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http://rosettacode.org/wiki/Classes | Classes | In object-oriented programming class is a set (a transitive closure) of types bound by the relation of inheritance. It is said that all types derived from some base type T and the type T itself form a class T.
The first type T from the class T sometimes is called the root type of the class.
A class of types itself, as a type, has the values and operations of its own.
The operations of are usually called methods of the root type.
Both operations and values are called polymorphic.
A polymorphic operation (method) selects an implementation depending on the actual specific type of the polymorphic argument.
The action of choice the type-specific implementation of a polymorphic operation is called dispatch. Correspondingly, polymorphic operations are often called dispatching or virtual.
Operations with multiple arguments and/or the results of the class are called multi-methods.
A further generalization of is the operation with arguments and/or results from different classes.
single-dispatch languages are those that allow only one argument or result to control the dispatch. Usually it is the first parameter, often hidden, so that a prefix notation x.f() is used instead of mathematical f(x).
multiple-dispatch languages allow many arguments and/or results to control the dispatch.
A polymorphic value has a type tag indicating its specific type from the class and the corresponding specific value of that type.
This type is sometimes called the most specific type of a [polymorphic] value.
The type tag of the value is used in order to resolve the dispatch.
The set of polymorphic values of a class is a transitive closure of the sets of values of all types from that class.
In many OO languages
the type of the class of T and T itself are considered equivalent.
In some languages they are distinct (like in Ada).
When class T and T are equivalent, there is no way to distinguish
polymorphic and specific values.
Task
Create a basic class with a method, a constructor, an instance variable and how to instantiate it.
| #E | E | def makeColor(name :String) {
def color {
to colorize(thing :String) {
return `$name $thing`
}
}
return color
} |
http://rosettacode.org/wiki/Classes | Classes | In object-oriented programming class is a set (a transitive closure) of types bound by the relation of inheritance. It is said that all types derived from some base type T and the type T itself form a class T.
The first type T from the class T sometimes is called the root type of the class.
A class of types itself, as a type, has the values and operations of its own.
The operations of are usually called methods of the root type.
Both operations and values are called polymorphic.
A polymorphic operation (method) selects an implementation depending on the actual specific type of the polymorphic argument.
The action of choice the type-specific implementation of a polymorphic operation is called dispatch. Correspondingly, polymorphic operations are often called dispatching or virtual.
Operations with multiple arguments and/or the results of the class are called multi-methods.
A further generalization of is the operation with arguments and/or results from different classes.
single-dispatch languages are those that allow only one argument or result to control the dispatch. Usually it is the first parameter, often hidden, so that a prefix notation x.f() is used instead of mathematical f(x).
multiple-dispatch languages allow many arguments and/or results to control the dispatch.
A polymorphic value has a type tag indicating its specific type from the class and the corresponding specific value of that type.
This type is sometimes called the most specific type of a [polymorphic] value.
The type tag of the value is used in order to resolve the dispatch.
The set of polymorphic values of a class is a transitive closure of the sets of values of all types from that class.
In many OO languages
the type of the class of T and T itself are considered equivalent.
In some languages they are distinct (like in Ada).
When class T and T are equivalent, there is no way to distinguish
polymorphic and specific values.
Task
Create a basic class with a method, a constructor, an instance variable and how to instantiate it.
| #EchoLisp | EchoLisp |
(lib 'gloops) ; load oo library
(define-class Person null (name (age :initform 66)))
(define-method tostring (Person) (lambda (p) ( format "🚶 %a " p.name)))
(define-method mailto (Person Person) (lambda( p o) (printf "From %a to️ %a : ..." p o)))
;; define a sub-class of Person with same methods
(define-class Writer (Person) (books))
(define-method tostring (Writer) (lambda (w)( format "🎩 %a" w.name)))
(define-method mailto (Person Writer)
(lambda (p w) (printf " From %a (age %d). Dear writer of %a ..." p p.age w.books )))
|
http://rosettacode.org/wiki/Cistercian_numerals | Cistercian numerals | Cistercian numerals were used across Europe by Cistercian monks during the Late Medieval Period as an alternative to Roman numerals. They were used to represent base 10 integers from 0 to 9999.
How they work
All Cistercian numerals begin with a vertical line segment, which by itself represents the number 0. Then, glyphs representing the digits 1 through 9 are optionally added to the four quadrants surrounding the vertical line segment. These glyphs are drawn with vertical and horizontal symmetry about the initial line segment. Each quadrant corresponds to a digit place in the number:
The upper-right quadrant represents the ones place.
The upper-left quadrant represents the tens place.
The lower-right quadrant represents the hundreds place.
The lower-left quadrant represents the thousands place.
Please consult the following image for examples of Cistercian numerals showing each glyph: [1]
Task
Write a function/procedure/routine to display any given Cistercian numeral. This could be done by drawing to the display, creating an image, or even as text (as long as it is a reasonable facsimile).
Use the routine to show the following Cistercian numerals:
0
1
20
300
4000
5555
6789
And a number of your choice!
Notes
Due to the inability to upload images to Rosetta Code as of this task's creation, showing output here on this page is not required. However, it is welcomed — especially for text output.
See also
Numberphile - The Forgotten Number System
dcode.fr - Online Cistercian numeral converter
| #Ruby | Ruby | def initN
n = Array.new(15){Array.new(11, ' ')}
for i in 1..15
n[i - 1][5] = 'x'
end
return n
end
def horiz(n, c1, c2, r)
for c in c1..c2
n[r][c] = 'x'
end
end
def verti(n, r1, r2, c)
for r in r1..r2
n[r][c] = 'x'
end
end
def diagd(n, c1, c2, r)
for c in c1..c2
n[r+c-c1][c] = 'x'
end
end
def diagu(n, c1, c2, r)
for c in c1..c2
n[r-c+c1][c] = 'x'
end
end
def initDraw
draw = []
draw[1] = lambda do |n| horiz(n, 6, 10, 0) end
draw[2] = lambda do |n| horiz(n, 6, 10, 4) end
draw[3] = lambda do |n| diagd(n, 6, 10, 0) end
draw[4] = lambda do |n| diagu(n, 6, 10, 4) end
draw[5] = lambda do |n|
draw[1].call(n)
draw[4].call(n)
end
draw[6] = lambda do |n| verti(n, 0, 4, 10) end
draw[7] = lambda do |n|
draw[1].call(n)
draw[6].call(n)
end
draw[8] = lambda do |n|
draw[2].call(n)
draw[6].call(n)
end
draw[9] = lambda do |n|
draw[1].call(n)
draw[8].call(n)
end
draw[10] = lambda do |n| horiz(n, 0, 4, 0) end
draw[20] = lambda do |n| horiz(n, 0, 4, 4) end
draw[30] = lambda do |n| diagu(n, 0, 4, 4) end
draw[40] = lambda do |n| diagd(n, 0, 4, 0) end
draw[50] = lambda do |n|
draw[10].call(n)
draw[40].call(n)
end
draw[60] = lambda do |n| verti(n, 0, 4, 0) end
draw[70] = lambda do |n|
draw[10].call(n)
draw[60].call(n)
end
draw[80] = lambda do |n|
draw[20].call(n)
draw[60].call(n)
end
draw[90] = lambda do |n|
draw[10].call(n)
draw[80].call(n)
end
draw[100] = lambda do |n| horiz(n, 6, 10, 14) end
draw[200] = lambda do |n| horiz(n, 6, 10, 10) end
draw[300] = lambda do |n| diagu(n, 6, 10, 14) end
draw[400] = lambda do |n| diagd(n, 6, 10, 10) end
draw[500] = lambda do |n|
draw[100].call(n)
draw[400].call(n)
end
draw[600] = lambda do |n| verti(n, 10, 14, 10) end
draw[700] = lambda do |n|
draw[100].call(n)
draw[600].call(n)
end
draw[800] = lambda do |n|
draw[200].call(n)
draw[600].call(n)
end
draw[900] = lambda do |n|
draw[100].call(n)
draw[800].call(n)
end
draw[1000] = lambda do |n| horiz(n, 0, 4, 14) end
draw[2000] = lambda do |n| horiz(n, 0, 4, 10) end
draw[3000] = lambda do |n| diagd(n, 0, 4, 10) end
draw[4000] = lambda do |n| diagu(n, 0, 4, 14) end
draw[5000] = lambda do |n|
draw[1000].call(n)
draw[4000].call(n)
end
draw[6000] = lambda do |n| verti(n, 10, 14, 0) end
draw[7000] = lambda do |n|
draw[1000].call(n)
draw[6000].call(n)
end
draw[8000] = lambda do |n|
draw[2000].call(n)
draw[6000].call(n)
end
draw[9000] = lambda do |n|
draw[1000].call(n)
draw[8000].call(n)
end
return draw
end
def printNumeral(n)
for a in n
for b in a
print b
end
print "\n"
end
print "\n"
end
draw = initDraw()
for number in [0, 1, 20, 300, 4000, 5555, 6789, 9999]
n = initN()
print number, ":\n"
thousands = (number / 1000).floor
number = number % 1000
hundreds = (number / 100).floor
number = number % 100
tens = (number / 10).floor
ones = number % 10
if thousands > 0 then
draw[thousands * 1000].call(n)
end
if hundreds > 0 then
draw[hundreds * 100].call(n)
end
if tens > 0 then
draw[tens * 10].call(n)
end
if ones > 0 then
draw[ones].call(n)
end
printNumeral(n)
end |
http://rosettacode.org/wiki/Closest-pair_problem | Closest-pair problem |
This page uses content from Wikipedia. The original article was at Closest pair of points problem. 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)
Task
Provide a function to find the closest two points among a set of given points in two dimensions, i.e. to solve the Closest pair of points problem in the planar case.
The straightforward solution is a O(n2) algorithm (which we can call brute-force algorithm); the pseudo-code (using indexes) could be simply:
bruteForceClosestPair of P(1), P(2), ... P(N)
if N < 2 then
return ∞
else
minDistance ← |P(1) - P(2)|
minPoints ← { P(1), P(2) }
foreach i ∈ [1, N-1]
foreach j ∈ [i+1, N]
if |P(i) - P(j)| < minDistance then
minDistance ← |P(i) - P(j)|
minPoints ← { P(i), P(j) }
endif
endfor
endfor
return minDistance, minPoints
endif
A better algorithm is based on the recursive divide&conquer approach, as explained also at Wikipedia's Closest pair of points problem, which is O(n log n); a pseudo-code could be:
closestPair of (xP, yP)
where xP is P(1) .. P(N) sorted by x coordinate, and
yP is P(1) .. P(N) sorted by y coordinate (ascending order)
if N ≤ 3 then
return closest points of xP using brute-force algorithm
else
xL ← points of xP from 1 to ⌈N/2⌉
xR ← points of xP from ⌈N/2⌉+1 to N
xm ← xP(⌈N/2⌉)x
yL ← { p ∈ yP : px ≤ xm }
yR ← { p ∈ yP : px > xm }
(dL, pairL) ← closestPair of (xL, yL)
(dR, pairR) ← closestPair of (xR, yR)
(dmin, pairMin) ← (dR, pairR)
if dL < dR then
(dmin, pairMin) ← (dL, pairL)
endif
yS ← { p ∈ yP : |xm - px| < dmin }
nS ← number of points in yS
(closest, closestPair) ← (dmin, pairMin)
for i from 1 to nS - 1
k ← i + 1
while k ≤ nS and yS(k)y - yS(i)y < dmin
if |yS(k) - yS(i)| < closest then
(closest, closestPair) ← (|yS(k) - yS(i)|, {yS(k), yS(i)})
endif
k ← k + 1
endwhile
endfor
return closest, closestPair
endif
References and further readings
Closest pair of points problem
Closest Pair (McGill)
Closest Pair (UCSB)
Closest pair (WUStL)
Closest pair (IUPUI)
| #Haskell | Haskell | import Data.List (minimumBy, tails, unfoldr, foldl1') --'
import System.Random (newStdGen, randomRs)
import Control.Arrow ((&&&))
import Data.Ord (comparing)
vecLeng [[a, b], [p, q]] = sqrt $ (a - p) ^ 2 + (b - q) ^ 2
findClosestPair =
foldl1'' ((minimumBy (comparing vecLeng) .) . (. return) . (:)) .
concatMap (\(x:xs) -> map ((x :) . return) xs) . init . tails
testCP = do
g <- newStdGen
let pts :: [[Double]]
pts = take 1000 . unfoldr (Just . splitAt 2) $ randomRs (-1, 1) g
print . (id &&& vecLeng) . findClosestPair $ pts
main = testCP
foldl1'' = foldl1'
|
http://rosettacode.org/wiki/Closures/Value_capture | Closures/Value capture | Task
Create a list of ten functions, in the simplest manner possible (anonymous functions are encouraged), such that the function at index i (you may choose to start i from either 0 or 1), when run, should return the square of the index, that is, i 2.
Display the result of running any but the last function, to demonstrate that the function indeed remembers its value.
Goal
Demonstrate how to create a series of independent closures based on the same template but maintain separate copies of the variable closed over.
In imperative languages, one would generally use a loop with a mutable counter variable.
For each function to maintain the correct number, it has to capture the value of the variable at the time it was created, rather than just a reference to the variable, which would have a different value by the time the function was run.
See also: Multiple distinct objects
| #Objective-C | Objective-C | NSMutableArray *funcs = [[NSMutableArray alloc] init];
for (int i = 0; i < 10; i++) {
[funcs addObject:[^ { return i * i; } copy]];
}
int (^foo)(void) = funcs[3];
NSLog(@"%d", foo()); // logs "9"
|
http://rosettacode.org/wiki/Closures/Value_capture | Closures/Value capture | Task
Create a list of ten functions, in the simplest manner possible (anonymous functions are encouraged), such that the function at index i (you may choose to start i from either 0 or 1), when run, should return the square of the index, that is, i 2.
Display the result of running any but the last function, to demonstrate that the function indeed remembers its value.
Goal
Demonstrate how to create a series of independent closures based on the same template but maintain separate copies of the variable closed over.
In imperative languages, one would generally use a loop with a mutable counter variable.
For each function to maintain the correct number, it has to capture the value of the variable at the time it was created, rather than just a reference to the variable, which would have a different value by the time the function was run.
See also: Multiple distinct objects
| #OCaml | OCaml | let () =
let cls = Array.init 10 (fun i -> (function () -> i * i)) in
Random.self_init ();
for i = 1 to 6 do
let x = Random.int 9 in
Printf.printf " fun.(%d) = %d\n" x (cls.(x) ());
done |
http://rosettacode.org/wiki/Circular_primes | Circular primes | Definitions
A circular prime is a prime number with the property that the number generated at each intermediate step when cyclically permuting its (base 10) digits will also be prime.
For example:
1193 is a circular prime, since 1931, 9311 and 3119 are all also prime.
Note that a number which is a cyclic permutation of a smaller circular prime is not considered to be itself a circular prime. So 13 is a circular prime, but 31 is not.
A repunit (denoted by R) is a number whose base 10 representation contains only the digit 1.
For example:
R(2) = 11 and R(5) = 11111 are repunits.
Task
Find the first 19 circular primes.
If your language has access to arbitrary precision integer arithmetic, given that they are all repunits, find the next 4 circular primes.
(Stretch) Determine which of the following repunits are probably circular primes: R(5003), R(9887), R(15073), R(25031), R(35317) and R(49081). The larger ones may take a long time to process so just do as many as you reasonably can.
See also
Wikipedia article - Circular primes.
Wikipedia article - Repunit.
OEIS sequence A016114 - Circular primes.
| #Sidef | Sidef | func is_circular_prime(n) {
n.is_prime || return false
var circular = n.digits
circular.min < circular.tail && return false
for k in (1 ..^ circular.len) {
with (circular.rotate(k).digits2num) {|p|
(p.is_prime && (p >= n)) || return false
}
}
return true
}
say "The first 19 circular primes are:"
say 19.by(is_circular_prime)
say "\nThe next 4 circular primes, in repunit format, are:"
{|n| (10**n - 1)/9 -> is_prob_prime }.first(4, 4..Inf).each {|n|
say "R(#{n})"
}
say "\nRepunit testing:"
[5003, 9887, 15073, 25031, 35317, 49081].each {|n|
var now = Time.micro
say ("R(#{n}) -> ", is_prob_prime((10**n - 1)/9) ? 'probably prime' : 'composite',
" (took: #{'%.3f' % Time.micro-now} sec)")
} |
http://rosettacode.org/wiki/Circles_of_given_radius_through_two_points | Circles of given radius through two points |
Given two points on a plane and a radius, usually two circles of given radius can be drawn through the points.
Exceptions
r==0.0 should be treated as never describing circles (except in the case where the points are coincident).
If the points are coincident then an infinite number of circles with the point on their circumference can be drawn, unless r==0.0 as well which then collapses the circles to a point.
If the points form a diameter then return two identical circles or return a single circle, according to which is the most natural mechanism for the implementation language.
If the points are too far apart then no circles can be drawn.
Task detail
Write a function/subroutine/method/... that takes two points and a radius and returns the two circles through those points, or some indication of special cases where two, possibly equal, circles cannot be returned.
Show here the output for the following inputs:
p1 p2 r
0.1234, 0.9876 0.8765, 0.2345 2.0
0.0000, 2.0000 0.0000, 0.0000 1.0
0.1234, 0.9876 0.1234, 0.9876 2.0
0.1234, 0.9876 0.8765, 0.2345 0.5
0.1234, 0.9876 0.1234, 0.9876 0.0
Related task
Total circles area.
See also
Finding the Center of a Circle from 2 Points and Radius from Math forum @ Drexel
| #C.23 | C# | using System;
public class CirclesOfGivenRadiusThroughTwoPoints
{
public static void Main()
{
double[][] values = new double[][] {
new [] { 0.1234, 0.9876, 0.8765, 0.2345, 2 },
new [] { 0.0, 2.0, 0.0, 0.0, 1 },
new [] { 0.1234, 0.9876, 0.1234, 0.9876, 2 },
new [] { 0.1234, 0.9876, 0.8765, 0.2345, 0.5 },
new [] { 0.1234, 0.9876, 0.1234, 0.9876, 0 }
};
foreach (var a in values) {
var p = new Point(a[0], a[1]);
var q = new Point(a[2], a[3]);
Console.WriteLine($"Points {p} and {q} with radius {a[4]}:");
try {
var centers = FindCircles(p, q, a[4]);
Console.WriteLine("\t" + string.Join(" and ", centers));
} catch (Exception ex) {
Console.WriteLine("\t" + ex.Message);
}
}
}
static Point[] FindCircles(Point p, Point q, double radius) {
if(radius < 0) throw new ArgumentException("Negative radius.");
if(radius == 0) {
if(p == q) return new [] { p };
else throw new InvalidOperationException("No circles.");
}
if (p == q) throw new InvalidOperationException("Infinite number of circles.");
double sqDistance = Point.SquaredDistance(p, q);
double sqDiameter = 4 * radius * radius;
if (sqDistance > sqDiameter) throw new InvalidOperationException("Points are too far apart.");
Point midPoint = new Point((p.X + q.X) / 2, (p.Y + q.Y) / 2);
if (sqDistance == sqDiameter) return new [] { midPoint };
double d = Math.Sqrt(radius * radius - sqDistance / 4);
double distance = Math.Sqrt(sqDistance);
double ox = d * (q.X - p.X) / distance, oy = d * (q.Y - p.Y) / distance;
return new [] {
new Point(midPoint.X - oy, midPoint.Y + ox),
new Point(midPoint.X + oy, midPoint.Y - ox)
};
}
public struct Point
{
public Point(double x, double y) : this() {
X = x;
Y = y;
}
public double X { get; }
public double Y { get; }
public static bool operator ==(Point p, Point q) => p.X == q.X && p.Y == q.Y;
public static bool operator !=(Point p, Point q) => p.X != q.X || p.Y != q.Y;
public static double SquaredDistance(Point p, Point q) {
double dx = q.X - p.X, dy = q.Y - p.Y;
return dx * dx + dy * dy;
}
public override string ToString() => $"({X}, {Y})";
}
} |
http://rosettacode.org/wiki/Chinese_zodiac | Chinese zodiac | Traditionally, the Chinese have counted years using two simultaneous cycles, one of length 10 (the "celestial stems") and one of length 12 (the "terrestrial branches"); the combination results in a repeating 60-year pattern. Mapping the branches to twelve traditional animal deities results in the well-known "Chinese zodiac", assigning each year to a given animal. For example, Tuesday, February 1, 2022 CE (in the common Gregorian calendar) will begin the lunisolar Year of the Tiger.
The celestial stems have no one-to-one mapping like that of the branches to animals; however, the five pairs of consecutive stems each belong to one of the five traditional Chinese elements (Wood, Fire, Earth, Metal, and Water). Further, one of the two years within each element's governance is associated with yin, the other with yang.
Thus, 2022 is also the yang year of Water. Note that since 12 is an even number, the association between animals and yin/yang doesn't change. Consecutive Years of the Rooster will cycle through the five elements, but will always be yin, despite the apparent conceptual mismatch between the specifically-male English animal name and the female aspect denoted by yin.
Task
Create a subroutine or program that will return or output the animal, yin/yang association, and element for the lunisolar year that begins in a given CE year.
You may optionally provide more information in the form of the year's numerical position within the 60-year cycle and/or its actual Chinese stem-branch name (in Han characters or Pinyin transliteration).
Requisite information
The animal cycle runs in this order: Rat, Ox, Tiger, Rabbit, Dragon, Snake, Horse, Goat, Monkey, Rooster, Dog, Pig.
The element cycle runs in this order: Wood, Fire, Earth, Metal, Water.
The yang year precedes the yin year within each element.
The current 60-year cycle began in 1984 CE; the first cycle of the Common Era began in 4 CE.
Thus, 1984 was the year of the Wood Rat (yang), 1985 was the year of the Wood Ox (yin), and 1986 the year of the Fire Tiger (yang); 2022 - which, as already noted, is the year of the Water Tiger (yang) - is the 39th year of the current cycle.
Information for optional task
The ten celestial stems are 甲 jiă, 乙 yĭ, 丙 bĭng, 丁 dīng, 戊 wù, 己 jĭ, 庚 gēng, 辛 xīn, 壬 rén, and 癸 gŭi. With the ASCII version of Pinyin tones, the names are written "jia3", "yi3", "bing3", "ding1", "wu4", "ji3", "geng1", "xin1", "ren2", and "gui3".
The twelve terrestrial branches are 子 zĭ, 丑 chŏu, 寅 yín, 卯 măo, 辰 chén, 巳 sì, 午 wŭ, 未 wèi, 申 shēn, 酉 yŏu, 戌 xū, 亥 hài. In ASCII Pinyin, those are "zi3", "chou3", "yin2", "mao3", "chen2", "si4", "wu3", "wei4", "shen1", "you3", "xu1", and "hai4".
Therefore 1984 was 甲子 (jiă-zĭ, or jia3-zi3). 2022 is 壬寅 (rén-yín or ren2-yin2).
| #BASIC256 | BASIC256 |
# Chinese zodiac
elementos = {"Wood", "Fire", "Earth", "Metal", "Water"}
animales = {"Rat", "Ox", "Tiger", "Rabbit", "Dragon", "Snake", "Horse", "Goat", "Monkey", "Rooster", "Dog", "Pig"}
aspectos = {"Yang","Yin"}
tallo_celestial = {'甲', '乙', '丙', '丁', '戊', '己', '庚', '辛', '壬', '癸'}
rama_terrestre = {'子','丑','寅','卯','辰','巳','午','未','申','酉','戌','亥'}
tallos_pinyin = {"jiă","yĭ","bĭng","dīng","wù","jĭ","gēng","xīn","rén","gŭi"}
ramas_pinyin = {"zĭ","chŏu","yín","măo","chén","sì","wŭ","wèi","shēn","yŏu","xū","hài"}
years = {1935, 1938, 1968, 1972, 1976, 1984, 2017}
For i = 0 To years[?]-1
xYear = years[i]
yElemento = elementos[((xYear - 4) % 10) \ 2]
yAnimal = animales[ (xYear - 4) % 12 ]
yAspectos = aspectos[ xYear % 2 ]
ytallo_celestial = tallo_celestial[((xYear - 4) % 10)]
yrama_terrestre = rama_terrestre[ (xYear - 4) % 12 ]
ytallos_pinyin = tallos_pinyin[ ((xYear - 4) % 10)]
yramas_pinyin = ramas_pinyin[ (xYear - 4) % 12 ]
ciclo = ((xYear - 4) % 60) + 1
Print xYear & ": " & ytallo_celestial & yrama_terrestre & " (" & ytallos_pinyin & "-" & yramas_pinyin & ", " & yElemento & " " & yAnimal & "; " & yAspectos & " - ciclo " &ciclo & "/60)"
Next i
End
|
http://rosettacode.org/wiki/Chinese_zodiac | Chinese zodiac | Traditionally, the Chinese have counted years using two simultaneous cycles, one of length 10 (the "celestial stems") and one of length 12 (the "terrestrial branches"); the combination results in a repeating 60-year pattern. Mapping the branches to twelve traditional animal deities results in the well-known "Chinese zodiac", assigning each year to a given animal. For example, Tuesday, February 1, 2022 CE (in the common Gregorian calendar) will begin the lunisolar Year of the Tiger.
The celestial stems have no one-to-one mapping like that of the branches to animals; however, the five pairs of consecutive stems each belong to one of the five traditional Chinese elements (Wood, Fire, Earth, Metal, and Water). Further, one of the two years within each element's governance is associated with yin, the other with yang.
Thus, 2022 is also the yang year of Water. Note that since 12 is an even number, the association between animals and yin/yang doesn't change. Consecutive Years of the Rooster will cycle through the five elements, but will always be yin, despite the apparent conceptual mismatch between the specifically-male English animal name and the female aspect denoted by yin.
Task
Create a subroutine or program that will return or output the animal, yin/yang association, and element for the lunisolar year that begins in a given CE year.
You may optionally provide more information in the form of the year's numerical position within the 60-year cycle and/or its actual Chinese stem-branch name (in Han characters or Pinyin transliteration).
Requisite information
The animal cycle runs in this order: Rat, Ox, Tiger, Rabbit, Dragon, Snake, Horse, Goat, Monkey, Rooster, Dog, Pig.
The element cycle runs in this order: Wood, Fire, Earth, Metal, Water.
The yang year precedes the yin year within each element.
The current 60-year cycle began in 1984 CE; the first cycle of the Common Era began in 4 CE.
Thus, 1984 was the year of the Wood Rat (yang), 1985 was the year of the Wood Ox (yin), and 1986 the year of the Fire Tiger (yang); 2022 - which, as already noted, is the year of the Water Tiger (yang) - is the 39th year of the current cycle.
Information for optional task
The ten celestial stems are 甲 jiă, 乙 yĭ, 丙 bĭng, 丁 dīng, 戊 wù, 己 jĭ, 庚 gēng, 辛 xīn, 壬 rén, and 癸 gŭi. With the ASCII version of Pinyin tones, the names are written "jia3", "yi3", "bing3", "ding1", "wu4", "ji3", "geng1", "xin1", "ren2", and "gui3".
The twelve terrestrial branches are 子 zĭ, 丑 chŏu, 寅 yín, 卯 măo, 辰 chén, 巳 sì, 午 wŭ, 未 wèi, 申 shēn, 酉 yŏu, 戌 xū, 亥 hài. In ASCII Pinyin, those are "zi3", "chou3", "yin2", "mao3", "chen2", "si4", "wu3", "wei4", "shen1", "you3", "xu1", and "hai4".
Therefore 1984 was 甲子 (jiă-zĭ, or jia3-zi3). 2022 is 壬寅 (rén-yín or ren2-yin2).
| #Befunge | Befunge | 0" :raeY">:#,_55+"< /8"&>+:66+%00p:55+v
v"Aspect: "0++88*5%2\0\+1%"<":p01++66/2%<
>00g5g7-0" :laminA"10g5g"<"+0" :tnemelE"v
v!:,+55$_v#!-*84,:g+5/< >:#,_$.,,.,@ >0#<
_>>:#,_$>>1+::"("%\"("^ ^"Cycle: " <<<<
$'-4;AGLS[_ %*06yang yin Rat Ox Tiger R |
abbit Dragon Snake Horse Goat Monkey Roo |
ster Dog Pig Wood Fire Earth Metal Water | |
http://rosettacode.org/wiki/Check_that_file_exists | Check that file exists | Task
Verify that a file called input.txt and a directory called docs exist.
This should be done twice:
once for the current working directory, and
once for a file and a directory in the filesystem root.
Optional criteria (May 2015): verify it works with:
zero-length files
an unusual filename: `Abdu'l-Bahá.txt
| #AArch64_Assembly | AArch64 Assembly |
/* ARM assembly AARCH64 Raspberry PI 3B */
/* program verifFic64.s */
/*******************************************/
/* Constantes file */
/*******************************************/
/* for this file see task include a file in language AArch64 assembly*/
.include "../includeConstantesARM64.inc"
.equ CHDIR, 49
.equ AT_FDCWD, -100
/*******************************************/
/* Initialized data */
/*******************************************/
.data
szMessFound1: .asciz "File 1 found.\n"
szMessFound2: .asciz "File 2 found.\n"
szMessNotFound1: .asciz "File 1 not found.\n"
szMessNotFound2: .asciz "File 2 not found.\n"
szMessDir2: .asciz "File 2 is a directory.\n"
szMessNotAuth2: .asciz "File 2 permission denied.\n"
szCarriageReturn: .asciz "\n"
/* areas strings */
szPath2: .asciz "/"
szFicName1: .asciz "test1.txt"
szFicName2: .asciz "root"
/*******************************************/
/* UnInitialized data */
/*******************************************/
.bss
/*******************************************/
/* code section */
/*******************************************/
.text
.global main
main: // entry of program
/*************************************
open file 1
************************************/
mov x0,AT_FDCWD // current directory
ldr x1,qAdrszFicName1 // file name
mov x2,#O_RDWR // flags
mov x3,#0 // mode
mov x8, #OPEN // call system OPEN
svc 0
cmp x0,#0 // error ?
ble 1f
mov x1,x0 // FD
ldr x0,qAdrszMessFound1
bl affichageMess
// close file
mov x0,x1 // Fd
mov x8, #CLOSE
svc 0
b 2f
1:
ldr x0,qAdrszMessNotFound1
bl affichageMess
2:
/*************************************
open file 2
************************************/
ldr x0,qAdrszPath2
mov x8,CHDIR // call system change directory
svc 0
mov x0,AT_FDCWD // current directory
ldr x1,qAdrszFicName2 // file name
mov x2,O_RDWR // flags
mov x3,0 // mode
mov x8,OPEN // call system OPEN
svc 0
cmp x0,-21 // is a directory ?
beq 4f
cmp x0,0 // error ?
ble 3f
mov x1,x0 // FD
ldr x0,qAdrszMessFound2
bl affichageMess
// close file
mov x0,x1 // Fd
mov x8, #CLOSE
svc 0
b 100f
3:
ldr x0,qAdrszMessNotFound2
bl affichageMess
b 100f
4:
ldr x0,qAdrszMessDir2
bl affichageMess
100: // standard end of the program
mov x0,0 // return code
mov x8,EXIT // request to exit program
svc 0 // perform the system call
qAdrszFicName1: .quad szFicName1
qAdrszFicName2: .quad szFicName2
qAdrszMessFound1: .quad szMessFound1
qAdrszMessFound2: .quad szMessFound2
qAdrszMessNotFound1: .quad szMessNotFound1
qAdrszMessNotFound2: .quad szMessNotFound2
qAdrszMessNotAuth2: .quad szMessNotAuth2
qAdrszPath2: .quad szPath2
qAdrszMessDir2: .quad szMessDir2
/********************************************************/
/* File Include fonctions */
/********************************************************/
/* for this file see task include a file in language AArch64 assembly */
.include "../includeARM64.inc"
|
http://rosettacode.org/wiki/Check_that_file_exists | Check that file exists | Task
Verify that a file called input.txt and a directory called docs exist.
This should be done twice:
once for the current working directory, and
once for a file and a directory in the filesystem root.
Optional criteria (May 2015): verify it works with:
zero-length files
an unusual filename: `Abdu'l-Bahá.txt
| #Action.21 | Action! | BYTE lastError
PROC MyError(BYTE errCode)
lastError=errCode
RETURN
BYTE FUNC FileExists(CHAR ARRAY fname)
DEFINE PTR="CARD"
PTR old
BYTE dev=[1]
lastError=0
old=Error
Error=MyError ;replace error handling to capture I/O error
Close(dev)
Open(dev,fname,4)
Close(dev)
Error=old ;restore the original error handling
IF lastError=0 THEN
RETURN (1)
FI
RETURN (0)
PROC Test(CHAR ARRAY fname)
BYTE exists
exists=FileExists(fname)
IF exists THEN
PrintF("File ""%S"" exists.%E",fname)
ELSE
PrintF("File ""%S"" does not exist.%E",fname)
FI
RETURN
PROC Main()
Test("D:INPUT.TXT")
Test("D:DOS.SYS")
RETURN |
http://rosettacode.org/wiki/Check_output_device_is_a_terminal | Check output device is a terminal | Task
Demonstrate how to check whether the output device is a terminal or not.
Related task
Check input device is a terminal
| #Factor | Factor |
IN: scratchpad USE: unix.ffi
IN: scratchpad 1 isatty
--- Data stack:
1
|
http://rosettacode.org/wiki/Check_output_device_is_a_terminal | Check output device is a terminal | Task
Demonstrate how to check whether the output device is a terminal or not.
Related task
Check input device is a terminal
| #FreeBASIC | FreeBASIC |
Open Cons For Output As #1
' Open Cons abre los flujos de entrada (stdin) o salida (stdout) estándar
' de la consola para leer o escribir.
If Err > 0 Then
Print #1, "stdout is not a tty."
Else
Print #1, "stdout is a tty."
End If
Close #1
Sleep
|
http://rosettacode.org/wiki/Check_output_device_is_a_terminal | Check output device is a terminal | Task
Demonstrate how to check whether the output device is a terminal or not.
Related task
Check input device is a terminal
| #Go | Go | package main
import (
"os"
"fmt"
)
func main() {
if fileInfo, _ := os.Stdout.Stat(); (fileInfo.Mode() & os.ModeCharDevice) != 0 {
fmt.Println("Hello terminal")
} else {
fmt.Println("Who are you? You're not a terminal")
}
} |
http://rosettacode.org/wiki/Check_output_device_is_a_terminal | Check output device is a terminal | Task
Demonstrate how to check whether the output device is a terminal or not.
Related task
Check input device is a terminal
| #Haskell | Haskell | module Main where
-- requires the unix package
-- https://hackage.haskell.org/package/unix
import System.Posix.Terminal (queryTerminal)
import System.Posix.IO (stdOutput)
main :: IO ()
main = do
istty <- queryTerminal stdOutput
putStrLn
(if istty
then "stdout is tty"
else "stdout is not tty") |
http://rosettacode.org/wiki/Cholesky_decomposition | Cholesky decomposition | Every symmetric, positive definite matrix A can be decomposed into a product of a unique lower triangular matrix L and its transpose:
A
=
L
L
T
{\displaystyle A=LL^{T}}
L
{\displaystyle L}
is called the Cholesky factor of
A
{\displaystyle A}
, and can be interpreted as a generalized square root of
A
{\displaystyle A}
, as described in Cholesky decomposition.
In a 3x3 example, we have to solve the following system of equations:
A
=
(
a
11
a
21
a
31
a
21
a
22
a
32
a
31
a
32
a
33
)
=
(
l
11
0
0
l
21
l
22
0
l
31
l
32
l
33
)
(
l
11
l
21
l
31
0
l
22
l
32
0
0
l
33
)
≡
L
L
T
=
(
l
11
2
l
21
l
11
l
31
l
11
l
21
l
11
l
21
2
+
l
22
2
l
31
l
21
+
l
32
l
22
l
31
l
11
l
31
l
21
+
l
32
l
22
l
31
2
+
l
32
2
+
l
33
2
)
{\displaystyle {\begin{aligned}A&={\begin{pmatrix}a_{11}&a_{21}&a_{31}\\a_{21}&a_{22}&a_{32}\\a_{31}&a_{32}&a_{33}\\\end{pmatrix}}\\&={\begin{pmatrix}l_{11}&0&0\\l_{21}&l_{22}&0\\l_{31}&l_{32}&l_{33}\\\end{pmatrix}}{\begin{pmatrix}l_{11}&l_{21}&l_{31}\\0&l_{22}&l_{32}\\0&0&l_{33}\end{pmatrix}}\equiv LL^{T}\\&={\begin{pmatrix}l_{11}^{2}&l_{21}l_{11}&l_{31}l_{11}\\l_{21}l_{11}&l_{21}^{2}+l_{22}^{2}&l_{31}l_{21}+l_{32}l_{22}\\l_{31}l_{11}&l_{31}l_{21}+l_{32}l_{22}&l_{31}^{2}+l_{32}^{2}+l_{33}^{2}\end{pmatrix}}\end{aligned}}}
We can see that for the diagonal elements (
l
k
k
{\displaystyle l_{kk}}
) of
L
{\displaystyle L}
there is a calculation pattern:
l
11
=
a
11
{\displaystyle l_{11}={\sqrt {a_{11}}}}
l
22
=
a
22
−
l
21
2
{\displaystyle l_{22}={\sqrt {a_{22}-l_{21}^{2}}}}
l
33
=
a
33
−
(
l
31
2
+
l
32
2
)
{\displaystyle l_{33}={\sqrt {a_{33}-(l_{31}^{2}+l_{32}^{2})}}}
or in general:
l
k
k
=
a
k
k
−
∑
j
=
1
k
−
1
l
k
j
2
{\displaystyle l_{kk}={\sqrt {a_{kk}-\sum _{j=1}^{k-1}l_{kj}^{2}}}}
For the elements below the diagonal (
l
i
k
{\displaystyle l_{ik}}
, where
i
>
k
{\displaystyle i>k}
) there is also a calculation pattern:
l
21
=
1
l
11
a
21
{\displaystyle l_{21}={\frac {1}{l_{11}}}a_{21}}
l
31
=
1
l
11
a
31
{\displaystyle l_{31}={\frac {1}{l_{11}}}a_{31}}
l
32
=
1
l
22
(
a
32
−
l
31
l
21
)
{\displaystyle l_{32}={\frac {1}{l_{22}}}(a_{32}-l_{31}l_{21})}
which can also be expressed in a general formula:
l
i
k
=
1
l
k
k
(
a
i
k
−
∑
j
=
1
k
−
1
l
i
j
l
k
j
)
{\displaystyle l_{ik}={\frac {1}{l_{kk}}}\left(a_{ik}-\sum _{j=1}^{k-1}l_{ij}l_{kj}\right)}
Task description
The task is to implement a routine which will return a lower Cholesky factor
L
{\displaystyle L}
for every given symmetric, positive definite nxn matrix
A
{\displaystyle A}
. You should then test it on the following two examples and include your output.
Example 1:
25 15 -5 5 0 0
15 18 0 --> 3 3 0
-5 0 11 -1 1 3
Example 2:
18 22 54 42 4.24264 0.00000 0.00000 0.00000
22 70 86 62 --> 5.18545 6.56591 0.00000 0.00000
54 86 174 134 12.72792 3.04604 1.64974 0.00000
42 62 134 106 9.89949 1.62455 1.84971 1.39262
Note
The Cholesky decomposition of a Pascal upper-triangle matrix is the Identity matrix of the same size.
The Cholesky decomposition of a Pascal symmetric matrix is the Pascal lower-triangle matrix of the same size. | #Clojure | Clojure | (defn cholesky
[matrix]
(let [n (count matrix)
A (to-array-2d matrix)
L (make-array Double/TYPE n n)]
(doseq [i (range n) j (range (inc i))]
(let [s (reduce + (for [k (range j)] (* (aget L i k) (aget L j k))))]
(aset L i j (if (= i j)
(Math/sqrt (- (aget A i i) s))
(* (/ 1.0 (aget L j j)) (- (aget A i j) s))))))
(vec (map vec L)))) |
http://rosettacode.org/wiki/Check_input_device_is_a_terminal | Check input device is a terminal | Task
Demonstrate how to check whether the input device is a terminal or not.
Related task
Check output device is a terminal
| #FreeBASIC | FreeBASIC |
Open Cons For Input As #1
' Open Cons abre los flujos de entrada (stdin) o salida (stdout) estándar
' de la consola para leer o escribir.
If Err Then
Print "Input doesn't come from tt."
Else
Print "Input comes from tty."
End If
Close #1
Sleep
|
http://rosettacode.org/wiki/Check_input_device_is_a_terminal | Check input device is a terminal | Task
Demonstrate how to check whether the input device is a terminal or not.
Related task
Check output device is a terminal
| #Go | Go | package main
import (
"golang.org/x/crypto/ssh/terminal"
"fmt"
"os"
)
func main() {
if terminal.IsTerminal(int(os.Stdin.Fd())) {
fmt.Println("Hello terminal")
} else {
fmt.Println("Who are you? You're not a terminal.")
}
} |
http://rosettacode.org/wiki/Check_input_device_is_a_terminal | Check input device is a terminal | Task
Demonstrate how to check whether the input device is a terminal or not.
Related task
Check output device is a terminal
| #Haskell | Haskell | module Main (main) where
import System.Posix.IO (stdInput)
import System.Posix.Terminal (queryTerminal)
main :: IO ()
main = do
isTTY <- queryTerminal stdInput
putStrLn $ if isTTY
then "stdin is TTY"
else "stdin is not TTY" |
http://rosettacode.org/wiki/Check_input_device_is_a_terminal | Check input device is a terminal | Task
Demonstrate how to check whether the input device is a terminal or not.
Related task
Check output device is a terminal
| #Jsish | Jsish | /* Check input device is a terminal, in Jsish */
;Interp.conf().subOpts.istty;
/*
=!EXPECTSTART!=
Interp.conf().subOpts.istty ==> false
=!EXPECTEND!=
*/ |
http://rosettacode.org/wiki/Check_input_device_is_a_terminal | Check input device is a terminal | Task
Demonstrate how to check whether the input device is a terminal or not.
Related task
Check output device is a terminal
| #Julia | Julia |
if isa(STDIN, Base.TTY)
println("This program sees STDIN as a TTY.")
else
println("This program does not see STDIN as a TTY.")
end
|
http://rosettacode.org/wiki/Cheryl%27s_birthday | Cheryl's birthday | Albert and Bernard just became friends with Cheryl, and they want to know when her birthday is.
Cheryl gave them a list of ten possible dates:
May 15, May 16, May 19
June 17, June 18
July 14, July 16
August 14, August 15, August 17
Cheryl then tells Albert the month of birth, and Bernard the day (of the month) of birth.
1) Albert: I don't know when Cheryl's birthday is, but I know that Bernard does not know too.
2) Bernard: At first I don't know when Cheryl's birthday is, but I know now.
3) Albert: Then I also know when Cheryl's birthday is.
Task
Write a computer program to deduce, by successive elimination, Cheryl's birthday.
Related task
Sum and Product Puzzle
References
Wikipedia article of the same name.
Tuple Relational Calculus
| #AutoHotkey | AutoHotkey | oDates:= {"May" : [ 15, 16, 19]
,"Jun" : [ 17, 18]
,"Jul" : [14, 16]
,"Aug" : [14, 15, 17]}
filter1(oDates)
filter2(oDates)
filter3(oDates)
MsgBox % result := checkAnswer(oDates)
return
filter1(ByRef oDates){ ; remove months that has a unique day in it.
for d, obj in MonthsOfDay(oDates)
if (obj.count() = 1)
for m, bool in obj
oDates.Remove(m)
}
filter2(ByRef oDates){ ; remove non-unique days from remaining months.
for d, obj in MonthsOfDay(oDates)
if (obj.count() > 1)
for m, bool in obj
for i, day in oDates[m]
if (day=d)
oDates[m].Remove(i)
}
filter3(ByRef oDates){ ; remove months that has multiple days from remaining months.
oRemove := []
for m, obj in oDates
if obj.count() > 1
oRemove.Push(m)
for i, m in oRemove
oDates.Remove(m)
}
MonthsOfDay(oDates){ ; create a list of months per day.
MonthsOfDay := []
for m, obj in oDates
for i, d in obj
MonthsOfDay[d, m] := 1
return MonthsOfDay
}
checkAnswer(oDates){ ; check unique answer if any.
if oDates.count()>1
return false
for m, obj in oDates
if obj.count() > 1
return false
else
return m " " obj.1
} |
http://rosettacode.org/wiki/Checkpoint_synchronization | Checkpoint synchronization | The checkpoint synchronization is a problem of synchronizing multiple tasks. Consider a workshop where several workers (tasks) assembly details of some mechanism. When each of them completes his work they put the details together. There is no store, so a worker who finished its part first must wait for others before starting another one. Putting details together is the checkpoint at which tasks synchronize themselves before going their paths apart.
The task
Implement checkpoint synchronization in your language.
Make sure that the solution is race condition-free. Note that a straightforward solution based on events is exposed to race condition. Let two tasks A and B need to be synchronized at a checkpoint. Each signals its event (EA and EB correspondingly), then waits for the AND-combination of the events (EA&EB) and resets its event. Consider the following scenario: A signals EA first and gets blocked waiting for EA&EB. Then B signals EB and loses the processor. Then A is released (both events are signaled) and resets EA. Now if B returns and enters waiting for EA&EB, it gets lost.
When a worker is ready it shall not continue before others finish. A typical implementation bug is when a worker is counted twice within one working cycle causing its premature completion. This happens when the quickest worker serves its cycle two times while the laziest one is lagging behind.
If you can, implement workers joining and leaving.
| #D | D | import std.stdio;
import std.parallelism: taskPool, defaultPoolThreads, totalCPUs;
void buildMechanism(uint nparts) {
auto details = new uint[nparts];
foreach (i, ref detail; taskPool.parallel(details)) {
writeln("Build detail ", i);
detail = i;
}
// This could be written more concisely via std.parallelism.reduce,
// but we want to see the checkpoint explicitly.
writeln("Checkpoint reached. Assemble details ...");
uint sum = 0;
foreach (immutable detail; details)
sum += detail;
writeln("Mechanism with ", nparts, " parts finished: ", sum);
}
void main() {
defaultPoolThreads = totalCPUs + 1; // totalCPUs - 1 on default.
buildMechanism(42);
buildMechanism(11);
} |
http://rosettacode.org/wiki/Checkpoint_synchronization | Checkpoint synchronization | The checkpoint synchronization is a problem of synchronizing multiple tasks. Consider a workshop where several workers (tasks) assembly details of some mechanism. When each of them completes his work they put the details together. There is no store, so a worker who finished its part first must wait for others before starting another one. Putting details together is the checkpoint at which tasks synchronize themselves before going their paths apart.
The task
Implement checkpoint synchronization in your language.
Make sure that the solution is race condition-free. Note that a straightforward solution based on events is exposed to race condition. Let two tasks A and B need to be synchronized at a checkpoint. Each signals its event (EA and EB correspondingly), then waits for the AND-combination of the events (EA&EB) and resets its event. Consider the following scenario: A signals EA first and gets blocked waiting for EA&EB. Then B signals EB and loses the processor. Then A is released (both events are signaled) and resets EA. Now if B returns and enters waiting for EA&EB, it gets lost.
When a worker is ready it shall not continue before others finish. A typical implementation bug is when a worker is counted twice within one working cycle causing its premature completion. This happens when the quickest worker serves its cycle two times while the laziest one is lagging behind.
If you can, implement workers joining and leaving.
| #E | E | /** A flagSet solves this problem: There are N things, each in a true or false
* state, and we want to know whether they are all true (or all false), and be
* able to bulk-change all of them, and all this without allowing double-
* counting -- setting a flag twice is idempotent.
*/
def makeFlagSet() {
# Each flag object is either in the true set or the false set.
def trues := [].asSet().diverge()
def falses := [].asSet().diverge()
return def flagSet {
/** Add a flag to the set. */
to join() {
def flag {
/** Get the value of this flag. */
to get() :boolean {
}
/** Set the value of this flag. */
to put(v :boolean) {
def [del,add] := if (v) { [falses,trues] } else { [trues,falses] }
if (del.contains(flag)) {
del.remove(flag)
add.addElement(flag)
}
}
/** Remove this flag from the set. */
to leave() :void {
trues.remove(flag)
falses.remove(flag)
}
}
falses.addElement(flag)
return flag
}
/** Are all the flags true (none false)? */
to allTrue() { return falses.size().isZero() }
/** Are all the flags false (none true)? */
to allFalse() { return trues.size().isZero() }
/** Set all the flags to the same value. */
to setAll(v :boolean) {
def [del,add] := if (v) { [falses,trues] } else { [trues,falses] }
add.addAll(del)
del.removeAll(del)
}
}
}
def makeCheckpoint() {
def [var continueSignal, var continueRes] := Ref.promise()
def readies := makeFlagSet()
/** Check whether all tasks have reached the checkpoint, and if so send the
* signal and go to the next round. */
def check() {
if (readies.allTrue()) {
readies.setAll(false)
continueRes.resolve(null) # send the continue signal
def [p, r] := Ref.promise() # prepare a new continue signal
continueSignal := p
continueRes := r
}
}
return def checkpoint {
to join() {
def &flag := readies.join()
return def membership {
to leave() {
(&flag).leave()
check <- ()
}
to deliver() {
flag := true
check <- ()
return continueSignal
}
}
}
}
}
def makeWorker(piece, checkpoint) {
def stops := timer.now() + 3000 + entropy.nextInt(2000)
var count := 0
def checkpointMember := checkpoint <- join()
def stopped
def run() {
# Pretend to do something lengthy; up to 1000 ms.
timer.whenPast(timer.now() + entropy.nextInt(1000), fn {
if (timer.now() >= stops) {
checkpointMember <- leave()
bind stopped := true
} else {
count += 1
println(`Delivering $piece#$count`)
when (checkpointMember <- deliver()) -> {
println(`Delivered $piece#$count`)
run()
}
}
})
}
run()
return stopped
}
def checkpoint := makeCheckpoint()
var waits := []
for piece in 1..5 {
waits with= makeWorker(piece, checkpoint)
}
interp.waitAtTop(promiseAllFulfilled(waits)) |
http://rosettacode.org/wiki/Collections | Collections | This task has been clarified. Its programming examples are in need of review to ensure that they still fit the requirements of the task.
Collections are abstractions to represent sets of values.
In statically-typed languages, the values are typically of a common data type.
Task
Create a collection, and add a few values to it.
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
| #Fancy | Fancy |
# creating an empty array and adding values
a = [] # => []
a[0]: 1 # => [1]
a[3]: 2 # => [1, nil, nil, 2]
# creating an array with the constructor
a = Array new # => []
|
http://rosettacode.org/wiki/Combinations | Combinations | Task
Given non-negative integers m and n, generate all size m combinations of the integers from 0 (zero) to n-1 in sorted order (each combination is sorted and the entire table is sorted).
Example
3 comb 5 is:
0 1 2
0 1 3
0 1 4
0 2 3
0 2 4
0 3 4
1 2 3
1 2 4
1 3 4
2 3 4
If it is more "natural" in your language to start counting from 1 (unity) instead of 0 (zero),
the combinations can be of the integers from 1 to n.
See also
The number of samples of size k from n objects.
With combinations and permutations generation tasks.
Order Unimportant
Order Important
Without replacement
(
n
k
)
=
n
C
k
=
n
(
n
−
1
)
…
(
n
−
k
+
1
)
k
(
k
−
1
)
…
1
{\displaystyle {\binom {n}{k}}=^{n}\operatorname {C} _{k}={\frac {n(n-1)\ldots (n-k+1)}{k(k-1)\dots 1}}}
n
P
k
=
n
⋅
(
n
−
1
)
⋅
(
n
−
2
)
⋯
(
n
−
k
+
1
)
{\displaystyle ^{n}\operatorname {P} _{k}=n\cdot (n-1)\cdot (n-2)\cdots (n-k+1)}
Task: Combinations
Task: Permutations
With replacement
(
n
+
k
−
1
k
)
=
n
+
k
−
1
C
k
=
(
n
+
k
−
1
)
!
(
n
−
1
)
!
k
!
{\displaystyle {\binom {n+k-1}{k}}=^{n+k-1}\operatorname {C} _{k}={(n+k-1)! \over (n-1)!k!}}
n
k
{\displaystyle n^{k}}
Task: Combinations with repetitions
Task: Permutations with repetitions
| #Lua | Lua |
function map(f, a, ...) if a then return f(a), map(f, ...) end end
function incr(k) return function(a) return k > a and a or a+1 end end
function combs(m, n)
if m * n == 0 then return {{}} end
local ret, old = {}, combs(m-1, n-1)
for i = 1, n do
for k, v in ipairs(old) do ret[#ret+1] = {i, map(incr(i), unpack(v))} end
end
return ret
end
for k, v in ipairs(combs(3, 5)) do print(unpack(v)) end
|
http://rosettacode.org/wiki/Conditional_structures | Conditional structures | Control Structures
These are examples of control structures. You may also be interested in:
Conditional structures
Exceptions
Flow-control structures
Loops
Task
List the conditional structures offered by a programming language. See Wikipedia: conditionals for descriptions.
Common conditional structures include if-then-else and switch.
Less common are arithmetic if, ternary operator and Hash-based conditionals.
Arithmetic if allows tight control over computed gotos, which optimizers have a hard time to figure out.
| #QB64 | QB64 |
Print "QB64/Qbasic conditional structures"
Dim k As String
Menu 1
View Print 13 To 23
Print "A menu example using the many options of IF statement"
k = " "
12: While k <> ""
k = UCase$(Input$(1))
If k = "O" GoTo O
If k = "F" Then 22
If k = "S" Then GoSub S: GoTo 12
If k = "C" Then GoSub 4: GoTo 12
If k = "E" Then GoSub 5: Exit While
Wend
Cls
Print "the same menu example with Select Case"
Sleep 2
While k <> ""
k = UCase$(Input$(1))
Select Case k
Case "O"
Print "You choose O"
Case "F"
Print "You choose F"
Case "S"
Print "You choose S"
Case "C"
Print "You choose C"
Case "E"
Print "You choose Exit"
_Delay 1
Exit While
Case Else
Print "Wrong choice"
End Select
Wend
View Print
Cls
Menu 2
View Print 13 To 23
Print "menu demonstration using ON value GOTO"
k = " "
While k <> ""
k = Input$(1)
On Val(k) GOSUB 1, 2, 3, 4, 5
Wend
End
1:
Print "Chosen O"
Return
2:
Print "Chosen F"
Return
3:
Print "Chosen S"
Return
4:
Print "Chosen C"
Return
5:
Print "Chosen E"
If k = "5" Then End
Return
O:
Print "You choose O"
GoTo 12
22:
Print "You choose F"
GoTo 12
S:
Print "You choose S"
Return
Sub Menu (Kind As Integer)
Locate 7, 33: Color 3, 4
Print "Choose the item"
Color 7, 0
Locate , 33
If Kind = 1 Then Print "Open a file"; Else Print "1) Open a file";
Color 14, 1
Locate , 33
If Kind = 1 Then Print "O" Else Print "1"
Color 7, 0
Locate , 33
If Kind = 1 Then Print "Find a file"; Else Print "2) Find a file";
Color 14, 1
Locate , 33
If Kind = 1 Then Print "F" Else Print "2"
Color 7, 0
Locate , 33
If Kind = 1 Then Print "Scan a file"; Else Print "3) Scan a file";
Color 14, 1
Locate , 33
If Kind = 1 Then Print "S" Else Print "3"
Color 7, 0
Locate , 33
If Kind = 1 Then Print "Copy a file"; Else Print "4) Copy a file";
Color 14, 1
Locate , 33
If Kind = 1 Then Print "C" Else Print "4"
Color 7, 0
Locate , 33
If Kind = 1 Then Print "Exit from Menu"; Else Print "5) Exit from Menu";
Color 14, 1
Locate , 33
If Kind = 1 Then Print "E" Else Print "5"
Color 7, 0
End Sub
|
http://rosettacode.org/wiki/Chinese_remainder_theorem | Chinese remainder theorem | Suppose
n
1
{\displaystyle n_{1}}
,
n
2
{\displaystyle n_{2}}
,
…
{\displaystyle \ldots }
,
n
k
{\displaystyle n_{k}}
are positive integers that are pairwise co-prime.
Then, for any given sequence of integers
a
1
{\displaystyle a_{1}}
,
a
2
{\displaystyle a_{2}}
,
…
{\displaystyle \dots }
,
a
k
{\displaystyle a_{k}}
, there exists an integer
x
{\displaystyle x}
solving the following system of simultaneous congruences:
x
≡
a
1
(
mod
n
1
)
x
≡
a
2
(
mod
n
2
)
⋮
x
≡
a
k
(
mod
n
k
)
{\displaystyle {\begin{aligned}x&\equiv a_{1}{\pmod {n_{1}}}\\x&\equiv a_{2}{\pmod {n_{2}}}\\&{}\ \ \vdots \\x&\equiv a_{k}{\pmod {n_{k}}}\end{aligned}}}
Furthermore, all solutions
x
{\displaystyle x}
of this system are congruent modulo the product,
N
=
n
1
n
2
…
n
k
{\displaystyle N=n_{1}n_{2}\ldots n_{k}}
.
Task
Write a program to solve a system of linear congruences by applying the Chinese Remainder Theorem.
If the system of equations cannot be solved, your program must somehow indicate this.
(It may throw an exception or return a special false value.)
Since there are infinitely many solutions, the program should return the unique solution
s
{\displaystyle s}
where
0
≤
s
≤
n
1
n
2
…
n
k
{\displaystyle 0\leq s\leq n_{1}n_{2}\ldots n_{k}}
.
Show the functionality of this program by printing the result such that the
n
{\displaystyle n}
's are
[
3
,
5
,
7
]
{\displaystyle [3,5,7]}
and the
a
{\displaystyle a}
's are
[
2
,
3
,
2
]
{\displaystyle [2,3,2]}
.
Algorithm: The following algorithm only applies if the
n
i
{\displaystyle n_{i}}
's are pairwise co-prime.
Suppose, as above, that a solution is required for the system of congruences:
x
≡
a
i
(
mod
n
i
)
f
o
r
i
=
1
,
…
,
k
{\displaystyle x\equiv a_{i}{\pmod {n_{i}}}\quad \mathrm {for} \;i=1,\ldots ,k}
Again, to begin, the product
N
=
n
1
n
2
…
n
k
{\displaystyle N=n_{1}n_{2}\ldots n_{k}}
is defined.
Then a solution
x
{\displaystyle x}
can be found as follows:
For each
i
{\displaystyle i}
, the integers
n
i
{\displaystyle n_{i}}
and
N
/
n
i
{\displaystyle N/n_{i}}
are co-prime.
Using the Extended Euclidean algorithm, we can find integers
r
i
{\displaystyle r_{i}}
and
s
i
{\displaystyle s_{i}}
such that
r
i
n
i
+
s
i
N
/
n
i
=
1
{\displaystyle r_{i}n_{i}+s_{i}N/n_{i}=1}
.
Then, one solution to the system of simultaneous congruences is:
x
=
∑
i
=
1
k
a
i
s
i
N
/
n
i
{\displaystyle x=\sum _{i=1}^{k}a_{i}s_{i}N/n_{i}}
and the minimal solution,
x
(
mod
N
)
{\displaystyle x{\pmod {N}}}
.
| #C.23 | C# | using System;
using System.Linq;
namespace ChineseRemainderTheorem
{
class Program
{
static void Main(string[] args)
{
int[] n = { 3, 5, 7 };
int[] a = { 2, 3, 2 };
int result = ChineseRemainderTheorem.Solve(n, a);
int counter = 0;
int maxCount = n.Length - 1;
while (counter <= maxCount)
{
Console.WriteLine($"{result} ≡ {a[counter]} (mod {n[counter]})");
counter++;
}
}
}
public static class ChineseRemainderTheorem
{
public static int Solve(int[] n, int[] a)
{
int prod = n.Aggregate(1, (i, j) => i * j);
int p;
int sm = 0;
for (int i = 0; i < n.Length; i++)
{
p = prod / n[i];
sm += a[i] * ModularMultiplicativeInverse(p, n[i]) * p;
}
return sm % prod;
}
private static int ModularMultiplicativeInverse(int a, int mod)
{
int b = a % mod;
for (int x = 1; x < mod; x++)
{
if ((b * x) % mod == 1)
{
return x;
}
}
return 1;
}
}
} |
http://rosettacode.org/wiki/Chernick%27s_Carmichael_numbers | Chernick's Carmichael numbers | In 1939, Jack Chernick proved that, for n ≥ 3 and m ≥ 1:
U(n, m) = (6m + 1) * (12m + 1) * Product_{i=1..n-2} (2^i * 9m + 1)
is a Carmichael number if all the factors are primes and, for n > 4, m is a multiple of 2^(n-4).
Example
U(3, m) = (6m + 1) * (12m + 1) * (18m + 1)
U(4, m) = U(3, m) * (2^2 * 9m + 1)
U(5, m) = U(4, m) * (2^3 * 9m + 1)
...
U(n, m) = U(n-1, m) * (2^(n-2) * 9m + 1)
The smallest Chernick's Carmichael number with 3 prime factors, is: U(3, 1) = 1729.
The smallest Chernick's Carmichael number with 4 prime factors, is: U(4, 1) = 63973.
The smallest Chernick's Carmichael number with 5 prime factors, is: U(5, 380) = 26641259752490421121.
For n = 5, the smallest number m that satisfy Chernick's conditions, is m = 380, therefore U(5, 380) is the smallest Chernick's Carmichael number with 5 prime factors.
U(5, 380) is a Chernick's Carmichael number because m = 380 is a multiple of 2^(n-4), where n = 5, and the factors { (6*380 + 1), (12*380 + 1), (18*380 + 1), (36*380 + 1), (72*380 + 1) } are all prime numbers.
Task
For n ≥ 3, let a(n) be the smallest Chernick's Carmichael number with n prime factors.
Compute a(n) for n = 3..9.
Optional: find a(10).
Note: it's perfectly acceptable to show the terms in factorized form:
a(3) = 7 * 13 * 19
a(4) = 7 * 13 * 19 * 37
a(5) = 2281 * 4561 * 6841 * 13681 * 27361
...
See also
Jack Chernick, On Fermat's simple theorem (PDF)
OEIS A318646: The least Chernick's "universal form" Carmichael number with n prime factors
Related tasks
Carmichael 3 strong pseudoprimes
| #Prolog | Prolog |
?- use_module(library(primality)).
u(3, M, A * B * C) :-
A is 6*M + 1, B is 12*M + 1, C is 18*M + 1, !.
u(N, M, U0 * D) :-
succ(Pn, N), u(Pn, M, U0),
D is 9*(1 << (N - 2))*M + 1.
prime_factorization(A*B) :- prime(B), prime_factorization(A), !.
prime_factorization(A) :- prime(A).
step(N, 1) :- N < 5, !.
step(5, 2) :- !.
step(N, K) :- K is 5*(1 << (N - 4)).
a(N, Factors) :- % due to backtracking nature of Prolog, a(n) will return all chernick-carmichael numbers.
N > 2, !,
step(N, I),
between(1, infinite, J), M is I * J,
u(N, M, Factors),
prime_factorization(Factors).
main :-
forall(
(between(3, 9, K), once(a(K, Factorization)), N is Factorization),
format("~w: ~w = ~w~n", [K, Factorization, N])),
halt.
?- main.
|
http://rosettacode.org/wiki/Chowla_numbers | Chowla numbers | Chowla numbers are also known as:
Chowla's function
chowla numbers
the chowla function
the chowla number
the chowla sequence
The chowla number of n is (as defined by Chowla's function):
the sum of the divisors of n excluding unity and n
where n is a positive integer
The sequence is named after Sarvadaman D. S. Chowla, (22 October 1907 ──► 10 December 1995),
a London born Indian American mathematician specializing in number theory.
German mathematician Carl Friedrich Gauss (1777─1855) said:
"Mathematics is the queen of the sciences ─ and number theory is the queen of mathematics".
Definitions
Chowla numbers can also be expressed as:
chowla(n) = sum of divisors of n excluding unity and n
chowla(n) = sum( divisors(n)) - 1 - n
chowla(n) = sum( properDivisors(n)) - 1
chowla(n) = sum(aliquotDivisors(n)) - 1
chowla(n) = aliquot(n) - 1
chowla(n) = sigma(n) - 1 - n
chowla(n) = sigmaProperDivisiors(n) - 1
chowla(a*b) = a + b, if a and b are distinct primes
if chowla(n) = 0, and n > 1, then n is prime
if chowla(n) = n - 1, and n > 1, then n is a perfect number
Task
create a chowla function that returns the chowla number for a positive integer n
Find and display (1 per line) for the 1st 37 integers:
the integer (the index)
the chowla number for that integer
For finding primes, use the chowla function to find values of zero
Find and display the count of the primes up to 100
Find and display the count of the primes up to 1,000
Find and display the count of the primes up to 10,000
Find and display the count of the primes up to 100,000
Find and display the count of the primes up to 1,000,000
Find and display the count of the primes up to 10,000,000
For finding perfect numbers, use the chowla function to find values of n - 1
Find and display all perfect numbers up to 35,000,000
use commas within appropriate numbers
show all output here
Related tasks
totient function
perfect numbers
Proper divisors
Sieve of Eratosthenes
See also
the OEIS entry for A48050 Chowla's function.
| #EasyLang | EasyLang | func chowla n . sum .
sum = 0
i = 2
while i * i <= n
if n mod i = 0
j = n div i
if i = j
sum += i
else
sum += i + j
.
.
i += 1
.
.
func sieve . c[] .
i = 3
while i * 3 < len c[]
if c[i] = 0
call chowla i h
if h = 0
j = 3 * i
while j < len c[]
c[j] = 1
j += 2 * i
.
.
.
i += 2
.
.
func commatize n . s$ .
s$[] = strchars n
s$ = ""
l = len s$[]
for i range len s$[]
if i > 0 and l mod 3 = 0
s$ &= ","
.
l -= 1
s$ &= s$[i]
.
.
print "chowla number from 1 to 37"
for i = 1 to 37
call chowla i h
print " " & i & ": " & h
.
func main . .
print ""
len c[] 10000000
count = 1
call sieve c[]
power = 100
i = 3
while i < len c[]
if c[i] = 0
count += 1
.
if i = power - 1
call commatize power p$
call commatize count c$
print "There are " & c$ & " primes up to " & p$
power *= 10
.
i += 2
.
print ""
limit = 35000000
count = 0
i = 2
k = 2
kk = 3
repeat
p = k * kk
until p > limit
call chowla p h
if h = p - 1
call commatize p s$
print s$ & " is a perfect number"
count += 1
.
k = kk + 1
kk += k
i += 1
.
call commatize limit s$
print "There are " & count & " perfect mumbers up to " & s$
.
call main |
http://rosettacode.org/wiki/Chowla_numbers | Chowla numbers | Chowla numbers are also known as:
Chowla's function
chowla numbers
the chowla function
the chowla number
the chowla sequence
The chowla number of n is (as defined by Chowla's function):
the sum of the divisors of n excluding unity and n
where n is a positive integer
The sequence is named after Sarvadaman D. S. Chowla, (22 October 1907 ──► 10 December 1995),
a London born Indian American mathematician specializing in number theory.
German mathematician Carl Friedrich Gauss (1777─1855) said:
"Mathematics is the queen of the sciences ─ and number theory is the queen of mathematics".
Definitions
Chowla numbers can also be expressed as:
chowla(n) = sum of divisors of n excluding unity and n
chowla(n) = sum( divisors(n)) - 1 - n
chowla(n) = sum( properDivisors(n)) - 1
chowla(n) = sum(aliquotDivisors(n)) - 1
chowla(n) = aliquot(n) - 1
chowla(n) = sigma(n) - 1 - n
chowla(n) = sigmaProperDivisiors(n) - 1
chowla(a*b) = a + b, if a and b are distinct primes
if chowla(n) = 0, and n > 1, then n is prime
if chowla(n) = n - 1, and n > 1, then n is a perfect number
Task
create a chowla function that returns the chowla number for a positive integer n
Find and display (1 per line) for the 1st 37 integers:
the integer (the index)
the chowla number for that integer
For finding primes, use the chowla function to find values of zero
Find and display the count of the primes up to 100
Find and display the count of the primes up to 1,000
Find and display the count of the primes up to 10,000
Find and display the count of the primes up to 100,000
Find and display the count of the primes up to 1,000,000
Find and display the count of the primes up to 10,000,000
For finding perfect numbers, use the chowla function to find values of n - 1
Find and display all perfect numbers up to 35,000,000
use commas within appropriate numbers
show all output here
Related tasks
totient function
perfect numbers
Proper divisors
Sieve of Eratosthenes
See also
the OEIS entry for A48050 Chowla's function.
| #Factor | Factor | USING: formatting fry grouping.extras io kernel math
math.primes.factors math.ranges math.statistics sequences
tools.memory.private ;
IN: rosetta-code.chowla-numbers
: chowla ( n -- m )
dup 1 = [ 1 - ] [ [ divisors sum ] [ - 1 - ] bi ] if ;
: show-chowla ( n -- )
[1,b] [ dup chowla "chowla(%02d) = %d\n" printf ] each ;
: count-primes ( seq -- )
dup 0 prefix [ [ 1 + ] dip 2 <range> ] 2clump-map
[ [ chowla zero? ] count ] map cum-sum
[ [ commas ] bi@ "Primes up to %s: %s\n" printf ] 2each ;
: show-perfect ( n -- )
[ 2 3 ] dip '[ 2dup * dup _ > ] [
dup [ chowla ] [ 1 - = ] bi
[ commas "%s is perfect\n" printf ] [ drop ] if
[ nip 1 + ] [ nip dupd + ] 2bi
] until 3drop ;
: chowla-demo ( -- )
37 show-chowla nl { 100 1000 10000 100000 1000000 10000000 }
count-primes nl 35e7 show-perfect ;
MAIN: chowla-demo |
http://rosettacode.org/wiki/Church_numerals | Church numerals | Task
In the Church encoding of natural numbers, the number N is encoded by a function that applies its first argument N times to its second argument.
Church zero always returns the identity function, regardless of its first argument. In other words, the first argument is not applied to the second argument at all.
Church one applies its first argument f just once to its second argument x, yielding f(x)
Church two applies its first argument f twice to its second argument x, yielding f(f(x))
and each successive Church numeral applies its first argument one additional time to its second argument, f(f(f(x))), f(f(f(f(x)))) ... The Church numeral 4, for example, returns a quadruple composition of the function supplied as its first argument.
Arithmetic operations on natural numbers can be similarly represented as functions on Church numerals.
In your language define:
Church Zero,
a Church successor function (a function on a Church numeral which returns the next Church numeral in the series),
functions for Addition, Multiplication and Exponentiation over Church numerals,
a function to convert integers to corresponding Church numerals,
and a function to convert Church numerals to corresponding integers.
You should:
Derive Church numerals three and four in terms of Church zero and a Church successor function.
use Church numeral arithmetic to obtain the the sum and the product of Church 3 and Church 4,
similarly obtain 4^3 and 3^4 in terms of Church numerals, using a Church numeral exponentiation function,
convert each result back to an integer, and return it or print it to the console.
| #Lua | Lua |
function churchZero()
return function(x) return x end
end
function churchSucc(c)
return function(f)
return function(x)
return f(c(f)(x))
end
end
end
function churchAdd(c, d)
return function(f)
return function(x)
return c(f)(d(f)(x))
end
end
end
function churchMul(c, d)
return function(f)
return c(d(f))
end
end
function churchExp(c, e)
return e(c)
end
function numToChurch(n)
local ret = churchZero
for i = 1, n do
ret = succ(ret)
end
return ret
end
function churchToNum(c)
return c(function(x) return x + 1 end)(0)
end
three = churchSucc(churchSucc(churchSucc(churchZero)))
four = churchSucc(churchSucc(churchSucc(churchSucc(churchZero))))
print("'three'\t=", churchToNum(three))
print("'four' \t=", churchToNum(four))
print("'three' * 'four' =", churchToNum(churchMul(three, four)))
print("'three' + 'four' =", churchToNum(churchAdd(three, four)))
print("'three' ^ 'four' =", churchToNum(churchExp(three, four)))
print("'four' ^ 'three' =", churchToNum(churchExp(four, three))) |
http://rosettacode.org/wiki/Classes | Classes | In object-oriented programming class is a set (a transitive closure) of types bound by the relation of inheritance. It is said that all types derived from some base type T and the type T itself form a class T.
The first type T from the class T sometimes is called the root type of the class.
A class of types itself, as a type, has the values and operations of its own.
The operations of are usually called methods of the root type.
Both operations and values are called polymorphic.
A polymorphic operation (method) selects an implementation depending on the actual specific type of the polymorphic argument.
The action of choice the type-specific implementation of a polymorphic operation is called dispatch. Correspondingly, polymorphic operations are often called dispatching or virtual.
Operations with multiple arguments and/or the results of the class are called multi-methods.
A further generalization of is the operation with arguments and/or results from different classes.
single-dispatch languages are those that allow only one argument or result to control the dispatch. Usually it is the first parameter, often hidden, so that a prefix notation x.f() is used instead of mathematical f(x).
multiple-dispatch languages allow many arguments and/or results to control the dispatch.
A polymorphic value has a type tag indicating its specific type from the class and the corresponding specific value of that type.
This type is sometimes called the most specific type of a [polymorphic] value.
The type tag of the value is used in order to resolve the dispatch.
The set of polymorphic values of a class is a transitive closure of the sets of values of all types from that class.
In many OO languages
the type of the class of T and T itself are considered equivalent.
In some languages they are distinct (like in Ada).
When class T and T are equivalent, there is no way to distinguish
polymorphic and specific values.
Task
Create a basic class with a method, a constructor, an instance variable and how to instantiate it.
| #Eiffel | Eiffel |
class MY_CLASS
end
|
http://rosettacode.org/wiki/Classes | Classes | In object-oriented programming class is a set (a transitive closure) of types bound by the relation of inheritance. It is said that all types derived from some base type T and the type T itself form a class T.
The first type T from the class T sometimes is called the root type of the class.
A class of types itself, as a type, has the values and operations of its own.
The operations of are usually called methods of the root type.
Both operations and values are called polymorphic.
A polymorphic operation (method) selects an implementation depending on the actual specific type of the polymorphic argument.
The action of choice the type-specific implementation of a polymorphic operation is called dispatch. Correspondingly, polymorphic operations are often called dispatching or virtual.
Operations with multiple arguments and/or the results of the class are called multi-methods.
A further generalization of is the operation with arguments and/or results from different classes.
single-dispatch languages are those that allow only one argument or result to control the dispatch. Usually it is the first parameter, often hidden, so that a prefix notation x.f() is used instead of mathematical f(x).
multiple-dispatch languages allow many arguments and/or results to control the dispatch.
A polymorphic value has a type tag indicating its specific type from the class and the corresponding specific value of that type.
This type is sometimes called the most specific type of a [polymorphic] value.
The type tag of the value is used in order to resolve the dispatch.
The set of polymorphic values of a class is a transitive closure of the sets of values of all types from that class.
In many OO languages
the type of the class of T and T itself are considered equivalent.
In some languages they are distinct (like in Ada).
When class T and T are equivalent, there is no way to distinguish
polymorphic and specific values.
Task
Create a basic class with a method, a constructor, an instance variable and how to instantiate it.
| #Elena | Elena | import extensions;
class MyClass
{
prop int Variable;
someMethod()
{
Variable := 1
}
constructor()
{
}
}
public program()
{
// instantiate the class
var instance := new MyClass();
// invoke the method
instance.someMethod();
// get the variable
console.printLine("Variable=",instance.Variable)
} |
http://rosettacode.org/wiki/Cistercian_numerals | Cistercian numerals | Cistercian numerals were used across Europe by Cistercian monks during the Late Medieval Period as an alternative to Roman numerals. They were used to represent base 10 integers from 0 to 9999.
How they work
All Cistercian numerals begin with a vertical line segment, which by itself represents the number 0. Then, glyphs representing the digits 1 through 9 are optionally added to the four quadrants surrounding the vertical line segment. These glyphs are drawn with vertical and horizontal symmetry about the initial line segment. Each quadrant corresponds to a digit place in the number:
The upper-right quadrant represents the ones place.
The upper-left quadrant represents the tens place.
The lower-right quadrant represents the hundreds place.
The lower-left quadrant represents the thousands place.
Please consult the following image for examples of Cistercian numerals showing each glyph: [1]
Task
Write a function/procedure/routine to display any given Cistercian numeral. This could be done by drawing to the display, creating an image, or even as text (as long as it is a reasonable facsimile).
Use the routine to show the following Cistercian numerals:
0
1
20
300
4000
5555
6789
And a number of your choice!
Notes
Due to the inability to upload images to Rosetta Code as of this task's creation, showing output here on this page is not required. However, it is welcomed — especially for text output.
See also
Numberphile - The Forgotten Number System
dcode.fr - Online Cistercian numeral converter
| #Wren | Wren | import "/fmt" for Fmt
var n
var init = Fn.new {
n = List.filled(15, null)
for (i in 0..14) {
n[i] = List.filled(11, " ")
n[i][5] = "x"
}
}
var horiz = Fn.new { |c1, c2, r| (c1..c2).each { |c| n[r][c] = "x" } }
var verti = Fn.new { |r1, r2, c| (r1..r2).each { |r| n[r][c] = "x" } }
var diagd = Fn.new { |c1, c2, r| (c1..c2).each { |c| n[r+c-c1][c] = "x" } }
var diagu = Fn.new { |c1, c2, r| (c1..c2).each { |c| n[r-c+c1][c] = "x" } }
var draw // map contains recursive closures
draw = {
1: Fn.new { horiz.call(6, 10, 0) },
2: Fn.new { horiz.call(6, 10, 4) },
3: Fn.new { diagd.call(6, 10, 0) },
4: Fn.new { diagu.call(6, 10, 4) },
5: Fn.new {
draw[1].call()
draw[4].call()
},
6: Fn.new { verti.call(0, 4, 10) },
7: Fn.new {
draw[1].call()
draw[6].call()
},
8: Fn.new {
draw[2].call()
draw[6].call()
},
9: Fn.new {
draw[1].call()
draw[8].call()
},
10: Fn.new { horiz.call(0, 4, 0) },
20: Fn.new { horiz.call(0, 4, 4) },
30: Fn.new { diagu.call(0, 4, 4) },
40: Fn.new { diagd.call(0, 4, 0) },
50: Fn.new {
draw[10].call()
draw[40].call()
},
60: Fn.new { verti.call(0, 4, 0) },
70: Fn.new {
draw[10].call()
draw[60].call()
},
80: Fn.new {
draw[20].call()
draw[60].call()
},
90: Fn.new {
draw[10].call()
draw[80].call()
},
100: Fn.new { horiz.call(6, 10, 14) },
200: Fn.new { horiz.call(6, 10, 10) },
300: Fn.new { diagu.call(6, 10, 14) },
400: Fn.new { diagd.call(6, 10, 10) },
500: Fn.new {
draw[100].call()
draw[400].call()
},
600: Fn.new { verti.call(10, 14, 10) },
700: Fn.new {
draw[100].call()
draw[600].call()
},
800: Fn.new {
draw[200].call()
draw[600].call()
},
900: Fn.new {
draw[100].call()
draw[800].call()
},
1000: Fn.new { horiz.call(0, 4, 14) },
2000: Fn.new { horiz.call(0, 4, 10) },
3000: Fn.new { diagd.call(0, 4, 10) },
4000: Fn.new { diagu.call(0, 4, 14) },
5000: Fn.new {
draw[1000].call()
draw[4000].call()
},
6000: Fn.new { verti.call(10, 14, 0) },
7000: Fn.new {
draw[1000].call()
draw[6000].call()
},
8000: Fn.new {
draw[2000].call()
draw[6000].call()
},
9000: Fn.new {
draw[1000].call()
draw[8000].call()
}
}
var numbers = [0, 1, 20, 300, 4000, 5555, 6789, 9999]
for (number in numbers) {
init.call()
System.print("%(number):")
var thousands = (number/1000).floor
number = number % 1000
var hundreds = (number/100).floor
number = number % 100
var tens = (number/10).floor
var ones = number % 10
if (thousands > 0) draw[thousands*1000].call()
if (hundreds > 0) draw[hundreds*100].call()
if (tens > 0) draw[tens*10].call()
if (ones > 0) draw[ones].call()
Fmt.mprint(n, 1, 0, "")
System.print()
} |
http://rosettacode.org/wiki/Closest-pair_problem | Closest-pair problem |
This page uses content from Wikipedia. The original article was at Closest pair of points problem. 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)
Task
Provide a function to find the closest two points among a set of given points in two dimensions, i.e. to solve the Closest pair of points problem in the planar case.
The straightforward solution is a O(n2) algorithm (which we can call brute-force algorithm); the pseudo-code (using indexes) could be simply:
bruteForceClosestPair of P(1), P(2), ... P(N)
if N < 2 then
return ∞
else
minDistance ← |P(1) - P(2)|
minPoints ← { P(1), P(2) }
foreach i ∈ [1, N-1]
foreach j ∈ [i+1, N]
if |P(i) - P(j)| < minDistance then
minDistance ← |P(i) - P(j)|
minPoints ← { P(i), P(j) }
endif
endfor
endfor
return minDistance, minPoints
endif
A better algorithm is based on the recursive divide&conquer approach, as explained also at Wikipedia's Closest pair of points problem, which is O(n log n); a pseudo-code could be:
closestPair of (xP, yP)
where xP is P(1) .. P(N) sorted by x coordinate, and
yP is P(1) .. P(N) sorted by y coordinate (ascending order)
if N ≤ 3 then
return closest points of xP using brute-force algorithm
else
xL ← points of xP from 1 to ⌈N/2⌉
xR ← points of xP from ⌈N/2⌉+1 to N
xm ← xP(⌈N/2⌉)x
yL ← { p ∈ yP : px ≤ xm }
yR ← { p ∈ yP : px > xm }
(dL, pairL) ← closestPair of (xL, yL)
(dR, pairR) ← closestPair of (xR, yR)
(dmin, pairMin) ← (dR, pairR)
if dL < dR then
(dmin, pairMin) ← (dL, pairL)
endif
yS ← { p ∈ yP : |xm - px| < dmin }
nS ← number of points in yS
(closest, closestPair) ← (dmin, pairMin)
for i from 1 to nS - 1
k ← i + 1
while k ≤ nS and yS(k)y - yS(i)y < dmin
if |yS(k) - yS(i)| < closest then
(closest, closestPair) ← (|yS(k) - yS(i)|, {yS(k), yS(i)})
endif
k ← k + 1
endwhile
endfor
return closest, closestPair
endif
References and further readings
Closest pair of points problem
Closest Pair (McGill)
Closest Pair (UCSB)
Closest pair (WUStL)
Closest pair (IUPUI)
| #Icon_and_Unicon | Icon and Unicon | record point(x,y)
procedure main()
minDist := 0
minPair := &null
every (points := [],p1 := readPoint()) do {
if *points == 1 then minDist := dSquared(p1,points[1])
every minDist >=:= dSquared(p1,p2 := !points) do minPair := [p1,p2]
push(points, p1)
}
if \minPair then {
write("(",minPair[1].x,",",minPair[1].y,") -> ",
"(",minPair[2].x,",",minPair[2].y,")")
}
else write("One or fewer points!")
end
procedure readPoint() # Skips lines that don't have two numbers on them
suspend !&input ? point(numeric(tab(upto(', '))), numeric((move(1),tab(0))))
end
procedure dSquared(p1,p2) # Compute the square of the distance
return (p2.x-p1.x)^2 + (p2.y-p1.y)^2 # (sufficient for closeness)
end |
http://rosettacode.org/wiki/Closures/Value_capture | Closures/Value capture | Task
Create a list of ten functions, in the simplest manner possible (anonymous functions are encouraged), such that the function at index i (you may choose to start i from either 0 or 1), when run, should return the square of the index, that is, i 2.
Display the result of running any but the last function, to demonstrate that the function indeed remembers its value.
Goal
Demonstrate how to create a series of independent closures based on the same template but maintain separate copies of the variable closed over.
In imperative languages, one would generally use a loop with a mutable counter variable.
For each function to maintain the correct number, it has to capture the value of the variable at the time it was created, rather than just a reference to the variable, which would have a different value by the time the function was run.
See also: Multiple distinct objects
| #Oforth | Oforth | : newClosure(i) #[ i sq ] ;
10 seq map(#newClosure) at(7) perform . |
http://rosettacode.org/wiki/Closures/Value_capture | Closures/Value capture | Task
Create a list of ten functions, in the simplest manner possible (anonymous functions are encouraged), such that the function at index i (you may choose to start i from either 0 or 1), when run, should return the square of the index, that is, i 2.
Display the result of running any but the last function, to demonstrate that the function indeed remembers its value.
Goal
Demonstrate how to create a series of independent closures based on the same template but maintain separate copies of the variable closed over.
In imperative languages, one would generally use a loop with a mutable counter variable.
For each function to maintain the correct number, it has to capture the value of the variable at the time it was created, rather than just a reference to the variable, which would have a different value by the time the function was run.
See also: Multiple distinct objects
| #PARI.2FGP | PARI/GP | vector(10,i,()->i^2)[5]() |
http://rosettacode.org/wiki/Circular_primes | Circular primes | Definitions
A circular prime is a prime number with the property that the number generated at each intermediate step when cyclically permuting its (base 10) digits will also be prime.
For example:
1193 is a circular prime, since 1931, 9311 and 3119 are all also prime.
Note that a number which is a cyclic permutation of a smaller circular prime is not considered to be itself a circular prime. So 13 is a circular prime, but 31 is not.
A repunit (denoted by R) is a number whose base 10 representation contains only the digit 1.
For example:
R(2) = 11 and R(5) = 11111 are repunits.
Task
Find the first 19 circular primes.
If your language has access to arbitrary precision integer arithmetic, given that they are all repunits, find the next 4 circular primes.
(Stretch) Determine which of the following repunits are probably circular primes: R(5003), R(9887), R(15073), R(25031), R(35317) and R(49081). The larger ones may take a long time to process so just do as many as you reasonably can.
See also
Wikipedia article - Circular primes.
Wikipedia article - Repunit.
OEIS sequence A016114 - Circular primes.
| #Wren | Wren | import "/math" for Int
import "/big" for BigInt
import "/str" for Str
var circs = []
var isCircular = Fn.new { |n|
var nn = n
var pow = 1 // will eventually contain 10 ^ d where d is number of digits in n
while (nn > 0) {
pow = pow * 10
nn = (nn/10).floor
}
nn = n
while (true) {
nn = nn * 10
var f = (nn/pow).floor // first digit
nn = nn + f * (1 - pow)
if (circs.contains(nn)) return false
if (nn == n) break
if (!Int.isPrime(nn)) return false
}
return true
}
var repunit = Fn.new { |n| BigInt.new(Str.repeat("1", n)) }
System.print("The first 19 circular primes are:")
var digits = [1, 3, 7, 9]
var q = [1, 2, 3, 5, 7, 9] // queue the numbers to be examined
var fq = [1, 2, 3, 5, 7, 9] // also queue the corresponding first digits
var count = 0
while (true) {
var f = q[0] // peek first element
var fd = fq[0] // peek first digit
if (Int.isPrime(f) && isCircular.call(f)) {
circs.add(f)
count = count + 1
if (count == 19) break
}
q.removeAt(0) // pop first element
fq.removeAt(0) // ditto for first digit queue
if (f != 2 && f != 5) { // if digits > 1 can't contain a 2 or 5
// add numbers with one more digit to queue
// only numbers whose last digit >= first digit need be added
for (d in digits) {
if (d >= fd) {
q.add(f*10+d)
fq.add(fd)
}
}
}
}
System.print(circs)
System.print("\nThe next 4 circular primes, in repunit format, are:")
count = 0
var rus = []
var primes = Int.primeSieve(10000)
for (p in primes[3..-1]) {
if (repunit.call(p).isProbablePrime(1)) {
rus.add("R(%(p))")
count = count + 1
if (count == 4) break
}
}
System.print(rus) |
http://rosettacode.org/wiki/Circles_of_given_radius_through_two_points | Circles of given radius through two points |
Given two points on a plane and a radius, usually two circles of given radius can be drawn through the points.
Exceptions
r==0.0 should be treated as never describing circles (except in the case where the points are coincident).
If the points are coincident then an infinite number of circles with the point on their circumference can be drawn, unless r==0.0 as well which then collapses the circles to a point.
If the points form a diameter then return two identical circles or return a single circle, according to which is the most natural mechanism for the implementation language.
If the points are too far apart then no circles can be drawn.
Task detail
Write a function/subroutine/method/... that takes two points and a radius and returns the two circles through those points, or some indication of special cases where two, possibly equal, circles cannot be returned.
Show here the output for the following inputs:
p1 p2 r
0.1234, 0.9876 0.8765, 0.2345 2.0
0.0000, 2.0000 0.0000, 0.0000 1.0
0.1234, 0.9876 0.1234, 0.9876 2.0
0.1234, 0.9876 0.8765, 0.2345 0.5
0.1234, 0.9876 0.1234, 0.9876 0.0
Related task
Total circles area.
See also
Finding the Center of a Circle from 2 Points and Radius from Math forum @ Drexel
| #C.2B.2B | C++ |
#include <iostream>
#include <cmath>
#include <tuple>
struct point { double x, y; };
bool operator==(const point& lhs, const point& rhs)
{ return std::tie(lhs.x, lhs.y) == std::tie(rhs.x, rhs.y); }
enum result_category { NONE, ONE_COINCEDENT, ONE_DIAMETER, TWO, INFINITE };
using result_t = std::tuple<result_category, point, point>;
double distance(point l, point r)
{ return std::hypot(l.x - r.x, l.y - r.y); }
result_t find_circles(point p1, point p2, double r)
{
point ans1 { 1/0., 1/0.}, ans2 { 1/0., 1/0.};
if (p1 == p2) {
if(r == 0.) return std::make_tuple(ONE_COINCEDENT, p1, p2 );
else return std::make_tuple(INFINITE, ans1, ans2);
}
point center { p1.x/2 + p2.x/2, p1.y/2 + p2.y/2};
double half_distance = distance(center, p1);
if(half_distance > r) return std::make_tuple(NONE, ans1, ans2);
if(half_distance - r == 0) return std::make_tuple(ONE_DIAMETER, center, ans2);
double root = std::hypot(r, half_distance) / distance(p1, p2);
ans1.x = center.x + root * (p1.y - p2.y);
ans1.y = center.y + root * (p2.x - p1.x);
ans2.x = center.x - root * (p1.y - p2.y);
ans2.y = center.y - root * (p2.x - p1.x);
return std::make_tuple(TWO, ans1, ans2);
}
void print(result_t result, std::ostream& out = std::cout)
{
point r1, r2; result_category res;
std::tie(res, r1, r2) = result;
switch(res) {
case NONE:
out << "There are no solutions, points are too far away\n"; break;
case ONE_COINCEDENT: case ONE_DIAMETER:
out << "Only one solution: " << r1.x << ' ' << r1.y << '\n'; break;
case INFINITE:
out << "Infinitely many circles can be drawn\n"; break;
case TWO:
out << "Two solutions: " << r1.x << ' ' << r1.y << " and " << r2.x << ' ' << r2.y << '\n'; break;
}
}
int main()
{
constexpr int size = 5;
const point points[size*2] = {
{0.1234, 0.9876}, {0.8765, 0.2345}, {0.0000, 2.0000}, {0.0000, 0.0000},
{0.1234, 0.9876}, {0.1234, 0.9876}, {0.1234, 0.9876}, {0.8765, 0.2345},
{0.1234, 0.9876}, {0.1234, 0.9876}
};
const double radius[size] = {2., 1., 2., .5, 0.};
for(int i = 0; i < size; ++i)
print(find_circles(points[i*2], points[i*2 + 1], radius[i]));
} |
http://rosettacode.org/wiki/Chinese_zodiac | Chinese zodiac | Traditionally, the Chinese have counted years using two simultaneous cycles, one of length 10 (the "celestial stems") and one of length 12 (the "terrestrial branches"); the combination results in a repeating 60-year pattern. Mapping the branches to twelve traditional animal deities results in the well-known "Chinese zodiac", assigning each year to a given animal. For example, Tuesday, February 1, 2022 CE (in the common Gregorian calendar) will begin the lunisolar Year of the Tiger.
The celestial stems have no one-to-one mapping like that of the branches to animals; however, the five pairs of consecutive stems each belong to one of the five traditional Chinese elements (Wood, Fire, Earth, Metal, and Water). Further, one of the two years within each element's governance is associated with yin, the other with yang.
Thus, 2022 is also the yang year of Water. Note that since 12 is an even number, the association between animals and yin/yang doesn't change. Consecutive Years of the Rooster will cycle through the five elements, but will always be yin, despite the apparent conceptual mismatch between the specifically-male English animal name and the female aspect denoted by yin.
Task
Create a subroutine or program that will return or output the animal, yin/yang association, and element for the lunisolar year that begins in a given CE year.
You may optionally provide more information in the form of the year's numerical position within the 60-year cycle and/or its actual Chinese stem-branch name (in Han characters or Pinyin transliteration).
Requisite information
The animal cycle runs in this order: Rat, Ox, Tiger, Rabbit, Dragon, Snake, Horse, Goat, Monkey, Rooster, Dog, Pig.
The element cycle runs in this order: Wood, Fire, Earth, Metal, Water.
The yang year precedes the yin year within each element.
The current 60-year cycle began in 1984 CE; the first cycle of the Common Era began in 4 CE.
Thus, 1984 was the year of the Wood Rat (yang), 1985 was the year of the Wood Ox (yin), and 1986 the year of the Fire Tiger (yang); 2022 - which, as already noted, is the year of the Water Tiger (yang) - is the 39th year of the current cycle.
Information for optional task
The ten celestial stems are 甲 jiă, 乙 yĭ, 丙 bĭng, 丁 dīng, 戊 wù, 己 jĭ, 庚 gēng, 辛 xīn, 壬 rén, and 癸 gŭi. With the ASCII version of Pinyin tones, the names are written "jia3", "yi3", "bing3", "ding1", "wu4", "ji3", "geng1", "xin1", "ren2", and "gui3".
The twelve terrestrial branches are 子 zĭ, 丑 chŏu, 寅 yín, 卯 măo, 辰 chén, 巳 sì, 午 wŭ, 未 wèi, 申 shēn, 酉 yŏu, 戌 xū, 亥 hài. In ASCII Pinyin, those are "zi3", "chou3", "yin2", "mao3", "chen2", "si4", "wu3", "wei4", "shen1", "you3", "xu1", and "hai4".
Therefore 1984 was 甲子 (jiă-zĭ, or jia3-zi3). 2022 is 壬寅 (rén-yín or ren2-yin2).
| #C | C | #include <math.h>
#include <stdio.h>
const char* animals[] = { "Rat","Ox","Tiger","Rabbit","Dragon","Snake","Horse","Goat","Monkey","Rooster","Dog","Pig" };
const char* elements[] = { "Wood","Fire","Earth","Metal","Water" };
const char* getElement(int year) {
int element = (int)floor((year - 4) % 10 / 2);
return elements[element];
}
const char* getAnimal(int year) {
return animals[(year - 4) % 12];
}
const char* getYY(int year) {
if (year % 2 == 0) {
return "yang";
} else {
return "yin";
}
}
int main() {
int years[] = { 1935, 1938, 1968, 1972, 1976, 2017 };
int i;
//the zodiac cycle didnt start until 4 CE, so years <4 shouldnt be valid
for (i = 0; i < 6; ++i) {
int year = years[i];
printf("%d is the year of the %s %s (%s).\n", year, getElement(year), getAnimal(year), getYY(year));
}
return 0;
} |
http://rosettacode.org/wiki/Check_that_file_exists | Check that file exists | Task
Verify that a file called input.txt and a directory called docs exist.
This should be done twice:
once for the current working directory, and
once for a file and a directory in the filesystem root.
Optional criteria (May 2015): verify it works with:
zero-length files
an unusual filename: `Abdu'l-Bahá.txt
| #Ada | Ada | with Ada.Text_IO; use Ada.Text_IO;
procedure File_Exists is
function Does_File_Exist (Name : String) return Boolean is
The_File : Ada.Text_IO.File_Type;
begin
Open (The_File, In_File, Name);
Close (The_File);
return True;
exception
when Name_Error =>
return False;
end Does_File_Exist;
begin
Put_Line (Boolean'Image (Does_File_Exist ("input.txt" )));
Put_Line (Boolean'Image (Does_File_Exist ("\input.txt")));
end File_Exists; |
http://rosettacode.org/wiki/Check_that_file_exists | Check that file exists | Task
Verify that a file called input.txt and a directory called docs exist.
This should be done twice:
once for the current working directory, and
once for a file and a directory in the filesystem root.
Optional criteria (May 2015): verify it works with:
zero-length files
an unusual filename: `Abdu'l-Bahá.txt
| #Aikido | Aikido |
function exists (filename) {
return stat (filename) != null
}
exists ("input.txt")
exists ("/input.txt")
exists ("docs")
exists ("/docs")
|
http://rosettacode.org/wiki/Check_output_device_is_a_terminal | Check output device is a terminal | Task
Demonstrate how to check whether the output device is a terminal or not.
Related task
Check input device is a terminal
| #J | J | 3=nc<'wd' |
http://rosettacode.org/wiki/Check_output_device_is_a_terminal | Check output device is a terminal | Task
Demonstrate how to check whether the output device is a terminal or not.
Related task
Check input device is a terminal
| #Javascript.2FNodeJS | Javascript/NodeJS | node -p -e "Boolean(process.stdout.isTTY)"
true |
http://rosettacode.org/wiki/Check_output_device_is_a_terminal | Check output device is a terminal | Task
Demonstrate how to check whether the output device is a terminal or not.
Related task
Check input device is a terminal
| #Julia | Julia |
if isa(STDOUT, Base.TTY)
println("This program sees STDOUT as a TTY.")
else
println("This program does not see STDOUT as a TTY.")
end
|
http://rosettacode.org/wiki/Check_output_device_is_a_terminal | Check output device is a terminal | Task
Demonstrate how to check whether the output device is a terminal or not.
Related task
Check input device is a terminal
| #Kotlin | Kotlin | // Kotlin Native version 0.5
import platform.posix.*
fun main(args: Array<String>) {
if (isatty(STDOUT_FILENO) != 0)
println("stdout is a terminal")
else
println("stdout is not a terminal")
} |
http://rosettacode.org/wiki/Check_output_device_is_a_terminal | Check output device is a terminal | Task
Demonstrate how to check whether the output device is a terminal or not.
Related task
Check input device is a terminal
| #Lua | Lua | local function isTTY ( fd )
fd = tonumber( fd ) or 1
local ok, exit, signal = os.execute( string.format( "test -t %d", fd ) )
return (ok and exit == "exit") and signal == 0 or false
end
print( "stdin", isTTY( 0 ) )
print( "stdout", isTTY( 1 ) )
print( "stderr", isTTY( 2 ) ) |
http://rosettacode.org/wiki/Cholesky_decomposition | Cholesky decomposition | Every symmetric, positive definite matrix A can be decomposed into a product of a unique lower triangular matrix L and its transpose:
A
=
L
L
T
{\displaystyle A=LL^{T}}
L
{\displaystyle L}
is called the Cholesky factor of
A
{\displaystyle A}
, and can be interpreted as a generalized square root of
A
{\displaystyle A}
, as described in Cholesky decomposition.
In a 3x3 example, we have to solve the following system of equations:
A
=
(
a
11
a
21
a
31
a
21
a
22
a
32
a
31
a
32
a
33
)
=
(
l
11
0
0
l
21
l
22
0
l
31
l
32
l
33
)
(
l
11
l
21
l
31
0
l
22
l
32
0
0
l
33
)
≡
L
L
T
=
(
l
11
2
l
21
l
11
l
31
l
11
l
21
l
11
l
21
2
+
l
22
2
l
31
l
21
+
l
32
l
22
l
31
l
11
l
31
l
21
+
l
32
l
22
l
31
2
+
l
32
2
+
l
33
2
)
{\displaystyle {\begin{aligned}A&={\begin{pmatrix}a_{11}&a_{21}&a_{31}\\a_{21}&a_{22}&a_{32}\\a_{31}&a_{32}&a_{33}\\\end{pmatrix}}\\&={\begin{pmatrix}l_{11}&0&0\\l_{21}&l_{22}&0\\l_{31}&l_{32}&l_{33}\\\end{pmatrix}}{\begin{pmatrix}l_{11}&l_{21}&l_{31}\\0&l_{22}&l_{32}\\0&0&l_{33}\end{pmatrix}}\equiv LL^{T}\\&={\begin{pmatrix}l_{11}^{2}&l_{21}l_{11}&l_{31}l_{11}\\l_{21}l_{11}&l_{21}^{2}+l_{22}^{2}&l_{31}l_{21}+l_{32}l_{22}\\l_{31}l_{11}&l_{31}l_{21}+l_{32}l_{22}&l_{31}^{2}+l_{32}^{2}+l_{33}^{2}\end{pmatrix}}\end{aligned}}}
We can see that for the diagonal elements (
l
k
k
{\displaystyle l_{kk}}
) of
L
{\displaystyle L}
there is a calculation pattern:
l
11
=
a
11
{\displaystyle l_{11}={\sqrt {a_{11}}}}
l
22
=
a
22
−
l
21
2
{\displaystyle l_{22}={\sqrt {a_{22}-l_{21}^{2}}}}
l
33
=
a
33
−
(
l
31
2
+
l
32
2
)
{\displaystyle l_{33}={\sqrt {a_{33}-(l_{31}^{2}+l_{32}^{2})}}}
or in general:
l
k
k
=
a
k
k
−
∑
j
=
1
k
−
1
l
k
j
2
{\displaystyle l_{kk}={\sqrt {a_{kk}-\sum _{j=1}^{k-1}l_{kj}^{2}}}}
For the elements below the diagonal (
l
i
k
{\displaystyle l_{ik}}
, where
i
>
k
{\displaystyle i>k}
) there is also a calculation pattern:
l
21
=
1
l
11
a
21
{\displaystyle l_{21}={\frac {1}{l_{11}}}a_{21}}
l
31
=
1
l
11
a
31
{\displaystyle l_{31}={\frac {1}{l_{11}}}a_{31}}
l
32
=
1
l
22
(
a
32
−
l
31
l
21
)
{\displaystyle l_{32}={\frac {1}{l_{22}}}(a_{32}-l_{31}l_{21})}
which can also be expressed in a general formula:
l
i
k
=
1
l
k
k
(
a
i
k
−
∑
j
=
1
k
−
1
l
i
j
l
k
j
)
{\displaystyle l_{ik}={\frac {1}{l_{kk}}}\left(a_{ik}-\sum _{j=1}^{k-1}l_{ij}l_{kj}\right)}
Task description
The task is to implement a routine which will return a lower Cholesky factor
L
{\displaystyle L}
for every given symmetric, positive definite nxn matrix
A
{\displaystyle A}
. You should then test it on the following two examples and include your output.
Example 1:
25 15 -5 5 0 0
15 18 0 --> 3 3 0
-5 0 11 -1 1 3
Example 2:
18 22 54 42 4.24264 0.00000 0.00000 0.00000
22 70 86 62 --> 5.18545 6.56591 0.00000 0.00000
54 86 174 134 12.72792 3.04604 1.64974 0.00000
42 62 134 106 9.89949 1.62455 1.84971 1.39262
Note
The Cholesky decomposition of a Pascal upper-triangle matrix is the Identity matrix of the same size.
The Cholesky decomposition of a Pascal symmetric matrix is the Pascal lower-triangle matrix of the same size. | #Common_Lisp | Common Lisp | ;; Calculates the Cholesky decomposition matrix L
;; for a positive-definite, symmetric nxn matrix A.
(defun chol (A)
(let* ((n (car (array-dimensions A)))
(L (make-array `(,n ,n) :initial-element 0)))
(do ((k 0 (incf k))) ((> k (- n 1)) nil)
;; First, calculate diagonal elements L_kk.
(setf (aref L k k)
(sqrt (- (aref A k k)
(do* ((j 0 (incf j))
(sum (expt (aref L k j) 2)
(incf sum (expt (aref L k j) 2))))
((> j (- k 1)) sum)))))
;; Then, all elements below a diagonal element, L_ik, i=k+1..n.
(do ((i (+ k 1) (incf i)))
((> i (- n 1)) nil)
(setf (aref L i k)
(/ (- (aref A i k)
(do* ((j 0 (incf j))
(sum (* (aref L i j) (aref L k j))
(incf sum (* (aref L i j) (aref L k j)))))
((> j (- k 1)) sum)))
(aref L k k)))))
;; Return the calculated matrix L.
L)) |
http://rosettacode.org/wiki/Check_input_device_is_a_terminal | Check input device is a terminal | Task
Demonstrate how to check whether the input device is a terminal or not.
Related task
Check output device is a terminal
| #Kotlin | Kotlin | // Kotlin Native version 0.5
import platform.posix.*
fun main(args: Array<String>) {
if (isatty(STDIN_FILENO) != 0)
println("stdin is a terminal")
else
println("stdin is not a terminal")
}
|
http://rosettacode.org/wiki/Check_input_device_is_a_terminal | Check input device is a terminal | Task
Demonstrate how to check whether the input device is a terminal or not.
Related task
Check output device is a terminal
| #Nemerle | Nemerle | def isTerm = System.Console.IsInputRedirected; |
http://rosettacode.org/wiki/Check_input_device_is_a_terminal | Check input device is a terminal | Task
Demonstrate how to check whether the input device is a terminal or not.
Related task
Check output device is a terminal
| #Nim | Nim | import terminal
echo if stdin.isatty: "stdin is a terminal" else: "stdin is not a terminal" |
http://rosettacode.org/wiki/Check_input_device_is_a_terminal | Check input device is a terminal | Task
Demonstrate how to check whether the input device is a terminal or not.
Related task
Check output device is a terminal
| #OCaml | OCaml | let () =
print_endline (
if Unix.isatty Unix.stdin
then "Input comes from tty."
else "Input doesn't come from tty."
) |
http://rosettacode.org/wiki/Check_input_device_is_a_terminal | Check input device is a terminal | Task
Demonstrate how to check whether the input device is a terminal or not.
Related task
Check output device is a terminal
| #Ol | Ol |
(define (isatty? fd) (syscall 16 fd 19))
(print (if (isatty? stdin)
"Input comes from tty."
"Input doesn't come from tty."))
|
http://rosettacode.org/wiki/Check_input_device_is_a_terminal | Check input device is a terminal | Task
Demonstrate how to check whether the input device is a terminal or not.
Related task
Check output device is a terminal
| #Perl | Perl | use strict;
use warnings;
use 5.010;
if (-t) {
say "Input comes from tty.";
}
else {
say "Input doesn't come from tty.";
} |
http://rosettacode.org/wiki/Cheryl%27s_birthday | Cheryl's birthday | Albert and Bernard just became friends with Cheryl, and they want to know when her birthday is.
Cheryl gave them a list of ten possible dates:
May 15, May 16, May 19
June 17, June 18
July 14, July 16
August 14, August 15, August 17
Cheryl then tells Albert the month of birth, and Bernard the day (of the month) of birth.
1) Albert: I don't know when Cheryl's birthday is, but I know that Bernard does not know too.
2) Bernard: At first I don't know when Cheryl's birthday is, but I know now.
3) Albert: Then I also know when Cheryl's birthday is.
Task
Write a computer program to deduce, by successive elimination, Cheryl's birthday.
Related task
Sum and Product Puzzle
References
Wikipedia article of the same name.
Tuple Relational Calculus
| #AWK | AWK |
# syntax: GAWK -f CHERYLS_BIRTHDAY.AWK [-v debug={0|1}]
#
# sorting:
# PROCINFO["sorted_in"] is used by GAWK
# SORTTYPE is used by Thompson Automation's TAWK
#
BEGIN {
debug += 0
PROCINFO["sorted_in"] = "@ind_num_asc" ; SORTTYPE = 1
n = split("05/15,05/16,05/19,06/17,06/18,07/14,07/16,08/14,08/15,08/17",arr,",")
for (i=1; i<=n; i++) { # move dates to a more friendly structure
mmdd_arr[arr[i]] = ""
}
print("Cheryl offers these ten MM/DD choices:")
cb_show_dates()
printf("Cheryl then tells Albert her birth 'month' and Bernard her birth 'day'.\n\n")
print("1. Albert: I don't know when Cheryl's birthday is, but I know that Bernard does not know too.")
cb_filter1()
print("2. Bernard: At first I don't know when Cheryl's birthday is, but I know now.")
cb_filter2()
print("3. Albert: Then I also know when Cheryl's birthday is.")
cb_filter3()
exit(0)
}
function cb_filter1( i,j) {
print("deduction: the month cannot have a unique day, leaving:")
cb_load_arrays(4)
for (j in arr1) {
if (arr1[j] == 1) {
if (debug) { printf("unique day %s\n",j) }
arr3[arr2[j]] = ""
}
}
cb_remove_dates()
}
function cb_filter2( i,j) {
print("deduction: the day must be unique, leaving:")
cb_load_arrays(4)
for (j in arr1) {
if (arr1[j] > 1) {
if (debug) { printf("non-unique day %s\n",j) }
arr3[j] = ""
}
}
cb_remove_dates("...")
}
function cb_filter3( i,j) {
print("deduction: the month must be unique, leaving:")
cb_load_arrays(1)
for (j in arr1) {
if (arr1[j] > 1) {
if (debug) { printf("non-unique month %s\n",j) }
arr3[j] = ""
}
}
cb_remove_dates()
}
function cb_load_arrays(col, i,key) {
delete arr1
delete arr2
delete arr3
for (i in mmdd_arr) {
key = substr(i,col,2)
arr1[key]++
arr2[key] = substr(i,1,2)
}
}
function cb_remove_dates(pattern, i,j) {
for (j in arr3) {
for (i in mmdd_arr) {
if (i ~ ("^" pattern j)) {
if (debug) { printf("removing %s\n",i) }
delete mmdd_arr[i]
}
}
}
cb_show_dates()
}
function cb_show_dates( i) {
if (debug) { printf("%d remaining\n",length(mmdd_arr)) }
for (i in mmdd_arr) {
printf("%s ",i)
}
printf("\n\n")
}
|
http://rosettacode.org/wiki/Checkpoint_synchronization | Checkpoint synchronization | The checkpoint synchronization is a problem of synchronizing multiple tasks. Consider a workshop where several workers (tasks) assembly details of some mechanism. When each of them completes his work they put the details together. There is no store, so a worker who finished its part first must wait for others before starting another one. Putting details together is the checkpoint at which tasks synchronize themselves before going their paths apart.
The task
Implement checkpoint synchronization in your language.
Make sure that the solution is race condition-free. Note that a straightforward solution based on events is exposed to race condition. Let two tasks A and B need to be synchronized at a checkpoint. Each signals its event (EA and EB correspondingly), then waits for the AND-combination of the events (EA&EB) and resets its event. Consider the following scenario: A signals EA first and gets blocked waiting for EA&EB. Then B signals EB and loses the processor. Then A is released (both events are signaled) and resets EA. Now if B returns and enters waiting for EA&EB, it gets lost.
When a worker is ready it shall not continue before others finish. A typical implementation bug is when a worker is counted twice within one working cycle causing its premature completion. This happens when the quickest worker serves its cycle two times while the laziest one is lagging behind.
If you can, implement workers joining and leaving.
| #Erlang | Erlang |
-module( checkpoint_synchronization ).
-export( [task/0] ).
task() ->
Pid = erlang:spawn( fun() -> checkpoint_loop([], []) end ),
[erlang:spawn(fun() -> random:seed(X, 1, 0), worker_loop(X, 3, Pid) end) || X <- lists:seq(1, 5)],
erlang:exit( Pid, normal ).
checkpoint_loop( Assemblings, Completes ) ->
receive
{starting, Worker} -> checkpoint_loop( [Worker | Assemblings], Completes );
{done, Worker} ->
New_assemblings = lists:delete( Worker, Assemblings ),
New_completes = checkpoint_loop_release( New_assemblings, [Worker | Completes] ),
checkpoint_loop( New_assemblings, New_completes )
end.
checkpoint_loop_release( [], Completes ) ->
[X ! all_complete || X <- Completes],
[];
checkpoint_loop_release( _Assemblings, Completes ) -> Completes.
worker_loop( _Worker, 0, _Checkpoint ) -> ok;
worker_loop( Worker, N, Checkpoint ) ->
Checkpoint ! {starting, erlang:self()},
io:fwrite( "Worker ~p ~p~n", [Worker, N] ),
timer:sleep( random:uniform(100) ),
Checkpoint ! {done, erlang:self()},
receive
all_complete -> ok
end,
worker_loop( Worker, N - 1, Checkpoint ).
|
http://rosettacode.org/wiki/Collections | Collections | This task has been clarified. Its programming examples are in need of review to ensure that they still fit the requirements of the task.
Collections are abstractions to represent sets of values.
In statically-typed languages, the values are typically of a common data type.
Task
Create a collection, and add a few values to it.
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
| #Forth | Forth | include ffl/car.fs
10 car-create ar \ create a dynamic array with initial size 10
2 0 ar car-set \ ar[0] = 2
3 1 ar car-set \ ar[1] = 3
1 0 ar car-insert \ ar[0] = 1 ar[1] = 2 ar[2] = 3
|
http://rosettacode.org/wiki/Combinations | Combinations | Task
Given non-negative integers m and n, generate all size m combinations of the integers from 0 (zero) to n-1 in sorted order (each combination is sorted and the entire table is sorted).
Example
3 comb 5 is:
0 1 2
0 1 3
0 1 4
0 2 3
0 2 4
0 3 4
1 2 3
1 2 4
1 3 4
2 3 4
If it is more "natural" in your language to start counting from 1 (unity) instead of 0 (zero),
the combinations can be of the integers from 1 to n.
See also
The number of samples of size k from n objects.
With combinations and permutations generation tasks.
Order Unimportant
Order Important
Without replacement
(
n
k
)
=
n
C
k
=
n
(
n
−
1
)
…
(
n
−
k
+
1
)
k
(
k
−
1
)
…
1
{\displaystyle {\binom {n}{k}}=^{n}\operatorname {C} _{k}={\frac {n(n-1)\ldots (n-k+1)}{k(k-1)\dots 1}}}
n
P
k
=
n
⋅
(
n
−
1
)
⋅
(
n
−
2
)
⋯
(
n
−
k
+
1
)
{\displaystyle ^{n}\operatorname {P} _{k}=n\cdot (n-1)\cdot (n-2)\cdots (n-k+1)}
Task: Combinations
Task: Permutations
With replacement
(
n
+
k
−
1
k
)
=
n
+
k
−
1
C
k
=
(
n
+
k
−
1
)
!
(
n
−
1
)
!
k
!
{\displaystyle {\binom {n+k-1}{k}}=^{n+k-1}\operatorname {C} _{k}={(n+k-1)! \over (n-1)!k!}}
n
k
{\displaystyle n^{k}}
Task: Combinations with repetitions
Task: Permutations with repetitions
| #M2000_Interpreter | M2000 Interpreter |
Module Checkit {
Global a$
Document a$
Module Combinations (m as long, n as long){
Module Level (n, s, h) {
If n=1 then {
while Len(s) {
Print h, car(s)
ToClipBoard()
s=cdr(s)
}
} Else {
While len(s) {
call Level n-1, cdr(s), cons(h, car(s))
s=cdr(s)
}
}
Sub ToClipBoard()
local m=each(h)
Local b$=""
While m {
b$+=If$(Len(b$)<>0->" ","")+Format$("{0::-10}",Array(m))
}
b$+=If$(Len(b$)<>0->" ","")+Format$("{0::-10}",Array(s,0))+{
}
a$<=b$ ' assign to global need <=
End Sub
}
If m<1 or n<1 then Error
s=(,)
for i=0 to n-1 {
s=cons(s, (i,))
}
Head=(,)
Call Level m, s, Head
}
Clear a$
Combinations 3, 5
ClipBoard a$
}
Checkit
|
http://rosettacode.org/wiki/Conditional_structures | Conditional structures | Control Structures
These are examples of control structures. You may also be interested in:
Conditional structures
Exceptions
Flow-control structures
Loops
Task
List the conditional structures offered by a programming language. See Wikipedia: conditionals for descriptions.
Common conditional structures include if-then-else and switch.
Less common are arithmetic if, ternary operator and Hash-based conditionals.
Arithmetic if allows tight control over computed gotos, which optimizers have a hard time to figure out.
| #Quackery | Quackery | /O> say "23 is greater than 42 is "
... 23 42 > not if [ say "not " ]
... say "true." cr
...
23 is greater than 42 is not true.
Stack empty.
/O> say "23 is less than 42 is "
... 23 42 < not if [ say "not " ]
... say "true." cr
...
23 is less than 42 is true.
Stack empty.
/O> 23 42 = iff
... [ say "23 is equal to 42." ]
... else
... [ say "23 is different to 42." ]
... cr
...
23 is different to 42.
Stack empty.
/O> 23 42 != iff
... [ say "23 is not equal to 42." ]
... else
... [ say "23 is the same as 42." ]
... cr
...
23 is not equal to 42.
Stack empty. |
http://rosettacode.org/wiki/Chinese_remainder_theorem | Chinese remainder theorem | Suppose
n
1
{\displaystyle n_{1}}
,
n
2
{\displaystyle n_{2}}
,
…
{\displaystyle \ldots }
,
n
k
{\displaystyle n_{k}}
are positive integers that are pairwise co-prime.
Then, for any given sequence of integers
a
1
{\displaystyle a_{1}}
,
a
2
{\displaystyle a_{2}}
,
…
{\displaystyle \dots }
,
a
k
{\displaystyle a_{k}}
, there exists an integer
x
{\displaystyle x}
solving the following system of simultaneous congruences:
x
≡
a
1
(
mod
n
1
)
x
≡
a
2
(
mod
n
2
)
⋮
x
≡
a
k
(
mod
n
k
)
{\displaystyle {\begin{aligned}x&\equiv a_{1}{\pmod {n_{1}}}\\x&\equiv a_{2}{\pmod {n_{2}}}\\&{}\ \ \vdots \\x&\equiv a_{k}{\pmod {n_{k}}}\end{aligned}}}
Furthermore, all solutions
x
{\displaystyle x}
of this system are congruent modulo the product,
N
=
n
1
n
2
…
n
k
{\displaystyle N=n_{1}n_{2}\ldots n_{k}}
.
Task
Write a program to solve a system of linear congruences by applying the Chinese Remainder Theorem.
If the system of equations cannot be solved, your program must somehow indicate this.
(It may throw an exception or return a special false value.)
Since there are infinitely many solutions, the program should return the unique solution
s
{\displaystyle s}
where
0
≤
s
≤
n
1
n
2
…
n
k
{\displaystyle 0\leq s\leq n_{1}n_{2}\ldots n_{k}}
.
Show the functionality of this program by printing the result such that the
n
{\displaystyle n}
's are
[
3
,
5
,
7
]
{\displaystyle [3,5,7]}
and the
a
{\displaystyle a}
's are
[
2
,
3
,
2
]
{\displaystyle [2,3,2]}
.
Algorithm: The following algorithm only applies if the
n
i
{\displaystyle n_{i}}
's are pairwise co-prime.
Suppose, as above, that a solution is required for the system of congruences:
x
≡
a
i
(
mod
n
i
)
f
o
r
i
=
1
,
…
,
k
{\displaystyle x\equiv a_{i}{\pmod {n_{i}}}\quad \mathrm {for} \;i=1,\ldots ,k}
Again, to begin, the product
N
=
n
1
n
2
…
n
k
{\displaystyle N=n_{1}n_{2}\ldots n_{k}}
is defined.
Then a solution
x
{\displaystyle x}
can be found as follows:
For each
i
{\displaystyle i}
, the integers
n
i
{\displaystyle n_{i}}
and
N
/
n
i
{\displaystyle N/n_{i}}
are co-prime.
Using the Extended Euclidean algorithm, we can find integers
r
i
{\displaystyle r_{i}}
and
s
i
{\displaystyle s_{i}}
such that
r
i
n
i
+
s
i
N
/
n
i
=
1
{\displaystyle r_{i}n_{i}+s_{i}N/n_{i}=1}
.
Then, one solution to the system of simultaneous congruences is:
x
=
∑
i
=
1
k
a
i
s
i
N
/
n
i
{\displaystyle x=\sum _{i=1}^{k}a_{i}s_{i}N/n_{i}}
and the minimal solution,
x
(
mod
N
)
{\displaystyle x{\pmod {N}}}
.
| #C.2B.2B | C++ | // Requires C++17
#include <iostream>
#include <numeric>
#include <vector>
#include <execution>
template<typename _Ty> _Ty mulInv(_Ty a, _Ty b) {
_Ty b0 = b;
_Ty x0 = 0;
_Ty x1 = 1;
if (b == 1) {
return 1;
}
while (a > 1) {
_Ty q = a / b;
_Ty amb = a % b;
a = b;
b = amb;
_Ty xqx = x1 - q * x0;
x1 = x0;
x0 = xqx;
}
if (x1 < 0) {
x1 += b0;
}
return x1;
}
template<typename _Ty> _Ty chineseRemainder(std::vector<_Ty> n, std::vector<_Ty> a) {
_Ty prod = std::reduce(std::execution::seq, n.begin(), n.end(), (_Ty)1, [](_Ty a, _Ty b) { return a * b; });
_Ty sm = 0;
for (int i = 0; i < n.size(); i++) {
_Ty p = prod / n[i];
sm += a[i] * mulInv(p, n[i]) * p;
}
return sm % prod;
}
int main() {
vector<int> n = { 3, 5, 7 };
vector<int> a = { 2, 3, 2 };
cout << chineseRemainder(n,a) << endl;
return 0;
} |
http://rosettacode.org/wiki/Chernick%27s_Carmichael_numbers | Chernick's Carmichael numbers | In 1939, Jack Chernick proved that, for n ≥ 3 and m ≥ 1:
U(n, m) = (6m + 1) * (12m + 1) * Product_{i=1..n-2} (2^i * 9m + 1)
is a Carmichael number if all the factors are primes and, for n > 4, m is a multiple of 2^(n-4).
Example
U(3, m) = (6m + 1) * (12m + 1) * (18m + 1)
U(4, m) = U(3, m) * (2^2 * 9m + 1)
U(5, m) = U(4, m) * (2^3 * 9m + 1)
...
U(n, m) = U(n-1, m) * (2^(n-2) * 9m + 1)
The smallest Chernick's Carmichael number with 3 prime factors, is: U(3, 1) = 1729.
The smallest Chernick's Carmichael number with 4 prime factors, is: U(4, 1) = 63973.
The smallest Chernick's Carmichael number with 5 prime factors, is: U(5, 380) = 26641259752490421121.
For n = 5, the smallest number m that satisfy Chernick's conditions, is m = 380, therefore U(5, 380) is the smallest Chernick's Carmichael number with 5 prime factors.
U(5, 380) is a Chernick's Carmichael number because m = 380 is a multiple of 2^(n-4), where n = 5, and the factors { (6*380 + 1), (12*380 + 1), (18*380 + 1), (36*380 + 1), (72*380 + 1) } are all prime numbers.
Task
For n ≥ 3, let a(n) be the smallest Chernick's Carmichael number with n prime factors.
Compute a(n) for n = 3..9.
Optional: find a(10).
Note: it's perfectly acceptable to show the terms in factorized form:
a(3) = 7 * 13 * 19
a(4) = 7 * 13 * 19 * 37
a(5) = 2281 * 4561 * 6841 * 13681 * 27361
...
See also
Jack Chernick, On Fermat's simple theorem (PDF)
OEIS A318646: The least Chernick's "universal form" Carmichael number with n prime factors
Related tasks
Carmichael 3 strong pseudoprimes
| #Python | Python |
"""
Python implementation of
http://rosettacode.org/wiki/Chernick%27s_Carmichael_numbers
"""
# use sympy for prime test
from sympy import isprime
# based on C version
def primality_pretest(k):
if not (k % 3) or not (k % 5) or not (k % 7) or not (k % 11) or not(k % 13) or not (k % 17) or not (k % 19) or not (k % 23):
return (k <= 23)
return True
def is_chernick(n, m):
t = 9 * m
if not primality_pretest(6 * m + 1):
return False
if not primality_pretest(12 * m + 1):
return False
for i in range(1,n-1):
if not primality_pretest((t << i) + 1):
return False
if not isprime(6 * m + 1):
return False
if not isprime(12 * m + 1):
return False
for i in range(1,n - 1):
if not isprime((t << i) + 1):
return False
return True
for n in range(3,10):
if n > 4:
multiplier = 1 << (n - 4)
else:
multiplier = 1
if n > 5:
multiplier *= 5
k = 1
while True:
m = k * multiplier
if is_chernick(n, m):
print("a("+str(n)+") has m = "+str(m))
break
k += 1
|
http://rosettacode.org/wiki/Chowla_numbers | Chowla numbers | Chowla numbers are also known as:
Chowla's function
chowla numbers
the chowla function
the chowla number
the chowla sequence
The chowla number of n is (as defined by Chowla's function):
the sum of the divisors of n excluding unity and n
where n is a positive integer
The sequence is named after Sarvadaman D. S. Chowla, (22 October 1907 ──► 10 December 1995),
a London born Indian American mathematician specializing in number theory.
German mathematician Carl Friedrich Gauss (1777─1855) said:
"Mathematics is the queen of the sciences ─ and number theory is the queen of mathematics".
Definitions
Chowla numbers can also be expressed as:
chowla(n) = sum of divisors of n excluding unity and n
chowla(n) = sum( divisors(n)) - 1 - n
chowla(n) = sum( properDivisors(n)) - 1
chowla(n) = sum(aliquotDivisors(n)) - 1
chowla(n) = aliquot(n) - 1
chowla(n) = sigma(n) - 1 - n
chowla(n) = sigmaProperDivisiors(n) - 1
chowla(a*b) = a + b, if a and b are distinct primes
if chowla(n) = 0, and n > 1, then n is prime
if chowla(n) = n - 1, and n > 1, then n is a perfect number
Task
create a chowla function that returns the chowla number for a positive integer n
Find and display (1 per line) for the 1st 37 integers:
the integer (the index)
the chowla number for that integer
For finding primes, use the chowla function to find values of zero
Find and display the count of the primes up to 100
Find and display the count of the primes up to 1,000
Find and display the count of the primes up to 10,000
Find and display the count of the primes up to 100,000
Find and display the count of the primes up to 1,000,000
Find and display the count of the primes up to 10,000,000
For finding perfect numbers, use the chowla function to find values of n - 1
Find and display all perfect numbers up to 35,000,000
use commas within appropriate numbers
show all output here
Related tasks
totient function
perfect numbers
Proper divisors
Sieve of Eratosthenes
See also
the OEIS entry for A48050 Chowla's function.
| #FreeBASIC | FreeBASIC |
' Chowla_numbers
#include "string.bi"
Dim Shared As Long limite
limite = 10000000
Dim Shared As Boolean c(limite)
Dim As Long count, topenumprimo, a
count = 1
topenumprimo = 100
Dim As Longint p, k, kk, limitenumperfect
limitenumperfect = 35000000
k = 2: kk = 3
Declare Function chowla(Byval n As Longint) As Longint
Declare Sub sieve(Byval limite As Long, c() As Boolean)
Function chowla(Byval n As Longint) As Longint
Dim As Long i, j, r
i = 2
Do While i * i <= n
j = n \ i
If n Mod i = 0 Then
r += i
If i <> j Then r += j
End If
i += 1
Loop
chowla = r
End Function
Sub sieve(Byval limite As Long, c() As Boolean)
Dim As Long i, j
Redim As Boolean c(limite - 1)
i = 3
Do While i * 3 < limite
If Not c(i) Then
If chowla(i) = false Then
j = 3 * i
Do While j < limite
c(j) = true
j += 2 * i
Loop
End If
End If
i += 2
Loop
End Sub
Print "Chowla numbers"
For a = 1 To 37
Print "chowla(" & Trim(Str(a)) & ") = " & Trim(Str(chowla(a)))
Next a
' Si chowla(n) = falso and n > 1 Entonces n es primo
Print: Print "Contando los numeros primos hasta: "
sieve(limite, c())
For a = 3 To limite - 1 Step 2
If Not c(a) Then count += 1
If a = topenumprimo - 1 Then
Print Using "########## hay"; topenumprimo;
Print count
topenumprimo *= 10
End If
Next a
' Si chowla(n) = n - 1 and n > 1 Entonces n es un número perfecto
Print: Print "Buscando numeros perfectos... "
count = 0
Do
p = k * kk : If p > limitenumperfect Then Exit Do
If chowla(p) = p - 1 Then
Print Using "##########,# es un numero perfecto"; p
count += 1
End If
k = kk + 1 : kk += k
Loop
Print: Print "Hay " & count & " numeros perfectos <= " & Format(limitenumperfect, "###############################,#")
Print: Print "Pulsa una tecla para salir"
Sleep
End
|
http://rosettacode.org/wiki/Church_numerals | Church numerals | Task
In the Church encoding of natural numbers, the number N is encoded by a function that applies its first argument N times to its second argument.
Church zero always returns the identity function, regardless of its first argument. In other words, the first argument is not applied to the second argument at all.
Church one applies its first argument f just once to its second argument x, yielding f(x)
Church two applies its first argument f twice to its second argument x, yielding f(f(x))
and each successive Church numeral applies its first argument one additional time to its second argument, f(f(f(x))), f(f(f(f(x)))) ... The Church numeral 4, for example, returns a quadruple composition of the function supplied as its first argument.
Arithmetic operations on natural numbers can be similarly represented as functions on Church numerals.
In your language define:
Church Zero,
a Church successor function (a function on a Church numeral which returns the next Church numeral in the series),
functions for Addition, Multiplication and Exponentiation over Church numerals,
a function to convert integers to corresponding Church numerals,
and a function to convert Church numerals to corresponding integers.
You should:
Derive Church numerals three and four in terms of Church zero and a Church successor function.
use Church numeral arithmetic to obtain the the sum and the product of Church 3 and Church 4,
similarly obtain 4^3 and 3^4 in terms of Church numerals, using a Church numeral exponentiation function,
convert each result back to an integer, and return it or print it to the console.
| #Nim | Nim | import macros, sugar
type
Fn = proc(p: pointer): pointer{.noSideEffect.}
Church = proc(f: Fn): Fn{.noSideEffect.}
MetaChurch = proc(c: Church): Church{.noSideEffect.}
#helpers:
template λfλx(exp): untyped = (f: Fn){.closure.}=>((x: pointer){.closure.}=>exp)
template λcλf(exp): untyped = (c: Church){.closure.}=>((f: Fn){.closure.}=>exp)
macro type_erase(body: untyped): untyped =
let
name = if body[0].kind == nnkPostFix: body[0][1] else: body[0]
typ = body[3][0]
quote do:
`body`
proc `name`(p: pointer): pointer =
template erased: untyped = cast[ptr `typ`](p)[]
erased = erased.`name`
p
macro type_erased(body: untyped): untyped =
let (id1, id2, id3) = (body[0][0][0], body[0][0][1], body[0][1])
quote do:
result = `id3`
result = cast[ptr typeof(`id3`)](
`id1`(`id2`)(result.addr)
)[]
#simple math
func zero*(): Church = λfλx: x
func succ*(c: Church): Church = λfλx: f (c f)x
func `+`*(c, d: Church): Church = λfλx: (c f) (d f)x
func `*`*(c, d: Church): Church = λfλx: c(d f)x
#exponentiation
func metazero(): MetaChurch = λcλf: f
func succ(m: MetaChurch): MetaChurch{.type_erase.} = λcλf: c (m c)f
converter toMeta*(c: Church): MetaChurch = type_erased: c(succ)(metazero())
func `^`*(c: Church, d: MetaChurch): Church = d c
#conversions to/from actual numbers
func incr(x: int): int{.type_erase.} = x+1
func toInt(c: Church): int = type_erased: c(incr)(0)
func toChurch*(x: int): Church = return if x <= 0: zero() else: toChurch(x-1).succ
func `$`*(c: Church): string = $c.toInt
when isMainModule:
let three = zero().succ.succ.succ
let four = 4.toChurch
echo [three+four, three*four, three^four, four^three]
|
http://rosettacode.org/wiki/Church_numerals | Church numerals | Task
In the Church encoding of natural numbers, the number N is encoded by a function that applies its first argument N times to its second argument.
Church zero always returns the identity function, regardless of its first argument. In other words, the first argument is not applied to the second argument at all.
Church one applies its first argument f just once to its second argument x, yielding f(x)
Church two applies its first argument f twice to its second argument x, yielding f(f(x))
and each successive Church numeral applies its first argument one additional time to its second argument, f(f(f(x))), f(f(f(f(x)))) ... The Church numeral 4, for example, returns a quadruple composition of the function supplied as its first argument.
Arithmetic operations on natural numbers can be similarly represented as functions on Church numerals.
In your language define:
Church Zero,
a Church successor function (a function on a Church numeral which returns the next Church numeral in the series),
functions for Addition, Multiplication and Exponentiation over Church numerals,
a function to convert integers to corresponding Church numerals,
and a function to convert Church numerals to corresponding integers.
You should:
Derive Church numerals three and four in terms of Church zero and a Church successor function.
use Church numeral arithmetic to obtain the the sum and the product of Church 3 and Church 4,
similarly obtain 4^3 and 3^4 in terms of Church numerals, using a Church numeral exponentiation function,
convert each result back to an integer, and return it or print it to the console.
| #OCaml | OCaml | (* Using type as suggested in https://stackoverflow.com/questions/43426709/does-ocamls-type-system-prevent-it-from-modeling-church-numerals
This is an explicitly polymorphic type : it says that f must be of type ('a -> 'a) -> 'a -> 'a for any possible a "at same time".
*)
type church_num = { f : 'a. ('a -> 'a) -> 'a -> 'a }
(* Zero means apply f 0 times to x, aka return x *)
let ch_zero : church_num = { f = fun _ -> fun x -> x }
(* One simplifies to just returning the function *)
let ch_one : church_num = { f = fun fn -> fn }
(* The next numeral of a church numeral would apply f one more time *)
let ch_succ (c : church_num) : church_num = { f = fun fn x -> fn (c.f fn x) }
(* Adding m and n is applying f m times and then also n times *)
let ch_add (m : church_num) (n : church_num) : church_num =
{ f = fun fn x -> n.f fn (m.f fn x) }
(* Multiplying is repeated addition : add n, m times *)
let ch_mul (m : church_num) (n : church_num) : church_num =
{ f = fun fn x -> m.f (n.f fn) x }
(* Exp is repeated multiplication : multiply by base, exp times.
However, Church numeral n is in some sense the n'th power of a function f applied to x
So exp base = apply function base to the exp'th power = base^exp.
*)
let ch_exp (base : church_num) (exp : church_num) : church_num =
{ f = fun fn x -> (exp.f base.f) fn x }
(* extended Church functions: *)
(* test function for church zero *)
let ch_is_zero (c : church_num) : church_num =
{ f = fun fn x -> c.f (fun _ -> fun _ -> fun xi -> xi) (* when argument is not ch_zero *)
(fun fi -> fi) (* when argument is ch_zero *) fn x }
(* church predecessor function; reduces function calls by one unless already church zero *)
let ch_pred (c : church_num) : church_num =
{ f = fun fn x -> c.f (fun g h -> h (g fn)) (fun _ -> x) (fun xi -> xi) }
(* church subtraction function; calls predecessor function second argument times on first *)
let ch_sub (m : church_num) (n : church_num) : church_num = n.f ch_pred m
(* church division function; counts number of times divisor can be recursively
subtracted from dividend *)
let ch_div (dvdnd : church_num) (dvsr : church_num) : church_num =
let rec divr n = (fun v -> v.f (fun _ -> (ch_succ (divr v)))
ch_zero) (ch_sub n dvsr)
in divr (ch_succ dvdnd)
(* conversion functions: *)
(* Convert a number to a church_num via recursion *)
let church_of_int (n : int) : church_num =
if n < 0
then raise (Invalid_argument (string_of_int n ^ " is not a natural number"))
else
(* Tail-recursed helper *)
let rec helper n acc =
if n = 0
then acc
else helper (n-1) (ch_succ acc)
in helper n ch_zero
(* Convert a church_num to an int is rather easy! Just +1 n times. *)
let int_of_church (n : church_num) : int = n.f succ 0
(* Now the tasks at hand: *)
(* Derive Church numerals three, four, eleven, and twelve,
in terms of Church zero and a Church successor function *)
let ch_three = church_of_int 3
let ch_four = ch_three |> ch_succ
let ch_eleven = church_of_int 11
let ch_twelve = ch_eleven |> ch_succ
(* Use Church numeral arithmetic to obtain the the sum and the product of Church 3 and Church 4 *)
let ch_7 = ch_add ch_three ch_four
let ch_12 = ch_mul ch_three ch_four
(* Similarly obtain 4^3 and 3^4 in terms of Church numerals, using a Church numeral exponentiation function *)
let ch_64 = ch_exp ch_four ch_three
let ch_81 = ch_exp ch_three ch_four
(* check that ch_is_zero works *)
let ch_1 = ch_is_zero ch_zero
let ch_0 = ch_is_zero ch_three
(* check church predecessor, subtraction, and division, functions work *)
let ch_2 = ch_pred ch_three
let ch_8 = ch_sub ch_eleven ch_three
let ch_3 = ch_div ch_eleven ch_three
let ch_4 = ch_div ch_twelve ch_three
(* Convert each result back to an integer, and return it as a string *)
let result = List.map (fun c -> string_of_int(int_of_church c))
[ ch_three; ch_four; ch_7; ch_12; ch_64; ch_81;
ch_eleven; ch_twelve; ch_1; ch_0; ch_2; ch_8; ch_3; ch_4 ]
|> String.concat "; " |> Printf.sprintf "[ %s ]"
;;
print_endline result |
http://rosettacode.org/wiki/Classes | Classes | In object-oriented programming class is a set (a transitive closure) of types bound by the relation of inheritance. It is said that all types derived from some base type T and the type T itself form a class T.
The first type T from the class T sometimes is called the root type of the class.
A class of types itself, as a type, has the values and operations of its own.
The operations of are usually called methods of the root type.
Both operations and values are called polymorphic.
A polymorphic operation (method) selects an implementation depending on the actual specific type of the polymorphic argument.
The action of choice the type-specific implementation of a polymorphic operation is called dispatch. Correspondingly, polymorphic operations are often called dispatching or virtual.
Operations with multiple arguments and/or the results of the class are called multi-methods.
A further generalization of is the operation with arguments and/or results from different classes.
single-dispatch languages are those that allow only one argument or result to control the dispatch. Usually it is the first parameter, often hidden, so that a prefix notation x.f() is used instead of mathematical f(x).
multiple-dispatch languages allow many arguments and/or results to control the dispatch.
A polymorphic value has a type tag indicating its specific type from the class and the corresponding specific value of that type.
This type is sometimes called the most specific type of a [polymorphic] value.
The type tag of the value is used in order to resolve the dispatch.
The set of polymorphic values of a class is a transitive closure of the sets of values of all types from that class.
In many OO languages
the type of the class of T and T itself are considered equivalent.
In some languages they are distinct (like in Ada).
When class T and T are equivalent, there is no way to distinguish
polymorphic and specific values.
Task
Create a basic class with a method, a constructor, an instance variable and how to instantiate it.
| #ERRE | ERRE | PROGRAM CLASS2_DEMO
CLASS QUADRATO
LOCAL LATO
PROCEDURE GETLATO(L)
LATO=L
END PROCEDURE
PROCEDURE AREA(->A)
A=LATO*LATO
END PROCEDURE
PROCEDURE PERIMETRO(->P)
P=4*LATO
END PROCEDURE
END CLASS
NEW P:QUADRATO,Q:QUADRATO
BEGIN
P_GETLATO(10)
P_AREA(->AREAP)
PRINT(AREAP)
Q_GETLATO(20)
Q_PERIMETRO(->PERIMETROQ)
PRINT(PERIMETROQ)
END PROGRAM
|
http://rosettacode.org/wiki/Classes | Classes | In object-oriented programming class is a set (a transitive closure) of types bound by the relation of inheritance. It is said that all types derived from some base type T and the type T itself form a class T.
The first type T from the class T sometimes is called the root type of the class.
A class of types itself, as a type, has the values and operations of its own.
The operations of are usually called methods of the root type.
Both operations and values are called polymorphic.
A polymorphic operation (method) selects an implementation depending on the actual specific type of the polymorphic argument.
The action of choice the type-specific implementation of a polymorphic operation is called dispatch. Correspondingly, polymorphic operations are often called dispatching or virtual.
Operations with multiple arguments and/or the results of the class are called multi-methods.
A further generalization of is the operation with arguments and/or results from different classes.
single-dispatch languages are those that allow only one argument or result to control the dispatch. Usually it is the first parameter, often hidden, so that a prefix notation x.f() is used instead of mathematical f(x).
multiple-dispatch languages allow many arguments and/or results to control the dispatch.
A polymorphic value has a type tag indicating its specific type from the class and the corresponding specific value of that type.
This type is sometimes called the most specific type of a [polymorphic] value.
The type tag of the value is used in order to resolve the dispatch.
The set of polymorphic values of a class is a transitive closure of the sets of values of all types from that class.
In many OO languages
the type of the class of T and T itself are considered equivalent.
In some languages they are distinct (like in Ada).
When class T and T are equivalent, there is no way to distinguish
polymorphic and specific values.
Task
Create a basic class with a method, a constructor, an instance variable and how to instantiate it.
| #F.23 | F# | type MyClass(init) = // constructor with one argument: init
let mutable var = init // a private instance variable
member x.Method() = // a simple method
var <- var + 1
printfn "%d" var
// create an instance and use it
let myObject = new MyClass(42)
myObject.Method() |
http://rosettacode.org/wiki/Closest-pair_problem | Closest-pair problem |
This page uses content from Wikipedia. The original article was at Closest pair of points problem. 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)
Task
Provide a function to find the closest two points among a set of given points in two dimensions, i.e. to solve the Closest pair of points problem in the planar case.
The straightforward solution is a O(n2) algorithm (which we can call brute-force algorithm); the pseudo-code (using indexes) could be simply:
bruteForceClosestPair of P(1), P(2), ... P(N)
if N < 2 then
return ∞
else
minDistance ← |P(1) - P(2)|
minPoints ← { P(1), P(2) }
foreach i ∈ [1, N-1]
foreach j ∈ [i+1, N]
if |P(i) - P(j)| < minDistance then
minDistance ← |P(i) - P(j)|
minPoints ← { P(i), P(j) }
endif
endfor
endfor
return minDistance, minPoints
endif
A better algorithm is based on the recursive divide&conquer approach, as explained also at Wikipedia's Closest pair of points problem, which is O(n log n); a pseudo-code could be:
closestPair of (xP, yP)
where xP is P(1) .. P(N) sorted by x coordinate, and
yP is P(1) .. P(N) sorted by y coordinate (ascending order)
if N ≤ 3 then
return closest points of xP using brute-force algorithm
else
xL ← points of xP from 1 to ⌈N/2⌉
xR ← points of xP from ⌈N/2⌉+1 to N
xm ← xP(⌈N/2⌉)x
yL ← { p ∈ yP : px ≤ xm }
yR ← { p ∈ yP : px > xm }
(dL, pairL) ← closestPair of (xL, yL)
(dR, pairR) ← closestPair of (xR, yR)
(dmin, pairMin) ← (dR, pairR)
if dL < dR then
(dmin, pairMin) ← (dL, pairL)
endif
yS ← { p ∈ yP : |xm - px| < dmin }
nS ← number of points in yS
(closest, closestPair) ← (dmin, pairMin)
for i from 1 to nS - 1
k ← i + 1
while k ≤ nS and yS(k)y - yS(i)y < dmin
if |yS(k) - yS(i)| < closest then
(closest, closestPair) ← (|yS(k) - yS(i)|, {yS(k), yS(i)})
endif
k ← k + 1
endwhile
endfor
return closest, closestPair
endif
References and further readings
Closest pair of points problem
Closest Pair (McGill)
Closest Pair (UCSB)
Closest pair (WUStL)
Closest pair (IUPUI)
| #IS-BASIC | IS-BASIC | 100 PROGRAM "Closestp.bas"
110 NUMERIC X(1 TO 10),Y(1 TO 10)
120 FOR I=1 TO 10
130 READ X(I),Y(I)
140 PRINT X(I),Y(I)
150 NEXT
160 LET MN=INF
170 FOR I=1 TO 9
180 FOR J=I+1 TO 10
190 LET DSQ=(X(I)-X(J))^2+(Y(I)-Y(J))^2
200 IF DSQ<MN THEN LET MN=DSQ:LET MINI=I:LET MINJ=J
210 NEXT
220 NEXT
230 PRINT "Closest pair is (";X(MINI);",";Y(MINI);") and (";X(MINJ);",";Y(MINJ);")":PRINT "at distance";SQR(MN)
240 DATA 0.654682,0.925557
250 DATA 0.409382,0.619391
260 DATA 0.891663,0.888594
270 DATA 0.716629,0.996200
280 DATA 0.477721,0.946355
290 DATA 0.925092,0.818220
300 DATA 0.624291,0.142924
310 DATA 0.211332,0.221507
320 DATA 0.293786,0.691701
330 DATA 0.839186,0.728260 |
http://rosettacode.org/wiki/Closures/Value_capture | Closures/Value capture | Task
Create a list of ten functions, in the simplest manner possible (anonymous functions are encouraged), such that the function at index i (you may choose to start i from either 0 or 1), when run, should return the square of the index, that is, i 2.
Display the result of running any but the last function, to demonstrate that the function indeed remembers its value.
Goal
Demonstrate how to create a series of independent closures based on the same template but maintain separate copies of the variable closed over.
In imperative languages, one would generally use a loop with a mutable counter variable.
For each function to maintain the correct number, it has to capture the value of the variable at the time it was created, rather than just a reference to the variable, which would have a different value by the time the function was run.
See also: Multiple distinct objects
| #Perl | Perl | my @f = map(sub { $_ * $_ }, 0 .. 9); # @f is an array of subs
print $f[$_](), "\n" for (0 .. 8); # call and print all but last |
http://rosettacode.org/wiki/Closures/Value_capture | Closures/Value capture | Task
Create a list of ten functions, in the simplest manner possible (anonymous functions are encouraged), such that the function at index i (you may choose to start i from either 0 or 1), when run, should return the square of the index, that is, i 2.
Display the result of running any but the last function, to demonstrate that the function indeed remembers its value.
Goal
Demonstrate how to create a series of independent closures based on the same template but maintain separate copies of the variable closed over.
In imperative languages, one would generally use a loop with a mutable counter variable.
For each function to maintain the correct number, it has to capture the value of the variable at the time it was created, rather than just a reference to the variable, which would have a different value by the time the function was run.
See also: Multiple distinct objects
| #Phix | Phix | with javascript_semantics
-- First some generic handling stuff, handles partial_args
-- of any mixture of any length and element types.
sequence closures = {}
function add_closure(integer rid, sequence partial_args)
closures = append(closures,{rid,partial_args})
return length(closures) -- (return an integer id)
end function
function call_closure(integer id, sequence args)
{integer rid, sequence partial_args} = closures[id]
return call_func(rid,partial_args&args)
end function
-- The test routine to be made into a closure, or ten
-- Note that all external references/captured variables must
-- be passed as arguments, and grouped together on the lhs
function square(integer i)
return i*i
end function
-- Create the ten closures as asked for.
-- Here, cids is just {1,2,3,4,5,6,7,8,9,10}, however ids would be more
-- useful for a mixed bag of closures, possibly stored all over the shop.
-- Likewise add_closure could have been a procedure for this demo, but
-- you would probably want the function in a real-world application.
sequence cids = {}
for i=1 to 10 do
--for i=11 to 20 do -- alternative test
cids &= add_closure(routine_id("square"),{i})
end for
-- And finally call em (this loop is blissfully unaware what function
-- it is actually calling, and what partial_arguments it is passing)
for i=1 to 10 do
printf(1," %d",call_closure(cids[i],{}))
end for
|
http://rosettacode.org/wiki/Circular_primes | Circular primes | Definitions
A circular prime is a prime number with the property that the number generated at each intermediate step when cyclically permuting its (base 10) digits will also be prime.
For example:
1193 is a circular prime, since 1931, 9311 and 3119 are all also prime.
Note that a number which is a cyclic permutation of a smaller circular prime is not considered to be itself a circular prime. So 13 is a circular prime, but 31 is not.
A repunit (denoted by R) is a number whose base 10 representation contains only the digit 1.
For example:
R(2) = 11 and R(5) = 11111 are repunits.
Task
Find the first 19 circular primes.
If your language has access to arbitrary precision integer arithmetic, given that they are all repunits, find the next 4 circular primes.
(Stretch) Determine which of the following repunits are probably circular primes: R(5003), R(9887), R(15073), R(25031), R(35317) and R(49081). The larger ones may take a long time to process so just do as many as you reasonably can.
See also
Wikipedia article - Circular primes.
Wikipedia article - Repunit.
OEIS sequence A016114 - Circular primes.
| #XPL0 | XPL0 | func IsPrime(N); \Return 'true' if N > 2 is a prime number
int N, I;
[if (N&1) = 0 \even number\ then return false;
for I:= 3 to sqrt(N) do
[if rem(N/I) = 0 then return false;
I:= I+1;
];
return true;
];
func CircPrime(N0); \Return 'true' if N0 is a circular prime
int N0, N, Digits, Rotation, I, R;
[N:= N0;
Digits:= 0; \count number of digits in N
repeat Digits:= Digits+1;
N:= N/10;
until N = 0;
N:= N0;
for Rotation:= 0 to Digits-1 do
[if not IsPrime(N) then return false;
N:= N/10; \rotate least sig digit into high end
R:= rem(0);
for I:= 0 to Digits-2 do
R:= R*10;
N:= N+R;
if N0 > N then \reject N0 if it has a smaller prime rotation
return false;
];
return true;
];
int Counter, N;
[IntOut(0, 2); ChOut(0, ^ ); \show first circular prime
Counter:= 1;
N:= 3; \remaining primes are odd
loop [if CircPrime(N) then
[IntOut(0, N); ChOut(0, ^ );
Counter:= Counter+1;
if Counter >= 19 then quit;
];
N:= N+2;
];
] |
http://rosettacode.org/wiki/Circular_primes | Circular primes | Definitions
A circular prime is a prime number with the property that the number generated at each intermediate step when cyclically permuting its (base 10) digits will also be prime.
For example:
1193 is a circular prime, since 1931, 9311 and 3119 are all also prime.
Note that a number which is a cyclic permutation of a smaller circular prime is not considered to be itself a circular prime. So 13 is a circular prime, but 31 is not.
A repunit (denoted by R) is a number whose base 10 representation contains only the digit 1.
For example:
R(2) = 11 and R(5) = 11111 are repunits.
Task
Find the first 19 circular primes.
If your language has access to arbitrary precision integer arithmetic, given that they are all repunits, find the next 4 circular primes.
(Stretch) Determine which of the following repunits are probably circular primes: R(5003), R(9887), R(15073), R(25031), R(35317) and R(49081). The larger ones may take a long time to process so just do as many as you reasonably can.
See also
Wikipedia article - Circular primes.
Wikipedia article - Repunit.
OEIS sequence A016114 - Circular primes.
| #Zig | Zig |
const std = @import("std");
const math = std.math;
const heap = std.heap;
const stdout = std.io.getStdOut().writer();
pub fn main() !void {
var arena = heap.ArenaAllocator.init(heap.page_allocator);
defer arena.deinit();
var candidates = std.PriorityQueue(u32).init(&arena.allocator, u32cmp);
defer candidates.deinit();
try stdout.print("The circular primes are:\n", .{});
try stdout.print("{:10}" ** 4, .{ 2, 3, 5, 7 });
var c: u32 = 4;
try candidates.add(0);
while (true) {
var n = candidates.remove();
if (n > 1_000_000)
break;
if (n > 10 and circular(n)) {
try stdout.print("{:10}", .{n});
c += 1;
if (c % 10 == 0)
try stdout.print("\n", .{});
}
try candidates.add(10 * n + 1);
try candidates.add(10 * n + 3);
try candidates.add(10 * n + 7);
try candidates.add(10 * n + 9);
}
try stdout.print("\n", .{});
}
fn u32cmp(a: u32, b: u32) math.Order {
return math.order(a, b);
}
fn circular(n0: u32) bool {
if (!isprime(n0))
return false
else {
var n = n0;
var d = @floatToInt(u32, @log10(@intToFloat(f32, n)));
return while (d > 0) : (d -= 1) {
n = rotate(n);
if (n < n0 or !isprime(n))
break false;
} else true;
}
}
fn rotate(n: u32) u32 {
if (n == 0)
return 0
else {
const d = @floatToInt(u32, @log10(@intToFloat(f32, n))); // digit count - 1
const m = math.pow(u32, 10, d);
return (n % m) * 10 + n / m;
}
}
fn isprime(n: u32) bool {
if (n < 2)
return false;
inline for ([3]u3{ 2, 3, 5 }) |p| {
if (n % p == 0)
return n == p;
}
const wheel235 = [_]u3{
6, 4, 2, 4, 2, 4, 6, 2,
};
var i: u32 = 1;
var f: u32 = 7;
return while (f * f <= n) {
if (n % f == 0)
break false;
f += wheel235[i];
i = (i + 1) & 0x07;
} else true;
}
|
http://rosettacode.org/wiki/Circles_of_given_radius_through_two_points | Circles of given radius through two points |
Given two points on a plane and a radius, usually two circles of given radius can be drawn through the points.
Exceptions
r==0.0 should be treated as never describing circles (except in the case where the points are coincident).
If the points are coincident then an infinite number of circles with the point on their circumference can be drawn, unless r==0.0 as well which then collapses the circles to a point.
If the points form a diameter then return two identical circles or return a single circle, according to which is the most natural mechanism for the implementation language.
If the points are too far apart then no circles can be drawn.
Task detail
Write a function/subroutine/method/... that takes two points and a radius and returns the two circles through those points, or some indication of special cases where two, possibly equal, circles cannot be returned.
Show here the output for the following inputs:
p1 p2 r
0.1234, 0.9876 0.8765, 0.2345 2.0
0.0000, 2.0000 0.0000, 0.0000 1.0
0.1234, 0.9876 0.1234, 0.9876 2.0
0.1234, 0.9876 0.8765, 0.2345 0.5
0.1234, 0.9876 0.1234, 0.9876 0.0
Related task
Total circles area.
See also
Finding the Center of a Circle from 2 Points and Radius from Math forum @ Drexel
| #D | D | import std.stdio, std.typecons, std.math;
class ValueException : Exception {
this(string msg_) pure { super(msg_); }
}
struct V2 { double x, y; }
struct Circle { double x, y, r; }
/**Following explanation at:
http://mathforum.org/library/drmath/view/53027.html
*/
Tuple!(Circle, Circle)
circlesFromTwoPointsAndRadius(in V2 p1, in V2 p2, in double r)
pure in {
assert(r >= 0, "radius can't be negative");
} body {
enum nBits = 40;
if (r.abs < (1.0 / (2.0 ^^ nBits)))
throw new ValueException("radius of zero");
if (feqrel(p1.x, p2.x) >= nBits &&
feqrel(p1.y, p2.y) >= nBits)
throw new ValueException("coincident points give" ~
" infinite number of Circles");
// Delta between points.
immutable d = V2(p2.x - p1.x, p2.y - p1.y);
// Distance between points.
immutable q = sqrt(d.x ^^ 2 + d.y ^^ 2);
if (q > 2.0 * r)
throw new ValueException("separation of points > diameter");
// Halfway point.
immutable h = V2((p1.x + p2.x) / 2, (p1.y + p2.y) / 2);
// Distance along the mirror line.
immutable dm = sqrt(r ^^ 2 - (q / 2) ^^ 2);
return typeof(return)(
Circle(h.x - dm * d.y / q, h.y + dm * d.x / q, r.abs),
Circle(h.x + dm * d.y / q, h.y - dm * d.x / q, r.abs));
}
void main() {
foreach (immutable t; [
tuple(V2(0.1234, 0.9876), V2(0.8765, 0.2345), 2.0),
tuple(V2(0.0000, 2.0000), V2(0.0000, 0.0000), 1.0),
tuple(V2(0.1234, 0.9876), V2(0.1234, 0.9876), 2.0),
tuple(V2(0.1234, 0.9876), V2(0.8765, 0.2345), 0.5),
tuple(V2(0.1234, 0.9876), V2(0.1234, 0.9876), 0.0)]) {
writefln("Through points:\n %s %s and radius %f\n" ~
"You can construct the following circles:", t[]);
try {
writefln(" %s\n %s\n",
circlesFromTwoPointsAndRadius(t[])[]);
} catch (ValueException v)
writefln(" ERROR: %s\n", v.msg);
}
} |
http://rosettacode.org/wiki/Chinese_zodiac | Chinese zodiac | Traditionally, the Chinese have counted years using two simultaneous cycles, one of length 10 (the "celestial stems") and one of length 12 (the "terrestrial branches"); the combination results in a repeating 60-year pattern. Mapping the branches to twelve traditional animal deities results in the well-known "Chinese zodiac", assigning each year to a given animal. For example, Tuesday, February 1, 2022 CE (in the common Gregorian calendar) will begin the lunisolar Year of the Tiger.
The celestial stems have no one-to-one mapping like that of the branches to animals; however, the five pairs of consecutive stems each belong to one of the five traditional Chinese elements (Wood, Fire, Earth, Metal, and Water). Further, one of the two years within each element's governance is associated with yin, the other with yang.
Thus, 2022 is also the yang year of Water. Note that since 12 is an even number, the association between animals and yin/yang doesn't change. Consecutive Years of the Rooster will cycle through the five elements, but will always be yin, despite the apparent conceptual mismatch between the specifically-male English animal name and the female aspect denoted by yin.
Task
Create a subroutine or program that will return or output the animal, yin/yang association, and element for the lunisolar year that begins in a given CE year.
You may optionally provide more information in the form of the year's numerical position within the 60-year cycle and/or its actual Chinese stem-branch name (in Han characters or Pinyin transliteration).
Requisite information
The animal cycle runs in this order: Rat, Ox, Tiger, Rabbit, Dragon, Snake, Horse, Goat, Monkey, Rooster, Dog, Pig.
The element cycle runs in this order: Wood, Fire, Earth, Metal, Water.
The yang year precedes the yin year within each element.
The current 60-year cycle began in 1984 CE; the first cycle of the Common Era began in 4 CE.
Thus, 1984 was the year of the Wood Rat (yang), 1985 was the year of the Wood Ox (yin), and 1986 the year of the Fire Tiger (yang); 2022 - which, as already noted, is the year of the Water Tiger (yang) - is the 39th year of the current cycle.
Information for optional task
The ten celestial stems are 甲 jiă, 乙 yĭ, 丙 bĭng, 丁 dīng, 戊 wù, 己 jĭ, 庚 gēng, 辛 xīn, 壬 rén, and 癸 gŭi. With the ASCII version of Pinyin tones, the names are written "jia3", "yi3", "bing3", "ding1", "wu4", "ji3", "geng1", "xin1", "ren2", and "gui3".
The twelve terrestrial branches are 子 zĭ, 丑 chŏu, 寅 yín, 卯 măo, 辰 chén, 巳 sì, 午 wŭ, 未 wèi, 申 shēn, 酉 yŏu, 戌 xū, 亥 hài. In ASCII Pinyin, those are "zi3", "chou3", "yin2", "mao3", "chen2", "si4", "wu3", "wei4", "shen1", "you3", "xu1", and "hai4".
Therefore 1984 was 甲子 (jiă-zĭ, or jia3-zi3). 2022 is 壬寅 (rén-yín or ren2-yin2).
| #C.23 | C# | using System;
namespace ChineseZodiac {
class Program {
static string[] animals = { "Rat", "Ox", "Tiger", "Rabbit", "Dragon", "Snake", "Horse", "Goat", "Monkey", "Rooster", "Dog", "Pig" };
static string[] elements = { "Wood", "Fire", "Earth", "Metal", "Water" };
static string[] animalChars = { "子", "丑", "寅", "卯", "辰", "巳", "午", "未", "申", "酉", "戌", "亥" };
static string[,] elementChars = { { "甲", "丙", "戊", "庚", "壬" }, { "乙", "丁", "己", "辛", "癸" } };
static string getYY(int year) {
if (year % 2 == 0) {
return "yang";
}
return "yin";
}
static void Main(string[] args) {
Console.OutputEncoding = System.Text.Encoding.UTF8;
int[] years = { 1935, 1938, 1968, 1972, 1976, 1984, 1985, 2017 };
for (int i = 0; i < years.Length; i++) {
int ei = (int)Math.Floor((years[i] - 4.0) % 10 / 2);
int ai = (years[i] - 4) % 12;
Console.WriteLine("{0} is the year of the {1} {2} ({3}). {4}{5}", years[i], elements[ei], animals[ai], getYY(years[i]), elementChars[years[i] % 2, ei], animalChars[(years[i] - 4) % 12]);
}
}
}
} |
http://rosettacode.org/wiki/Check_that_file_exists | Check that file exists | Task
Verify that a file called input.txt and a directory called docs exist.
This should be done twice:
once for the current working directory, and
once for a file and a directory in the filesystem root.
Optional criteria (May 2015): verify it works with:
zero-length files
an unusual filename: `Abdu'l-Bahá.txt
| #ALGOL_68 | ALGOL 68 | # Check files and directories exist #
# check a file exists by attempting to open it for input #
# returns TRUE if the file exists, FALSE otherwise #
PROC file exists = ( STRING file name )BOOL:
IF FILE f;
open( f, file name, stand in channel ) = 0
THEN
# file opened OK so must exist #
close( f );
TRUE
ELSE
# file cannot be opened - assume it does not exist #
FALSE
FI # file exists # ;
# print a suitable messages if the specified file exists #
PROC test file exists = ( STRING name )VOID:
print( ( "file: "
, name
, IF file exists( name ) THEN " does" ELSE " does not" FI
, " exist"
, newline
)
);
# print a suitable messages if the specified directory exists #
PROC test directory exists = ( STRING name )VOID:
print( ( "dir: "
, name
, IF file is directory( name ) THEN " does" ELSE " does not" FI
, " exist"
, newline
)
);
# test the flies and directories mentioned in the task exist or not #
test file exists( "input.txt" );
test file exists( "\input.txt");
test directory exists( "docs" );
test directory exists( "\docs" ) |
http://rosettacode.org/wiki/Check_output_device_is_a_terminal | Check output device is a terminal | Task
Demonstrate how to check whether the output device is a terminal or not.
Related task
Check input device is a terminal
| #Nemerle | Nemerle | def isTerm = System.Console.IsOutputRedirected; |
http://rosettacode.org/wiki/Check_output_device_is_a_terminal | Check output device is a terminal | Task
Demonstrate how to check whether the output device is a terminal or not.
Related task
Check input device is a terminal
| #Nim | Nim | import terminal
stderr.write if stdout.isatty: "stdout is a terminal\n" else: "stdout is not a terminal\n" |
http://rosettacode.org/wiki/Check_output_device_is_a_terminal | Check output device is a terminal | Task
Demonstrate how to check whether the output device is a terminal or not.
Related task
Check input device is a terminal
| #OCaml | OCaml | let () =
print_endline (
if Unix.isatty Unix.stdout
then "Output goes to tty."
else "Output doesn't go to tty."
) |
http://rosettacode.org/wiki/Check_output_device_is_a_terminal | Check output device is a terminal | Task
Demonstrate how to check whether the output device is a terminal or not.
Related task
Check input device is a terminal
| #Ol | Ol |
(define (isatty? fd) (syscall 16 fd 19))
(print (if (isatty? stdout)
"stdout is a tty."
"stdout is not a tty."))
|
http://rosettacode.org/wiki/Character_codes | Character codes |
Task
Given a character value in your language, print its code (could be ASCII code, Unicode code, or whatever your language uses).
Example
The character 'a' (lowercase letter A) has a code of 97 in ASCII (as well as Unicode, as ASCII forms the beginning of Unicode).
Conversely, given a code, print out the corresponding character.
| #11l | 11l | print(‘a’.code) // prints "97"
print(Char(code' 97)) // prints "a" |
http://rosettacode.org/wiki/Cholesky_decomposition | Cholesky decomposition | Every symmetric, positive definite matrix A can be decomposed into a product of a unique lower triangular matrix L and its transpose:
A
=
L
L
T
{\displaystyle A=LL^{T}}
L
{\displaystyle L}
is called the Cholesky factor of
A
{\displaystyle A}
, and can be interpreted as a generalized square root of
A
{\displaystyle A}
, as described in Cholesky decomposition.
In a 3x3 example, we have to solve the following system of equations:
A
=
(
a
11
a
21
a
31
a
21
a
22
a
32
a
31
a
32
a
33
)
=
(
l
11
0
0
l
21
l
22
0
l
31
l
32
l
33
)
(
l
11
l
21
l
31
0
l
22
l
32
0
0
l
33
)
≡
L
L
T
=
(
l
11
2
l
21
l
11
l
31
l
11
l
21
l
11
l
21
2
+
l
22
2
l
31
l
21
+
l
32
l
22
l
31
l
11
l
31
l
21
+
l
32
l
22
l
31
2
+
l
32
2
+
l
33
2
)
{\displaystyle {\begin{aligned}A&={\begin{pmatrix}a_{11}&a_{21}&a_{31}\\a_{21}&a_{22}&a_{32}\\a_{31}&a_{32}&a_{33}\\\end{pmatrix}}\\&={\begin{pmatrix}l_{11}&0&0\\l_{21}&l_{22}&0\\l_{31}&l_{32}&l_{33}\\\end{pmatrix}}{\begin{pmatrix}l_{11}&l_{21}&l_{31}\\0&l_{22}&l_{32}\\0&0&l_{33}\end{pmatrix}}\equiv LL^{T}\\&={\begin{pmatrix}l_{11}^{2}&l_{21}l_{11}&l_{31}l_{11}\\l_{21}l_{11}&l_{21}^{2}+l_{22}^{2}&l_{31}l_{21}+l_{32}l_{22}\\l_{31}l_{11}&l_{31}l_{21}+l_{32}l_{22}&l_{31}^{2}+l_{32}^{2}+l_{33}^{2}\end{pmatrix}}\end{aligned}}}
We can see that for the diagonal elements (
l
k
k
{\displaystyle l_{kk}}
) of
L
{\displaystyle L}
there is a calculation pattern:
l
11
=
a
11
{\displaystyle l_{11}={\sqrt {a_{11}}}}
l
22
=
a
22
−
l
21
2
{\displaystyle l_{22}={\sqrt {a_{22}-l_{21}^{2}}}}
l
33
=
a
33
−
(
l
31
2
+
l
32
2
)
{\displaystyle l_{33}={\sqrt {a_{33}-(l_{31}^{2}+l_{32}^{2})}}}
or in general:
l
k
k
=
a
k
k
−
∑
j
=
1
k
−
1
l
k
j
2
{\displaystyle l_{kk}={\sqrt {a_{kk}-\sum _{j=1}^{k-1}l_{kj}^{2}}}}
For the elements below the diagonal (
l
i
k
{\displaystyle l_{ik}}
, where
i
>
k
{\displaystyle i>k}
) there is also a calculation pattern:
l
21
=
1
l
11
a
21
{\displaystyle l_{21}={\frac {1}{l_{11}}}a_{21}}
l
31
=
1
l
11
a
31
{\displaystyle l_{31}={\frac {1}{l_{11}}}a_{31}}
l
32
=
1
l
22
(
a
32
−
l
31
l
21
)
{\displaystyle l_{32}={\frac {1}{l_{22}}}(a_{32}-l_{31}l_{21})}
which can also be expressed in a general formula:
l
i
k
=
1
l
k
k
(
a
i
k
−
∑
j
=
1
k
−
1
l
i
j
l
k
j
)
{\displaystyle l_{ik}={\frac {1}{l_{kk}}}\left(a_{ik}-\sum _{j=1}^{k-1}l_{ij}l_{kj}\right)}
Task description
The task is to implement a routine which will return a lower Cholesky factor
L
{\displaystyle L}
for every given symmetric, positive definite nxn matrix
A
{\displaystyle A}
. You should then test it on the following two examples and include your output.
Example 1:
25 15 -5 5 0 0
15 18 0 --> 3 3 0
-5 0 11 -1 1 3
Example 2:
18 22 54 42 4.24264 0.00000 0.00000 0.00000
22 70 86 62 --> 5.18545 6.56591 0.00000 0.00000
54 86 174 134 12.72792 3.04604 1.64974 0.00000
42 62 134 106 9.89949 1.62455 1.84971 1.39262
Note
The Cholesky decomposition of a Pascal upper-triangle matrix is the Identity matrix of the same size.
The Cholesky decomposition of a Pascal symmetric matrix is the Pascal lower-triangle matrix of the same size. | #D | D | import std.stdio, std.math, std.numeric;
T[][] cholesky(T)(in T[][] A) pure nothrow /*@safe*/ {
auto L = new T[][](A.length, A.length);
foreach (immutable r, row; L)
row[r + 1 .. $] = 0;
foreach (immutable i; 0 .. A.length)
foreach (immutable j; 0 .. i + 1) {
auto t = dotProduct(L[i][0 .. j], L[j][0 .. j]);
L[i][j] = (i == j) ? (A[i][i] - t) ^^ 0.5 :
(1.0 / L[j][j] * (A[i][j] - t));
}
return L;
}
void main() {
immutable double[][] m1 = [[25, 15, -5],
[15, 18, 0],
[-5, 0, 11]];
writefln("%(%(%2.0f %)\n%)\n", m1.cholesky);
immutable double[][] m2 = [[18, 22, 54, 42],
[22, 70, 86, 62],
[54, 86, 174, 134],
[42, 62, 134, 106]];
writefln("%(%(%2.3f %)\n%)", m2.cholesky);
} |
http://rosettacode.org/wiki/Check_input_device_is_a_terminal | Check input device is a terminal | Task
Demonstrate how to check whether the input device is a terminal or not.
Related task
Check output device is a terminal
| #Phix | Phix | without js -- (no input redirection in a browser!)
printf(1,"stdin:%t, stdout:%t, stderr:%t\n",{isatty(0),isatty(1),isatty(2)})
|
http://rosettacode.org/wiki/Check_input_device_is_a_terminal | Check input device is a terminal | Task
Demonstrate how to check whether the input device is a terminal or not.
Related task
Check output device is a terminal
| #Pike | Pike | void main()
{
if(Stdio.Terminfo.is_tty())
write("Input comes from tty.\n");
else
write("Input doesn't come from tty.\n");
} |
http://rosettacode.org/wiki/Check_input_device_is_a_terminal | Check input device is a terminal | Task
Demonstrate how to check whether the input device is a terminal or not.
Related task
Check output device is a terminal
| #Python | Python | from sys import stdin
if stdin.isatty():
print("Input comes from tty.")
else:
print("Input doesn't come from tty.") |
http://rosettacode.org/wiki/Check_input_device_is_a_terminal | Check input device is a terminal | Task
Demonstrate how to check whether the input device is a terminal or not.
Related task
Check output device is a terminal
| #Quackery | Quackery | [ $ |from sys import stdin
to_stack( 1 if stdin.isatty() else 0)|
python ] is ttyin ( --> b )
ttyin if
[ say "Looks like a teletype." ]
else
[ say "Not a teletype." ] |
http://rosettacode.org/wiki/Check_input_device_is_a_terminal | Check input device is a terminal | Task
Demonstrate how to check whether the input device is a terminal or not.
Related task
Check output device is a terminal
| #Racket | Racket |
(terminal-port? (current-input-port))
|
http://rosettacode.org/wiki/Check_input_device_is_a_terminal | Check input device is a terminal | Task
Demonstrate how to check whether the input device is a terminal or not.
Related task
Check output device is a terminal
| #Raku | Raku | say $*IN.t ?? "Input comes from tty." !! "Input doesn't come from tty."; |
http://rosettacode.org/wiki/Check_input_device_is_a_terminal | Check input device is a terminal | Task
Demonstrate how to check whether the input device is a terminal or not.
Related task
Check output device is a terminal
| #REXX | REXX | /*REXX program determines if input comes from terminal or standard input*/
if queued() then say 'input comes from the terminal.'
else say 'input comes from the (stacked) terminal queue.'
/*stick a fork in it, we're done.*/
|
http://rosettacode.org/wiki/Cheryl%27s_birthday | Cheryl's birthday | Albert and Bernard just became friends with Cheryl, and they want to know when her birthday is.
Cheryl gave them a list of ten possible dates:
May 15, May 16, May 19
June 17, June 18
July 14, July 16
August 14, August 15, August 17
Cheryl then tells Albert the month of birth, and Bernard the day (of the month) of birth.
1) Albert: I don't know when Cheryl's birthday is, but I know that Bernard does not know too.
2) Bernard: At first I don't know when Cheryl's birthday is, but I know now.
3) Albert: Then I also know when Cheryl's birthday is.
Task
Write a computer program to deduce, by successive elimination, Cheryl's birthday.
Related task
Sum and Product Puzzle
References
Wikipedia article of the same name.
Tuple Relational Calculus
| #C | C | #include <stdbool.h>
#include <stdio.h>
char *months[] = {
"ERR", "Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec"
};
struct Date {
int month, day;
bool active;
} dates[] = {
{5,15,true}, {5,16,true}, {5,19,true},
{6,17,true}, {6,18,true},
{7,14,true}, {7,16,true},
{8,14,true}, {8,15,true}, {8,17,true}
};
#define UPPER_BOUND (sizeof(dates) / sizeof(struct Date))
void printRemaining() {
int i, c;
for (i = 0, c = 0; i < UPPER_BOUND; i++) {
if (dates[i].active) {
c++;
}
}
printf("%d remaining.\n", c);
}
void printAnswer() {
int i;
for (i = 0; i < UPPER_BOUND; i++) {
if (dates[i].active) {
printf("%s, %d\n", months[dates[i].month], dates[i].day);
}
}
}
void firstPass() {
// the month cannot have a unique day
int i, j, c;
for (i = 0; i < UPPER_BOUND; i++) {
c = 0;
for (j = 0; j < UPPER_BOUND; j++) {
if (dates[j].day == dates[i].day) {
c++;
}
}
if (c == 1) {
for (j = 0; j < UPPER_BOUND; j++) {
if (!dates[j].active) continue;
if (dates[j].month == dates[i].month) {
dates[j].active = false;
}
}
}
}
}
void secondPass() {
// the day must now be unique
int i, j, c;
for (i = 0; i < UPPER_BOUND; i++) {
if (!dates[i].active) continue;
c = 0;
for (j = 0; j < UPPER_BOUND; j++) {
if (!dates[j].active) continue;
if (dates[j].day == dates[i].day) {
c++;
}
}
if (c > 1) {
for (j = 0; j < UPPER_BOUND; j++) {
if (!dates[j].active) continue;
if (dates[j].day == dates[i].day) {
dates[j].active = false;
}
}
}
}
}
void thirdPass() {
// the month must now be unique
int i, j, c;
for (i = 0; i < UPPER_BOUND; i++) {
if (!dates[i].active) continue;
c = 0;
for (j = 0; j < UPPER_BOUND; j++) {
if (!dates[j].active) continue;
if (dates[j].month == dates[i].month) {
c++;
}
}
if (c > 1) {
for (j = 0; j < UPPER_BOUND; j++) {
if (!dates[j].active) continue;
if (dates[j].month == dates[i].month) {
dates[j].active = false;
}
}
}
}
}
int main() {
printRemaining();
// the month cannot have a unique day
firstPass();
printRemaining();
// the day must now be unique
secondPass();
printRemaining();
// the month must now be unique
thirdPass();
printAnswer();
return 0;
} |
http://rosettacode.org/wiki/Checkpoint_synchronization | Checkpoint synchronization | The checkpoint synchronization is a problem of synchronizing multiple tasks. Consider a workshop where several workers (tasks) assembly details of some mechanism. When each of them completes his work they put the details together. There is no store, so a worker who finished its part first must wait for others before starting another one. Putting details together is the checkpoint at which tasks synchronize themselves before going their paths apart.
The task
Implement checkpoint synchronization in your language.
Make sure that the solution is race condition-free. Note that a straightforward solution based on events is exposed to race condition. Let two tasks A and B need to be synchronized at a checkpoint. Each signals its event (EA and EB correspondingly), then waits for the AND-combination of the events (EA&EB) and resets its event. Consider the following scenario: A signals EA first and gets blocked waiting for EA&EB. Then B signals EB and loses the processor. Then A is released (both events are signaled) and resets EA. Now if B returns and enters waiting for EA&EB, it gets lost.
When a worker is ready it shall not continue before others finish. A typical implementation bug is when a worker is counted twice within one working cycle causing its premature completion. This happens when the quickest worker serves its cycle two times while the laziest one is lagging behind.
If you can, implement workers joining and leaving.
| #FreeBASIC | FreeBASIC | #include "ontimer.bi"
Randomize Timer
Dim Shared As Uinteger nWorkers = 3
Dim Shared As Uinteger tID(nWorkers)
Dim Shared As Integer cnt(nWorkers)
Dim Shared As Integer checked = 0
Sub checkpoint()
Dim As Boolean sync
If checked = 0 Then sync = False
checked += 1
If (sync = False) And (checked = nWorkers) Then
sync = True
Color 14 : Print "--Sync Point--"
checked = 0
End If
End Sub
Sub task(worker As Uinteger)
Redim Preserve cnt(nWorkers)
Select Case cnt(worker)
Case 0
cnt(worker) = Rnd * 3
Color 15 : Print "Worker " & worker & " starting (" & cnt(worker) & " ticks)"
Case -1
Exit Select
Case Else
cnt(worker) -= 1
If cnt(worker) = 0 Then
Color 7 : Print "Worker "; worker; " ready and waiting"
cnt(worker) = -1
checkpoint
cnt(worker) = 0
End If
End Select
End Sub
Sub worker1
task(1)
End Sub
Sub worker2
task(2)
End Sub
Sub worker3
task(3)
End Sub
Do
OnTimer(500, @worker1, 1)
OnTimer(100, @worker2, 1)
OnTimer(900, @worker3, 1)
Sleep 1000
Loop |
http://rosettacode.org/wiki/Collections | Collections | This task has been clarified. Its programming examples are in need of review to ensure that they still fit the requirements of the task.
Collections are abstractions to represent sets of values.
In statically-typed languages, the values are typically of a common data type.
Task
Create a collection, and add a few values to it.
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
| #Fortran | Fortran | REAL A(36) !Declares a one-dimensional array A(1), A(2), ... A(36)
A(1) = 1 !Assigns a value to the first element.
A(2) = 3*A(1) + 5 !The second element gets 8. |
http://rosettacode.org/wiki/Combinations | Combinations | Task
Given non-negative integers m and n, generate all size m combinations of the integers from 0 (zero) to n-1 in sorted order (each combination is sorted and the entire table is sorted).
Example
3 comb 5 is:
0 1 2
0 1 3
0 1 4
0 2 3
0 2 4
0 3 4
1 2 3
1 2 4
1 3 4
2 3 4
If it is more "natural" in your language to start counting from 1 (unity) instead of 0 (zero),
the combinations can be of the integers from 1 to n.
See also
The number of samples of size k from n objects.
With combinations and permutations generation tasks.
Order Unimportant
Order Important
Without replacement
(
n
k
)
=
n
C
k
=
n
(
n
−
1
)
…
(
n
−
k
+
1
)
k
(
k
−
1
)
…
1
{\displaystyle {\binom {n}{k}}=^{n}\operatorname {C} _{k}={\frac {n(n-1)\ldots (n-k+1)}{k(k-1)\dots 1}}}
n
P
k
=
n
⋅
(
n
−
1
)
⋅
(
n
−
2
)
⋯
(
n
−
k
+
1
)
{\displaystyle ^{n}\operatorname {P} _{k}=n\cdot (n-1)\cdot (n-2)\cdots (n-k+1)}
Task: Combinations
Task: Permutations
With replacement
(
n
+
k
−
1
k
)
=
n
+
k
−
1
C
k
=
(
n
+
k
−
1
)
!
(
n
−
1
)
!
k
!
{\displaystyle {\binom {n+k-1}{k}}=^{n+k-1}\operatorname {C} _{k}={(n+k-1)! \over (n-1)!k!}}
n
k
{\displaystyle n^{k}}
Task: Combinations with repetitions
Task: Permutations with repetitions
| #M4 | M4 | divert(-1)
define(`set',`define(`$1[$2]',`$3')')
define(`get',`defn(`$1[$2]')')
define(`setrange',`ifelse(`$3',`',$2,`define($1[$2],$3)`'setrange($1,
incr($2),shift(shift(shift($@))))')')
define(`for',
`ifelse($#,0,``$0'',
`ifelse(eval($2<=$3),1,
`pushdef(`$1',$2)$4`'popdef(`$1')$0(`$1',incr($2),$3,`$4')')')')
define(`show',
`for(`k',0,decr($1),`get(a,k) ')')
define(`chklim',
`ifelse(get(`a',$3),eval($2-($1-$3)),
`chklim($1,$2,decr($3))',
`set(`a',$3,incr(get(`a',$3)))`'for(`k',incr($3),decr($2),
`set(`a',k,incr(get(`a',decr(k))))')`'nextcomb($1,$2)')')
define(`nextcomb',
`show($1)
ifelse(eval(get(`a',0)<$2-$1),1,
`chklim($1,$2,decr($1))')')
define(`comb',
`for(`j',0,decr($1),`set(`a',j,j)')`'nextcomb($1,$2)')
divert
comb(3,5) |
http://rosettacode.org/wiki/Conditional_structures | Conditional structures | Control Structures
These are examples of control structures. You may also be interested in:
Conditional structures
Exceptions
Flow-control structures
Loops
Task
List the conditional structures offered by a programming language. See Wikipedia: conditionals for descriptions.
Common conditional structures include if-then-else and switch.
Less common are arithmetic if, ternary operator and Hash-based conditionals.
Arithmetic if allows tight control over computed gotos, which optimizers have a hard time to figure out.
| #R | R | x <- 0
if(x == 0) print("foo")
x <- 1
if(x == 0) print("foo")
if(x == 0) print("foo") else print("bar") |
http://rosettacode.org/wiki/Chinese_remainder_theorem | Chinese remainder theorem | Suppose
n
1
{\displaystyle n_{1}}
,
n
2
{\displaystyle n_{2}}
,
…
{\displaystyle \ldots }
,
n
k
{\displaystyle n_{k}}
are positive integers that are pairwise co-prime.
Then, for any given sequence of integers
a
1
{\displaystyle a_{1}}
,
a
2
{\displaystyle a_{2}}
,
…
{\displaystyle \dots }
,
a
k
{\displaystyle a_{k}}
, there exists an integer
x
{\displaystyle x}
solving the following system of simultaneous congruences:
x
≡
a
1
(
mod
n
1
)
x
≡
a
2
(
mod
n
2
)
⋮
x
≡
a
k
(
mod
n
k
)
{\displaystyle {\begin{aligned}x&\equiv a_{1}{\pmod {n_{1}}}\\x&\equiv a_{2}{\pmod {n_{2}}}\\&{}\ \ \vdots \\x&\equiv a_{k}{\pmod {n_{k}}}\end{aligned}}}
Furthermore, all solutions
x
{\displaystyle x}
of this system are congruent modulo the product,
N
=
n
1
n
2
…
n
k
{\displaystyle N=n_{1}n_{2}\ldots n_{k}}
.
Task
Write a program to solve a system of linear congruences by applying the Chinese Remainder Theorem.
If the system of equations cannot be solved, your program must somehow indicate this.
(It may throw an exception or return a special false value.)
Since there are infinitely many solutions, the program should return the unique solution
s
{\displaystyle s}
where
0
≤
s
≤
n
1
n
2
…
n
k
{\displaystyle 0\leq s\leq n_{1}n_{2}\ldots n_{k}}
.
Show the functionality of this program by printing the result such that the
n
{\displaystyle n}
's are
[
3
,
5
,
7
]
{\displaystyle [3,5,7]}
and the
a
{\displaystyle a}
's are
[
2
,
3
,
2
]
{\displaystyle [2,3,2]}
.
Algorithm: The following algorithm only applies if the
n
i
{\displaystyle n_{i}}
's are pairwise co-prime.
Suppose, as above, that a solution is required for the system of congruences:
x
≡
a
i
(
mod
n
i
)
f
o
r
i
=
1
,
…
,
k
{\displaystyle x\equiv a_{i}{\pmod {n_{i}}}\quad \mathrm {for} \;i=1,\ldots ,k}
Again, to begin, the product
N
=
n
1
n
2
…
n
k
{\displaystyle N=n_{1}n_{2}\ldots n_{k}}
is defined.
Then a solution
x
{\displaystyle x}
can be found as follows:
For each
i
{\displaystyle i}
, the integers
n
i
{\displaystyle n_{i}}
and
N
/
n
i
{\displaystyle N/n_{i}}
are co-prime.
Using the Extended Euclidean algorithm, we can find integers
r
i
{\displaystyle r_{i}}
and
s
i
{\displaystyle s_{i}}
such that
r
i
n
i
+
s
i
N
/
n
i
=
1
{\displaystyle r_{i}n_{i}+s_{i}N/n_{i}=1}
.
Then, one solution to the system of simultaneous congruences is:
x
=
∑
i
=
1
k
a
i
s
i
N
/
n
i
{\displaystyle x=\sum _{i=1}^{k}a_{i}s_{i}N/n_{i}}
and the minimal solution,
x
(
mod
N
)
{\displaystyle x{\pmod {N}}}
.
| #Clojure | Clojure | (ns test-p.core
(:require [clojure.math.numeric-tower :as math]))
(defn extended-gcd
"The extended Euclidean algorithm
Returns a list containing the GCD and the Bézout coefficients
corresponding to the inputs. "
[a b]
(cond (zero? a) [(math/abs b) 0 1]
(zero? b) [(math/abs a) 1 0]
:else (loop [s 0
s0 1
t 1
t0 0
r (math/abs b)
r0 (math/abs a)]
(if (zero? r)
[r0 s0 t0]
(let [q (quot r0 r)]
(recur (- s0 (* q s)) s
(- t0 (* q t)) t
(- r0 (* q r)) r))))))
(defn chinese_remainder
" Main routine to return the chinese remainder "
[n a]
(let [prod (apply * n)
reducer (fn [sum [n_i a_i]]
(let [p (quot prod n_i) ; p = prod / n_i
egcd (extended-gcd p n_i) ; Extended gcd
inv_p (second egcd)] ; Second item is the inverse
(+ sum (* a_i inv_p p))))
sum-prod (reduce reducer 0 (map vector n a))] ; Replaces the Python for loop to sum
; (map vector n a) is same as
; ; Python's version Zip (n, a)
(mod sum-prod prod))) ; Result line
(def n [3 5 7])
(def a [2 3 2])
(println (chinese_remainder n a))
|
http://rosettacode.org/wiki/Chernick%27s_Carmichael_numbers | Chernick's Carmichael numbers | In 1939, Jack Chernick proved that, for n ≥ 3 and m ≥ 1:
U(n, m) = (6m + 1) * (12m + 1) * Product_{i=1..n-2} (2^i * 9m + 1)
is a Carmichael number if all the factors are primes and, for n > 4, m is a multiple of 2^(n-4).
Example
U(3, m) = (6m + 1) * (12m + 1) * (18m + 1)
U(4, m) = U(3, m) * (2^2 * 9m + 1)
U(5, m) = U(4, m) * (2^3 * 9m + 1)
...
U(n, m) = U(n-1, m) * (2^(n-2) * 9m + 1)
The smallest Chernick's Carmichael number with 3 prime factors, is: U(3, 1) = 1729.
The smallest Chernick's Carmichael number with 4 prime factors, is: U(4, 1) = 63973.
The smallest Chernick's Carmichael number with 5 prime factors, is: U(5, 380) = 26641259752490421121.
For n = 5, the smallest number m that satisfy Chernick's conditions, is m = 380, therefore U(5, 380) is the smallest Chernick's Carmichael number with 5 prime factors.
U(5, 380) is a Chernick's Carmichael number because m = 380 is a multiple of 2^(n-4), where n = 5, and the factors { (6*380 + 1), (12*380 + 1), (18*380 + 1), (36*380 + 1), (72*380 + 1) } are all prime numbers.
Task
For n ≥ 3, let a(n) be the smallest Chernick's Carmichael number with n prime factors.
Compute a(n) for n = 3..9.
Optional: find a(10).
Note: it's perfectly acceptable to show the terms in factorized form:
a(3) = 7 * 13 * 19
a(4) = 7 * 13 * 19 * 37
a(5) = 2281 * 4561 * 6841 * 13681 * 27361
...
See also
Jack Chernick, On Fermat's simple theorem (PDF)
OEIS A318646: The least Chernick's "universal form" Carmichael number with n prime factors
Related tasks
Carmichael 3 strong pseudoprimes
| #Raku | Raku | use Inline::Perl5;
use ntheory:from<Perl5> <:all>;
sub chernick-factors ($n, $m) {
6×$m + 1, 12×$m + 1, |((1 .. $n-2).map: { (1 +< $_) × 9×$m + 1 } )
}
sub chernick-carmichael-number ($n) {
my $multiplier = 1 +< (($n-4) max 0);
my $iterator = $n < 5 ?? (1 .. *) !! (1 .. *).map: * × 5;
$multiplier × $iterator.first: -> $m {
[&&] chernick-factors($n, $m × $multiplier).map: { is_prime($_) }
}
}
for 3 .. 9 -> $n {
my $m = chernick-carmichael-number($n);
my @f = chernick-factors($n, $m);
say "U($n, $m): {[×] @f} = {@f.join(' ⨉ ')}";
} |
http://rosettacode.org/wiki/Chowla_numbers | Chowla numbers | Chowla numbers are also known as:
Chowla's function
chowla numbers
the chowla function
the chowla number
the chowla sequence
The chowla number of n is (as defined by Chowla's function):
the sum of the divisors of n excluding unity and n
where n is a positive integer
The sequence is named after Sarvadaman D. S. Chowla, (22 October 1907 ──► 10 December 1995),
a London born Indian American mathematician specializing in number theory.
German mathematician Carl Friedrich Gauss (1777─1855) said:
"Mathematics is the queen of the sciences ─ and number theory is the queen of mathematics".
Definitions
Chowla numbers can also be expressed as:
chowla(n) = sum of divisors of n excluding unity and n
chowla(n) = sum( divisors(n)) - 1 - n
chowla(n) = sum( properDivisors(n)) - 1
chowla(n) = sum(aliquotDivisors(n)) - 1
chowla(n) = aliquot(n) - 1
chowla(n) = sigma(n) - 1 - n
chowla(n) = sigmaProperDivisiors(n) - 1
chowla(a*b) = a + b, if a and b are distinct primes
if chowla(n) = 0, and n > 1, then n is prime
if chowla(n) = n - 1, and n > 1, then n is a perfect number
Task
create a chowla function that returns the chowla number for a positive integer n
Find and display (1 per line) for the 1st 37 integers:
the integer (the index)
the chowla number for that integer
For finding primes, use the chowla function to find values of zero
Find and display the count of the primes up to 100
Find and display the count of the primes up to 1,000
Find and display the count of the primes up to 10,000
Find and display the count of the primes up to 100,000
Find and display the count of the primes up to 1,000,000
Find and display the count of the primes up to 10,000,000
For finding perfect numbers, use the chowla function to find values of n - 1
Find and display all perfect numbers up to 35,000,000
use commas within appropriate numbers
show all output here
Related tasks
totient function
perfect numbers
Proper divisors
Sieve of Eratosthenes
See also
the OEIS entry for A48050 Chowla's function.
| #Go | Go | package main
import "fmt"
func chowla(n int) int {
if n < 1 {
panic("argument must be a positive integer")
}
sum := 0
for i := 2; i*i <= n; i++ {
if n%i == 0 {
j := n / i
if i == j {
sum += i
} else {
sum += i + j
}
}
}
return sum
}
func sieve(limit int) []bool {
// True denotes composite, false denotes prime.
// Only interested in odd numbers >= 3
c := make([]bool, limit)
for i := 3; i*3 < limit; i += 2 {
if !c[i] && chowla(i) == 0 {
for j := 3 * i; j < limit; j += 2 * i {
c[j] = true
}
}
}
return c
}
func commatize(n int) string {
s := fmt.Sprintf("%d", n)
le := len(s)
for i := le - 3; i >= 1; i -= 3 {
s = s[0:i] + "," + s[i:]
}
return s
}
func main() {
for i := 1; i <= 37; i++ {
fmt.Printf("chowla(%2d) = %d\n", i, chowla(i))
}
fmt.Println()
count := 1
limit := int(1e7)
c := sieve(limit)
power := 100
for i := 3; i < limit; i += 2 {
if !c[i] {
count++
}
if i == power-1 {
fmt.Printf("Count of primes up to %-10s = %s\n", commatize(power), commatize(count))
power *= 10
}
}
fmt.Println()
count = 0
limit = 35000000
for i := uint(2); ; i++ {
p := 1 << (i - 1) * (1<<i - 1) // perfect numbers must be of this form
if p > limit {
break
}
if chowla(p) == p-1 {
fmt.Printf("%s is a perfect number\n", commatize(p))
count++
}
}
fmt.Println("There are", count, "perfect numbers <= 35,000,000")
} |
http://rosettacode.org/wiki/Church_numerals | Church numerals | Task
In the Church encoding of natural numbers, the number N is encoded by a function that applies its first argument N times to its second argument.
Church zero always returns the identity function, regardless of its first argument. In other words, the first argument is not applied to the second argument at all.
Church one applies its first argument f just once to its second argument x, yielding f(x)
Church two applies its first argument f twice to its second argument x, yielding f(f(x))
and each successive Church numeral applies its first argument one additional time to its second argument, f(f(f(x))), f(f(f(f(x)))) ... The Church numeral 4, for example, returns a quadruple composition of the function supplied as its first argument.
Arithmetic operations on natural numbers can be similarly represented as functions on Church numerals.
In your language define:
Church Zero,
a Church successor function (a function on a Church numeral which returns the next Church numeral in the series),
functions for Addition, Multiplication and Exponentiation over Church numerals,
a function to convert integers to corresponding Church numerals,
and a function to convert Church numerals to corresponding integers.
You should:
Derive Church numerals three and four in terms of Church zero and a Church successor function.
use Church numeral arithmetic to obtain the the sum and the product of Church 3 and Church 4,
similarly obtain 4^3 and 3^4 in terms of Church numerals, using a Church numeral exponentiation function,
convert each result back to an integer, and return it or print it to the console.
| #Octave | Octave |
zero = @(f) @(x) x;
succ = @(n) @(f) @(x) f(n(f)(x));
add = @(m, n) @(f) @(x) m(f)(n(f)(x));
mul = @(m, n) @(f) @(x) m(n(f))(x);
pow = @(b, e) e(b);
% Need a short-circuiting ternary
iif = @(varargin) varargin{3 - varargin{1}}();
% Helper for anonymous recursion
% The branches are thunked to prevent infinite recursion
to_church_ = @(f, i) iif(i == 0, @() zero, @() succ(f(f, i - 1)));
to_church = @(i) to_church_(to_church_, i);
to_int = @(c) c(@(n) n + 1)(0);
three = succ(succ(succ(zero)));
four = succ(succ(succ(succ(zero))));
cellfun(to_int, {
add(three, four),
mul(three, four),
pow(three, four),
pow(four, three)}) |
http://rosettacode.org/wiki/Classes | Classes | In object-oriented programming class is a set (a transitive closure) of types bound by the relation of inheritance. It is said that all types derived from some base type T and the type T itself form a class T.
The first type T from the class T sometimes is called the root type of the class.
A class of types itself, as a type, has the values and operations of its own.
The operations of are usually called methods of the root type.
Both operations and values are called polymorphic.
A polymorphic operation (method) selects an implementation depending on the actual specific type of the polymorphic argument.
The action of choice the type-specific implementation of a polymorphic operation is called dispatch. Correspondingly, polymorphic operations are often called dispatching or virtual.
Operations with multiple arguments and/or the results of the class are called multi-methods.
A further generalization of is the operation with arguments and/or results from different classes.
single-dispatch languages are those that allow only one argument or result to control the dispatch. Usually it is the first parameter, often hidden, so that a prefix notation x.f() is used instead of mathematical f(x).
multiple-dispatch languages allow many arguments and/or results to control the dispatch.
A polymorphic value has a type tag indicating its specific type from the class and the corresponding specific value of that type.
This type is sometimes called the most specific type of a [polymorphic] value.
The type tag of the value is used in order to resolve the dispatch.
The set of polymorphic values of a class is a transitive closure of the sets of values of all types from that class.
In many OO languages
the type of the class of T and T itself are considered equivalent.
In some languages they are distinct (like in Ada).
When class T and T are equivalent, there is no way to distinguish
polymorphic and specific values.
Task
Create a basic class with a method, a constructor, an instance variable and how to instantiate it.
| #Factor | Factor | TUPLE: my-class foo bar baz ;
M: my-class quux foo>> 20 + ;
C: <my-class> my-class
10 20 30 <my-class> quux ! result: 30
TUPLE: my-child-class < my-class quxx ;
C: <my-child-class> my-child-class
M: my-child-class foobar 20 >>quux ;
20 20 30 <my-child-class> foobar quux ! result: 30 |
http://rosettacode.org/wiki/Classes | Classes | In object-oriented programming class is a set (a transitive closure) of types bound by the relation of inheritance. It is said that all types derived from some base type T and the type T itself form a class T.
The first type T from the class T sometimes is called the root type of the class.
A class of types itself, as a type, has the values and operations of its own.
The operations of are usually called methods of the root type.
Both operations and values are called polymorphic.
A polymorphic operation (method) selects an implementation depending on the actual specific type of the polymorphic argument.
The action of choice the type-specific implementation of a polymorphic operation is called dispatch. Correspondingly, polymorphic operations are often called dispatching or virtual.
Operations with multiple arguments and/or the results of the class are called multi-methods.
A further generalization of is the operation with arguments and/or results from different classes.
single-dispatch languages are those that allow only one argument or result to control the dispatch. Usually it is the first parameter, often hidden, so that a prefix notation x.f() is used instead of mathematical f(x).
multiple-dispatch languages allow many arguments and/or results to control the dispatch.
A polymorphic value has a type tag indicating its specific type from the class and the corresponding specific value of that type.
This type is sometimes called the most specific type of a [polymorphic] value.
The type tag of the value is used in order to resolve the dispatch.
The set of polymorphic values of a class is a transitive closure of the sets of values of all types from that class.
In many OO languages
the type of the class of T and T itself are considered equivalent.
In some languages they are distinct (like in Ada).
When class T and T are equivalent, there is no way to distinguish
polymorphic and specific values.
Task
Create a basic class with a method, a constructor, an instance variable and how to instantiate it.
| #Falcon | Falcon | class class_name[ ( param_list ) ] [ from inh1[, inh2, ..., inhN] ]
[ static block ]
[ properties declaration ]
[init block]
[method list]
end |
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