n - k - 1
where
(ys, zs) = partition (< x) xs
k = length ys
medianMay :: (Fractional a, Ord a) => [a] -> Maybe a
medianMay xs
| n < 1 = Nothing
| even n = Just ((nth xs (div n 2) + nth xs (div n 2 - 1)) / 2.0)
| otherwise = Just (nth xs (div n 2))
where
n = length xs
main :: IO ()
main =
mapM_
(printMay . medianMay)
[[], [7], [5, 3, 4], [5, 4, 2, 3], [3, 4, 1, -8.4, 7.2, 4, 1, 1.2]]
where
printMay = maybe (putStrLn "(not defined)") print |
http://rosettacode.org/wiki/Averages/Pythagorean_means | Averages/Pythagorean means | Task[edit]
Compute all three of the Pythagorean means of the set of integers 1 through 10 (inclusive).
Show that
A
(
x
1
,
…
,
x
n
)
≥
G
(
x
1
,
…
,
x
n
)
≥
H
(
x
1
,
…
,
x
n
)
{\displaystyle A(x_{1},\ldots ,x_{n})\geq G(x_{1},\ldots ,x_{n})\geq H(x_{1},\ldots ,x_{n})}
for this set of positive integers.
The most common of the three means, the arithmetic mean, is the sum of the list divided by its length:
A
(
x
1
,
…
,
x
n
)
=
x
1
+
⋯
+
x
n
n
{\displaystyle A(x_{1},\ldots ,x_{n})={\frac {x_{1}+\cdots +x_{n}}{n}}}
The geometric mean is the
n
{\displaystyle n}
th root of the product of the list:
G
(
x
1
,
…
,
x
n
)
=
x
1
⋯
x
n
n
{\displaystyle G(x_{1},\ldots ,x_{n})={\sqrt[{n}]{x_{1}\cdots x_{n}}}}
The harmonic mean is
n
{\displaystyle n}
divided by the sum of the reciprocal of each item in the list:
H
(
x
1
,
…
,
x
n
)
=
n
1
x
1
+
⋯
+
1
x
n
{\displaystyle H(x_{1},\ldots ,x_{n})={\frac {n}{{\frac {1}{x_{1}}}+\cdots +{\frac {1}{x_{n}}}}}}
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #Logo | Logo | to compute_means :count
local "sum
make "sum 0
local "product
make "product 1
local "reciprocal_sum
make "reciprocal_sum 0
repeat :count [
make "sum sum :sum repcount
make "product product :product repcount
make "reciprocal_sum sum :reciprocal_sum (quotient repcount)
]
output (sentence (quotient :sum :count) (power :product (quotient :count))
(quotient :count :reciprocal_sum))
end
make "means compute_means 10
print sentence [Arithmetic mean is] item 1 :means
print sentence [Geometric mean is] item 2 :means
print sentence [Harmonic mean is] item 3 :means
bye |
http://rosettacode.org/wiki/Balanced_ternary | Balanced ternary | Balanced ternary is a way of representing numbers. Unlike the prevailing binary representation, a balanced ternary integer is in base 3, and each digit can have the values 1, 0, or −1.
Examples
Decimal 11 = 32 + 31 − 30, thus it can be written as "++−"
Decimal 6 = 32 − 31 + 0 × 30, thus it can be written as "+−0"
Task
Implement balanced ternary representation of integers with the following:
Support arbitrarily large integers, both positive and negative;
Provide ways to convert to and from text strings, using digits '+', '-' and '0' (unless you are already using strings to represent balanced ternary; but see requirement 5).
Provide ways to convert to and from native integer type (unless, improbably, your platform's native integer type is balanced ternary). If your native integers can't support arbitrary length, overflows during conversion must be indicated.
Provide ways to perform addition, negation and multiplication directly on balanced ternary integers; do not convert to native integers first.
Make your implementation efficient, with a reasonable definition of "efficient" (and with a reasonable definition of "reasonable").
Test case With balanced ternaries a from string "+-0++0+", b from native integer -436, c "+-++-":
write out a, b and c in decimal notation;
calculate a × (b − c), write out the result in both ternary and decimal notations.
Note: The pages generalised floating point addition and generalised floating point multiplication have code implementing arbitrary precision floating point balanced ternary.
| #Scala | Scala |
object TernaryBit {
val P = TernaryBit(+1)
val M = TernaryBit(-1)
val Z = TernaryBit( 0)
implicit def asChar(t: TernaryBit): Char = t.charValue
implicit def valueOf(c: Char): TernaryBit = {
c match {
case '0' => 0
case '+' => 1
case '-' => -1
case nc => throw new IllegalArgumentException("Illegal ternary symbol " + nc)
}
}
implicit def asInt(t: TernaryBit): Int = t.intValue
implicit def valueOf(i: Int): TernaryBit = TernaryBit(i)
}
case class TernaryBit(val intValue: Int) {
def inverse: TernaryBit = TernaryBit(-intValue)
def charValue = intValue match {
case 0 => '0'
case 1 => '+'
case -1 => '-'
}
}
class Ternary(val bits: List[TernaryBit]) {
def + (b: Ternary) = {
val sumBits: List[Int] = bits.map(_.intValue).zipAll(b.bits.map(_.intValue), 0, 0).map(p => p._1 + p._2)
// normalize
val iv: Tuple2[List[Int], Int] = (List(), 0)
val (revBits, carry) = sumBits.foldLeft(iv)((accu: Tuple2[List[Int], Int], e: Int) => {
val s = e + accu._2
(((s + 1 + 3 * 100) % 3 - 1) :: accu._1 , (s + 1 + 3 * 100) / 3 - 100)
})
new Ternary(( TernaryBit(carry) :: revBits.map(TernaryBit(_))).reverse )
}
def - (b: Ternary) = {this + (-b)}
def <<<(a: Int): Ternary = { List.fill(a)(TernaryBit.Z) ++ bits}
def >>>(a: Int): Ternary = { bits.drop(a) }
def unary_- = { bits.map(_.inverse) }
def ** (b: TernaryBit): Ternary = {
b match {
case TernaryBit.P => this
case TernaryBit.M => - this
case TernaryBit.Z => 0
}
}
def * (mul: Ternary): Ternary = {
// might be done more efficiently - perform normalize only once
mul.bits.reverse.foldLeft(new Ternary(Nil))((a: Ternary, b: TernaryBit) => (a <<< 1) + (this ** b))
}
def intValue = bits.foldRight(0)((c, a) => a*3 + c.intValue)
override def toString = new String(bits.reverse.map(_.charValue).toArray)
}
object Ternary {
implicit def asString(t: Ternary): String = t.toString()
implicit def valueOf(s: String): Ternary = new Ternary(s.toList.reverse.map(TernaryBit.valueOf(_)))
implicit def asBits(t: Ternary): List[TernaryBit] = t.bits
implicit def valueOf(l: List[TernaryBit]): Ternary = new Ternary(l)
implicit def asInt(t: Ternary): BigInt = t.intValue
// XXX not tail recursive
implicit def valueOf(i: BigInt): Ternary = {
if (i < 0) -valueOf(-i)
else if (i == 0) new Ternary(List())
else if (i % 3 == 0) TernaryBit.Z :: valueOf(i / 3)
else if (i % 3 == 1) TernaryBit.P :: valueOf(i / 3)
else /*(i % 3 == 2)*/ TernaryBit.M :: valueOf((i + 1) / 3)
}
implicit def intToTernary(i: Int): Ternary = valueOf(i)
}
|
http://rosettacode.org/wiki/Babbage_problem | Babbage problem |
Charles Babbage, looking ahead to the sorts of problems his Analytical Engine would be able to solve, gave this example:
What is the smallest positive integer whose square ends in the digits 269,696?
— Babbage, letter to Lord Bowden, 1837; see Hollingdale and Tootill, Electronic Computers, second edition, 1970, p. 125.
He thought the answer might be 99,736, whose square is 9,947,269,696; but he couldn't be certain.
Task[edit]
The task is to find out if Babbage had the right answer — and to do so, as far as your language allows it, in code that Babbage himself would have been able to read and understand.
As Babbage evidently solved the task with pencil and paper, a similar efficient solution is preferred.
For these purposes, Charles Babbage may be taken to be an intelligent person, familiar with mathematics and with the idea of a computer; he has written the first drafts of simple computer programmes in tabular form. [Babbage Archive Series L].
Motivation
The aim of the task is to write a program that is sufficiently clear and well-documented for such a person to be able to read it and be confident that it does indeed solve the specified problem.
| #SenseTalk | SenseTalk |
(*
Answer Charles Babbage's question:
What is the smallest positive integer whose square ends in the digits 269,696?
*)
put 1 into int
repeat forever
if int squared ends with "269696" then
put "The smallest positive integer whose square ends in the digits 269696 is" && int
exit all -- don't keep repeating forever!
end if
add 1 to int
end repeat
|
http://rosettacode.org/wiki/Babbage_problem | Babbage problem |
Charles Babbage, looking ahead to the sorts of problems his Analytical Engine would be able to solve, gave this example:
What is the smallest positive integer whose square ends in the digits 269,696?
— Babbage, letter to Lord Bowden, 1837; see Hollingdale and Tootill, Electronic Computers, second edition, 1970, p. 125.
He thought the answer might be 99,736, whose square is 9,947,269,696; but he couldn't be certain.
Task[edit]
The task is to find out if Babbage had the right answer — and to do so, as far as your language allows it, in code that Babbage himself would have been able to read and understand.
As Babbage evidently solved the task with pencil and paper, a similar efficient solution is preferred.
For these purposes, Charles Babbage may be taken to be an intelligent person, familiar with mathematics and with the idea of a computer; he has written the first drafts of simple computer programmes in tabular form. [Babbage Archive Series L].
Motivation
The aim of the task is to write a program that is sufficiently clear and well-documented for such a person to be able to read it and be confident that it does indeed solve the specified problem.
| #SequenceL | SequenceL | main() := babbage(0);
babbage(current) :=
current when current * current mod 1000000 = 269696
else
babbage(current + 1); |
http://rosettacode.org/wiki/Approximate_equality | Approximate equality | Sometimes, when testing whether the solution to a task (for example, here on Rosetta Code) is correct, the
difference in floating point calculations between different language implementations becomes significant.
For example, a difference between 32 bit and 64 bit floating point calculations may appear by
about the 8th significant digit in base 10 arithmetic.
Task
Create a function which returns true if two floating point numbers are approximately equal.
The function should allow for differences in the magnitude of numbers, so that, for example,
100000000000000.01 may be approximately equal to 100000000000000.011,
even though 100.01 is not approximately equal to 100.011.
If the language has such a feature in its standard library, this may be used instead of a custom function.
Show the function results with comparisons on the following pairs of values:
100000000000000.01, 100000000000000.011 (note: should return true)
100.01, 100.011 (note: should return false)
10000000000000.001 / 10000.0, 1000000000.0000001000
0.001, 0.0010000001
0.000000000000000000000101, 0.0
sqrt(2) * sqrt(2), 2.0
-sqrt(2) * sqrt(2), -2.0
3.14159265358979323846, 3.14159265358979324
Answers should be true for the first example and false in the second, so that just rounding the numbers to a fixed number of decimals should not be enough. Otherwise answers may vary and still be correct. See the Python code for one type of solution.
| #Smalltalk | Smalltalk | { #(100000000000000.01 100000000000000.011) .
#(100.01 100.011) .
{10000000000000.001 / 10000.0 . 1000000000.0000001000} .
#(0.001 0.0010000001) .
#(0.000000000000000000000101 0.0) .
{ 2 sqrt * 2 sqrt . 2.0} .
{ 2 sqrt negated * 2 sqrt . -2.0} .
#(3.14159265358979323846 3.14159265358979324)
} pairsDo:[:val1 :val2 |
Stdout printCR: e'{val1} =~= {val2} -> {val1 isAlmostEqualTo:val2 nEpsilon:2}'
] |
http://rosettacode.org/wiki/Approximate_equality | Approximate equality | Sometimes, when testing whether the solution to a task (for example, here on Rosetta Code) is correct, the
difference in floating point calculations between different language implementations becomes significant.
For example, a difference between 32 bit and 64 bit floating point calculations may appear by
about the 8th significant digit in base 10 arithmetic.
Task
Create a function which returns true if two floating point numbers are approximately equal.
The function should allow for differences in the magnitude of numbers, so that, for example,
100000000000000.01 may be approximately equal to 100000000000000.011,
even though 100.01 is not approximately equal to 100.011.
If the language has such a feature in its standard library, this may be used instead of a custom function.
Show the function results with comparisons on the following pairs of values:
100000000000000.01, 100000000000000.011 (note: should return true)
100.01, 100.011 (note: should return false)
10000000000000.001 / 10000.0, 1000000000.0000001000
0.001, 0.0010000001
0.000000000000000000000101, 0.0
sqrt(2) * sqrt(2), 2.0
-sqrt(2) * sqrt(2), -2.0
3.14159265358979323846, 3.14159265358979324
Answers should be true for the first example and false in the second, so that just rounding the numbers to a fixed number of decimals should not be enough. Otherwise answers may vary and still be correct. See the Python code for one type of solution.
| #Swift | Swift | import Foundation
extension FloatingPoint {
@inlinable
public func isAlmostEqual(
to other: Self,
tolerance: Self = Self.ulpOfOne.squareRoot()
) -> Bool {
// tolerances outside of [.ulpOfOne,1) yield well-defined but useless results,
// so this is enforced by an assert rathern than a precondition.
assert(tolerance >= .ulpOfOne && tolerance < 1, "tolerance should be in [.ulpOfOne, 1).")
// The simple computation below does not necessarily give sensible
// results if one of self or other is infinite; we need to rescale
// the computation in that case.
guard self.isFinite && other.isFinite else {
return rescaledAlmostEqual(to: other, tolerance: tolerance)
}
// This should eventually be rewritten to use a scaling facility to be
// defined on FloatingPoint suitable for hypot and scaled sums, but the
// following is good enough to be useful for now.
let scale = max(abs(self), abs(other), .leastNormalMagnitude)
return abs(self - other) < scale*tolerance
}
@usableFromInline
internal func rescaledAlmostEqual(to other: Self, tolerance: Self) -> Bool {
// NaN is considered to be not approximately equal to anything, not even
// itself.
if self.isNaN || other.isNaN { return false }
if self.isInfinite {
if other.isInfinite { return self == other }
// Self is infinite and other is finite. Replace self with the binade
// of the greatestFiniteMagnitude, and reduce the exponent of other by
// one to compensate.
let scaledSelf = Self(sign: self.sign,
exponent: Self.greatestFiniteMagnitude.exponent,
significand: 1)
let scaledOther = Self(sign: .plus,
exponent: -1,
significand: other)
// Now both values are finite, so re-run the naive comparison.
return scaledSelf.isAlmostEqual(to: scaledOther, tolerance: tolerance)
}
// If self is finite and other is infinite, flip order and use scaling
// defined above, since this relation is symmetric.
return other.rescaledAlmostEqual(to: self, tolerance: tolerance)
}
}
let testCases = [
(100000000000000.01, 100000000000000.011),
(100.01, 100.011),
(10000000000000.001 / 10000.0, 1000000000.0000001000),
(0.001, 0.0010000001),
(0.000000000000000000000101, 0.0),
(sqrt(2) * sqrt(2), 2.0),
(-sqrt(2) * sqrt(2), -2.0),
(3.14159265358979323846, 3.14159265358979324)
]
for testCase in testCases {
print("\(testCase.0), \(testCase.1) => \(testCase.0.isAlmostEqual(to: testCase.1))")
} |
http://rosettacode.org/wiki/Approximate_equality | Approximate equality | Sometimes, when testing whether the solution to a task (for example, here on Rosetta Code) is correct, the
difference in floating point calculations between different language implementations becomes significant.
For example, a difference between 32 bit and 64 bit floating point calculations may appear by
about the 8th significant digit in base 10 arithmetic.
Task
Create a function which returns true if two floating point numbers are approximately equal.
The function should allow for differences in the magnitude of numbers, so that, for example,
100000000000000.01 may be approximately equal to 100000000000000.011,
even though 100.01 is not approximately equal to 100.011.
If the language has such a feature in its standard library, this may be used instead of a custom function.
Show the function results with comparisons on the following pairs of values:
100000000000000.01, 100000000000000.011 (note: should return true)
100.01, 100.011 (note: should return false)
10000000000000.001 / 10000.0, 1000000000.0000001000
0.001, 0.0010000001
0.000000000000000000000101, 0.0
sqrt(2) * sqrt(2), 2.0
-sqrt(2) * sqrt(2), -2.0
3.14159265358979323846, 3.14159265358979324
Answers should be true for the first example and false in the second, so that just rounding the numbers to a fixed number of decimals should not be enough. Otherwise answers may vary and still be correct. See the Python code for one type of solution.
| #Tcl | Tcl | catch {namespace delete test_almost_equal_decimal} ;# Start with a clean namespace
namespace eval test_almost_equal_decimal {
package require Tcl 8.5 ;# required by tcllib
package require math::decimal ;# from tcllib
namespace import ::math::decimal::* ;# for: setVariable, fromstr, and compare
array set yesno {0 Yes -1 No 1 No} ;# For nice output
# More info here: http://speleotrove.com/decimal/dax3274.html
# This puts the library into "simplified" mode. Which
# rounds the "decimal digits" in the coefficient to the
# number of digits that "precision" is set to.
setVariable extended 0
setVariable precision 9
set data {
{100000000000000.01 100000000000000.011}
{100.01 100.011}
{[expr {10000000000000.001 / 10000.0}] 1000000000.0000001000}
{0.001 0.0010000001}
{0.000000000000000000000101 0.0}
{[expr { sqrt(2) * sqrt(2)}] 2.0}
{[expr {-sqrt(2) * sqrt(2)}] -2.0}
{3.14159265358979323846 3.14159265358979324}
}
set data [subst $data] ;# resolves expressions in the list
foreach {a b} [join $data] {
set a_d [fromstr $a]
set b_d [fromstr $b]
puts [format "Is %26s ≈ %21s ? %4s." $a $b $yesno([compare $a_d $b_d])]
}
}
|
http://rosettacode.org/wiki/Balanced_brackets | Balanced brackets | Task:
Generate a string with N opening brackets [ and with N closing brackets ], in some arbitrary order.
Determine whether the generated string is balanced; that is, whether it consists entirely of pairs of opening/closing brackets (in that order), none of which mis-nest.
Examples
(empty) OK
[] OK
[][] OK
[[][]] OK
][ NOT OK
][][ NOT OK
[]][[] NOT OK
| #Erlang | Erlang |
-module( balanced_brackets ).
-export( [generate/1, is_balanced/1, task/0] ).
generate( N ) ->
[generate_bracket(random:uniform()) || _X <- lists:seq(1, 2*N)].
is_balanced( String ) -> is_balanced_loop( String, 0 ).
task() ->
lists:foreach( fun (N) ->
String = generate( N ),
Result = is_balanced( String ),
io:fwrite( "~s is ~s~n", [String, task_balanced(Result)] )
end,
lists:seq(0, 5) ).
is_balanced_loop( _String, N ) when N < 0 -> false;
is_balanced_loop( [], 0 ) -> true;
is_balanced_loop( [], _N ) -> false;
is_balanced_loop( [$[ | T], N ) -> is_balanced_loop( T, N + 1 );
is_balanced_loop( [$] | T], N ) -> is_balanced_loop( T, N - 1 ).
generate_bracket( N ) when N =< 0.5 -> $[;
generate_bracket( N ) when N > 0.5 -> $].
task_balanced( true ) -> "OK";
task_balanced( false ) -> "NOT OK".
|
http://rosettacode.org/wiki/Append_a_record_to_the_end_of_a_text_file | Append a record to the end of a text file | Many systems offer the ability to open a file for writing, such that any data written will be appended to the end of the file. Further, the file operations will always adjust the position pointer to guarantee the end of the file, even in a multitasking environment.
This feature is most useful in the case of log files, where many jobs may be appending to the log file at the same time, or where care must be taken to avoid concurrently overwriting the same record from another job.
Task
Given a two record sample for a mythical "passwd" file:
Write these records out in the typical system format.
Ideally these records will have named fields of various types.
Close the file, then reopen the file for append.
Append a new record to the file and close the file again.
Take appropriate care to avoid concurrently overwrites from another job.
Open the file and demonstrate the new record has indeed written to the end.
Source record field types and contents.
account
password
UID
GID
fullname,office,extension,homephone,email
directory
shell
string
string
int
int
struct(string,string,string,string,string)
string
string
jsmith
x
1001
1000
Joe Smith,Room 1007,(234)555-8917,(234)555-0077,[email protected]
/home/jsmith
/bin/bash
jdoe
x
1002
1000
Jane Doe,Room 1004,(234)555-8914,(234)555-0044,[email protected]
/home/jdoe
/bin/bash
Record to be appended.
account
password
UID
GID
fullname,office,extension,homephone,email
directory
shell
string
string
int
int
struct(string,string,string,string,string)
string
string
xyz
x
1003
1000
X Yz,Room 1003,(234)555-8913,(234)555-0033,[email protected]
/home/xyz
/bin/bash
Resulting file format: should mimic Linux's /etc/passwd file format with particular attention to the "," separator used in the GECOS field. But if the specific language has a particular or unique format of storing records in text file, then this format should be named and demonstrated with an additional example.
Expected output:
Appended record: xyz:x:1003:1000:X Yz,Room 1003,(234)555-8913,(234)555-0033,[email protected]:/home/xyz:/bin/bash
Finally: Provide a summary of the language's "append record" capabilities in a table. eg.
Append Capabilities.
Data Representation
IO
Library
Append
Possible
Automatic
Append
Multi-tasking
Safe
In core
On disk
C struct
CSV text file
glibc/stdio
☑
☑
☑ (Not all, eg NFS)
Alternatively: If the language's appends can not guarantee its writes will always append, then note this restriction in the table. If possible, provide an actual code example (possibly using file/record locking) to guarantee correct concurrent appends.
| #Icon_and_Unicon | Icon and Unicon | procedure main()
orig := [
"jsmith:x:1001:1000:Joe Smith,Room 1007,(234)555-8917,(234)555-0077,[email protected]:/home/jsmith:/bin/bash",
"jdoe:x:1002:1000:Jane Doe,Room 1004,(234)555-8914,(234)555-0044,[email protected]:/home/jsmith:/bin/bash"
]
new := [
"xyz:x:1003:1000:X:Yz,Room 1003,(234)555-8913,(234)555-0033,[email protected]:/home/xyz:/bin/bash"
]
fName := !open("mktemp","rp")
every (f := open(fName,"w")) | write(f,!orig) | close(f)
every (f := open(fName,"a")) | write(f,!new) | close(f)
every (f := open(fName,"r")) | write(!f) | close(f)
end |
http://rosettacode.org/wiki/Associative_array/Creation | Associative array/Creation | Task
The goal is to create an associative array (also known as a dictionary, map, or hash).
Related tasks:
Associative arrays/Iteration
Hash from two arrays
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #Babel | Babel |
(("foo" 13)
("bar" 42)
("baz" 77)) ls2map ! |
http://rosettacode.org/wiki/Anti-primes | Anti-primes | The anti-primes
(or highly composite numbers, sequence A002182 in the OEIS)
are the natural numbers with more factors than any smaller than itself.
Task
Generate and show here, the first twenty anti-primes.
Related tasks
Factors of an integer
Sieve of Eratosthenes
| #Elixir | Elixir | defmodule AntiPrimes do
def divcount(n) when is_integer(n), do: divcount(n, 1, 0)
def divcount(n, d, count) when d * d > n, do: count
def divcount(n, d, count) do
divs = case rem(n, d) do
0 ->
case n - d * d do
0 -> 1
_ -> 2
end
_ -> 0
end
divcount(n, d + 1, count + divs)
end
def antiprimes(n), do: antiprimes(n, 1, 0, [])
def antiprimes(0, _, _, l), do: Enum.reverse(l)
def antiprimes(n, m, max, l) do
count = divcount(m)
case count > max do
true -> antiprimes(n-1, m+1, count, [m|l])
false -> antiprimes(n, m+1, max, l)
end
end
def main() do
:io.format("The first 20 anti-primes are ~w~n", [antiprimes(20)])
end
end |
http://rosettacode.org/wiki/Anti-primes | Anti-primes | The anti-primes
(or highly composite numbers, sequence A002182 in the OEIS)
are the natural numbers with more factors than any smaller than itself.
Task
Generate and show here, the first twenty anti-primes.
Related tasks
Factors of an integer
Sieve of Eratosthenes
| #Erlang | Erlang | divcount(N) -> divcount(N, 1, 0).
divcount(N, D, Count) when D*D > N -> Count;
divcount(N, D, Count) ->
Divs = case N rem D of
0 ->
case N - D*D of
0 -> 1;
_ -> 2
end;
_ -> 0
end,
divcount(N, D + 1, Count + Divs).
antiprimes(N) -> antiprimes(N, 1, 0, []).
antiprimes(0, _, _, L) -> lists:reverse(L);
antiprimes(N, M, Max, L) ->
Count = divcount(M),
case Count > Max of
true -> antiprimes(N-1, M+1, Count, [M|L]);
false -> antiprimes(N, M+1, Max, L)
end.
main(_) ->
io:format("The first 20 anti-primes are ~w~n", [antiprimes(20)]).
|
http://rosettacode.org/wiki/Atomic_updates | Atomic updates |
Task
Define a data type consisting of a fixed number of 'buckets', each containing a nonnegative integer value, which supports operations to:
get the current value of any bucket
remove a specified amount from one specified bucket and add it to another, preserving the total of all bucket values, and clamping the transferred amount to ensure the values remain non-negative
In order to exercise this data type, create one set of buckets, and start three concurrent tasks:
As often as possible, pick two buckets and make their values closer to equal.
As often as possible, pick two buckets and arbitrarily redistribute their values.
At whatever rate is convenient, display (by any means) the total value and, optionally, the individual values of each bucket.
The display task need not be explicit; use of e.g. a debugger or trace tool is acceptable provided it is simple to set up to provide the display.
This task is intended as an exercise in atomic operations. The sum of the bucket values must be preserved even if the two tasks attempt to perform transfers simultaneously, and a straightforward solution is to ensure that at any time, only one transfer is actually occurring — that the transfer operation is atomic.
| #Phix | Phix | without js -- (no threads or critical sections in JavaScript)
constant nBuckets = 20
sequence buckets = tagset(nBuckets) -- {1,2,3,..,20}
constant bucket_cs = init_cs() -- critical section
atom equals = 0, rands = 0 -- operation counts
integer terminate = 0 -- control flag
procedure mythreads(integer eq)
-- if eq then equalise else randomise
integer b1,b2,amt
while not terminate do
b1 = rand(nBuckets)
b2 = rand(nBuckets)
if b1!=b2 then -- (test not actually needed)
enter_cs(bucket_cs)
if eq then
amt = floor((buckets[b1]-buckets[b2])/2)
equals += 1
else
amt = rand(buckets[b1]+1)-1
rands += 1
end if
buckets[b1] -= amt
buckets[b2] += amt
leave_cs(bucket_cs)
end if
end while
exit_thread(0)
end procedure
procedure display()
enter_cs(bucket_cs)
?{sum(buckets),equals,rands,buckets}
leave_cs(bucket_cs)
end procedure
display()
constant threads = {create_thread(routine_id("mythreads"),{1}), -- equalise
create_thread(routine_id("mythreads"),{0})} -- randomise
constant ESC = #1B
while not find(get_key(),{ESC,'q','Q'}) do
sleep(1)
display()
end while
terminate = 1
wait_thread(threads)
delete_cs(bucket_cs)
|
http://rosettacode.org/wiki/Assertions | Assertions | Assertions are a way of breaking out of code when there is an error or an unexpected input.
Some languages throw exceptions and some treat it as a break point.
Task
Show an assertion in your language by asserting that an integer variable is equal to 42.
| #Modula-3 | Modula-3 | <*ASSERT a = 42*> |
http://rosettacode.org/wiki/Assertions | Assertions | Assertions are a way of breaking out of code when there is an error or an unexpected input.
Some languages throw exceptions and some treat it as a break point.
Task
Show an assertion in your language by asserting that an integer variable is equal to 42.
| #Nanoquery | Nanoquery | a = 5
assert (a = 42) |
http://rosettacode.org/wiki/Assertions | Assertions | Assertions are a way of breaking out of code when there is an error or an unexpected input.
Some languages throw exceptions and some treat it as a break point.
Task
Show an assertion in your language by asserting that an integer variable is equal to 42.
| #Nemerle | Nemerle | assert (foo == 42, $"foo == $foo, not 42.") |
http://rosettacode.org/wiki/Assertions | Assertions | Assertions are a way of breaking out of code when there is an error or an unexpected input.
Some languages throw exceptions and some treat it as a break point.
Task
Show an assertion in your language by asserting that an integer variable is equal to 42.
| #NGS | NGS | a = 42
assert(a==42)
assert(a, 42)
assert(a==42, "Not 42!")
assert(a, 42, "Not 42!") |
http://rosettacode.org/wiki/Apply_a_callback_to_an_array | Apply a callback to an array | Task
Take a combined set of elements and apply a function to each element.
| #Common_Lisp | Common Lisp | (map nil #'print #(1 2 3 4 5)) |
http://rosettacode.org/wiki/Apply_a_callback_to_an_array | Apply a callback to an array | Task
Take a combined set of elements and apply a function to each element.
| #Component_Pascal | Component Pascal |
MODULE Callback;
IMPORT StdLog;
TYPE
Callback = PROCEDURE (x: INTEGER;OUT doubled: INTEGER);
Callback2 = PROCEDURE (x: INTEGER): INTEGER;
PROCEDURE Apply(proc: Callback; VAR x: ARRAY OF INTEGER);
VAR
i: INTEGER;
BEGIN
FOR i := 0 TO LEN(x) - 1 DO;
proc(x[i],x[i]);
END
END Apply;
PROCEDURE Apply2(func: Callback2; VAR x: ARRAY OF INTEGER);
VAR
i: INTEGER;
BEGIN
FOR i := 0 TO LEN(x) - 1 DO;
x[i] := func(x[i]);
END
END Apply2;
PROCEDURE Double(x: INTEGER; OUT y: INTEGER);
BEGIN
y := x * x;
END Double;
PROCEDURE Double2(x: INTEGER): INTEGER;
BEGIN
RETURN x * x
END Double2;
PROCEDURE Do*;
VAR
i: INTEGER;
ary: ARRAY 10 OF INTEGER;
BEGIN
FOR i := 0 TO LEN(ary) - 1 DO ary[i] := i END;
Apply(Double,ary);
FOR i := 0 TO LEN(ary) - 1 DO
StdLog.Int(ary[i]);StdLog.Ln
END;
StdLog.Ln;
Apply2(Double2,ary);
FOR i := 0 TO LEN(ary) - 1 DO
StdLog.Int(ary[i]);StdLog.Ln
END
END Do;
END Callback.
|
http://rosettacode.org/wiki/Averages/Mode | Averages/Mode | Task[edit]
Write a program to find the mode value of a collection.
The case where the collection is empty may be ignored. Care must be taken to handle the case where the mode is non-unique.
If it is not appropriate or possible to support a general collection, use a vector (array), if possible. If it is not appropriate or possible to support an unspecified value type, use integers.
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #Perl | Perl | use strict;
use List::Util qw(max);
sub mode
{
my %c;
foreach my $e ( @_ ) {
$c{$e}++;
}
my $best = max(values %c);
return grep { $c{$_} == $best } keys %c;
} |
http://rosettacode.org/wiki/Associative_array/Iteration | Associative array/Iteration | Show how to iterate over the key-value pairs of an associative array, and print each pair out.
Also show how to iterate just over the keys, or the values, if there is a separate way to do that in your language.
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #EchoLisp | EchoLisp |
(lib 'hash) ;; load hash.lib
(define H (make-hash))
;; fill hash table
(hash-set H 'Simon 42)
(hash-set H 'Albert 666)
(hash-set H 'Antoinette 33)
;; iterate over (key . value ) pairs
(for ([kv H]) (writeln kv))
(Simon . 42)
(Albert . 666)
(Antoinette . 33)
;; iterate over keys
(for ([k (hash-keys H)]) (writeln 'key-> k))
key-> Simon
key-> Albert
key-> Antoinette
;; iterate over values
(for ([v (hash-values H)]) (writeln 'value-> v))
value-> 42
value-> 666
value-> 33
|
http://rosettacode.org/wiki/Apply_a_digital_filter_(direct_form_II_transposed) | Apply a digital filter (direct form II transposed) | Digital filters are used to apply a mathematical operation to a sampled signal. One of the common formulations is the "direct form II transposed" which can represent both infinite impulse response (IIR) and finite impulse response (FIR) filters, as well as being more numerically stable than other forms. [1]
Task
Filter a signal using an order 3 low-pass Butterworth filter. The coefficients for the filter are a=[1.00000000, -2.77555756e-16, 3.33333333e-01, -1.85037171e-17] and b = [0.16666667, 0.5, 0.5, 0.16666667]
The signal that needs filtering is the following vector: [-0.917843918645, 0.141984778794, 1.20536903482, 0.190286794412, -0.662370894973, -1.00700480494, -0.404707073677 ,0.800482325044, 0.743500089861, 1.01090520172, 0.741527555207, 0.277841675195, 0.400833448236, -0.2085993586, -0.172842103641, -0.134316096293, 0.0259303398477, 0.490105989562, 0.549391221511, 0.9047198589]
| #Phixmonti | Phixmonti | include ..\Utilitys.pmt
( 1.00000000 -2.77555756e-16 3.33333333e-01 -1.85037171e-17 ) var a
( 0.16666667 0.5 0.5 0.16666667 ) var b
( -0.917843918645 0.141984778794 1.20536903482 0.190286794412 -0.662370894973
-1.00700480494 -0.404707073677 0.800482325044 0.743500089861 1.01090520172
0.741527555207 0.277841675195 0.400833448236 -0.2085993586 -0.172842103641
-0.134316096293 0.0259303398477 0.490105989562 0.549391221511 0.9047198589 )
len dup 0 swap repeat >ps
for var i
0 >ps
b len i min for var j
j get rot i j - 1 + get rot * ps> + >ps swap
endfor
drop
a len i min for var j
j get ps> tps i j - 1 + get nip rot * - >ps
endfor
drop
ps> a 1 get nip / ps> swap i set >ps
endfor
drop ps> ? |
http://rosettacode.org/wiki/Apply_a_digital_filter_(direct_form_II_transposed) | Apply a digital filter (direct form II transposed) | Digital filters are used to apply a mathematical operation to a sampled signal. One of the common formulations is the "direct form II transposed" which can represent both infinite impulse response (IIR) and finite impulse response (FIR) filters, as well as being more numerically stable than other forms. [1]
Task
Filter a signal using an order 3 low-pass Butterworth filter. The coefficients for the filter are a=[1.00000000, -2.77555756e-16, 3.33333333e-01, -1.85037171e-17] and b = [0.16666667, 0.5, 0.5, 0.16666667]
The signal that needs filtering is the following vector: [-0.917843918645, 0.141984778794, 1.20536903482, 0.190286794412, -0.662370894973, -1.00700480494, -0.404707073677 ,0.800482325044, 0.743500089861, 1.01090520172, 0.741527555207, 0.277841675195, 0.400833448236, -0.2085993586, -0.172842103641, -0.134316096293, 0.0259303398477, 0.490105989562, 0.549391221511, 0.9047198589]
| #Python | Python | #!/bin/python
from __future__ import print_function
from scipy import signal
import matplotlib.pyplot as plt
if __name__=="__main__":
sig = [-0.917843918645,0.141984778794,1.20536903482,0.190286794412,-0.662370894973,-1.00700480494,
-0.404707073677,0.800482325044,0.743500089861,1.01090520172,0.741527555207,
0.277841675195,0.400833448236,-0.2085993586,-0.172842103641,-0.134316096293,
0.0259303398477,0.490105989562,0.549391221511,0.9047198589]
#Create an order 3 lowpass butterworth filter
#Generated using b, a = signal.butter(3, 0.5)
a = [1.00000000, -2.77555756e-16, 3.33333333e-01, -1.85037171e-17]
b = [0.16666667, 0.5, 0.5, 0.16666667]
#Apply the filter to signal
filt = signal.lfilter(b, a, sig)
print (filt)
plt.plot(sig, 'b')
plt.plot(filt, 'r--')
plt.show() |
http://rosettacode.org/wiki/Averages/Arithmetic_mean | Averages/Arithmetic mean | Task[edit]
Write a program to find the mean (arithmetic average) of a numeric vector.
In case of a zero-length input, since the mean of an empty set of numbers is ill-defined, the program may choose to behave in any way it deems appropriate, though if the programming language has an established convention for conveying math errors or undefined values, it's preferable to follow it.
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #Erlang | Erlang | mean([]) -> 0;
mean(L) -> lists:sum(L)/erlang:length(L). |
http://rosettacode.org/wiki/Average_loop_length | Average loop length | Let f be a uniformly-randomly chosen mapping from the numbers 1..N to the numbers 1..N (note: not necessarily a permutation of 1..N; the mapping could produce a number in more than one way or not at all). At some point, the sequence 1, f(1), f(f(1))... will contain a repetition, a number that occurring for the second time in the sequence.
Task
Write a program or a script that estimates, for each N, the average length until the first such repetition.
Also calculate this expected length using an analytical formula, and optionally compare the simulated result with the theoretical one.
This problem comes from the end of Donald Knuth's Christmas tree lecture 2011.
Example of expected output:
N average analytical (error)
=== ========= ============ =========
1 1.0000 1.0000 ( 0.00%)
2 1.4992 1.5000 ( 0.05%)
3 1.8784 1.8889 ( 0.56%)
4 2.2316 2.2188 ( 0.58%)
5 2.4982 2.5104 ( 0.49%)
6 2.7897 2.7747 ( 0.54%)
7 3.0153 3.0181 ( 0.09%)
8 3.2429 3.2450 ( 0.07%)
9 3.4536 3.4583 ( 0.14%)
10 3.6649 3.6602 ( 0.13%)
11 3.8091 3.8524 ( 1.12%)
12 3.9986 4.0361 ( 0.93%)
13 4.2074 4.2123 ( 0.12%)
14 4.3711 4.3820 ( 0.25%)
15 4.5275 4.5458 ( 0.40%)
16 4.6755 4.7043 ( 0.61%)
17 4.8877 4.8579 ( 0.61%)
18 4.9951 5.0071 ( 0.24%)
19 5.1312 5.1522 ( 0.41%)
20 5.2699 5.2936 ( 0.45%)
| #Simula | Simula | BEGIN
REAL PROCEDURE FACTORIAL(N); INTEGER N;
BEGIN
REAL RESULT;
INTEGER I;
RESULT := 1.0;
FOR I := 2 STEP 1 UNTIL N DO
RESULT := RESULT * I;
FACTORIAL := RESULT;
END FACTORIAL;
REAL PROCEDURE ANALYTICAL (N); INTEGER N;
BEGIN
REAL SUM, RN;
INTEGER I;
RN := N;
FOR I := 1 STEP 1 UNTIL N DO
BEGIN
SUM := SUM + FACTORIAL(N) / FACTORIAL(N - I) / RN ** I;
END;
ANALYTICAL := SUM;
END ANALYTICAL;
REAL PROCEDURE EXPERIMENTAL(N); INTEGER N;
BEGIN
INTEGER NUM;
INTEGER COUNT;
INTEGER RUN;
FOR RUN := 1 STEP 1 UNTIL TESTS DO
BEGIN
BOOLEAN ARRAY BITS(1:N);
INTEGER I;
FOR I := 1 STEP 1 UNTIL N DO
BEGIN
NUM := RANDINT(1,N,SEED);
IF BITS(NUM) THEN GOTO L;
BITS(NUM) := TRUE;
COUNT := COUNT + 1;
END FOR I;
L:
END FOR RUN;
EXPERIMENTAL := COUNT / TESTS;
END EXPERIMENTAL;
INTEGER SEED, TESTS;
SEED := ININT;
TESTS := 1000000;
BEGIN
REAL A, E, ERR;
INTEGER I;
OUTTEXT(" N AVG CALC %DIFF"); OUTIMAGE;
FOR I := 1 STEP 1 UNTIL 20 DO
BEGIN
A := ANALYTICAL(I);
E := EXPERIMENTAL(I);
ERR := (ABS(E-A)/A)*100.0;
OUTINT(I, 2);
OUTFIX(E, 4, 7);
OUTFIX(A, 4, 10);
OUTFIX(ERR, 4, 10);
OUTIMAGE;
END FOR I;
END;
END |
http://rosettacode.org/wiki/Average_loop_length | Average loop length | Let f be a uniformly-randomly chosen mapping from the numbers 1..N to the numbers 1..N (note: not necessarily a permutation of 1..N; the mapping could produce a number in more than one way or not at all). At some point, the sequence 1, f(1), f(f(1))... will contain a repetition, a number that occurring for the second time in the sequence.
Task
Write a program or a script that estimates, for each N, the average length until the first such repetition.
Also calculate this expected length using an analytical formula, and optionally compare the simulated result with the theoretical one.
This problem comes from the end of Donald Knuth's Christmas tree lecture 2011.
Example of expected output:
N average analytical (error)
=== ========= ============ =========
1 1.0000 1.0000 ( 0.00%)
2 1.4992 1.5000 ( 0.05%)
3 1.8784 1.8889 ( 0.56%)
4 2.2316 2.2188 ( 0.58%)
5 2.4982 2.5104 ( 0.49%)
6 2.7897 2.7747 ( 0.54%)
7 3.0153 3.0181 ( 0.09%)
8 3.2429 3.2450 ( 0.07%)
9 3.4536 3.4583 ( 0.14%)
10 3.6649 3.6602 ( 0.13%)
11 3.8091 3.8524 ( 1.12%)
12 3.9986 4.0361 ( 0.93%)
13 4.2074 4.2123 ( 0.12%)
14 4.3711 4.3820 ( 0.25%)
15 4.5275 4.5458 ( 0.40%)
16 4.6755 4.7043 ( 0.61%)
17 4.8877 4.8579 ( 0.61%)
18 4.9951 5.0071 ( 0.24%)
19 5.1312 5.1522 ( 0.41%)
20 5.2699 5.2936 ( 0.45%)
| #Tcl | Tcl | # Generate a list of the numbers increasing from $a to $b
proc range {a b} {
for {set result {}} {$a <= $b} {incr a} {lappend result $a}
return $result
}
# Computing the expected value analytically
proc tcl::mathfunc::factorial n {
::tcl::mathop::* {*}[range 2 $n]
}
proc Analytical {n} {
set sum 0.0
foreach x [range 1 $n] {
set sum [expr {$sum + factorial($n) / factorial($n-$x) / double($n)**$x}]
}
return $sum
}
# Determining an approximation to the value experimentally
proc Experimental {n numTests} {
set count 0
set u0 [lrepeat $n 1]
foreach run [range 1 $numTests] {
set unseen $u0
for {set i 0} {[lindex $unseen $i]} {incr count} {
lset unseen $i 0
set i [expr {int(rand()*$n)}]
}
}
return [expr {$count / double($numTests)}]
}
# Tabulate the results in exactly the original format
puts " N average analytical (error)"
puts "=== ========= ============ ========="
foreach n [range 1 20] {
set a [Analytical $n]
set e [Experimental $n 100000]
puts [format "%3d %9.4f %12.4f (%6.2f%%)" $n $e $a [expr {abs($e-$a)/$a*100.0}]]
} |
http://rosettacode.org/wiki/Averages/Simple_moving_average | Averages/Simple moving average | Computing the simple moving average of a series of numbers.
Task[edit]
Create a stateful function/class/instance that takes a period and returns a routine that takes a number as argument and returns a simple moving average of its arguments so far.
Description
A simple moving average is a method for computing an average of a stream of numbers by only averaging the last P numbers from the stream, where P is known as the period.
It can be implemented by calling an initialing routine with P as its argument, I(P), which should then return a routine that when called with individual, successive members of a stream of numbers, computes the mean of (up to), the last P of them, lets call this SMA().
The word stateful in the task description refers to the need for SMA() to remember certain information between calls to it:
The period, P
An ordered container of at least the last P numbers from each of its individual calls.
Stateful also means that successive calls to I(), the initializer, should return separate routines that do not share saved state so they could be used on two independent streams of data.
Pseudo-code for an implementation of SMA is:
function SMA(number: N):
stateful integer: P
stateful list: stream
number: average
stream.append_last(N)
if stream.length() > P:
# Only average the last P elements of the stream
stream.delete_first()
if stream.length() == 0:
average = 0
else:
average = sum( stream.values() ) / stream.length()
return average
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #TI-89_BASIC | TI-89 BASIC | movinavg(list,p)
Func
Local r, i, z
For i,1,dim(list)
max(i-p,0)→z
sum(mid(list,z+1,i-z))/(i-z)→r[i]
EndFor
r
EndFunc
|
http://rosettacode.org/wiki/Attractive_numbers | Attractive numbers | A number is an attractive number if the number of its prime factors (whether distinct or not) is also prime.
Example
The number 20, whose prime decomposition is 2 × 2 × 5, is an attractive number because the number of its prime factors (3) is also prime.
Task
Show sequence items up to 120.
Reference
The OEIS entry: A063989: Numbers with a prime number of prime divisors.
| #Nim | Nim | import strformat
const MAX = 120
proc isPrime(n: int): bool =
var d = 5
if n < 2:
return false
if n mod 2 == 0:
return n == 2
if n mod 3 == 0:
return n == 3
while d * d <= n:
if n mod d == 0:
return false
inc d, 2
if n mod d == 0:
return false
inc d, 4
return true
proc countPrimeFactors(n_in: int): int =
var count = 0
var f = 2
var n = n_in
if n == 1:
return 0
if isPrime(n):
return 1
while true:
if n mod f == 0:
inc count
n = n div f
if n == 1:
return count
if isPrime(n):
f = n
elif (f >= 3):
inc f, 2
else:
f = 3
proc main() =
var n, count: int = 0
echo fmt"The attractive numbers up to and including {MAX} are:"
for i in 1..MAX:
n = countPrimeFactors(i)
if isPrime(n):
write(stdout, fmt"{i:4d}")
inc count
if count mod 20 == 0:
write(stdout, "\n")
write(stdout, "\n")
main()
|
http://rosettacode.org/wiki/Attractive_numbers | Attractive numbers | A number is an attractive number if the number of its prime factors (whether distinct or not) is also prime.
Example
The number 20, whose prime decomposition is 2 × 2 × 5, is an attractive number because the number of its prime factors (3) is also prime.
Task
Show sequence items up to 120.
Reference
The OEIS entry: A063989: Numbers with a prime number of prime divisors.
| #Objeck | Objeck | class AttractiveNumber {
function : Main(args : String[]) ~ Nil {
max := 120;
"The attractive numbers up to and including {$max} are:"->PrintLine();
count := 0;
for(i := 1; i <= max; i += 1;) {
n := CountPrimeFactors(i);
if(IsPrime(n)) {
" {$i}"->Print();
if(++count % 20 = 0) {
""->PrintLine();
};
};
};
""->PrintLine();
}
function : IsPrime(n : Int) ~ Bool {
if(n < 2) {
return false;
};
if(n % 2 = 0) {
return n = 2;
};
if(n % 3 = 0) {
return n = 3;
};
d := 5;
while(d *d <= n) {
if(n % d = 0) {
return false;
};
d += 2;
if(n % d = 0) {
return false;
};
d += 4;
};
return true;
}
function : CountPrimeFactors(n : Int) ~ Int {
if(n = 1) {
return 0;
};
if(IsPrime(n)) {
return 1;
};
count := 0;
f := 2;
while(true) {
if(n % f = 0) {
count++;
n /= f;
if(n = 1) { return count; };
if(IsPrime(n)) { f := n; };
}
else if(f >= 3) {
f += 2;
}
else {
f := 3;
};
};
return -1;
}
} |
http://rosettacode.org/wiki/Averages/Mean_time_of_day | Averages/Mean time of day | Task[edit]
A particular activity of bats occurs at these times of the day:
23:00:17, 23:40:20, 00:12:45, 00:17:19
Using the idea that there are twenty-four hours in a day,
which is analogous to there being 360 degrees in a circle,
map times of day to and from angles;
and using the ideas of Averages/Mean angle
compute and show the average time of the nocturnal activity
to an accuracy of one second of time.
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #Racket | Racket |
#lang racket
(define (mean-angle/radians as)
(define n (length as))
(atan (* (/ 1 n) (for/sum ([αj as]) (sin αj)))
(* (/ 1 n) (for/sum ([αj as]) (cos αj)))))
(define (mean-time times)
(define secs/day (* 60 60 24))
(define (time->deg time)
(/ (for/fold ([sum 0]) ([t (map string->number (string-split time ":"))])
(+ (* 60 sum) t))
secs/day 1/360 (/ 180 pi)))
(define secs
(modulo (inexact->exact (round (* (mean-angle/radians (map time->deg times))
(/ 180 pi) 1/360 secs/day)))
secs/day))
(let loop ([s secs] [ts '()])
(if (zero? s) (string-join ts ":")
(let-values ([(q r) (quotient/remainder s 60)])
(loop q (cons (~r r #:min-width 2 #:pad-string "0") ts))))))
(mean-time '("23:00:17" "23:40:20" "00:12:45" "00:17:19"))
|
http://rosettacode.org/wiki/Averages/Mean_time_of_day | Averages/Mean time of day | Task[edit]
A particular activity of bats occurs at these times of the day:
23:00:17, 23:40:20, 00:12:45, 00:17:19
Using the idea that there are twenty-four hours in a day,
which is analogous to there being 360 degrees in a circle,
map times of day to and from angles;
and using the ideas of Averages/Mean angle
compute and show the average time of the nocturnal activity
to an accuracy of one second of time.
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #Raku | Raku | sub tod2rad($_) { [+](.comb(/\d+/) Z* 3600,60,1) * tau / 86400 }
sub rad2tod ($r) {
my $x = $r * 86400 / tau;
(($x xx 3 Z/ 3600,60,1) Z% 24,60,60).fmt('%02d',':');
}
sub phase ($c) { $c.polar[1] }
sub mean-time (@t) { rad2tod phase [+] map { cis tod2rad $_ }, @t }
my @times = ["23:00:17", "23:40:20", "00:12:45", "00:17:19"];
say "{ mean-time(@times) } is the mean time of @times[]"; |
http://rosettacode.org/wiki/AVL_tree | AVL tree |
This page uses content from Wikipedia. The original article was at AVL tree. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance)
In computer science, an AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; at no time do they differ by more than one because rebalancing is done ensure this is the case. Lookup, insertion, and deletion all take O(log n) time in both the average and worst cases, where n is the number of nodes in the tree prior to the operation. Insertions and deletions may require the tree to be rebalanced by one or more tree rotations. Note the tree of nodes comprise a set, so duplicate node keys are not allowed.
AVL trees are often compared with red-black trees because they support the same set of operations and because red-black trees also take O(log n) time for the basic operations. Because AVL trees are more rigidly balanced, they are faster than red-black trees for lookup-intensive applications. Similar to red-black trees, AVL trees are height-balanced, but in general not weight-balanced nor μ-balanced; that is, sibling nodes can have hugely differing numbers of descendants.
Task
Implement an AVL tree in the language of choice, and provide at least basic operations.
| #TypeScript | TypeScript | /** A single node in an AVL tree */
class AVLnode <T> {
balance: number
left: AVLnode<T>
right: AVLnode<T>
constructor(public key: T, public parent: AVLnode<T> = null) {
this.balance = 0
this.left = null
this.right = null
}
}
/** The balanced AVL tree */
class AVLtree <T> {
// public members organized here
constructor() {
this.root = null
}
insert(key: T): boolean {
if (this.root === null) {
this.root = new AVLnode<T>(key)
} else {
let n: AVLnode<T> = this.root,
parent: AVLnode<T> = null
while (true) {
if(n.key === key) {
return false
}
parent = n
let goLeft: boolean = n.key > key
n = goLeft ? n.left : n.right
if (n === null) {
if (goLeft) {
parent.left = new AVLnode<T>(key, parent)
} else {
parent.right = new AVLnode<T>(key, parent)
}
this.rebalance(parent)
break
}
}
}
return true
}
deleteKey(delKey: T): void {
if (this.root === null) {
return
}
let n: AVLnode<T> = this.root,
parent: AVLnode<T> = this.root,
delNode: AVLnode<T> = null,
child: AVLnode<T> = this.root
while (child !== null) {
parent = n
n = child
child = delKey >= n.key ? n.right : n.left
if (delKey === n.key) {
delNode = n
}
}
if (delNode !== null) {
delNode.key = n.key
child = n.left !== null ? n.left : n.right
if (this.root.key === delKey) {
this.root = child
} else {
if (parent.left === n) {
parent.left = child
} else {
parent.right = child
}
this.rebalance(parent)
}
}
}
treeBalanceString(n: AVLnode<T> = this.root): string {
if (n !== null) {
return `${this.treeBalanceString(n.left)} ${n.balance} ${this.treeBalanceString(n.right)}`
}
return ""
}
toString(n: AVLnode<T> = this.root): string {
if (n !== null) {
return `${this.toString(n.left)} ${n.key} ${this.toString(n.right)}`
}
return ""
}
// private members organized here
private root: AVLnode<T>
private rotateLeft(a: AVLnode<T>): AVLnode<T> {
let b: AVLnode<T> = a.right
b.parent = a.parent
a.right = b.left
if (a.right !== null) {
a.right.parent = a
}
b.left = a
a.parent = b
if (b.parent !== null) {
if (b.parent.right === a) {
b.parent.right = b
} else {
b.parent.left = b
}
}
this.setBalance(a)
this.setBalance(b)
return b
}
private rotateRight(a: AVLnode<T>): AVLnode<T> {
let b: AVLnode<T> = a.left
b.parent = a.parent
a.left = b.right
if (a.left !== null) {
a.left.parent = a
}
b.right = a
a.parent = b
if (b.parent !== null) {
if (b.parent.right === a) {
b.parent.right = b
} else {
b.parent.left = b
}
}
this.setBalance(a)
this.setBalance(b)
return b
}
private rotateLeftThenRight(n: AVLnode<T>): AVLnode<T> {
n.left = this.rotateLeft(n.left)
return this.rotateRight(n)
}
private rotateRightThenLeft(n: AVLnode<T>): AVLnode<T> {
n.right = this.rotateRight(n.right)
return this.rotateLeft(n)
}
private rebalance(n: AVLnode<T>): void {
this.setBalance(n)
if (n.balance === -2) {
if(this.height(n.left.left) >= this.height(n.left.right)) {
n = this.rotateRight(n)
} else {
n = this.rotateLeftThenRight(n)
}
} else if (n.balance === 2) {
if(this.height(n.right.right) >= this.height(n.right.left)) {
n = this.rotateLeft(n)
} else {
n = this.rotateRightThenLeft(n)
}
}
if (n.parent !== null) {
this.rebalance(n.parent)
} else {
this.root = n
}
}
private height(n: AVLnode<T>): number {
if (n === null) {
return -1
}
return 1 + Math.max(this.height(n.left), this.height(n.right))
}
private setBalance(n: AVLnode<T>): void {
n.balance = this.height(n.right) - this.height(n.left)
}
public showNodeBalance(n: AVLnode<T>): string {
if (n !== null) {
return `${this.showNodeBalance(n.left)} ${n.balance} ${this.showNodeBalance(n.right)}`
}
return ""
}
}
|
http://rosettacode.org/wiki/Averages/Mean_angle | Averages/Mean angle | When calculating the average or mean of an angle one has to take into account how angles wrap around so that any angle in degrees plus any integer multiple of 360 degrees is a measure of the same angle.
If one wanted an average direction of the wind over two readings where the first reading was of 350 degrees and the second was of 10 degrees then the average of the numbers is 180 degrees, whereas if you can note that 350 degrees is equivalent to -10 degrees and so you have two readings at 10 degrees either side of zero degrees leading to a more fitting mean angle of zero degrees.
To calculate the mean angle of several angles:
Assume all angles are on the unit circle and convert them to complex numbers expressed in real and imaginary form.
Compute the mean of the complex numbers.
Convert the complex mean to polar coordinates whereupon the phase of the complex mean is the required angular mean.
(Note that, since the mean is the sum divided by the number of numbers, and division by a positive real number does not affect the angle, you can also simply compute the sum for step 2.)
You can alternatively use this formula:
Given the angles
α
1
,
…
,
α
n
{\displaystyle \alpha _{1},\dots ,\alpha _{n}}
the mean is computed by
α
¯
=
atan2
(
1
n
⋅
∑
j
=
1
n
sin
α
j
,
1
n
⋅
∑
j
=
1
n
cos
α
j
)
{\displaystyle {\bar {\alpha }}=\operatorname {atan2} \left({\frac {1}{n}}\cdot \sum _{j=1}^{n}\sin \alpha _{j},{\frac {1}{n}}\cdot \sum _{j=1}^{n}\cos \alpha _{j}\right)}
Task[edit]
write a function/method/subroutine/... that given a list of angles in degrees returns their mean angle.
(You should use a built-in function if you have one that does this for degrees or radians).
Use the function to compute the means of these lists of angles (in degrees):
[350, 10]
[90, 180, 270, 360]
[10, 20, 30]
Show your output here.
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #Python | Python | >>> from cmath import rect, phase
>>> from math import radians, degrees
>>> def mean_angle(deg):
... return degrees(phase(sum(rect(1, radians(d)) for d in deg)/len(deg)))
...
>>> for angles in [[350, 10], [90, 180, 270, 360], [10, 20, 30]]:
... print('The mean angle of', angles, 'is:', round(mean_angle(angles), 12), 'degrees')
...
The mean angle of [350, 10] is: -0.0 degrees
The mean angle of [90, 180, 270, 360] is: -90.0 degrees
The mean angle of [10, 20, 30] is: 20.0 degrees
>>> |
http://rosettacode.org/wiki/Averages/Mean_angle | Averages/Mean angle | When calculating the average or mean of an angle one has to take into account how angles wrap around so that any angle in degrees plus any integer multiple of 360 degrees is a measure of the same angle.
If one wanted an average direction of the wind over two readings where the first reading was of 350 degrees and the second was of 10 degrees then the average of the numbers is 180 degrees, whereas if you can note that 350 degrees is equivalent to -10 degrees and so you have two readings at 10 degrees either side of zero degrees leading to a more fitting mean angle of zero degrees.
To calculate the mean angle of several angles:
Assume all angles are on the unit circle and convert them to complex numbers expressed in real and imaginary form.
Compute the mean of the complex numbers.
Convert the complex mean to polar coordinates whereupon the phase of the complex mean is the required angular mean.
(Note that, since the mean is the sum divided by the number of numbers, and division by a positive real number does not affect the angle, you can also simply compute the sum for step 2.)
You can alternatively use this formula:
Given the angles
α
1
,
…
,
α
n
{\displaystyle \alpha _{1},\dots ,\alpha _{n}}
the mean is computed by
α
¯
=
atan2
(
1
n
⋅
∑
j
=
1
n
sin
α
j
,
1
n
⋅
∑
j
=
1
n
cos
α
j
)
{\displaystyle {\bar {\alpha }}=\operatorname {atan2} \left({\frac {1}{n}}\cdot \sum _{j=1}^{n}\sin \alpha _{j},{\frac {1}{n}}\cdot \sum _{j=1}^{n}\cos \alpha _{j}\right)}
Task[edit]
write a function/method/subroutine/... that given a list of angles in degrees returns their mean angle.
(You should use a built-in function if you have one that does this for degrees or radians).
Use the function to compute the means of these lists of angles (in degrees):
[350, 10]
[90, 180, 270, 360]
[10, 20, 30]
Show your output here.
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #R | R |
deg2rad <- function(x) {
x * pi/180
}
rad2deg <- function(x) {
x * 180/pi
}
deg2vec <- function(x) {
c(sin(deg2rad(x)), cos(deg2rad(x)))
}
vec2deg <- function(x) {
res <- rad2deg(atan2(x[1], x[2]))
if (res < 0) {
360 + res
} else {
res
}
}
mean_vec <- function(x) {
y <- lapply(x, deg2vec)
Reduce(`+`, y)/length(y)
}
mean_deg <- function(x) {
vec2deg(mean_vec(x))
}
mean_deg(c(350, 10))
mean_deg(c(90, 180, 270, 360))
mean_deg(c(10, 20, 30))
|
http://rosettacode.org/wiki/Averages/Median | Averages/Median | Task[edit]
Write a program to find the median value of a vector of floating-point numbers.
The program need not handle the case where the vector is empty, but must handle the case where there are an even number of elements. In that case, return the average of the two middle values.
There are several approaches to this. One is to sort the elements, and then pick the element(s) in the middle.
Sorting would take at least O(n logn). Another approach would be to build a priority queue from the elements, and then extract half of the elements to get to the middle element(s). This would also take O(n logn). The best solution is to use the selection algorithm to find the median in O(n) time.
See also
Quickselect_algorithm
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #HicEst | HicEst | REAL :: n=10, vec(n)
vec = RAN(1)
SORT(Vector=vec, Sorted=vec) ! in-place Merge-Sort
IF( MOD(n,2) ) THEN ! odd n
median = vec( CEILING(n/2) )
ELSE
median = ( vec(n/2) + vec(n/2 + 1) ) / 2
ENDIF |
http://rosettacode.org/wiki/Averages/Pythagorean_means | Averages/Pythagorean means | Task[edit]
Compute all three of the Pythagorean means of the set of integers 1 through 10 (inclusive).
Show that
A
(
x
1
,
…
,
x
n
)
≥
G
(
x
1
,
…
,
x
n
)
≥
H
(
x
1
,
…
,
x
n
)
{\displaystyle A(x_{1},\ldots ,x_{n})\geq G(x_{1},\ldots ,x_{n})\geq H(x_{1},\ldots ,x_{n})}
for this set of positive integers.
The most common of the three means, the arithmetic mean, is the sum of the list divided by its length:
A
(
x
1
,
…
,
x
n
)
=
x
1
+
⋯
+
x
n
n
{\displaystyle A(x_{1},\ldots ,x_{n})={\frac {x_{1}+\cdots +x_{n}}{n}}}
The geometric mean is the
n
{\displaystyle n}
th root of the product of the list:
G
(
x
1
,
…
,
x
n
)
=
x
1
⋯
x
n
n
{\displaystyle G(x_{1},\ldots ,x_{n})={\sqrt[{n}]{x_{1}\cdots x_{n}}}}
The harmonic mean is
n
{\displaystyle n}
divided by the sum of the reciprocal of each item in the list:
H
(
x
1
,
…
,
x
n
)
=
n
1
x
1
+
⋯
+
1
x
n
{\displaystyle H(x_{1},\ldots ,x_{n})={\frac {n}{{\frac {1}{x_{1}}}+\cdots +{\frac {1}{x_{n}}}}}}
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #Lua | Lua | function fsum(f, a, ...) return a and f(a) + fsum(f, ...) or 0 end
function pymean(t, f, finv) return finv(fsum(f, unpack(t)) / #t) end
nums = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
--arithmetic
a = pymean(nums, function(n) return n end, function(n) return n end)
--geometric
g = pymean(nums, math.log, math.exp)
--harmonic
h = pymean(nums, function(n) return 1/n end, function(n) return 1/n end)
print(a, g, h)
assert(a >= g and g >= h) |
http://rosettacode.org/wiki/Balanced_ternary | Balanced ternary | Balanced ternary is a way of representing numbers. Unlike the prevailing binary representation, a balanced ternary integer is in base 3, and each digit can have the values 1, 0, or −1.
Examples
Decimal 11 = 32 + 31 − 30, thus it can be written as "++−"
Decimal 6 = 32 − 31 + 0 × 30, thus it can be written as "+−0"
Task
Implement balanced ternary representation of integers with the following:
Support arbitrarily large integers, both positive and negative;
Provide ways to convert to and from text strings, using digits '+', '-' and '0' (unless you are already using strings to represent balanced ternary; but see requirement 5).
Provide ways to convert to and from native integer type (unless, improbably, your platform's native integer type is balanced ternary). If your native integers can't support arbitrary length, overflows during conversion must be indicated.
Provide ways to perform addition, negation and multiplication directly on balanced ternary integers; do not convert to native integers first.
Make your implementation efficient, with a reasonable definition of "efficient" (and with a reasonable definition of "reasonable").
Test case With balanced ternaries a from string "+-0++0+", b from native integer -436, c "+-++-":
write out a, b and c in decimal notation;
calculate a × (b − c), write out the result in both ternary and decimal notations.
Note: The pages generalised floating point addition and generalised floating point multiplication have code implementing arbitrary precision floating point balanced ternary.
| #Tcl | Tcl | package require Tcl 8.5
proc bt-int b {
set n 0
foreach c [split $b ""] {
set n [expr {$n * 3}]
switch -- $c {
+ { incr n 1 }
- { incr n -1 }
}
}
return $n
}
proc int-bt n {
if {$n == 0} {
return "0"
}
while {$n != 0} {
lappend result [lindex {0 + -} [expr {$n % 3}]]
set n [expr {$n / 3 + ($n%3 == 2)}]
}
return [join [lreverse $result] ""]
}
proc bt-neg b {
string map {+ - - +} $b
}
proc bt-sub {a b} {
bt-add $a [bt-neg $b]
}
proc bt-add-digits {a b c} {
if {$a eq ""} {set a 0}
if {$b eq ""} {set b 0}
if {$a ne 0} {append a 1}
if {$b ne 0} {append b 1}
lindex {{0 -1} {+ -1} {- 0} {0 0} {+ 0} {- 1} {0 1}} [expr {$a+$b+$c+3}]
}
proc bt-add {a b} {
set c 0
set result {}
foreach ca [lreverse [split $a ""]] cb [lreverse [split $b ""]] {
lassign [bt-add-digits $ca $cb $c] d c
lappend result $d
}
if {$c ne "0"} {lappend result [lindex {0 + -} $c]}
if {![llength $result]} {return "0"}
string trimleft [join [lreverse $result] ""] 0
}
proc bt-mul {a b} {
if {$a eq "0" || $a eq "" || $b eq "0"} {return "0"}
set sub [bt-mul [string range $a 0 end-1] $b]0
switch -- [string index $a end] {
0 { return $sub }
+ { return [bt-add $sub $b] }
- { return [bt-sub $sub $b] }
}
} |
http://rosettacode.org/wiki/Babbage_problem | Babbage problem |
Charles Babbage, looking ahead to the sorts of problems his Analytical Engine would be able to solve, gave this example:
What is the smallest positive integer whose square ends in the digits 269,696?
— Babbage, letter to Lord Bowden, 1837; see Hollingdale and Tootill, Electronic Computers, second edition, 1970, p. 125.
He thought the answer might be 99,736, whose square is 9,947,269,696; but he couldn't be certain.
Task[edit]
The task is to find out if Babbage had the right answer — and to do so, as far as your language allows it, in code that Babbage himself would have been able to read and understand.
As Babbage evidently solved the task with pencil and paper, a similar efficient solution is preferred.
For these purposes, Charles Babbage may be taken to be an intelligent person, familiar with mathematics and with the idea of a computer; he has written the first drafts of simple computer programmes in tabular form. [Babbage Archive Series L].
Motivation
The aim of the task is to write a program that is sufficiently clear and well-documented for such a person to be able to read it and be confident that it does indeed solve the specified problem.
| #Shen | Shen | (define babbage
N -> N where (= 269696 (shen.mod (* N N) 1000000)))
N -> (babbage (+ N 1))
(babbage 1) |
http://rosettacode.org/wiki/Babbage_problem | Babbage problem |
Charles Babbage, looking ahead to the sorts of problems his Analytical Engine would be able to solve, gave this example:
What is the smallest positive integer whose square ends in the digits 269,696?
— Babbage, letter to Lord Bowden, 1837; see Hollingdale and Tootill, Electronic Computers, second edition, 1970, p. 125.
He thought the answer might be 99,736, whose square is 9,947,269,696; but he couldn't be certain.
Task[edit]
The task is to find out if Babbage had the right answer — and to do so, as far as your language allows it, in code that Babbage himself would have been able to read and understand.
As Babbage evidently solved the task with pencil and paper, a similar efficient solution is preferred.
For these purposes, Charles Babbage may be taken to be an intelligent person, familiar with mathematics and with the idea of a computer; he has written the first drafts of simple computer programmes in tabular form. [Babbage Archive Series L].
Motivation
The aim of the task is to write a program that is sufficiently clear and well-documented for such a person to be able to read it and be confident that it does indeed solve the specified problem.
| #Sidef | Sidef | var n = 0
while (n*n % 1000000 != 269696) {
n += 2
}
say n |
http://rosettacode.org/wiki/Approximate_equality | Approximate equality | Sometimes, when testing whether the solution to a task (for example, here on Rosetta Code) is correct, the
difference in floating point calculations between different language implementations becomes significant.
For example, a difference between 32 bit and 64 bit floating point calculations may appear by
about the 8th significant digit in base 10 arithmetic.
Task
Create a function which returns true if two floating point numbers are approximately equal.
The function should allow for differences in the magnitude of numbers, so that, for example,
100000000000000.01 may be approximately equal to 100000000000000.011,
even though 100.01 is not approximately equal to 100.011.
If the language has such a feature in its standard library, this may be used instead of a custom function.
Show the function results with comparisons on the following pairs of values:
100000000000000.01, 100000000000000.011 (note: should return true)
100.01, 100.011 (note: should return false)
10000000000000.001 / 10000.0, 1000000000.0000001000
0.001, 0.0010000001
0.000000000000000000000101, 0.0
sqrt(2) * sqrt(2), 2.0
-sqrt(2) * sqrt(2), -2.0
3.14159265358979323846, 3.14159265358979324
Answers should be true for the first example and false in the second, so that just rounding the numbers to a fixed number of decimals should not be enough. Otherwise answers may vary and still be correct. See the Python code for one type of solution.
| #Visual_Basic_.NET | Visual Basic .NET | Imports System.Runtime.CompilerServices
Module Module1
<Extension()>
Function ApproxEquals(ByVal value As Double, other As Double, epsilon As Double)
Return Math.Abs(value - other) < epsilon
End Function
Sub Test(a As Double, b As Double)
Dim epsilon = 1.0E-18
Console.WriteLine($"{a}, {b} => {a.ApproxEquals(b, epsilon)}")
End Sub
Sub Main()
Test(100000000000000.02, 100000000000000.02)
Test(100.01, 100.011)
Test(10000000000000.002 / 10000.0, 1000000000.0000001)
Test(0.001, 0.0010000001)
Test(1.01E-22, 0.0)
Test(Math.Sqrt(2) * Math.Sqrt(2), 2.0)
Test(-Math.Sqrt(2) * Math.Sqrt(2), -2.0)
Test(3.1415926535897931, 3.1415926535897931)
End Sub
End Module |
http://rosettacode.org/wiki/Approximate_equality | Approximate equality | Sometimes, when testing whether the solution to a task (for example, here on Rosetta Code) is correct, the
difference in floating point calculations between different language implementations becomes significant.
For example, a difference between 32 bit and 64 bit floating point calculations may appear by
about the 8th significant digit in base 10 arithmetic.
Task
Create a function which returns true if two floating point numbers are approximately equal.
The function should allow for differences in the magnitude of numbers, so that, for example,
100000000000000.01 may be approximately equal to 100000000000000.011,
even though 100.01 is not approximately equal to 100.011.
If the language has such a feature in its standard library, this may be used instead of a custom function.
Show the function results with comparisons on the following pairs of values:
100000000000000.01, 100000000000000.011 (note: should return true)
100.01, 100.011 (note: should return false)
10000000000000.001 / 10000.0, 1000000000.0000001000
0.001, 0.0010000001
0.000000000000000000000101, 0.0
sqrt(2) * sqrt(2), 2.0
-sqrt(2) * sqrt(2), -2.0
3.14159265358979323846, 3.14159265358979324
Answers should be true for the first example and false in the second, so that just rounding the numbers to a fixed number of decimals should not be enough. Otherwise answers may vary and still be correct. See the Python code for one type of solution.
| #Wren | Wren | var tol = 1e-16
var pairs = [
[100000000000000.01, 100000000000000.011],
[100.01, 100.011],
[10000000000000.001 / 10000.0, 1000000000.0000001000],
[0.001, 0.0010000001],
[0.000000000000000000000101, 0.0],
[2.sqrt * 2.sqrt, 2.0],
[-2.sqrt * 2.sqrt, -2.0],
[3.14159265358979323846, 3.14159265358979324]
]
System.print("Approximate equality of test cases for a tolerance of %(tol):")
var i = 0
for (pair in pairs) {
i = i + 1
System.print(" %(i) -> %((pair[0] - pair[1]).abs < tol)")
} |
http://rosettacode.org/wiki/Balanced_brackets | Balanced brackets | Task:
Generate a string with N opening brackets [ and with N closing brackets ], in some arbitrary order.
Determine whether the generated string is balanced; that is, whether it consists entirely of pairs of opening/closing brackets (in that order), none of which mis-nest.
Examples
(empty) OK
[] OK
[][] OK
[[][]] OK
][ NOT OK
][][ NOT OK
[]][[] NOT OK
| #Euphoria | Euphoria | function check_brackets(sequence s)
integer level
level = 0
for i = 1 to length(s) do
if s[i] = '[' then
level += 1
elsif s[i] = ']' then
level -= 1
if level < 0 then
return 0
end if
end if
end for
return level = 0
end function
function generate_brackets(integer n)
integer opened,closed,r
sequence s
opened = n
closed = n
s = ""
for i = 1 to n*2 do
r = rand(opened+closed)
if r<=opened then
s &= '['
opened -= 1
else
s &= ']'
closed -= 1
end if
end for
return s
end function
sequence s
for i = 1 to 10 do
s = generate_brackets(3)
puts(1,s)
if check_brackets(s) then
puts(1," OK\n")
else
puts(1," NOT OK\n")
end if
end for |
http://rosettacode.org/wiki/Append_a_record_to_the_end_of_a_text_file | Append a record to the end of a text file | Many systems offer the ability to open a file for writing, such that any data written will be appended to the end of the file. Further, the file operations will always adjust the position pointer to guarantee the end of the file, even in a multitasking environment.
This feature is most useful in the case of log files, where many jobs may be appending to the log file at the same time, or where care must be taken to avoid concurrently overwriting the same record from another job.
Task
Given a two record sample for a mythical "passwd" file:
Write these records out in the typical system format.
Ideally these records will have named fields of various types.
Close the file, then reopen the file for append.
Append a new record to the file and close the file again.
Take appropriate care to avoid concurrently overwrites from another job.
Open the file and demonstrate the new record has indeed written to the end.
Source record field types and contents.
account
password
UID
GID
fullname,office,extension,homephone,email
directory
shell
string
string
int
int
struct(string,string,string,string,string)
string
string
jsmith
x
1001
1000
Joe Smith,Room 1007,(234)555-8917,(234)555-0077,[email protected]
/home/jsmith
/bin/bash
jdoe
x
1002
1000
Jane Doe,Room 1004,(234)555-8914,(234)555-0044,[email protected]
/home/jdoe
/bin/bash
Record to be appended.
account
password
UID
GID
fullname,office,extension,homephone,email
directory
shell
string
string
int
int
struct(string,string,string,string,string)
string
string
xyz
x
1003
1000
X Yz,Room 1003,(234)555-8913,(234)555-0033,[email protected]
/home/xyz
/bin/bash
Resulting file format: should mimic Linux's /etc/passwd file format with particular attention to the "," separator used in the GECOS field. But if the specific language has a particular or unique format of storing records in text file, then this format should be named and demonstrated with an additional example.
Expected output:
Appended record: xyz:x:1003:1000:X Yz,Room 1003,(234)555-8913,(234)555-0033,[email protected]:/home/xyz:/bin/bash
Finally: Provide a summary of the language's "append record" capabilities in a table. eg.
Append Capabilities.
Data Representation
IO
Library
Append
Possible
Automatic
Append
Multi-tasking
Safe
In core
On disk
C struct
CSV text file
glibc/stdio
☑
☑
☑ (Not all, eg NFS)
Alternatively: If the language's appends can not guarantee its writes will always append, then note this restriction in the table. If possible, provide an actual code example (possibly using file/record locking) to guarantee correct concurrent appends.
| #J | J | require'strings ~system/packages/misc/xenos.ijs'
record=: [:|: <@deb;._1@(':',]);._2@do bind '0 :0'
passfields=: <;._1':username:password:gid:uid:gecos:home:shell'
passrec=: LF,~ [: }.@;@ , ':';"0 (passfields i. {. ) { a:,~ {:
R1=: passrec record''
username: jsmith
password: x
gid: 1001
uid: 1000
gecos: Joe Smith,Room 1007,(234)555-8917,(234)555-0077,[email protected]
home: /home/jsmith
shell: /bin/bash
)
R2=: passrec record''
username: jdoe
password: x
gid: 1002
uid: 1000
gecos: Jane Doe,Room 1004,(234)555-8914,(234)555-0044,[email protected]
home: /home/jdoe
shell: /bin/bash
)
R3=: passrec record''
username: xyz
password: x
gid: 1003
uid: 1000
gecos: X Yz,Room 1003,(234)555-8913,(234)555-0033,[email protected]
home: /home/xyz
shell: /bin/bash
)
passwd=: <'/tmp/passwd.txt' NB. file needs to be writable on implementation machine
(R1,R2) fwrite passwd
R3 fappend passwd
assert 1 e. R3 E. fread passwd |
http://rosettacode.org/wiki/Associative_array/Creation | Associative array/Creation | Task
The goal is to create an associative array (also known as a dictionary, map, or hash).
Related tasks:
Associative arrays/Iteration
Hash from two arrays
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #BaCon | BaCon | DECLARE associative ASSOC STRING
associative("abc") = "first three"
associative("xyz") = "last three"
PRINT associative("xyz") |
http://rosettacode.org/wiki/Anti-primes | Anti-primes | The anti-primes
(or highly composite numbers, sequence A002182 in the OEIS)
are the natural numbers with more factors than any smaller than itself.
Task
Generate and show here, the first twenty anti-primes.
Related tasks
Factors of an integer
Sieve of Eratosthenes
| #F.23 | F# |
// Find Antı-Primes. Nigel Galloway: Secember 10th., 2018
let N=200000000000000000000000000I
let fI,_=Seq.scan(fun (_,g) e->(e,e*g)) (2I,4I) (primes|>Seq.skip 1|>Seq.map bigint)|>Seq.takeWhile(fun(_,n)->n<N)|>List.ofSeq|>List.unzip
let fG g=Seq.unfold(fun ((n,i,e) as z)->Some(z,(n+1,i+1,(e*g)))) (1,2,g)|>Seq.takeWhile(fun(_,_,n)->n<N)
let fE n i=n|>Seq.collect(fun(n,e,g)->Seq.map(fun(a,c,b)->(a,c*e,g*b)) (i|>Seq.takeWhile(fun(g,_,_)->g<=n)) |> Seq.takeWhile(fun(_,_,n)->n<N))
let fL,_=Seq.concat(Seq.scan(fun n g->fE n (fG g)) (seq[(2147483647,1,1I)]) fI)|>List.ofSeq|>List.sortBy(fun(_,_,n)->n)|>List.fold(fun ((a,b) as z) (_,n,g)->if n>b then ((n,g)::a,n) else z) ([],0)
|
http://rosettacode.org/wiki/Atomic_updates | Atomic updates |
Task
Define a data type consisting of a fixed number of 'buckets', each containing a nonnegative integer value, which supports operations to:
get the current value of any bucket
remove a specified amount from one specified bucket and add it to another, preserving the total of all bucket values, and clamping the transferred amount to ensure the values remain non-negative
In order to exercise this data type, create one set of buckets, and start three concurrent tasks:
As often as possible, pick two buckets and make their values closer to equal.
As often as possible, pick two buckets and arbitrarily redistribute their values.
At whatever rate is convenient, display (by any means) the total value and, optionally, the individual values of each bucket.
The display task need not be explicit; use of e.g. a debugger or trace tool is acceptable provided it is simple to set up to provide the display.
This task is intended as an exercise in atomic operations. The sum of the bucket values must be preserved even if the two tasks attempt to perform transfers simultaneously, and a straightforward solution is to ensure that at any time, only one transfer is actually occurring — that the transfer operation is atomic.
| #PicoLisp | PicoLisp | (seed (in "/dev/urandom" (rd 8)))
(de *Buckets . 15) # Number of buckets
# E/R model
(class +Bucket +Entity)
(rel key (+Key +Number)) # Key 1 .. *Buckets
(rel val (+Number)) # Value 1 .. 999
# Create new DB file
(pool (tmp "buckets.db"))
# Create *Buckets buckets with values between 1 and 999
(for K *Buckets
(new T '(+Bucket) 'key K 'val (rand 1 999)) )
(commit)
# Pick a random bucket
(de pickBucket ()
(db 'key '+Bucket (rand 1 *Buckets)) )
# First process
(unless (fork)
(seed *Pid) # Ensure local random sequence
(loop
(let (B1 (pickBucket) B2 (pickBucket)) # Pick two buckets 'B1' and 'B2'
(dbSync) # Atomic DB operation
(let (V1 (; B1 val) V2 (; B2 val)) # Get current values
(cond
((> V1 V2)
(dec> B1 'val) # Make them closer to equal
(inc> B2 'val) )
((> V2 V1)
(dec> B2 'val)
(inc> B1 'val) ) ) )
(commit 'upd) # Close transaction
(wait 1) ) ) )
# Second process
(unless (fork)
(seed *Pid) # Ensure local random sequence
(loop
(let (B1 (pickBucket) B2 (pickBucket)) # Pick two buckets 'B1' and 'B2'
(unless (== B1 B2) # Found two different ones?
(dbSync) # Atomic DB operation
(let (V1 (; B1 val) V2 (; B2 val)) # Get current values
(cond
((> V1 V2 0)
(inc> B1 'val) # Redistribute them
(dec> B2 'val) )
((> V2 V1 0)
(inc> B2 'val)
(dec> B1 'val) ) ) )
(commit 'upd) # Close transaction
(wait 1) ) ) ) )
# Third process
(unless (fork)
(loop
(let Lst (collect 'key '+Bucket) # Get all buckets
(for This Lst # Print current values
(printsp (: val)) )
(prinl # and total sum
"-- Total: "
(sum '((This) (: val)) Lst) ) )
(wait 2000) ) ) # Sleep two seconds
(wait) |
http://rosettacode.org/wiki/Atomic_updates | Atomic updates |
Task
Define a data type consisting of a fixed number of 'buckets', each containing a nonnegative integer value, which supports operations to:
get the current value of any bucket
remove a specified amount from one specified bucket and add it to another, preserving the total of all bucket values, and clamping the transferred amount to ensure the values remain non-negative
In order to exercise this data type, create one set of buckets, and start three concurrent tasks:
As often as possible, pick two buckets and make their values closer to equal.
As often as possible, pick two buckets and arbitrarily redistribute their values.
At whatever rate is convenient, display (by any means) the total value and, optionally, the individual values of each bucket.
The display task need not be explicit; use of e.g. a debugger or trace tool is acceptable provided it is simple to set up to provide the display.
This task is intended as an exercise in atomic operations. The sum of the bucket values must be preserved even if the two tasks attempt to perform transfers simultaneously, and a straightforward solution is to ensure that at any time, only one transfer is actually occurring — that the transfer operation is atomic.
| #PureBasic | PureBasic | #Buckets=9
#TotalAmount=200
Global Dim Buckets(#Buckets)
Global BMutex=CreateMutex()
Global Quit=#False
Procedure max(x,y)
If x>=y: ProcedureReturn x
Else: ProcedureReturn y
EndIf
EndProcedure
Procedure Move(WantedAmount, From, Dest)
Protected RealAmount
If from<>Dest
LockMutex(BMutex)
RealAmount=max(0, Buckets(from)-WantedAmount)
Buckets(From)-RealAmount
Buckets(Dest)+RealAmount
UnlockMutex(BMutex)
EndIf
ProcedureReturn RealAmount
EndProcedure
Procedure Level(A,B)
Protected i, j, t
If A<>B
LockMutex(BMutex)
t=Buckets(A)+Buckets(B)
i=t/2: j=t-i
Buckets(A)=i
Buckets(B)=j
UnlockMutex(BMutex)
EndIf
EndProcedure
Procedure DoInvent(Array A(1))
Protected i, sum
LockMutex(BMutex)
For i=0 To ArraySize(Buckets())
A(i)=Buckets(i)
sum+A(i)
Next i
UnlockMutex(BMutex)
ProcedureReturn sum
EndProcedure
Procedure MixingThread(arg)
Repeat
Move(Random(#TotalAmount),Random(#Buckets),Random(#Buckets))
Until Quit
EndProcedure
Procedure LevelingThread(arg)
Repeat
Level(Random(#Buckets),Random(#Buckets))
Until Quit
EndProcedure
If OpenWindow(0,0,0,100,150,"Atomic updates",#PB_Window_SystemMenu)
Define Thread1=CreateThread(@MixingThread(),0)
Define Thread2=CreateThread(@MixingThread(),0)
Define i, Event
Dim Inventory(#Buckets)
; Set up a small GUI
For i=0 To 9
TextGadget(i, 0,i*15,50, 15,"Bucket #"+Str(i))
Next i
TextGadget(10,55,135,40,15,"=")
AddWindowTimer(0,0,500)
Buckets(0)=#TotalAmount
Repeat
Event=WaitWindowEvent()
If Event=#PB_Event_Timer
i=DoInvent(Inventory())
SetGadgetText(10,"="+Str(i))
For i=0 To #Buckets
SetGadgetText(i, Str(Inventory(i)))
Next i
EndIf
Until Event=#PB_Event_CloseWindow
Quit=#True ; Tell threads to shut down
WaitThread(Thread1): WaitThread(Thread2)
EndIf |
http://rosettacode.org/wiki/Assertions | Assertions | Assertions are a way of breaking out of code when there is an error or an unexpected input.
Some languages throw exceptions and some treat it as a break point.
Task
Show an assertion in your language by asserting that an integer variable is equal to 42.
| #Nim | Nim | var a = 42
assert(a == 42, "Not 42!") |
http://rosettacode.org/wiki/Assertions | Assertions | Assertions are a way of breaking out of code when there is an error or an unexpected input.
Some languages throw exceptions and some treat it as a break point.
Task
Show an assertion in your language by asserting that an integer variable is equal to 42.
| #Oberon-2 | Oberon-2 |
MODULE Assertions;
VAR
a: INTEGER;
BEGIN
a := 40;
ASSERT(a = 42);
END Assertions.
|
http://rosettacode.org/wiki/Assertions | Assertions | Assertions are a way of breaking out of code when there is an error or an unexpected input.
Some languages throw exceptions and some treat it as a break point.
Task
Show an assertion in your language by asserting that an integer variable is equal to 42.
| #Objeck | Objeck | class Test {
function : Main(args : String[]) ~ Nil {
if(args->Size() = 1) {
a := args[0]->ToInt();
Runtime->Assert(a = 42);
};
}
}
|
http://rosettacode.org/wiki/Apply_a_callback_to_an_array | Apply a callback to an array | Task
Take a combined set of elements and apply a function to each element.
| #Crystal | Crystal | values = [1, 2, 3]
new_values = values.map do |number|
number * 2
end
puts new_values #=> [2, 4, 6] |
http://rosettacode.org/wiki/Apply_a_callback_to_an_array | Apply a callback to an array | Task
Take a combined set of elements and apply a function to each element.
| #D | D | import std.stdio, std.algorithm;
void main() {
auto items = [1, 2, 3, 4, 5];
auto m = items.map!(x => x + 5)();
writeln(m);
} |
http://rosettacode.org/wiki/Averages/Mode | Averages/Mode | Task[edit]
Write a program to find the mode value of a collection.
The case where the collection is empty may be ignored. Care must be taken to handle the case where the mode is non-unique.
If it is not appropriate or possible to support a general collection, use a vector (array), if possible. If it is not appropriate or possible to support an unspecified value type, use integers.
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #Phix | Phix | function mode(sequence s)
-- returns a list of the most common values, each of which occurs the same number of times
integer nxt = 1, count = 1, maxc = 1
sequence res = {}
if length(s)!=0 then
s = sort(s)
object prev = s[1]
for i=2 to length(s) do
if s[i]!=prev then
s[nxt] = {count,prev}
nxt += 1
prev = s[i]
count = 1
else
count += 1
if count>maxc then
maxc = count
end if
end if
end for
s[nxt] = {count,prev}
res = ""
for i=1 to nxt do
if s[i][1]=maxc then
res = append(res,s[i][2])
end if
end for
end if
return res
end function
?mode({1, 2, 5, -5, -9.5, 3.14159})
?mode({ 1, "blue", 2, 7.5, 5, "green", "red", 5, 2, "blue", "white" })
?mode({1, 2, 3, 1, 2, 4, 2, 5, 2, 3, 3, 1, 3, 6})
?mode({.2, .7, .1, .8, .2})
?mode({"two", 7, 1, 8, "two", 8})
?mode("Hello there world")
?mode({})
{} = wait_key()
|
http://rosettacode.org/wiki/Associative_array/Iteration | Associative array/Iteration | Show how to iterate over the key-value pairs of an associative array, and print each pair out.
Also show how to iterate just over the keys, or the values, if there is a separate way to do that in your language.
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #Elena | Elena | import system'collections;
import system'routines;
import extensions;
public program()
{
// 1. Create
var map := Dictionary.new();
map["key"] := "foox";
map["key"] := "foo";
map["key2"]:= "foo2";
map["key3"]:= "foo3";
map["key4"]:= "foo4";
// Enumerate
map.forEach:
(keyValue){ console.printLine(keyValue.Key," : ",keyValue.Value) }
} |
http://rosettacode.org/wiki/Associative_array/Iteration | Associative array/Iteration | Show how to iterate over the key-value pairs of an associative array, and print each pair out.
Also show how to iterate just over the keys, or the values, if there is a separate way to do that in your language.
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #Elixir | Elixir | IO.inspect d = Map.new([foo: 1, bar: 2, baz: 3])
Enum.each(d, fn kv -> IO.inspect kv end)
Enum.each(d, fn {k,v} -> IO.puts "#{inspect k} => #{v}" end)
Enum.each(Map.keys(d), fn key -> IO.inspect key end)
Enum.each(Map.values(d), fn value -> IO.inspect value end) |
http://rosettacode.org/wiki/Apply_a_digital_filter_(direct_form_II_transposed) | Apply a digital filter (direct form II transposed) | Digital filters are used to apply a mathematical operation to a sampled signal. One of the common formulations is the "direct form II transposed" which can represent both infinite impulse response (IIR) and finite impulse response (FIR) filters, as well as being more numerically stable than other forms. [1]
Task
Filter a signal using an order 3 low-pass Butterworth filter. The coefficients for the filter are a=[1.00000000, -2.77555756e-16, 3.33333333e-01, -1.85037171e-17] and b = [0.16666667, 0.5, 0.5, 0.16666667]
The signal that needs filtering is the following vector: [-0.917843918645, 0.141984778794, 1.20536903482, 0.190286794412, -0.662370894973, -1.00700480494, -0.404707073677 ,0.800482325044, 0.743500089861, 1.01090520172, 0.741527555207, 0.277841675195, 0.400833448236, -0.2085993586, -0.172842103641, -0.134316096293, 0.0259303398477, 0.490105989562, 0.549391221511, 0.9047198589]
| #Racket | Racket | #lang racket
(define a (vector 1.00000000E0 -2.77555756E-16 3.33333333E-01 -1.85037171E-17))
(define b (vector 0.16666667E0 0.50000000E0 0.50000000E0 0.16666667E0))
(define s (vector -0.917843918645 0.141984778794 1.20536903482 0.190286794412 -0.662370894973
-1.00700480494 -0.404707073677 0.800482325044 0.743500089861 1.01090520172
0.741527555207 0.277841675195 0.400833448236 -0.2085993586 -0.172842103641
-0.134316096293 0.0259303398477 0.490105989562 0.549391221511 0.9047198589))
(define (filter-signal-direct-form-ii-transposed coeff1 coeff2 signal)
(define signal-size (vector-length signal))
(define filtered-signal (make-vector signal-size 0))
(for ((i signal-size))
(vector-set! filtered-signal
i
(/ (for/fold ((s (for/fold ((s 0)) ((j (vector-length coeff2)) #:when (>= i j))
(+ s (* (vector-ref coeff2 j) (vector-ref signal (- i j)))))))
((j (vector-length coeff1)) #:when (>= i j))
(- s (* (vector-ref coeff1 j) (vector-ref filtered-signal (- i j)))))
(vector-ref coeff1 0))))
filtered-signal)
(filter-signal-direct-form-ii-transposed a b s) |
http://rosettacode.org/wiki/Apply_a_digital_filter_(direct_form_II_transposed) | Apply a digital filter (direct form II transposed) | Digital filters are used to apply a mathematical operation to a sampled signal. One of the common formulations is the "direct form II transposed" which can represent both infinite impulse response (IIR) and finite impulse response (FIR) filters, as well as being more numerically stable than other forms. [1]
Task
Filter a signal using an order 3 low-pass Butterworth filter. The coefficients for the filter are a=[1.00000000, -2.77555756e-16, 3.33333333e-01, -1.85037171e-17] and b = [0.16666667, 0.5, 0.5, 0.16666667]
The signal that needs filtering is the following vector: [-0.917843918645, 0.141984778794, 1.20536903482, 0.190286794412, -0.662370894973, -1.00700480494, -0.404707073677 ,0.800482325044, 0.743500089861, 1.01090520172, 0.741527555207, 0.277841675195, 0.400833448236, -0.2085993586, -0.172842103641, -0.134316096293, 0.0259303398477, 0.490105989562, 0.549391221511, 0.9047198589]
| #Raku | Raku | sub TDF-II-filter ( @signal, @a, @b ) {
my @out = 0 xx @signal;
for ^@signal -> $i {
my $this;
$this += @b[$_] * @signal[$i-$_] if $i-$_ >= 0 for ^@b;
$this -= @a[$_] * @out[$i-$_] if $i-$_ >= 0 for ^@a;
@out[$i] = $this / @a[0];
}
@out
}
my @signal = [
-0.917843918645, 0.141984778794, 1.20536903482, 0.190286794412,
-0.662370894973, -1.00700480494, -0.404707073677, 0.800482325044,
0.743500089861, 1.01090520172, 0.741527555207, 0.277841675195,
0.400833448236, -0.2085993586, -0.172842103641, -0.134316096293,
0.0259303398477, 0.490105989562, 0.549391221511, 0.9047198589
];
my @a = [ 1.00000000, -2.77555756e-16, 3.33333333e-01, -1.85037171e-17 ];
my @b = [ 0.16666667, 0.5, 0.5, 0.16666667 ];
say TDF-II-filter(@signal, @a, @b)».fmt("% 0.8f")
Z~ flat (', ' xx 4, ",\n") xx *; |
http://rosettacode.org/wiki/Averages/Arithmetic_mean | Averages/Arithmetic mean | Task[edit]
Write a program to find the mean (arithmetic average) of a numeric vector.
In case of a zero-length input, since the mean of an empty set of numbers is ill-defined, the program may choose to behave in any way it deems appropriate, though if the programming language has an established convention for conveying math errors or undefined values, it's preferable to follow it.
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #Euphoria | Euphoria | function mean(sequence s)
atom sum
if length(s) = 0 then
return 0
else
sum = 0
for i = 1 to length(s) do
sum += s[i]
end for
return sum/length(s)
end if
end function
sequence test
test = {1.0, 2.0, 5.0, -5.0, 9.5, 3.14159}
? mean(test) |
http://rosettacode.org/wiki/Averages/Arithmetic_mean | Averages/Arithmetic mean | Task[edit]
Write a program to find the mean (arithmetic average) of a numeric vector.
In case of a zero-length input, since the mean of an empty set of numbers is ill-defined, the program may choose to behave in any way it deems appropriate, though if the programming language has an established convention for conveying math errors or undefined values, it's preferable to follow it.
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #Excel | Excel | =AVERAGE(A1:A10) |
http://rosettacode.org/wiki/Average_loop_length | Average loop length | Let f be a uniformly-randomly chosen mapping from the numbers 1..N to the numbers 1..N (note: not necessarily a permutation of 1..N; the mapping could produce a number in more than one way or not at all). At some point, the sequence 1, f(1), f(f(1))... will contain a repetition, a number that occurring for the second time in the sequence.
Task
Write a program or a script that estimates, for each N, the average length until the first such repetition.
Also calculate this expected length using an analytical formula, and optionally compare the simulated result with the theoretical one.
This problem comes from the end of Donald Knuth's Christmas tree lecture 2011.
Example of expected output:
N average analytical (error)
=== ========= ============ =========
1 1.0000 1.0000 ( 0.00%)
2 1.4992 1.5000 ( 0.05%)
3 1.8784 1.8889 ( 0.56%)
4 2.2316 2.2188 ( 0.58%)
5 2.4982 2.5104 ( 0.49%)
6 2.7897 2.7747 ( 0.54%)
7 3.0153 3.0181 ( 0.09%)
8 3.2429 3.2450 ( 0.07%)
9 3.4536 3.4583 ( 0.14%)
10 3.6649 3.6602 ( 0.13%)
11 3.8091 3.8524 ( 1.12%)
12 3.9986 4.0361 ( 0.93%)
13 4.2074 4.2123 ( 0.12%)
14 4.3711 4.3820 ( 0.25%)
15 4.5275 4.5458 ( 0.40%)
16 4.6755 4.7043 ( 0.61%)
17 4.8877 4.8579 ( 0.61%)
18 4.9951 5.0071 ( 0.24%)
19 5.1312 5.1522 ( 0.41%)
20 5.2699 5.2936 ( 0.45%)
| #Unicon | Unicon | link printf, factors
$define MAX_N 20
$define TIMES 1000000
$define RAND_MAX 2147483647
procedure expected(n)
local sum := 0
every i := 1 to n do
sum +:= factorial(n) / (n ^ i) / factorial(n - i)
return sum
end
procedure test(n, times)
local i, count := 0, x, bits
every i := 0 to times-1 do {
x := 1
bits := 0
while iand(bits, x)=0 do {
count +:= 1
bits := ior(bits, x)
x := ishift(1 , ?n-1)
}
}
return count
end
procedure main(void)
local n, cnt, avg, theory, diff
write(" n\tavg\texp.\tdiff\n", repl("-",29))
every n := 1 to MAX_N do {
cnt := test(n, TIMES)
avg := real(cnt) / TIMES
theory := expected(n)
diff := (avg / theory - 1) * 100
printf("%2d %8.4r %8.4r %6.3r%%\n", n, avg, theory, diff)
}
return 0
end |
http://rosettacode.org/wiki/Average_loop_length | Average loop length | Let f be a uniformly-randomly chosen mapping from the numbers 1..N to the numbers 1..N (note: not necessarily a permutation of 1..N; the mapping could produce a number in more than one way or not at all). At some point, the sequence 1, f(1), f(f(1))... will contain a repetition, a number that occurring for the second time in the sequence.
Task
Write a program or a script that estimates, for each N, the average length until the first such repetition.
Also calculate this expected length using an analytical formula, and optionally compare the simulated result with the theoretical one.
This problem comes from the end of Donald Knuth's Christmas tree lecture 2011.
Example of expected output:
N average analytical (error)
=== ========= ============ =========
1 1.0000 1.0000 ( 0.00%)
2 1.4992 1.5000 ( 0.05%)
3 1.8784 1.8889 ( 0.56%)
4 2.2316 2.2188 ( 0.58%)
5 2.4982 2.5104 ( 0.49%)
6 2.7897 2.7747 ( 0.54%)
7 3.0153 3.0181 ( 0.09%)
8 3.2429 3.2450 ( 0.07%)
9 3.4536 3.4583 ( 0.14%)
10 3.6649 3.6602 ( 0.13%)
11 3.8091 3.8524 ( 1.12%)
12 3.9986 4.0361 ( 0.93%)
13 4.2074 4.2123 ( 0.12%)
14 4.3711 4.3820 ( 0.25%)
15 4.5275 4.5458 ( 0.40%)
16 4.6755 4.7043 ( 0.61%)
17 4.8877 4.8579 ( 0.61%)
18 4.9951 5.0071 ( 0.24%)
19 5.1312 5.1522 ( 0.41%)
20 5.2699 5.2936 ( 0.45%)
| #VBA | VBA | Const MAX = 20
Const ITER = 1000000
Function expected(n As Long) As Double
Dim sum As Double
For i = 1 To n
sum = sum + WorksheetFunction.Fact(n) / n ^ i / WorksheetFunction.Fact(n - i)
Next i
expected = sum
End Function
Function test(n As Long) As Double
Dim count As Long
Dim x As Long, bits As Long
For i = 1 To ITER
x = 1
bits = 0
Do While Not bits And x
count = count + 1
bits = bits Or x
x = 2 ^ (Int(n * Rnd()))
Loop
Next i
test = count / ITER
End Function
Public Sub main()
Dim n As Long
Debug.Print " n avg. exp. (error%)"
Debug.Print "== ====== ====== ========"
For n = 1 To MAX
av = test(n)
ex = expected(n)
Debug.Print Format(n, "@@"); " "; Format(av, "0.0000"); " ";
Debug.Print Format(ex, "0.0000"); " ("; Format(Abs(1 - av / ex), "0.000%"); ")"
Next n
End Sub |
http://rosettacode.org/wiki/Averages/Simple_moving_average | Averages/Simple moving average | Computing the simple moving average of a series of numbers.
Task[edit]
Create a stateful function/class/instance that takes a period and returns a routine that takes a number as argument and returns a simple moving average of its arguments so far.
Description
A simple moving average is a method for computing an average of a stream of numbers by only averaging the last P numbers from the stream, where P is known as the period.
It can be implemented by calling an initialing routine with P as its argument, I(P), which should then return a routine that when called with individual, successive members of a stream of numbers, computes the mean of (up to), the last P of them, lets call this SMA().
The word stateful in the task description refers to the need for SMA() to remember certain information between calls to it:
The period, P
An ordered container of at least the last P numbers from each of its individual calls.
Stateful also means that successive calls to I(), the initializer, should return separate routines that do not share saved state so they could be used on two independent streams of data.
Pseudo-code for an implementation of SMA is:
function SMA(number: N):
stateful integer: P
stateful list: stream
number: average
stream.append_last(N)
if stream.length() > P:
# Only average the last P elements of the stream
stream.delete_first()
if stream.length() == 0:
average = 0
else:
average = sum( stream.values() ) / stream.length()
return average
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #VBA | VBA | Class sma
'to be stored in a class module with name "sma"
Private n As Integer 'period
Private arr() As Double 'circular list
Private index As Integer 'pointer into arr
Private oldsma As Double
Public Sub init(size As Integer)
n = size
ReDim arr(n - 1)
index = 0
End Sub
Public Function sma(number As Double) As Double
sma = oldsma + (-arr(index) + number) / n
oldsma = sma
arr(index) = number
index = (index + 1) Mod n
End Function
Normal module
Public Sub main()
s = [{1,2,3,4,5,5,4,3,2,1}]
Dim sma3 As New sma
Dim sma5 As New sma
sma3.init 3
sma5.init 5
For i = 1 To UBound(s)
Debug.Print i, Format(sma3.sma(CDbl(s(i))), "0.00000"),
Debug.Print Format(sma5.sma(CDbl(s(i))), "0.00000")
Next i
End Sub |
http://rosettacode.org/wiki/Attractive_numbers | Attractive numbers | A number is an attractive number if the number of its prime factors (whether distinct or not) is also prime.
Example
The number 20, whose prime decomposition is 2 × 2 × 5, is an attractive number because the number of its prime factors (3) is also prime.
Task
Show sequence items up to 120.
Reference
The OEIS entry: A063989: Numbers with a prime number of prime divisors.
| #Pascal | Pascal | program AttractiveNumbers;
{ numbers with count of factors = prime
* using modified sieve of erathosthes
* by adding the power of the prime to multiples
* of the composite number }
{$IFDEF FPC}
{$MODE DELPHI}
{$ELSE}
{$APPTYPE CONSOLE}
{$ENDIF}
uses
sysutils;//timing
const
cTextMany = ' with many factors ';
cText2 = ' with only two factors ';
cText1 = ' with only one factor ';
type
tValue = LongWord;
tpValue = ^tValue;
tPower = array[0..63] of tValue;//2^64
var
power : tPower;
sieve : array of byte;
function NextPotCnt(p: tValue):tValue;
//return the first power <> 0
//power == n to base prim
var
i : NativeUint;
begin
result := 0;
repeat
i := power[result];
Inc(i);
IF i < p then
BREAK
else
begin
i := 0;
power[result] := 0;
inc(result);
end;
until false;
power[result] := i;
inc(result);
end;
procedure InitSieveWith2;
//the prime 2, because its the first one, is the one,
//which can can be speed up tremendously, by moving
var
pSieve : pByte;
CopyWidth,lmt : NativeInt;
Begin
pSieve := @sieve[0];
Lmt := High(sieve);
sieve[1] := 0;
sieve[2] := 1; // aka 2^1 -> one factor
CopyWidth := 2;
while CopyWidth*2 <= Lmt do
Begin
// copy idx 1,2 to 3,4 | 1..4 to 5..8 | 1..8 to 9..16
move(pSieve[1],pSieve[CopyWidth+1],CopyWidth);
// 01 -> 0101 -> 01020102-> 0102010301020103
inc(CopyWidth,CopyWidth);//*2
//increment the factor of last element by one.
inc(pSieve[CopyWidth]);
//idx 12 1234 12345678
//value 01 -> 0102 -> 01020103-> 0102010301020104
end;
//copy the rest
move(pSieve[1],pSieve[CopyWidth+1],Lmt-CopyWidth);
//mark 0,1 not prime, 255 factors are today not possible 2^255 >> Uint64
sieve[0]:= 255;
sieve[1]:= 255;
sieve[2]:= 0; // make prime again
end;
procedure OutCntTime(T:TDateTime;txt:String;cnt:NativeInt);
Begin
writeln(cnt:12,txt,T*86400:10:3,' s');
end;
procedure sievefactors;
var
T0 : TDateTime;
pSieve : pByte;
i,j,i2,k,lmt,cnt : NativeUInt;
Begin
InitSieveWith2;
pSieve := @sieve[0];
Lmt := High(sieve);
//Divide into 3 section
//first i*i*i<= lmt with time expensive NextPotCnt
T0 := now;
cnt := 0;
//third root of limit calculate only once, no comparison ala while i*i*i<= lmt do
k := trunc(exp(ln(Lmt)/3));
For i := 3 to k do
if pSieve[i] = 0 then
Begin
inc(cnt);
j := 2*i;
fillChar(Power,Sizeof(Power),#0);
Power[0] := 1;
repeat
inc(pSieve[j],NextPotCnt(i));
inc(j,i);
until j > lmt;
end;
OutCntTime(now-T0,cTextMany,cnt);
T0 := now;
//second i*i <= lmt
cnt := 0;
i := k+1;
k := trunc(sqrt(Lmt));
For i := i to k do
if pSieve[i] = 0 then
Begin
//first increment all multiples of prime by one
inc(cnt);
j := 2*i;
repeat
inc(pSieve[j]);
inc(j,i);
until j>lmt;
//second increment all multiples prime*prime by one
i2 := i*i;
j := i2;
repeat
inc(pSieve[j]);
inc(j,i2);
until j>lmt;
end;
OutCntTime(now-T0,cText2,cnt);
T0 := now;
//third i*i > lmt -> only one new factor
cnt := 0;
inc(k);
For i := k to Lmt shr 1 do
if pSieve[i] = 0 then
Begin
inc(cnt);
j := 2*i;
repeat
inc(pSieve[j]);
inc(j,i);
until j>lmt;
end;
OutCntTime(now-T0,cText1,cnt);
end;
const
smallLmt = 120;
//needs 1e10 Byte = 10 Gb maybe someone got 128 Gb :-) nearly linear time
BigLimit = 10*1000*1000*1000;
var
T0,T : TDateTime;
i,cnt,lmt : NativeInt;
Begin
setlength(sieve,smallLmt+1);
sievefactors;
cnt := 0;
For i := 2 to smallLmt do
Begin
if sieve[sieve[i]] = 0 then
Begin
write(i:4);
inc(cnt);
if cnt>19 then
Begin
writeln;
cnt := 0;
end;
end;
end;
writeln;
writeln;
T0 := now;
setlength(sieve,BigLimit+1);
T := now;
writeln('time allocating : ',(T-T0) *86400 :8:3,' s');
sievefactors;
T := now-T;
writeln('time sieving : ',T*86400 :8:3,' s');
T:= now;
cnt := 0;
i := 0;
lmt := 10;
repeat
repeat
inc(i);
{IF sieve[sieve[i]] = 0 then inc(cnt); takes double time is not relevant}
inc(cnt,ORD(sieve[sieve[i]] = 0));
until i = lmt;
writeln(lmt:11,cnt:12);
lmt := 10*lmt;
until lmt >High(sieve);
T := now-T;
writeln('time counting : ',T*86400 :8:3,' s');
writeln('time total : ',(now-T0)*86400 :8:3,' s');
end. |
http://rosettacode.org/wiki/Averages/Mean_time_of_day | Averages/Mean time of day | Task[edit]
A particular activity of bats occurs at these times of the day:
23:00:17, 23:40:20, 00:12:45, 00:17:19
Using the idea that there are twenty-four hours in a day,
which is analogous to there being 360 degrees in a circle,
map times of day to and from angles;
and using the ideas of Averages/Mean angle
compute and show the average time of the nocturnal activity
to an accuracy of one second of time.
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #REXX | REXX | /* REXX ---------------------------------------------------------------
* 25.06.2014 Walter Pachl
* taken from ooRexx using my very aged sin/cos/artan functions
*--------------------------------------------------------------------*/
times='23:00:17 23:40:20 00:12:45 00:17:19'
sum=0
day=86400
pi=3.14159265358979323846264
x=0
y=0
Do i=1 To words(times) /* loop over times */
time.i=word(times,i) /* pick a time */
alpha.i=t2a(time.i) /* convert to angle (radians) */
/* Say time.i format(alpha.i,6,9) */
x=x+sin(alpha.i) /* accumulate sines */
y=y+cos(alpha.i) /* accumulate cosines */
End
ww=arctan(x/y) /* compute average angle */
ss=ww*86400/(2*pi) /* convert to seconds */
If ss<0 Then ss=ss+day /* avoid negative value */
m=ss%60 /* split into hh mm ss */
s=ss-m*60
h=m%60
m=m-h*60
Say f2(h)':'f2(m)':'f2(s) /* show the mean time */
Exit
t2a: Procedure Expose day pi /* convert time to angle */
Parse Arg hh ':' mm ':' ss
sec=(hh*60+mm)*60+ss
If sec>(day/2) Then
sec=sec-day
a=2*pi*sec/day
Return a
f2: return right(format(arg(1),2,0),2,0)
sin: Procedure Expose pi
Parse Arg x
prec=digits()
Numeric Digits (2*prec)
Do While x>pi
x=x-pi
End
Do While x<-pi
x=x+pi
End
o=x
u=1
r=x
Do i=3 By 2
ra=r
o=-o*x*x
u=u*i*(i-1)
r=r+(o/u)
If r=ra Then Leave
End
Numeric Digits prec
Return r+0
cos: Procedure Expose pi
Parse Arg x
prec=digits()
Numeric Digits (2*prec)
Numeric Fuzz 3
o=1
u=1
r=1
Do i=1 By 2
ra=r
o=-o*x*x
u=u*i*(i+1)
r=r+(o/u)
If r=ra Then Leave
End
Numeric Digits prec
Return r+0
arctan: Procedure
Parse Arg x
prec=digits()
Numeric Digits (2*prec)
Numeric Fuzz 3
o=x
u=1
r=x
k=0
Do i=3 By 2
ra=r
o=-o*x*x
r=r+(o/i)
If r=ra Then
Leave
k=k+1
If k//1000=0 Then
Say i left(r,40) format(abs(o/i),15,5)
End
Numeric Digits (prec)
Return r+0 |
http://rosettacode.org/wiki/AVL_tree | AVL tree |
This page uses content from Wikipedia. The original article was at AVL tree. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance)
In computer science, an AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; at no time do they differ by more than one because rebalancing is done ensure this is the case. Lookup, insertion, and deletion all take O(log n) time in both the average and worst cases, where n is the number of nodes in the tree prior to the operation. Insertions and deletions may require the tree to be rebalanced by one or more tree rotations. Note the tree of nodes comprise a set, so duplicate node keys are not allowed.
AVL trees are often compared with red-black trees because they support the same set of operations and because red-black trees also take O(log n) time for the basic operations. Because AVL trees are more rigidly balanced, they are faster than red-black trees for lookup-intensive applications. Similar to red-black trees, AVL trees are height-balanced, but in general not weight-balanced nor μ-balanced; that is, sibling nodes can have hugely differing numbers of descendants.
Task
Implement an AVL tree in the language of choice, and provide at least basic operations.
| #Wren | Wren | class Node {
construct new(key, parent) {
_key = key
_parent = parent
_balance = 0
_left = null
_right = null
}
key { _key }
parent { _parent }
balance { _balance }
left { _left }
right { _right }
key=(k) { _key = k }
parent=(p) { _parent = p }
balance=(v) { _balance = v }
left=(n) { _left = n }
right= (n) { _right = n }
}
class AvlTree {
construct new() {
_root = null
}
insert(key) {
if (!_root) {
_root = Node.new(key, null)
} else {
var n = _root
while (true) {
if (n.key == key) return false
var parent = n
var goLeft = n.key > key
n = goLeft ? n.left : n.right
if (!n) {
if (goLeft) {
parent.left = Node.new(key, parent)
} else {
parent.right = Node.new(key, parent)
}
rebalance(parent)
break
}
}
}
return true
}
delete(delKey) {
if (!_root) return
var n = _root
var parent = _root
var delNode = null
var child = _root
while (child) {
parent = n
n = child
child = (delKey >= n.key) ? n.right : n.left
if (delKey == n.key) delNode = n
}
if (delNode) {
delNode.key = n.key
child = n.left ? n.left : n.right
if (_root.key == delKey) {
_root = child
if (_root) _root.parent = null
} else {
if (parent.left == n) {
parent.left = child
} else {
parent.right = child
}
if (child) child.parent = parent
rebalance(parent)
}
}
}
rebalance(n) {
setBalance([n])
var nn = n
if (nn.balance == -2) {
if (height(nn.left.left) >= height(nn.left.right)) {
nn = rotateRight(nn)
} else {
nn = rotateLeftThenRight(nn)
}
} else if (nn.balance == 2) {
if (height(nn.right.right) >= height(nn.right.left)) {
nn = rotateLeft(nn)
} else {
nn = rotateRightThenLeft(nn)
}
}
if (nn.parent) rebalance(nn.parent) else _root = nn
}
rotateLeft(a) {
var b = a.right
b.parent = a.parent
a.right = b.left
if (a.right) a.right.parent = a
b.left = a
a.parent = b
if (b.parent) {
if (b.parent.right == a) {
b.parent.right = b
} else {
b.parent.left = b
}
}
setBalance([a, b])
return b
}
rotateRight(a) {
var b = a.left
b.parent = a.parent
a.left = b.right
if (a.left) a.left.parent = a
b.right = a
a.parent = b
if (b.parent) {
if (b.parent.right == a) {
b.parent.right = b
} else {
b.parent.left = b
}
}
setBalance([a, b])
return b
}
rotateLeftThenRight(n) {
n.left = rotateLeft(n.left)
return rotateRight(n)
}
rotateRightThenLeft(n) {
n.right = rotateRight(n.right)
return rotateLeft(n)
}
height(n) {
if (!n) return -1
return 1 + height(n.left).max(height(n.right))
}
setBalance(nodes) {
for (n in nodes) n.balance = height(n.right) - height(n.left)
}
printKey() {
printKey(_root)
System.print()
}
printKey(n) {
if (n) {
printKey(n.left)
System.write("%(n.key) ")
printKey(n.right)
}
}
printBalance() {
printBalance(_root)
System.print()
}
printBalance(n) {
if (n) {
printBalance(n.left)
System.write("%(n.balance) ")
printBalance(n.right)
}
}
}
var tree = AvlTree.new()
System.print("Inserting values 1 to 10")
for (i in 1..10) tree.insert(i)
System.write("Printing key : ")
tree.printKey()
System.write("Printing balance : ")
tree.printBalance() |
http://rosettacode.org/wiki/Averages/Mean_angle | Averages/Mean angle | When calculating the average or mean of an angle one has to take into account how angles wrap around so that any angle in degrees plus any integer multiple of 360 degrees is a measure of the same angle.
If one wanted an average direction of the wind over two readings where the first reading was of 350 degrees and the second was of 10 degrees then the average of the numbers is 180 degrees, whereas if you can note that 350 degrees is equivalent to -10 degrees and so you have two readings at 10 degrees either side of zero degrees leading to a more fitting mean angle of zero degrees.
To calculate the mean angle of several angles:
Assume all angles are on the unit circle and convert them to complex numbers expressed in real and imaginary form.
Compute the mean of the complex numbers.
Convert the complex mean to polar coordinates whereupon the phase of the complex mean is the required angular mean.
(Note that, since the mean is the sum divided by the number of numbers, and division by a positive real number does not affect the angle, you can also simply compute the sum for step 2.)
You can alternatively use this formula:
Given the angles
α
1
,
…
,
α
n
{\displaystyle \alpha _{1},\dots ,\alpha _{n}}
the mean is computed by
α
¯
=
atan2
(
1
n
⋅
∑
j
=
1
n
sin
α
j
,
1
n
⋅
∑
j
=
1
n
cos
α
j
)
{\displaystyle {\bar {\alpha }}=\operatorname {atan2} \left({\frac {1}{n}}\cdot \sum _{j=1}^{n}\sin \alpha _{j},{\frac {1}{n}}\cdot \sum _{j=1}^{n}\cos \alpha _{j}\right)}
Task[edit]
write a function/method/subroutine/... that given a list of angles in degrees returns their mean angle.
(You should use a built-in function if you have one that does this for degrees or radians).
Use the function to compute the means of these lists of angles (in degrees):
[350, 10]
[90, 180, 270, 360]
[10, 20, 30]
Show your output here.
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #Racket | Racket |
#lang racket
(define (mean-angle αs)
(radians->degrees
(mean-angle/radians
(map degrees->radians αs))))
(define (mean-angle/radians αs)
(define n (length αs))
(atan (* (/ 1 n) (for/sum ([α_j αs]) (sin α_j)))
(* (/ 1 n) (for/sum ([α_j αs]) (cos α_j)))))
(mean-angle '(350 0 10))
(mean-angle '(90 180 270 360))
(mean-angle '(10 20 30))
|
http://rosettacode.org/wiki/Averages/Mean_angle | Averages/Mean angle | When calculating the average or mean of an angle one has to take into account how angles wrap around so that any angle in degrees plus any integer multiple of 360 degrees is a measure of the same angle.
If one wanted an average direction of the wind over two readings where the first reading was of 350 degrees and the second was of 10 degrees then the average of the numbers is 180 degrees, whereas if you can note that 350 degrees is equivalent to -10 degrees and so you have two readings at 10 degrees either side of zero degrees leading to a more fitting mean angle of zero degrees.
To calculate the mean angle of several angles:
Assume all angles are on the unit circle and convert them to complex numbers expressed in real and imaginary form.
Compute the mean of the complex numbers.
Convert the complex mean to polar coordinates whereupon the phase of the complex mean is the required angular mean.
(Note that, since the mean is the sum divided by the number of numbers, and division by a positive real number does not affect the angle, you can also simply compute the sum for step 2.)
You can alternatively use this formula:
Given the angles
α
1
,
…
,
α
n
{\displaystyle \alpha _{1},\dots ,\alpha _{n}}
the mean is computed by
α
¯
=
atan2
(
1
n
⋅
∑
j
=
1
n
sin
α
j
,
1
n
⋅
∑
j
=
1
n
cos
α
j
)
{\displaystyle {\bar {\alpha }}=\operatorname {atan2} \left({\frac {1}{n}}\cdot \sum _{j=1}^{n}\sin \alpha _{j},{\frac {1}{n}}\cdot \sum _{j=1}^{n}\cos \alpha _{j}\right)}
Task[edit]
write a function/method/subroutine/... that given a list of angles in degrees returns their mean angle.
(You should use a built-in function if you have one that does this for degrees or radians).
Use the function to compute the means of these lists of angles (in degrees):
[350, 10]
[90, 180, 270, 360]
[10, 20, 30]
Show your output here.
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #Raku | Raku | # Of course, you can still use pi and 180.
sub deg2rad { $^d * tau / 360 }
sub rad2deg { $^r * 360 / tau }
sub phase ($c) {
my ($mag,$ang) = $c.polar;
return NaN if $mag < 1e-16;
$ang;
}
sub meanAngle { rad2deg phase [+] map { cis deg2rad $_ }, @^angles }
say meanAngle($_).fmt("%.2f\tis the mean angle of "), $_ for
[350, 10],
[90, 180, 270, 360],
[10, 20, 30]; |
http://rosettacode.org/wiki/Averages/Median | Averages/Median | Task[edit]
Write a program to find the median value of a vector of floating-point numbers.
The program need not handle the case where the vector is empty, but must handle the case where there are an even number of elements. In that case, return the average of the two middle values.
There are several approaches to this. One is to sort the elements, and then pick the element(s) in the middle.
Sorting would take at least O(n logn). Another approach would be to build a priority queue from the elements, and then extract half of the elements to get to the middle element(s). This would also take O(n logn). The best solution is to use the selection algorithm to find the median in O(n) time.
See also
Quickselect_algorithm
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #Icon_and_Unicon | Icon and Unicon | procedure main(args)
write(median(args))
end
procedure median(A)
A := sort(A)
n := *A
return if n % 2 = 1 then A[n/2+1]
else (A[n/2]+A[n/2+1])/2.0 | 0 # 0 if empty list
end |
http://rosettacode.org/wiki/Averages/Median | Averages/Median | Task[edit]
Write a program to find the median value of a vector of floating-point numbers.
The program need not handle the case where the vector is empty, but must handle the case where there are an even number of elements. In that case, return the average of the two middle values.
There are several approaches to this. One is to sort the elements, and then pick the element(s) in the middle.
Sorting would take at least O(n logn). Another approach would be to build a priority queue from the elements, and then extract half of the elements to get to the middle element(s). This would also take O(n logn). The best solution is to use the selection algorithm to find the median in O(n) time.
See also
Quickselect_algorithm
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #J | J | require 'stats/base'
median 1 9 2 4
3 |
http://rosettacode.org/wiki/Averages/Pythagorean_means | Averages/Pythagorean means | Task[edit]
Compute all three of the Pythagorean means of the set of integers 1 through 10 (inclusive).
Show that
A
(
x
1
,
…
,
x
n
)
≥
G
(
x
1
,
…
,
x
n
)
≥
H
(
x
1
,
…
,
x
n
)
{\displaystyle A(x_{1},\ldots ,x_{n})\geq G(x_{1},\ldots ,x_{n})\geq H(x_{1},\ldots ,x_{n})}
for this set of positive integers.
The most common of the three means, the arithmetic mean, is the sum of the list divided by its length:
A
(
x
1
,
…
,
x
n
)
=
x
1
+
⋯
+
x
n
n
{\displaystyle A(x_{1},\ldots ,x_{n})={\frac {x_{1}+\cdots +x_{n}}{n}}}
The geometric mean is the
n
{\displaystyle n}
th root of the product of the list:
G
(
x
1
,
…
,
x
n
)
=
x
1
⋯
x
n
n
{\displaystyle G(x_{1},\ldots ,x_{n})={\sqrt[{n}]{x_{1}\cdots x_{n}}}}
The harmonic mean is
n
{\displaystyle n}
divided by the sum of the reciprocal of each item in the list:
H
(
x
1
,
…
,
x
n
)
=
n
1
x
1
+
⋯
+
1
x
n
{\displaystyle H(x_{1},\ldots ,x_{n})={\frac {n}{{\frac {1}{x_{1}}}+\cdots +{\frac {1}{x_{n}}}}}}
See also
Tasks for calculating statistical measures
in one go
moving (sliding window)
moving (cumulative)
Mean
Arithmetic
Statistics/Basic
Averages/Arithmetic mean
Averages/Pythagorean means
Averages/Simple moving average
Geometric
Averages/Pythagorean means
Harmonic
Averages/Pythagorean means
Quadratic
Averages/Root mean square
Circular
Averages/Mean angle
Averages/Mean time of day
Median
Averages/Median
Mode
Averages/Mode
Standard deviation
Statistics/Basic
Cumulative standard deviation
| #M2000_Interpreter | M2000 Interpreter |
Module CheckIt {
sum=lambda -> {
Read m as array
if len(m)=0 then =0 : exit
sum=Array(m, Dimension(m,0))
If len(m)=1 then =sum : exit
k=each(m,2,-1)
While k {
sum+=Array(k)
}
=sum
}
mean=lambda sum (a as array) ->{
=sum(a)/len(a)
}
prod=lambda -> {
m=array
if len(m)=0 then =0 : exit
prod=Array(m, Dimension(m,0))
If len(m)=1 then =prod : exit
k=each(m,2,-1)
While k {
prod*=Array(k)
}
=prod
}
geomean=lambda prod (a as array) -> {
=prod(a)^(1/len(a))
}
harmomean=lambda (a as array) -> {
if len(a)=0 then =0 : exit
sum=1/Array(a, Dimension(a,0))
If len(a)=1 then =1/sum : exit
k=each(a,2,-1)
While k {
sum+=1/Array(k)
}
=len(a)/sum
}
Print sum((1,2,3,4,5))=15
Print prod((1,2,3,4,5))=120
Print mean((1,2,3,4,5))==3
\\ use == to apply rounding before comparison
Print geomean((1,2,3,4,5))==2.60517108469735
Print harmomean((1,2,3,4,5))==2.18978102189784
Generator =lambda x=1 ->{=x : x++}
dim a(10)<<Generator()
Print mean(a())==5.5
Print geomean(a())==4.52872868811677
Print harmomean(a())==3.41417152147412
}
CheckIt
|
http://rosettacode.org/wiki/Balanced_ternary | Balanced ternary | Balanced ternary is a way of representing numbers. Unlike the prevailing binary representation, a balanced ternary integer is in base 3, and each digit can have the values 1, 0, or −1.
Examples
Decimal 11 = 32 + 31 − 30, thus it can be written as "++−"
Decimal 6 = 32 − 31 + 0 × 30, thus it can be written as "+−0"
Task
Implement balanced ternary representation of integers with the following:
Support arbitrarily large integers, both positive and negative;
Provide ways to convert to and from text strings, using digits '+', '-' and '0' (unless you are already using strings to represent balanced ternary; but see requirement 5).
Provide ways to convert to and from native integer type (unless, improbably, your platform's native integer type is balanced ternary). If your native integers can't support arbitrary length, overflows during conversion must be indicated.
Provide ways to perform addition, negation and multiplication directly on balanced ternary integers; do not convert to native integers first.
Make your implementation efficient, with a reasonable definition of "efficient" (and with a reasonable definition of "reasonable").
Test case With balanced ternaries a from string "+-0++0+", b from native integer -436, c "+-++-":
write out a, b and c in decimal notation;
calculate a × (b − c), write out the result in both ternary and decimal notations.
Note: The pages generalised floating point addition and generalised floating point multiplication have code implementing arbitrary precision floating point balanced ternary.
| #Visual_Basic_.NET | Visual Basic .NET | Imports System.Text
Module Module1
Sub Main()
Dim a As New BalancedTernary("+-0++0+")
Console.WriteLine("a: {0} = {1}", a, a.ToLong)
Dim b As New BalancedTernary(-436)
Console.WriteLine("b: {0} = {1}", b, b.ToLong)
Dim c As New BalancedTernary("+-++-")
Console.WriteLine("c: {0} = {1}", c, c.ToLong)
Dim d = a * (b - c)
Console.WriteLine("a * (b - c): {0} = {1}", d, d.ToLong)
End Sub
Class BalancedTernary
Private Enum BalancedTernaryDigit
MINUS = -1
ZERO = 0
PLUS = 1
End Enum
Private ReadOnly value() As BalancedTernaryDigit
' empty = 0
Public Sub New()
ReDim value(-1)
End Sub
' create from string
Public Sub New(str As String)
ReDim value(str.Length - 1)
For i = 1 To str.Length
If str(i - 1) = "-" Then
value(i - 1) = BalancedTernaryDigit.MINUS
ElseIf str(i - 1) = "0" Then
value(i - 1) = BalancedTernaryDigit.ZERO
ElseIf str(i - 1) = "+" Then
value(i - 1) = BalancedTernaryDigit.PLUS
Else
Throw New ArgumentException("Unknown Digit: " + str(i - 1))
End If
Next
Array.Reverse(value)
End Sub
' convert integer
Public Sub New(l As Long)
Dim value As New List(Of BalancedTernaryDigit)
Dim sign = Math.Sign(l)
l = Math.Abs(l)
While l <> 0
Dim remainder = CType(l Mod 3, Byte)
If remainder = 0 OrElse remainder = 1 Then
value.Add(remainder)
l /= 3
ElseIf remainder = 2 Then
value.Add(BalancedTernaryDigit.MINUS)
l = (l + 1) / 3
End If
End While
Me.value = value.ToArray
If sign < 0 Then
Invert()
End If
End Sub
' copy constructor
Public Sub New(origin As BalancedTernary)
ReDim value(origin.value.Length - 1)
Array.Copy(origin.value, value, origin.value.Length)
End Sub
' only for internal use
Private Sub New(value() As BalancedTernaryDigit)
Dim endi = value.Length - 1
While endi > 0 AndAlso value(endi) = BalancedTernaryDigit.ZERO
endi -= 1
End While
ReDim Me.value(endi)
Array.Copy(value, Me.value, endi + 1)
End Sub
' invert the values
Private Sub Invert()
For i = 1 To value.Length
value(i - 1) = CType(-CType(value(i - 1), Integer), BalancedTernaryDigit)
Next
End Sub
' convert to string
Public Overrides Function ToString() As String
Dim result As New StringBuilder
Dim i = value.Length - 1
While i >= 0
If value(i) = BalancedTernaryDigit.MINUS Then
result.Append("-")
ElseIf value(i) = BalancedTernaryDigit.ZERO Then
result.Append("0")
ElseIf value(i) = BalancedTernaryDigit.PLUS Then
result.Append("+")
End If
i -= 1
End While
Return result.ToString
End Function
' convert to long
Public Function ToLong() As Long
Dim result = 0L
For i = 1 To value.Length
result += value(i - 1) * Math.Pow(3.0, i - 1)
Next
Return result
End Function
' unary minus
Public Shared Operator -(origin As BalancedTernary) As BalancedTernary
Dim result As New BalancedTernary(origin)
result.Invert()
Return result
End Operator
' addition of digits
Private Shared carry = BalancedTernaryDigit.ZERO
Private Shared Function Add(a As BalancedTernaryDigit, b As BalancedTernaryDigit) As BalancedTernaryDigit
If a <> b Then
carry = BalancedTernaryDigit.ZERO
Return a + b
Else
carry = a
Return -CType(b, Integer)
End If
End Function
' addition of balanced ternary numbers
Public Shared Operator +(a As BalancedTernary, b As BalancedTernary) As BalancedTernary
Dim maxLength = Math.Max(a.value.Length, b.value.Length)
Dim resultValue(maxLength) As BalancedTernaryDigit
For i = 1 To maxLength
If i - 1 < a.value.Length Then
resultValue(i - 1) = Add(resultValue(i - 1), a.value(i - 1))
resultValue(i) = carry
Else
carry = BalancedTernaryDigit.ZERO
End If
If i - 1 < b.value.Length Then
resultValue(i - 1) = Add(resultValue(i - 1), b.value(i - 1))
resultValue(i) = Add(resultValue(i), carry)
End If
Next
Return New BalancedTernary(resultValue)
End Operator
' subtraction of balanced ternary numbers
Public Shared Operator -(a As BalancedTernary, b As BalancedTernary) As BalancedTernary
Return a + (-b)
End Operator
' multiplication of balanced ternary numbers
Public Shared Operator *(a As BalancedTernary, b As BalancedTernary) As BalancedTernary
Dim longValue = a.value
Dim shortValue = b.value
Dim result As New BalancedTernary
If a.value.Length < b.value.Length Then
longValue = b.value
shortValue = a.value
End If
For i = 1 To shortValue.Length
If shortValue(i - 1) <> BalancedTernaryDigit.ZERO Then
Dim temp(i + longValue.Length - 2) As BalancedTernaryDigit
For j = 1 To longValue.Length
temp(i + j - 2) = CType(shortValue(i - 1) * longValue(j - 1), BalancedTernaryDigit)
Next
result += New BalancedTernary(temp)
End If
Next
Return result
End Operator
End Class
End Module |
http://rosettacode.org/wiki/Babbage_problem | Babbage problem |
Charles Babbage, looking ahead to the sorts of problems his Analytical Engine would be able to solve, gave this example:
What is the smallest positive integer whose square ends in the digits 269,696?
— Babbage, letter to Lord Bowden, 1837; see Hollingdale and Tootill, Electronic Computers, second edition, 1970, p. 125.
He thought the answer might be 99,736, whose square is 9,947,269,696; but he couldn't be certain.
Task[edit]
The task is to find out if Babbage had the right answer — and to do so, as far as your language allows it, in code that Babbage himself would have been able to read and understand.
As Babbage evidently solved the task with pencil and paper, a similar efficient solution is preferred.
For these purposes, Charles Babbage may be taken to be an intelligent person, familiar with mathematics and with the idea of a computer; he has written the first drafts of simple computer programmes in tabular form. [Babbage Archive Series L].
Motivation
The aim of the task is to write a program that is sufficiently clear and well-documented for such a person to be able to read it and be confident that it does indeed solve the specified problem.
| #Simula | Simula | BEGIN
INTEGER PROBE, SQUARE;
BOOLEAN DONE;
WHILE NOT DONE DO BEGIN
PROBE := PROBE + 1;
SQUARE := PROBE * PROBE;
IF MOD(SQUARE, 1000000) = 269696 THEN BEGIN
OUTTEXT("THE SMALLEST NUMBER: ");
OUTINT(PROBE,0);
OUTIMAGE;
OUTTEXT("THE SQUARE : ");
OUTINT(SQUARE,0);
OUTIMAGE;
DONE := TRUE;
END;
END;
END |
http://rosettacode.org/wiki/Babbage_problem | Babbage problem |
Charles Babbage, looking ahead to the sorts of problems his Analytical Engine would be able to solve, gave this example:
What is the smallest positive integer whose square ends in the digits 269,696?
— Babbage, letter to Lord Bowden, 1837; see Hollingdale and Tootill, Electronic Computers, second edition, 1970, p. 125.
He thought the answer might be 99,736, whose square is 9,947,269,696; but he couldn't be certain.
Task[edit]
The task is to find out if Babbage had the right answer — and to do so, as far as your language allows it, in code that Babbage himself would have been able to read and understand.
As Babbage evidently solved the task with pencil and paper, a similar efficient solution is preferred.
For these purposes, Charles Babbage may be taken to be an intelligent person, familiar with mathematics and with the idea of a computer; he has written the first drafts of simple computer programmes in tabular form. [Babbage Archive Series L].
Motivation
The aim of the task is to write a program that is sufficiently clear and well-documented for such a person to be able to read it and be confident that it does indeed solve the specified problem.
| #Smalltalk | Smalltalk | "We use one variable, called n. Let it initially be equal to 1. Then keep increasing it by 1 for only as long as the remainder after dividing by a million is not equal to 269,696; finally, show the value of n."
| n |
n := 1.
[ n squared \\ 1000000 = 269696 ] whileFalse: [ n := n + 1 ].
n |
http://rosettacode.org/wiki/Approximate_equality | Approximate equality | Sometimes, when testing whether the solution to a task (for example, here on Rosetta Code) is correct, the
difference in floating point calculations between different language implementations becomes significant.
For example, a difference between 32 bit and 64 bit floating point calculations may appear by
about the 8th significant digit in base 10 arithmetic.
Task
Create a function which returns true if two floating point numbers are approximately equal.
The function should allow for differences in the magnitude of numbers, so that, for example,
100000000000000.01 may be approximately equal to 100000000000000.011,
even though 100.01 is not approximately equal to 100.011.
If the language has such a feature in its standard library, this may be used instead of a custom function.
Show the function results with comparisons on the following pairs of values:
100000000000000.01, 100000000000000.011 (note: should return true)
100.01, 100.011 (note: should return false)
10000000000000.001 / 10000.0, 1000000000.0000001000
0.001, 0.0010000001
0.000000000000000000000101, 0.0
sqrt(2) * sqrt(2), 2.0
-sqrt(2) * sqrt(2), -2.0
3.14159265358979323846, 3.14159265358979324
Answers should be true for the first example and false in the second, so that just rounding the numbers to a fixed number of decimals should not be enough. Otherwise answers may vary and still be correct. See the Python code for one type of solution.
| #XPL0 | XPL0 | func ApproxEqual(A, B); \Return 'true' if approximately equal
real A, B;
real Epsilon;
[Epsilon:= abs(A) * 1E-15;
return abs(A-B) < Epsilon;
];
real Data;
int I;
[Format(0, 16);
Data:=[ [100000000000000.01, 100000000000000.011], \should return true
[100.01, 100.011], \should return false
[10000000000000.001 / 10000.0, 1000000000.0000001000],
[0.001, 0.0010000001],
[0.000000000000000000000101, 0.0], \is undefined
[sqrt(2.0) * sqrt(2.0), 2.0],
[-1.0 * sqrt(2.0) * sqrt(2.0), -2.0], \-sqrt doesn't compile!
[3.14159265358979323846, 3.14159265358979324] ];
for I:= 0 to 7 do
[IntOut(0, I+1); Text(0, ". ");
RlOut(0, Data(I,0)); ChOut(0, ^ ); RlOut(0, Data(I,1));
Text(0, if ApproxEqual(Data(I,0), Data(I,1)) then " true" else " false");
CrLf(0);
];
] |
http://rosettacode.org/wiki/Approximate_equality | Approximate equality | Sometimes, when testing whether the solution to a task (for example, here on Rosetta Code) is correct, the
difference in floating point calculations between different language implementations becomes significant.
For example, a difference between 32 bit and 64 bit floating point calculations may appear by
about the 8th significant digit in base 10 arithmetic.
Task
Create a function which returns true if two floating point numbers are approximately equal.
The function should allow for differences in the magnitude of numbers, so that, for example,
100000000000000.01 may be approximately equal to 100000000000000.011,
even though 100.01 is not approximately equal to 100.011.
If the language has such a feature in its standard library, this may be used instead of a custom function.
Show the function results with comparisons on the following pairs of values:
100000000000000.01, 100000000000000.011 (note: should return true)
100.01, 100.011 (note: should return false)
10000000000000.001 / 10000.0, 1000000000.0000001000
0.001, 0.0010000001
0.000000000000000000000101, 0.0
sqrt(2) * sqrt(2), 2.0
-sqrt(2) * sqrt(2), -2.0
3.14159265358979323846, 3.14159265358979324
Answers should be true for the first example and false in the second, so that just rounding the numbers to a fixed number of decimals should not be enough. Otherwise answers may vary and still be correct. See the Python code for one type of solution.
| #zkl | zkl | testValues:=T(
T(100000000000000.01,100000000000000.011),
T(100.01, 100.011),
T(10000000000000.001 / 10000.0, 1000000000.0000001),
T(0.001, 0.0010000001),
T(0.00000000000000000101, 0.0),
T( (2.0).sqrt()*(2.0).sqrt(), 2.0),
T( -(2.0).sqrt()*(2.0).sqrt(), -2.0),
T(100000000000000003.0, 100000000000000004.0),
T(3.14159265358979323846, 3.14159265358979324)
);
tolerance:=-1e-9; // <0 for relative comparison
foreach x,y in (testValues){
maybeNot:=( if(x.closeTo(y,tolerance)) " \u2248" else "!\u2248" );
println("% 25.19g %s %- 25.19g %g".fmt(x,maybeNot,y, (x-y).abs()));
} |
http://rosettacode.org/wiki/Balanced_brackets | Balanced brackets | Task:
Generate a string with N opening brackets [ and with N closing brackets ], in some arbitrary order.
Determine whether the generated string is balanced; that is, whether it consists entirely of pairs of opening/closing brackets (in that order), none of which mis-nest.
Examples
(empty) OK
[] OK
[][] OK
[[][]] OK
][ NOT OK
][][ NOT OK
[]][[] NOT OK
| #Excel | Excel | bracketReport
=LAMBDA(bracketPair,
LAMBDA(s,
LET(
depths, SCANLCOLS(
LAMBDA(depth, charDelta,
depth + charDelta
)
)(0)(
codesFromBrackets(bracketPair)(
MID(s,
SEQUENCE(1, LEN(s), 1, 1),
1
)
)
),
overClosePosn, IFNA(MATCH(-1, depths, 0), 0),
IF(0 <> overClosePosn,
"Overclosed at char " & (overClosePosn - 1),
IF(0 <> LASTCOL(depths),
"Underclosed by end",
"OK"
)
)
)
)
)
testRandomBrackets
=LAMBDA(bracketPair,
LAMBDA(n,
LET(
sample, CONCAT(
bracketFromCode(bracketPair)(
RANDARRAY(1, n, -1, 1, TRUE)
)
),
APPENDCOLS(sample)(
bracketReport(bracketPair)(sample)
)
)
)
)
bracketFromCode
=LAMBDA(bracketPairString,
LAMBDA(c,
IF(0 <> c,
IF(0 < c,
MID(bracketPairString, 1, 1),
MID(bracketPairString, 2, 1)
),
"x"
)
)
)
codesFromBrackets
=LAMBDA(bracketPair,
LAMBDA(s,
LAMBDA(c,
LET(
posn, IFERROR(
FIND(c, bracketPair),
0
),
rem, "0 for non-bracket chars, (1, -1) for (open, close)",
IF(0 <> posn,
IF(1 < posn, -1, 1),
0
)
)
)(
MID(s,
SEQUENCE(1, LEN(s), 1, 1),
1
)
)
)
) |
http://rosettacode.org/wiki/Append_a_record_to_the_end_of_a_text_file | Append a record to the end of a text file | Many systems offer the ability to open a file for writing, such that any data written will be appended to the end of the file. Further, the file operations will always adjust the position pointer to guarantee the end of the file, even in a multitasking environment.
This feature is most useful in the case of log files, where many jobs may be appending to the log file at the same time, or where care must be taken to avoid concurrently overwriting the same record from another job.
Task
Given a two record sample for a mythical "passwd" file:
Write these records out in the typical system format.
Ideally these records will have named fields of various types.
Close the file, then reopen the file for append.
Append a new record to the file and close the file again.
Take appropriate care to avoid concurrently overwrites from another job.
Open the file and demonstrate the new record has indeed written to the end.
Source record field types and contents.
account
password
UID
GID
fullname,office,extension,homephone,email
directory
shell
string
string
int
int
struct(string,string,string,string,string)
string
string
jsmith
x
1001
1000
Joe Smith,Room 1007,(234)555-8917,(234)555-0077,[email protected]
/home/jsmith
/bin/bash
jdoe
x
1002
1000
Jane Doe,Room 1004,(234)555-8914,(234)555-0044,[email protected]
/home/jdoe
/bin/bash
Record to be appended.
account
password
UID
GID
fullname,office,extension,homephone,email
directory
shell
string
string
int
int
struct(string,string,string,string,string)
string
string
xyz
x
1003
1000
X Yz,Room 1003,(234)555-8913,(234)555-0033,[email protected]
/home/xyz
/bin/bash
Resulting file format: should mimic Linux's /etc/passwd file format with particular attention to the "," separator used in the GECOS field. But if the specific language has a particular or unique format of storing records in text file, then this format should be named and demonstrated with an additional example.
Expected output:
Appended record: xyz:x:1003:1000:X Yz,Room 1003,(234)555-8913,(234)555-0033,[email protected]:/home/xyz:/bin/bash
Finally: Provide a summary of the language's "append record" capabilities in a table. eg.
Append Capabilities.
Data Representation
IO
Library
Append
Possible
Automatic
Append
Multi-tasking
Safe
In core
On disk
C struct
CSV text file
glibc/stdio
☑
☑
☑ (Not all, eg NFS)
Alternatively: If the language's appends can not guarantee its writes will always append, then note this restriction in the table. If possible, provide an actual code example (possibly using file/record locking) to guarantee correct concurrent appends.
| #Java | Java | import static java.util.Objects.requireNonNull;
import java.io.IOException;
import java.nio.file.Files;
import java.nio.file.Path;
import java.nio.file.Paths;
import java.nio.file.StandardOpenOption;
import java.util.Arrays;
import java.util.Collections;
import java.util.List;
import java.util.stream.Collectors;
import java.util.stream.Stream;
public class RecordAppender {
static class Record {
private final String account;
private final String password;
private final int uid;
private final int gid;
private final List<String> gecos;
private final String directory;
private final String shell;
public Record(String account, String password, int uid, int gid, List<String> gecos, String directory, String shell) {
this.account = requireNonNull(account);
this.password = requireNonNull(password);
this.uid = uid;
this.gid = gid;
this.gecos = requireNonNull(gecos);
this.directory = requireNonNull(directory);
this.shell = requireNonNull(shell);
}
@Override
public String toString() {
return account + ':' + password + ':' + uid + ':' + gid + ':' + String.join(",", gecos) + ':' + directory + ':' + shell;
}
public static Record parse(String text) {
String[] tokens = text.split(":");
return new Record(
tokens[0],
tokens[1],
Integer.parseInt(tokens[2]),
Integer.parseInt(tokens[3]),
Arrays.asList(tokens[4].split(",")),
tokens[5],
tokens[6]);
}
}
public static void main(String[] args) throws IOException {
List<String> rawData = Arrays.asList(
"jsmith:x:1001:1000:Joe Smith,Room 1007,(234)555-8917,(234)555-0077,[email protected]:/home/jsmith:/bin/bash",
"jdoe:x:1002:1000:Jane Doe,Room 1004,(234)555-8914,(234)555-0044,[email protected]:/home/jdoe:/bin/bash",
"xyz:x:1003:1000:X Yz,Room 1003,(234)555-8913,(234)555-0033,[email protected]:/home/xyz:/bin/bash"
);
List<Record> records = rawData.stream().map(Record::parse).collect(Collectors.toList());
Path tmp = Paths.get("_rosetta", ".passwd");
Files.createDirectories(tmp.getParent());
Files.write(tmp, (Iterable<String>) records.stream().limit(2).map(Record::toString)::iterator);
Files.write(tmp, Collections.singletonList(records.get(2).toString()), StandardOpenOption.APPEND);
try (Stream<String> lines = Files.lines(tmp)) {
lines.map(Record::parse).forEach(System.out::println);
}
}
} |
http://rosettacode.org/wiki/Associative_array/Creation | Associative array/Creation | Task
The goal is to create an associative array (also known as a dictionary, map, or hash).
Related tasks:
Associative arrays/Iteration
Hash from two arrays
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #BASIC256 | BASIC256 | global values$, keys$
dim values$[1]
dim keys$[1]
call updateKey("a","apple")
call updateKey("b","banana")
call updateKey("c","cucumber")
gosub show
print "I like to eat a " + getValue$("c") + " on my salad."
call deleteKey("b")
call updateKey("c","carrot")
call updateKey("e","endive")
gosub show
end
show:
for t = 0 to countKeys()-1
print getKeyByIndex$(t) + " " + getValueByIndex$(t)
next t
print
return
subroutine updateKey(key$, value$)
# update or add an item
i=findKey(key$)
if i=-1 then
i = freeKey()
keys$[i] = key$
end if
values$[i] = value$
end subroutine
subroutine deleteKey(key$)
# delete by clearing the key
i=findKey(key$)
if i<>-1 then
keys$[i] = ""
end if
end subroutine
function freeKey()
# find index of a free element in the array
for n = 0 to keys$[?]-1
if keys$[n]="" then return n
next n
redim keys$[n+1]
redim values$[n+1]
return n
end function
function findKey(key$)
# return index or -1 if not found
for n = 0 to keys$[?]-1
if key$=keys$[n] then return n
next n
return -1
end function
function getValue$(key$)
# return a value by the key or "" if not existing
i=findKey(key$)
if i=-1 then
return ""
end if
return values$[i]
end function
function countKeys()
# return number of items
# remember to skip the empty keys (deleted ones)
k = 0
for n = 0 to keys$[?] -1
if keys$[n]<>"" then k++
next n
return k
end function
function getValueByIndex$(i)
# get a value by the index
# remember to skip the empty keys (deleted ones)
k = 0
for n = 0 to keys$[?] -1
if keys$[n]<>"" then
if k=i then return values$[k]
k++
endif
next n
return ""
end function
function getKeyByIndex$(i)
# get a key by the index
# remember to skip the empty keys (deleted ones)
k = 0
for n = 0 to keys$[?] -1
if keys$[n]<>"" then
if k=i then return keys$[k]
k++
endif
next n
return ""
end function |
http://rosettacode.org/wiki/Anti-primes | Anti-primes | The anti-primes
(or highly composite numbers, sequence A002182 in the OEIS)
are the natural numbers with more factors than any smaller than itself.
Task
Generate and show here, the first twenty anti-primes.
Related tasks
Factors of an integer
Sieve of Eratosthenes
| #Factor | Factor | USING: assocs formatting kernel locals make math
math.primes.factors sequences.extras ;
IN: rosetta-code.anti-primes
<PRIVATE
: count-divisors ( n -- m )
dup 1 = [ group-factors values [ 1 + ] map-product ] unless ;
: (n-anti-primes) ( md n count -- ?md' n' ?count' )
dup 0 >
[| max-div! n count! |
n count-divisors :> d
d max-div > [ d max-div! n , count 1 - count! ] when
max-div n dup 60 >= 20 1 ? + count (n-anti-primes)
] when ;
PRIVATE>
: n-anti-primes ( n -- seq )
[ 0 1 ] dip [ (n-anti-primes) 3drop ] { } make ;
: anti-primes-demo ( -- )
20 n-anti-primes "First 20 anti-primes:\n%[%d, %]\n" printf ;
MAIN: anti-primes-demo |
http://rosettacode.org/wiki/Anti-primes | Anti-primes | The anti-primes
(or highly composite numbers, sequence A002182 in the OEIS)
are the natural numbers with more factors than any smaller than itself.
Task
Generate and show here, the first twenty anti-primes.
Related tasks
Factors of an integer
Sieve of Eratosthenes
| #Forth | Forth |
include ./factors.fs
: max-count ( n1 n2 -- n f )
\ n is max(n1, factor-count n2); if n is new maximum then f = true.
\
count-factors 2dup <
if nip true
else drop false
then ;
: .anti-primes ( n -- )
0 1 rot \ stack: max, candidate, count
begin
>r dup >r max-count
if r> dup . r> 1-
else r> r>
then swap 1+ swap
dup 0= until drop 2drop ;
." The first 20 anti-primes are: " 20 .anti-primes cr
bye
|
http://rosettacode.org/wiki/Atomic_updates | Atomic updates |
Task
Define a data type consisting of a fixed number of 'buckets', each containing a nonnegative integer value, which supports operations to:
get the current value of any bucket
remove a specified amount from one specified bucket and add it to another, preserving the total of all bucket values, and clamping the transferred amount to ensure the values remain non-negative
In order to exercise this data type, create one set of buckets, and start three concurrent tasks:
As often as possible, pick two buckets and make their values closer to equal.
As often as possible, pick two buckets and arbitrarily redistribute their values.
At whatever rate is convenient, display (by any means) the total value and, optionally, the individual values of each bucket.
The display task need not be explicit; use of e.g. a debugger or trace tool is acceptable provided it is simple to set up to provide the display.
This task is intended as an exercise in atomic operations. The sum of the bucket values must be preserved even if the two tasks attempt to perform transfers simultaneously, and a straightforward solution is to ensure that at any time, only one transfer is actually occurring — that the transfer operation is atomic.
| #Python | Python | from __future__ import with_statement # required for Python 2.5
import threading
import random
import time
terminate = threading.Event()
class Buckets:
def __init__(self, nbuckets):
self.nbuckets = nbuckets
self.values = [random.randrange(10) for i in range(nbuckets)]
self.lock = threading.Lock()
def __getitem__(self, i):
return self.values[i]
def transfer(self, src, dst, amount):
with self.lock:
amount = min(amount, self.values[src])
self.values[src] -= amount
self.values[dst] += amount
def snapshot(self):
# copy of the current state (synchronized)
with self.lock:
return self.values[:]
def randomize(buckets):
nbuckets = buckets.nbuckets
while not terminate.isSet():
src = random.randrange(nbuckets)
dst = random.randrange(nbuckets)
if dst!=src:
amount = random.randrange(20)
buckets.transfer(src, dst, amount)
def equalize(buckets):
nbuckets = buckets.nbuckets
while not terminate.isSet():
src = random.randrange(nbuckets)
dst = random.randrange(nbuckets)
if dst!=src:
amount = (buckets[src] - buckets[dst]) // 2
if amount>=0: buckets.transfer(src, dst, amount)
else: buckets.transfer(dst, src, -amount)
def print_state(buckets):
snapshot = buckets.snapshot()
for value in snapshot:
print '%2d' % value,
print '=', sum(snapshot)
# create 15 buckets
buckets = Buckets(15)
# the randomize thread
t1 = threading.Thread(target=randomize, args=[buckets])
t1.start()
# the equalize thread
t2 = threading.Thread(target=equalize, args=[buckets])
t2.start()
# main thread, display
try:
while True:
print_state(buckets)
time.sleep(1)
except KeyboardInterrupt: # ^C to finish
terminate.set()
# wait until all worker threads finish
t1.join()
t2.join() |
http://rosettacode.org/wiki/Assertions | Assertions | Assertions are a way of breaking out of code when there is an error or an unexpected input.
Some languages throw exceptions and some treat it as a break point.
Task
Show an assertion in your language by asserting that an integer variable is equal to 42.
| #Objective-C | Objective-C | NSAssert(a == 42, @"Error message"); |
http://rosettacode.org/wiki/Assertions | Assertions | Assertions are a way of breaking out of code when there is an error or an unexpected input.
Some languages throw exceptions and some treat it as a break point.
Task
Show an assertion in your language by asserting that an integer variable is equal to 42.
| #OCaml | OCaml | let a = get_some_value () in
assert (a = 42); (* throws Assert_failure when a is not 42 *)
(* evaluate stuff to return here when a is 42 *) |
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