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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.
| #Oz | Oz | declare
class Something
feat
name %% immutable, public attribute (called a "feature")
attr
count %% mutable, private attribute
%% public method which is used as an initializer
meth init(N)
self.name = N
count := 0
end
%% public method
meth increase
count := @count + 1
end
end
in
%% create an instance
Object = {New Something init("object")}
%% call a method
{Object increase} |
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)
| #Raku | Raku | sub MAIN ($N = 5000) {
my @points = (^$N).map: { [rand × 20 - 10, rand × 20 - 10] }
printf "%.8f between (%.5f, %.5f), (%.5f, %.5f)\n", $_[2], @($_[1]), @($_[0]) for
closest-pair(@points), closest-pair-simple(@points)
}
sub dist-squared(@a, @b) { (@a[0] - @b[0])² + (@a[1] - @b[1])² }
sub closest-pair-simple(@points is copy) {
return ∞ if @points < 2;
my ($a, $b, $d) = |@points[0,1], dist-squared(|@points[0,1]);
while @points {
my \p = pop @points;
for @points -> \l {
($a, $b, $d) = p, l, $_ if $_ < $d given dist-squared(p, l);
}
}
$a, $b, $d.sqrt
}
sub closest-pair(@r) {
closest-pair-real (@r.sort: *.[0]), (@r.sort: *.[1])
}
sub closest-pair-real(@rx, @ry) {
return closest-pair-simple(@rx) if @rx ≤ 3;
my \N = @rx;
my \midx = ceiling(N/2) - 1;
my @PL := @rx[ 0 .. midx];
my @PR := @rx[midx+1 ..^ N ];
my \xm = @rx[midx;0];
(.[0] ≤ xm ?? my @yR !! my @yL).push: @$_ for @ry;
my (\al, \bl, \dL) = closest-pair-real(@PL, @yR);
my (\ar, \br, \dR) = closest-pair-real(@PR, @yL);
my ($w1, $w2, $closest) = dR < dL ?? (ar, br, dR) !! (al, bl, dL);
my @yS = @ry.grep: { (xm - .[0]).abs < $closest }
for 0 ..^ @yS -> \i {
for i+1 ..^ @yS -> \k {
next unless @yS[k;1] - @yS[i;1] < $closest;
($w1, $w2, $closest) = |@yS[k, i], $_ if $_ < $closest given dist-squared(|@yS[k, i]).sqrt;
}
}
$w1, $w2, $closest
} |
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
| #OCaml | OCaml |
(* Task : Circles of given radius through two points *)
(* Types to make code even more readable *)
type point = float * float
type radius = float
type circle = Circle of radius * point
type circ_output =
NoSolution
| OneSolution of circle
| TwoSolutions of circle * circle
| InfiniteSolutions
;;
(* Actual function *)
let circles_2points_radius (x1, y1 : point) (x2, y2 : point) (r : radius) =
let (dx, dy) = (x2 -. x1, y2 -. y1) in
let dist_sq = dx *. dx +. dy *. dy in
match dist_sq, r with
(* Edge case - point circles *)
| 0., 0. -> OneSolution (Circle (r, (x1, y1)))
(* Edge case - coinciding points *)
| 0., _ -> InfiniteSolutions
| _ ->
let side_len_sq = r *. r -. dist_sq /. 4. in
let midp = ((x1 +. x2) *. 0.5, (y1 +. y2) *. 0.5) in
(* Points are too far apart; same whether r = 0 or not *)
if side_len_sq < 0. then NoSolution
(* Points are on diameter *)
else if side_len_sq = 0. then OneSolution (Circle (r, midp))
else
(* A right-angle triangle is made with the radius as hyp, dist/2 as one side *)
let side_len = sqrt (r *. r -. dist_sq /. 4.) in
let dist = sqrt dist_sq in
(* A 90-deg rotation of a vector (x, y) is obtained by either (y, -x) or (-y, x)
We need both, so pick one and the other is its negative.
*)
let (vx, vy) = (-. dy *. side_len /. dist, dx *. side_len /. dist) in
let c1 = Circle (r, (fst midp +. vx, snd midp +. vy)) in
let c2 = Circle (r, (fst midp -. vx, snd midp -. vy)) in
TwoSolutions (c1, c2)
;;
(* Relevant tests and printing *)
let tests = [
(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;
] ;;
let format_output (out : circ_output) = match out with
| NoSolution -> print_endline "No solution"
| OneSolution (Circle (_, (x, y))) -> Printf.printf "One solution: (%.6f, %.6f)\n" x y
| TwoSolutions (Circle (_, (x1, y1)), Circle (_, (x2, y2))) ->
Printf.printf "Two solutions: (%.6f, %.6f) and (%.6f, %.6f)\n" x1 y1 x2 y2
| InfiniteSolutions -> print_endline "Infinite solutions"
;;
let _ =
List.iter
(fun (a, b, c) -> circles_2points_radius a b c |> format_output)
tests
;;
|
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).
| #Prolog | Prolog |
:- initialization(main).
animals(['Rat', 'Ox', 'Tiger', 'Rabbit', 'Dragon', 'Snake', 'Horse', 'Goat', 'Monkey', 'Rooster', 'Dog', 'Pig']).
elements(['Wood', 'Fire', 'Earth', 'Metal', 'Water']).
animal_chars(['子','丑','寅','卯','辰','巳','午','未','申','酉','戌','亥']).
element_chars([['甲', '丙', '戊', '庚', '壬'], ['乙', '丁', '己', '辛', '癸']]).
years([1935, 1938, 1968, 1972, 1976, 1984, 1985, 2017]).
year_animal(Year, Animal) :-
I is ((Year - 4) mod 12) + 1,
animals(Animals),
nth(I, Animals, Animal).
year_element(Year, Element) :-
I is ((Year - 4) mod 10) div 2 + 1,
elements(Elements),
nth(I, Elements, Element).
year_animal_char(Year, AnimalChar) :-
I is (Year - 4) mod 12 + 1,
animal_chars(AnimalChars),
nth(I, AnimalChars, AnimalChar).
year_element_char(Year, ElementChar) :-
I1 is Year mod 2 + 1,
element_chars(ElementChars),
nth(I1, ElementChars, ElementChars1),
I2 is (Year - 4) mod 10 div 2 + 1,
nth(I2, ElementChars1, ElementChar).
year_yin_yang(Year, YinYang) :-
Year mod 2 =:= 0 -> YinYang = 'yang' ; YinYang = 'yin'.
main :-
years(Years),
forall(member(Year, Years), (
write(Year),
write(' is the year of the '),
year_element(Year, Element),
write(Element),
write(' '),
year_animal(Year, Animal),
write(Animal),
write(' '),
year_yin_yang(Year, YinYang),
write('('),
write(YinYang),
write('). '),
year_element_char(Year, ElementChar),
write(ElementChar),
year_animal_char(Year, AnimalChar),
write(AnimalChar),
nl
)).
|
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
| #Haskell | Haskell | import System.Directory (doesFileExist, doesDirectoryExist)
check :: (FilePath -> IO Bool) -> FilePath -> IO ()
check p s = do
result <- p s
putStrLn $
s ++
if result
then " does exist"
else " does not exist"
main :: IO ()
main = do
check doesFileExist "input.txt"
check doesDirectoryExist "docs"
check doesFileExist "/input.txt"
check doesDirectoryExist "/docs" |
http://rosettacode.org/wiki/Chaos_game | Chaos game | The Chaos Game is a method of generating the attractor of an iterated function system (IFS).
One of the best-known and simplest examples creates a fractal, using a polygon and an initial point selected at random.
Task
Play the Chaos Game using the corners of an equilateral triangle as the reference points. Add a starting point at random (preferably inside the triangle). Then add the next point halfway between the starting point and one of the reference points. This reference point is chosen at random.
After a sufficient number of iterations, the image of a Sierpinski Triangle should emerge.
See also
The Game of Chaos
| #PARI.2FGP | PARI/GP |
\\ Chaos Game (Sierpinski triangle) 2/15/17 aev
pChaosGameS3(size,lim)={
my(sz1=size\2,sz2=sz1*sqrt(3),M=matrix(size,size),x,y,xf,yf,v);
x=random(size); y=random(sz2);
for(i=1,lim, v=random(3);
if(v==0, x/=2; y/=2;);
if(v==1, x=sz1+(sz1-x)/2; y=sz2-(sz2-y)/2;);
if(v==2, x=size-(size-x)/2; y/=2;);
xf=floor(x); yf=floor(y); if(xf<1||xf>size||yf<1||yf>size, next);
M[xf,yf]=1;
);\\fend
plotmat(M);
}
\\ Test:
pChaosGameS3(600,30000); \\ SierpTri1.png
|
http://rosettacode.org/wiki/Chat_server | Chat server | Task
Write a server for a minimal text based chat.
People should be able to connect via ‘telnet’, sign on with a nickname, and type messages which will then be seen by all other connected users. Arrivals and departures of chat members should generate appropriate notification messages.
| #Racket | Racket |
#lang racket
(define outs (list (current-output-port)))
(define ((tell-all who o) line)
(for ([c outs] #:unless (eq? o c)) (displayln (~a who ": " line) c)))
(define ((client i o))
(define nick (begin (display "Nick: " o) (read-line i)))
(define tell (tell-all nick o))
(let loop ([line "(joined)"])
(if (eof-object? line)
(begin (tell "(left)") (set! outs (remq o outs)) (close-output-port o))
(begin (tell line) (loop (read-line i))))))
(define (chat-server listener)
(define-values [i o] (tcp-accept listener))
(for ([p (list i o)]) (file-stream-buffer-mode p 'none))
(thread (client i o)) (set! outs (cons o outs)) (chat-server listener))
(void (thread (λ() (chat-server (tcp-listen 12321)))))
((client (current-input-port) (current-output-port)))
|
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.
| #Elixir | Elixir | iex(1)> code = ?a
97
iex(2)> to_string([code])
"a" |
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.
| #Emacs_Lisp | Emacs Lisp | (string-to-char "a") ;=> 97
(format "%c" 97) ;=> "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. | #PL.2FI | PL/I | (subscriptrange):
decompose: procedure options (main); /* 31 October 2013 */
declare a(*,*) float controlled;
allocate a(3,3) initial (25, 15, -5,
15, 18, 0,
-5, 0, 11);
put skip list ('Original matrix:');
put edit (a) (skip, 3 f(4));
call cholesky(a);
put skip list ('Decomposed matrix');
put edit (a) (skip, 3 f(4));
free a;
allocate a(4,4) initial (18, 22, 54, 42,
22, 70, 86, 62,
54, 86, 174, 134,
42, 62, 134, 106);
put skip list ('Original matrix:');
put edit (a) (skip, (hbound(a,1)) f(12) );
call cholesky(a);
put skip list ('Decomposed matrix');
put edit (a) (skip, (hbound(a,1)) f(12,5) );
cholesky: procedure(a);
declare a(*,*) float;
declare L(hbound(a,1), hbound(a,2)) float;
declare s float;
declare (i, j, k) fixed binary;
L = 0;
do i = lbound(a,1) to hbound(a,1);
do j = lbound(a,2) to i;
s = 0;
do k = lbound(a,2) to j-1;
s = s + L(i,k) * L(j,k);
end;
if i = j then
L(i,j) = sqrt(a(i,i) - s);
else
L(i,j) = (a(i,j) - s) / L(j,j);
end;
end;
a = L;
end cholesky;
end decompose; |
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
| #Vlang | Vlang | import time
struct Birthday {
month int
day int
}
fn (b Birthday) str() string {
return "${time.long_months[b.month-1]} $b.day"
}
fn (b Birthday) month_uniquie_in(bds []Birthday) bool {
mut count := 0
for bd in bds {
if bd.month == b.month {
count++
}
}
if count == 1 {
return true
}
return false
}
fn (b Birthday) day_unique_in(bds []Birthday) bool {
mut count := 0
for bd in bds {
if bd.day == b.day {
count++
}
}
if count == 1 {
return true
}
return false
}
fn (b Birthday) month_with_unique_day_in(bds []Birthday) bool {
for bd in bds {
if bd.month == b.month && bd.day_unique_in(bds) {
return true
}
}
return false
}
fn main() {
choices := [
Birthday{5, 15}, Birthday{5, 16}, Birthday{5, 19}, Birthday{6, 17}, Birthday{6, 18},
Birthday{7, 14}, Birthday{7, 16}, Birthday{8, 14}, Birthday{8, 15}, Birthday{8, 17},
]
// Albert knows the month but doesn't know the day.
// So the month can't be unique within the choices.
mut filtered := []Birthday{}
for bd in choices {
if !bd.month_uniquie_in(choices) {
filtered << bd
}
}
// Albert also knows that Bernard doesn't know the answer.
// So the month can't have a unique day.
mut filtered2 := []Birthday{}
for bd in filtered {
if !bd.month_with_unique_day_in(filtered) {
filtered2 << bd
}
}
// Bernard now knows the answer.
// So the day must be unique within the remaining choices.
mut filtered3 := []Birthday{}
for bd in filtered2 {
if bd.day_unique_in(filtered2) {
filtered3 << bd
}
}
// Albert now knows the answer too.
// So the month must be unique within the remaining choices.
mut filtered4 := []Birthday{}
for bd in filtered3 {
if bd.month_uniquie_in(filtered3) {
filtered4 << bd
}
}
if filtered4.len == 1 {
println("Cheryl's Birthday is ${filtered4[0]}")
} else {
println("Something went wrong!")
}
} |
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
| #Wren | Wren | var Months = [
"January", "February", "March", "April", "May", "June",
"July", "August", "September", "October", "November", "December"
]
class Birthday {
construct new(month, day) {
_month = month
_day = day
}
month { _month }
day { _day }
toString { "%(Months[_month-1]) %(day)" }
monthUniqueIn(bds) { bds.count { |bd| _month == bd.month } == 1 }
dayUniqueIn(bds) { bds.count { |bd| _day == bd.day } == 1 }
monthWithUniqueDayIn(bds) { bds.any { |bd| (_month == bd.month) && bd.dayUniqueIn(bds) } }
}
var choices = [
Birthday.new(5, 15), Birthday.new(5, 16), Birthday.new(5, 19), Birthday.new(6, 17),
Birthday.new(6, 18), Birthday.new(7, 14), Birthday.new(7, 16), Birthday.new(8, 14),
Birthday.new(8, 15), Birthday.new(8, 17)
]
// Albert knows the month but doesn't know the day.
// So the month can't be unique within the choices.
var filtered = choices.where { |bd| !bd.monthUniqueIn(choices) }.toList
// Albert also knows that Bernard doesn't know the answer.
// So the month can't have a unique day.
filtered = filtered.where { |bd| !bd.monthWithUniqueDayIn(filtered) }.toList
// Bernard now knows the answer.
// So the day must be unique within the remaining choices.
filtered = filtered.where { |bd| bd.dayUniqueIn(filtered) }.toList
// Albert now knows the answer too.
// So the month must be unique within the remaining choices.
filtered = filtered.where { |bd| bd.monthUniqueIn(filtered) }.toList
if (filtered.count == 1) {
System.print("Cheryl's birthday is %(filtered[0])")
} else {
System.print("Something went wrong!")
} |
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
| #OCaml | OCaml | [1; 2; 3; 4; 5] |
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
| #R | R | print(combn(0:4, 3)) |
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.
| #Tailspin | Tailspin |
templates foo
when <=0> do 'zero' -> !OUT::write
when <..0> do
'negative ' -> !OUT::write
-$ -> #
when <?($ mod 2 <=0>)> do 'even' -> !OUT::write
otherwise 'odd' -> !OUT::write
end foo
|
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.
| #Tcl | Tcl | if {$foo == 3} {
puts "foo is three"
} elseif {$foo == 4} {
puts "foo is four"
} else {
puts "foo is neither three nor four"
} |
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}}}
.
| #PicoLisp | PicoLisp | (de modinv (A B)
(let (B0 B X0 0 X1 1 Q 0 T1 0)
(while (< 1 A)
(setq
Q (/ A B)
T1 B
B (% A B)
A T1
T1 X0
X0 (- X1 (* Q X0))
X1 T1 ) )
(if (lt0 X1) (+ X1 B0) X1) ) )
(de chinrem (N A)
(let P (apply * N)
(%
(sum
'((N A)
(setq T1 (/ P N))
(* A (modinv T1 N) T1) )
N
A )
P ) ) )
(println
(chinrem (3 5 7) (2 3 2))
(chinrem (11 12 13) (10 4 12)) )
(bye) |
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}}}
.
| #Prolog | Prolog |
product(A, B, C) :- C is A*B.
pair(X, Y, X-Y).
egcd(_, 0, 1, 0) :- !.
egcd(A, B, X, Y) :-
divmod(A, B, Q, R),
egcd(B, R, S, X),
Y is S - Q*X.
modinv(A, B, X) :-
egcd(A, B, X, Y),
A*X + B*Y =:= 1.
crt_fold(A, M, P, R0, R1) :- % system of equations of (x = a) (mod m); p = M/m
modinv(P, M, Inv),
R1 is R0 + A*Inv*P.
crt(Pairs, N) :-
maplist(pair, As, Ms, Pairs),
foldl(product, Ms, 1, M),
maplist(divmod(M), Ms, Ps, _), % p(n) <- M/m(n)
foldl(crt_fold, As, Ms, Ps, 0, N0),
N is N0 mod M.
|
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.
| #Pascal | Pascal | {
# a class is a package (i.e. a namespace) with methods in it
package MyClass;
# a constructor is a function that returns a blessed reference
sub new {
my $class = shift;
bless {variable => 0}, $class;
# the instance object is a hashref in disguise.
# (it can be a ref to anything.)
}
# an instance method is a function that takes an object as first argument.
# the -> invocation syntax takes care of that nicely, see Usage paragraph below.
sub some_method {
my $self = shift;
$self->{variable} = 1;
}
} |
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.
| #Perl | Perl | {
# a class is a package (i.e. a namespace) with methods in it
package MyClass;
# a constructor is a function that returns a blessed reference
sub new {
my $class = shift;
bless {variable => 0}, $class;
# the instance object is a hashref in disguise.
# (it can be a ref to anything.)
}
# an instance method is a function that takes an object as first argument.
# the -> invocation syntax takes care of that nicely, see Usage paragraph below.
sub some_method {
my $self = shift;
$self->{variable} = 1;
}
} |
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)
| #REXX | REXX | /*REXX program solves the closest pair of points problem (in two dimensions). */
parse arg N LO HI seed . /*obtain optional arguments from the CL*/
if N=='' | N=="," then N= 100 /*Not specified? Then use the default.*/
if LO=='' | LO=="," then LO= 0 /* " " " " " " */
if HI=='' | HI=="," then HI= 20000 /* " " " " " " */
if datatype(seed, 'W') then call random ,,seed /*seed for RANDOM (BIF) repeatability.*/
w= length(HI); w= w + (w//2==0) /*W: for aligning the output columns.*/
/*╔══════════════════════╗*/ do j=1 for N /*generate N random points*/
/*║ generate N points. ║*/ @x.j= random(LO, HI) /* " a " X */
/*╚══════════════════════╝*/ @y.j= random(LO, HI) /* " a " Y */
end /*j*/ /*X & Y make the point.*/
A= 1; B= 2 /* [↓] MIND is actually the squared */
minD= (@x.A - @x.B)**2 + (@y.A - @y.B)**2 /* distance between the 1st two points.*/
/* [↓] use of XJ & YJ speed things up.*/
do j=1 for N-1; xj= @x.j; yj= @y.j /*find min distance between a point ···*/
do k=j+1 for N-j-1 /* ··· and all other (higher) points. */
sd= (xj - @x.k)**2 + (yj - @y.k)**2 /*compute squared distance from points.*/
if sd<minD then parse value sd j k with minD A B
end /*k*/ /* [↑] needn't take SQRT of SD (yet).*/
end /*j*/ /* [↑] when done, A & B are the points*/
$= 'For ' N " points, the minimum distance between the two points: "
say $ center("x", w, '═')" " center('y', w, "═") ' is: ' sqrt( abs(minD)) / 1
say left('', length($) - 1) "["right(@x.A, w)',' right(@y.A, w)"]"
say left('', length($) - 1) "["right(@x.B, w)',' right(@y.B, w)"]"
exit 0 /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
sqrt: procedure; parse arg x; if x=0 then return 0; d=digits(); m.=9; numeric form; h=d+6
numeric digits; parse value format(x,2,1,,0) 'E0' with g 'E' _ .; g= g *.5'e'_ % 2
do j=0 while h>9; m.j= h; h= h % 2 + 1; end /*j*/
do k=j+5 to 0 by -1; numeric digits m.k; g= (g+x/g)*.5; end /*k*/; return g |
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
| #Oforth | Oforth | : circleCenter(x1, y1, x2, y2, r)
| d xmid ymid r1 md |
x2 x1 - sq y2 y1 - sq + sqrt -> d
x1 x2 + 2 / -> xmid
y1 y2 + 2 / -> ymid
2 r * -> r1
d 0.0 == ifTrue: [ "Infinite number of circles" . return ]
d r1 == ifTrue: [ System.Out "One circle: (" << xmid << ", " << ymid << ")" << cr return ]
d r1 > ifTrue: [ "No circle" . return ]
r sq d 2 / sq - sqrt ->md
System.Out "C1 : (" << xmid y1 y2 - md * d / + << ", " << ymid x2 x1 - md * d / + << ")" << cr
System.Out "C2 : (" << xmid y1 y2 - md * d / - << ", " << ymid x2 x1 - md * d / - << ")" << cr
; |
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).
| #PureBasic | PureBasic | EnableExplicit
#BASE=4
#SPC=Chr(32)
Procedure.s ChineseZodiac(n.i)
Define cycle_year.i=n-#BASE,
stem_number.i = cycle_year%10+1,
element_number.i = Round(stem_number/2,#PB_Round_Nearest),
branch_number.i = cycle_year%12+1,
aspect_number.i = cycle_year%2+1,
index.i = cycle_year%60+1,
celestial$ = Chr(PeekU(?Celestial_stem+SizeOf(Character)*(stem_number-1))),
c_pinyin$ = StringField(PeekS(?Stem),stem_number,"\"),
element$ = StringField(PeekS(?Element),element_number,"\"),
branch_han$ = Chr(PeekU(?Terrestrial_branch+SizeOf(Character)*(branch_number-1))),
b_pinyin$ = StringField(PeekS(?Branch),branch_number,"\"),
animal$ = StringField(PeekS(?Zodiac_animal),branch_number,"\"),
aspect$ = StringField(PeekS(?Aspect),aspect_number,"\"),
YearOfCycle$ = Str(index)
ProcedureReturn Str(n)+#SPC+
LSet(element$,7,#SPC)+#SPC+
LSet(animal$,7,#SPC)+#SPC+
LSet(aspect$,6,#SPC)+#SPC+
RSet(YearOfCycle$,11)+#SPC+
LSet(c_pinyin$+"-"+b_pinyin$,9,#SPC)+#SPC+
celestial$+branch_han$
EndProcedure
LoadFont(0,"Consolas",12)
If OpenWindow(0,#PB_Ignore,#PB_Ignore,600,400,"Chinese Zodiac",#PB_Window_ScreenCentered|#PB_Window_SystemMenu)
EditorGadget(0, 8, 8, 600-16, 400-16) : SetGadgetFont(0,FontID(0))
Define header$="Year Element Animal Aspect YearOfCycle ASCII Chinese"
AddGadgetItem(0,-1,header$)
AddGadgetItem(0,-1,ChineseZodiac(1935))
AddGadgetItem(0,-1,ChineseZodiac(1938))
AddGadgetItem(0,-1,ChineseZodiac(1968))
AddGadgetItem(0,-1,ChineseZodiac(1972))
AddGadgetItem(0,-1,ChineseZodiac(1976))
AddGadgetItem(0,-1,ChineseZodiac(1984))
AddGadgetItem(0,-1,ChineseZodiac(Year(Date())))
Repeat : Until WaitWindowEvent() = #PB_Event_CloseWindow
EndIf
DataSection
Celestial_stem: : Data.u $7532, $4E59, $4E19, $4E01, $620A, $5DF1, $5E9A, $8F9B, $58EC, $7678
Terrestrial_branch: : Data.u $5B50, $4E11, $5BC5, $536F, $8FB0, $5DF3, $5348, $672A, $7533, $9149, $620C, $4EA5
Zodiac_animal: : Data.s "Rat\Ox\Tiger\Rabbit\Dragon\Snake\Horse\Goat\Monkey\Rooster\Dog\Pig"
Element: : Data.s "Wood\Fire\Earth\Metal\Water"
Aspect: : Data.s "yang\yin"
Stem: : Data.s "jiă\yĭ\bĭng\dīng\wù\jĭ\gēng\xīn\rén\gŭi"
Branch: : Data.s "zĭ\chŏu\yín\măo\chén\sì\wŭ\wèi\shēn\yŏu\xū\hài"
EndDataSection |
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
| #hexiscript | hexiscript | println "File \"input.txt\"? " + (exists "input.txt")
println "Dir \"docs\"? " + (exists "docs/")
println "File \"/input.txt\"? " + (exists "/input.txt")
println "Dir \"/docs\"? " + (exists "/docs/") |
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
| #HicEst | HicEst | OPEN(FIle= 'input.txt', OLD, IOStat=ios, ERror=99)
OPEN(FIle='C:\input.txt', OLD, IOStat=ios, ERror=99)
! ...
99 WRITE(Messagebox='!') 'File does not exist. Error message ', ios |
http://rosettacode.org/wiki/Chaos_game | Chaos game | The Chaos Game is a method of generating the attractor of an iterated function system (IFS).
One of the best-known and simplest examples creates a fractal, using a polygon and an initial point selected at random.
Task
Play the Chaos Game using the corners of an equilateral triangle as the reference points. Add a starting point at random (preferably inside the triangle). Then add the next point halfway between the starting point and one of the reference points. This reference point is chosen at random.
After a sufficient number of iterations, the image of a Sierpinski Triangle should emerge.
See also
The Game of Chaos
| #Pascal | Pascal |
program ChaosGame;
// FPC 3.0.2
uses
Graph, windows, math;
// Return a point on a circle defined by angle and the circles radius
// Angle 0 = Radius points to the left
// Angle 90 = Radius points upwards
Function PointOfCircle(Angle: SmallInt; Radius: integer): TPoint;
var Ia: Double;
begin
Ia:=DegToRad(-Angle);
result.x:=round(cos(Ia)*Radius);
result.y:=round(sin(Ia)*Radius);
end;
{ Main }
var
GraphDev,GraphMode: smallint;
Triangle: array[0..2] of Tpoint; // Corners of the triangle
TriPnt: Byte; // Point in ^^^^
Origin: TPoint; // Defines center of triangle
Itterations: integer; // Number of Itterations
Radius: Integer;
View: viewPorttype;
CurPnt: TPoint;
Rect: TRect;
Counter: integer;
begin
Repeat {forever}
// Get the Itteration count 0=exit
Write('Itterations: ');
ReadLn(Itterations);
if Itterations=0 then halt;
// Set Up Graphics screen (everythings Auto detect)
GraphDev:=Detect;
GraphMode:=0;
InitGraph(GraphDev,GraphMode,'');
if GraphResult<>grok then
begin
Writeln('Graphics doesn''t work');
Halt;
end;
// set Origin to center of the _Triangle_ (Not the creen)
GetViewSettings(View);
Rect.Create(View.x1,View.y1+10,View.x2,View.y2-10);
Origin:=Rect.CenterPoint;
Origin.Offset(0,Rect.Height div 6); // Center Triangle on screen
// Define Equilateral triangle,
Radius:=Origin.y; // Radius of Circumscribed circle
for Counter:=0 to 2 do
Triangle[Counter]:=PointOfCircle((Counter*120)+90,Radius)+Origin;
// Choose random starting point, in the incsribed circle of the triangle
Radius:=Radius div 2; // Radius of inscribed circle
CurPnt:=PointOfCircle(random(360),random(Radius div 2))+Origin;
// Play the Chaos Game
for Counter:=0 to Itterations do
begin
TriPnt:=Random(3); // Select Triangle Point
Rect.Create(Triangle[TriPnt],CurPnt);; // Def. rect. between TriPnt and CurPnt
CurPnt:=Rect.CenterPoint; // New CurPnt is center of rectangle
putPixel(CurPnt.x,CurPnt.y,cyan); // Plot the new CurPnt
end;
until False;
end.
|
http://rosettacode.org/wiki/Chat_server | Chat server | Task
Write a server for a minimal text based chat.
People should be able to connect via ‘telnet’, sign on with a nickname, and type messages which will then be seen by all other connected users. Arrivals and departures of chat members should generate appropriate notification messages.
| #Raku | Raku | react {
my %connections;
whenever IO::Socket::Async.listen('localhost', 4004) -> $conn {
my $name;
$conn.print: "Please enter your name: ";
whenever $conn.Supply.lines -> $message {
if !$name {
if %connections{$message} {
$conn.print: "Name already taken, choose another one: ";
}
else {
$name = $message;
%connections{$name} = $conn;
broadcast "+++ %s arrived +++", $name;
}
}
else {
broadcast "%s> %s", $name, $message;
}
LAST {
broadcast "--- %s left ---", $name;
%connections{$name}:delete;
$conn.close ;
}
QUIT {
default {
say "oh no, $_";
}
}
}
}
sub broadcast ($format, $from, *@message) {
my $text = sprintf $format, $from, |@message;
say $text;
for %connections.kv -> $name, $conn {
$conn.print: "$text\n" if $name ne $from;
}
}
} |
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.
| #Erlang | Erlang | 1> F = fun([X]) -> X end.
#Fun<erl_eval.6.13229925>
2> F("a").
97 |
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.
| #Euphoria | Euphoria | printf(1,"%d\n", 'a') -- prints "97"
printf(1,"%s\n", 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. | #PowerShell | PowerShell |
function cholesky ($a) {
$l = @()
if ($a) {
$n = $a.count
$end = $n - 1
$l = 1..$n | foreach {$row = @(0) * $n; ,$row}
foreach ($k in 0..$end) {
$m = $k - 1
$sum = 0
if(0 -lt $k) {
foreach ($j in 0..$m) {$sum += $l[$k][$j]*$l[$k][$j]}
}
$l[$k][$k] = [Math]::Sqrt($a[$k][$k] - $sum)
if ($k -lt $end) {
foreach ($i in ($k+1)..$end) {
$sum = 0
if (0 -lt $k) {
foreach ($j in 0..$m) {$sum += $l[$i][$j]*$l[$k][$j]}
}
$l[$i][$k] = ($a[$i][$k] - $sum)/$l[$k][$k]
}
}
}
}
$l
}
function show($a) {$a | foreach {"$_"}}
$a1 = @(
@(25, 15, -5),
@(15, 18, 0),
@(-5, 0, 11)
)
"a1 ="
show $a1
""
"l1 ="
show (cholesky $a1)
""
$a2 = @(
@(18, 22, 54, 42),
@(22, 70, 86, 62),
@(54, 86, 174, 134),
@(42, 62, 134, 106)
)
"a2 ="
show $a2
""
"l2 ="
show (cholesky $a2)
|
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
| #zkl | zkl | dates:=T(T("May", 15), T("May", 16), T("May", 19),
T("June", 17), T("June", 18),
T("July", 14), T("July", 16),
T("August",14), T("August",15), T("August",17) );
mDs:=dates.pump(Dictionary().appendKV); // "June":(15,16,19), ...
dMs:=dates.pump(Dictionary().appendKV,"reverse"); // 15:"May", 16:"May", 19:"May", ...
// remove unique days (18,19) --> "July":(14,16),"August":(14,15,17)
dMs.values.apply2('wrap(ms){ if(ms.len()==1) mDs.del(ms[0]) });
// find intersection of above days --> (14)
fcn intersection(l1,l2){ l1.pump(List,l2.holds,'==(True),Void.Filter) }
badDs:=mDs.values.reduce(intersection);
// --> July:(16),August:(15,17) --> ( ("July",(16)) )
theDay:=mDs.filter('wrap([(m,ds)]){ ds.removeEach(badDs).len()==1 });
// print birthday such that muliples are shown, if any
println("Cheryl's birthday is ",theDay.flatten().flatten().concat(" ")); |
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
| #Oforth | Oforth | Buffer A collection of bytes
Mem A mutable collection of bytes
Interval A first value, a last value and a step.
Pair A collection of 2 elements (with key/value features).
List A collection of n elements
ListBuffer A mutable collection of n elements that can grow when necessary
String A collection of n characters
Symbol A collection of n characters that are identity (if they are equal, they are the same object).
StringBuffer A mutable collection of n charaters that can grow when necessary
|
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
| #Racket | Racket |
(define sublists
(match-lambda**
[(0 _) '(())]
[(_ '()) '()]
[(m (cons x xs)) (append (map (curry cons x) (sublists (- m 1) xs))
(sublists m xs))]))
(define (combinations n m)
(sublists n (range m)))
|
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.
| #Tern | Tern | if(a > b)
println(a); |
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}}}
.
| #PureBasic | PureBasic | EnableExplicit
DisableDebugger
DataSection
LBL_n1:
Data.i 3,5,7
LBL_a1:
Data.i 2,3,2
LBL_n2:
Data.i 11,12,13
LBL_a2:
Data.i 10,4,12
LBL_n3:
Data.i 10,4,9
LBL_a3:
Data.i 11,22,19
EndDataSection
Procedure ErrorHdl()
Print(ErrorMessage())
Input()
EndProcedure
Macro PrintData(n,a)
Define Idx.i=0
Print("[")
While n+SizeOf(Integer)*Idx<a
Print("( ")
Print(Str(PeekI(a+SizeOf(Integer)*Idx)))
Print(" . ")
Print(Str(PeekI(n+SizeOf(Integer)*Idx)))
Print(" )")
Idx+1
Wend
Print(~"]\nx = ")
EndMacro
Procedure.i Produkt_n(n_Adr.i,a_Adr.i)
Define p.i=1
While n_Adr<a_Adr
p*PeekI(n_Adr)
n_Adr+SizeOf(Integer)
Wend
ProcedureReturn p
EndProcedure
Procedure.i Eval_x1(a.i,b.i)
Define b0.i=b, x0.i=0, x1.i=1, q.i, t.i
If b=1 : ProcedureReturn x1 : EndIf
While a>1
q=Int(a/b)
t=b : b=a%b : a=t
t=x0 : x0=x1-q*x0 : x1=t
Wend
If x1<0 : ProcedureReturn x1+b0 : EndIf
ProcedureReturn x1
EndProcedure
Procedure.i ChineseRem(n_Adr.i,a_Adr.i)
Define prod.i=Produkt_n(n_Adr,a_Adr), a.i, b.i, p.i, Idx.i=0, sum.i
While n_Adr+SizeOf(Integer)*Idx<a_Adr
b=PeekI(n_Adr+SizeOf(Integer)*Idx)
p=Int(prod/b) : a=p
sum+PeekI(a_Adr+SizeOf(Integer)*Idx)*Eval_x1(a,b)*p
Idx+1
Wend
ProcedureReturn sum%prod
EndProcedure
OnErrorCall(@ErrorHdl())
OpenConsole("Chinese remainder theorem")
PrintData(?LBL_n1,?LBL_a1)
PrintN(Str(ChineseRem(?LBL_n1,?LBL_a1)))
PrintData(?LBL_n2,?LBL_a2)
PrintN(Str(ChineseRem(?LBL_n2,?LBL_a2)))
PrintData(?LBL_n3,?LBL_a3)
PrintN(Str(ChineseRem(?LBL_n3,?LBL_a3)))
Input() |
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.
| #Phix | Phix | without js -- (else cffi namespaces error, classes not supported by pwa/p2js anyway)
requires ("1.0.2") -- (free up the temp used in the v.show() call)
class five nullable
private integer n = 3
function get_n()
return n
end function
procedure set_n(integer n)
this.n = n
end procedure
procedure show()
printf(1,"show: n is %d\n",{n})
end procedure
function five(integer n = 4)
printf(1,"constructor five(%d) called\n",n)
this.n = n
return this
end function
procedure ~five()
printf(1,"destructor ~five(%d) called\n",n)
end procedure
end class
five v = new({5})
assert(v.n=5)
v.n = 6
v.show()
v=NULL
|
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)
| #Ring | Ring |
decimals(10)
x = list(10)
y = list(10)
x[1] = 0.654682
y[1] = 0.925557
x[2] = 0.409382
y[2] = 0.619391
x[3] = 0.891663
y[3] = 0.888594
x[4] = 0.716629
y[4] = 0.996200
x[5] = 0.477721
y[5] = 0.946355
x[6] = 0.925092
y[6] = 0.818220
x[7] = 0.624291
y[7] = 0.142924
x[8] = 0.211332
y[8] = 0.221507
x[9] = 0.293786
y[9] = 0.691701
x[10] = 0.839186
y[10] = 0.728260
min = 10000
for i = 1 to 9
for j = i+1 to 10
dsq = pow((x[i] - x[j]),2) + pow((y[i] - y[j]),2)
if dsq < min min = dsq mini = i minj = j ok
next
next
see "closest pair is : " + mini + " and " + minj + " at distance " + sqrt(min)
|
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
| #ooRexx | ooRexx | /*REXX pgm finds 2 circles with a specific radius given two (X,Y) points*/
a.=''
a.1=0.1234 0.9876 0.8765 0.2345 2
a.2=0.0000 2.0000 0.0000 0.0000 1
a.3=0.1234 0.9876 0.1234 0.9876 2
a.4=0.1234 0.9876 0.8765 0.2345 0.5
a.5=0.1234 0.9876 0.1234 0.9876 0
Say ' x1 y1 x2 y2 radius cir1x cir1y cir2x cir2y'
Say ' ------ ------ ------ ------ ------ ------ ------ ------ ------'
Do j=1 By 1 While a.j<>''
Do k=1 For 4
w.k=f(word(a.j,k))
End
Say w.1 w.2 w.3 w.4 format(word(a.j,5),5,1) twocircles(a.j)
End
Exit
twocircles: Procedure
Parse Arg px py qx qy r .
If r=0 Then
Return ' radius of zero gives no circles.'
x=(qx-px)/2
y=(qy-py)/2
bx=px+x
by=py+y
pb=rxCalcsqrt(x**2+y**2)
If pb=0 Then
Return ' coincident points give infinite circles'
If pb>r Then
Return ' points are too far apart for the given radius'
cb=rxCalcsqrt(r**2-pb**2)
x1=y*cb/pb
y1=x*cb/pb
Return f(bx-x1) f(by+y1) f(bx+x1) f(by-y1)
f: Return format(arg(1),2,4) /* format a number with 4 dec dig.*/
::requires 'rxMath' library |
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).
| #Python | Python |
# coding: utf-8
from __future__ import print_function
from datetime import datetime
pinyin = {
'甲': 'jiă',
'乙': 'yĭ',
'丙': 'bĭng',
'丁': 'dīng',
'戊': 'wù',
'己': 'jĭ',
'庚': 'gēng',
'辛': 'xīn',
'壬': 'rén',
'癸': 'gŭi',
'子': 'zĭ',
'丑': 'chŏu',
'寅': 'yín',
'卯': 'măo',
'辰': 'chén',
'巳': 'sì',
'午': 'wŭ',
'未': 'wèi',
'申': 'shēn',
'酉': 'yŏu',
'戌': 'xū',
'亥': 'hài'
}
animals = ['Rat', 'Ox', 'Tiger', 'Rabbit', 'Dragon', 'Snake',
'Horse', 'Goat', 'Monkey', 'Rooster', 'Dog', 'Pig']
elements = ['Wood', 'Fire', 'Earth', 'Metal', 'Water']
celestial = ['甲', '乙', '丙', '丁', '戊', '己', '庚', '辛', '壬', '癸']
terrestrial = ['子', '丑', '寅', '卯', '辰', '巳', '午', '未', '申', '酉', '戌', '亥']
aspects = ['yang', 'yin']
def calculate(year):
BASE = 4
year = int(year)
cycle_year = year - BASE
stem_number = cycle_year % 10
stem_han = celestial[stem_number]
stem_pinyin = pinyin[stem_han]
element_number = stem_number // 2
element = elements[element_number]
branch_number = cycle_year % 12
branch_han = terrestrial[branch_number]
branch_pinyin = pinyin[branch_han]
animal = animals[branch_number]
aspect_number = cycle_year % 2
aspect = aspects[aspect_number]
index = cycle_year % 60 + 1
print("{}: {}{} ({}-{}, {} {}; {} - year {} of the cycle)"
.format(year, stem_han, branch_han,
stem_pinyin, branch_pinyin, element, animal, aspect, index))
current_year = datetime.now().year
years = [1935, 1938, 1968, 1972, 1976, current_year]
for year in years:
calculate(year) |
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
| #HolyC | HolyC | U0 FileExists(U8 *f) {
if (FileFind(f) && !IsDir(f)) {
Print("'%s' file exists.\n", f);
} else {
Print("'%s' file does not exist.\n", f);
}
}
U0 DirExists(U8 *d) {
if (IsDir(d)) {
Print("'%s' directory exists.\n", d);
} else {
Print("'%s' directory does not exist.\n", d);
}
}
FileExists("input.txt");
FileExists("::/input.txt");
DirExists("docs");
DirExists("::/docs"); |
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
| #i | i | concept exists(path) {
open(path)
errors {
if error.DoesNotExist()
print(path, " does not exist!")
end
return
}
print(path, " exists!")
}
software {
exists("input.txt")
exists("/input.txt")
exists("docs")
exists("/docs")
exists("docs/Abdu'l-Bahá.txt")
} |
http://rosettacode.org/wiki/Chaos_game | Chaos game | The Chaos Game is a method of generating the attractor of an iterated function system (IFS).
One of the best-known and simplest examples creates a fractal, using a polygon and an initial point selected at random.
Task
Play the Chaos Game using the corners of an equilateral triangle as the reference points. Add a starting point at random (preferably inside the triangle). Then add the next point halfway between the starting point and one of the reference points. This reference point is chosen at random.
After a sufficient number of iterations, the image of a Sierpinski Triangle should emerge.
See also
The Game of Chaos
| #Perl | Perl | use Imager;
my $width = 1000;
my $height = 1000;
my @points = (
[ $width/2, 0],
[ 0, $height-1],
[$height-1, $height-1],
);
my $img = Imager->new(
xsize => $width,
ysize => $height,
channels => 3,
);
my $color = Imager::Color->new('#ff0000');
my $r = [int(rand($width)), int(rand($height))];
foreach my $i (1 .. 100000) {
my $p = $points[rand @points];
my $h = [
int(($p->[0] + $r->[0]) / 2),
int(($p->[1] + $r->[1]) / 2),
];
$img->setpixel(
x => $h->[0],
y => $h->[1],
color => $color,
);
$r = $h;
}
$img->write(file => 'chaos_game_triangle.png'); |
http://rosettacode.org/wiki/Chat_server | Chat server | Task
Write a server for a minimal text based chat.
People should be able to connect via ‘telnet’, sign on with a nickname, and type messages which will then be seen by all other connected users. Arrivals and departures of chat members should generate appropriate notification messages.
| #Ruby | Ruby | require 'gserver'
class ChatServer < GServer
def initialize *args
super
#Keep a list for broadcasting messages
@chatters = []
#We'll need this for thread safety
@mutex = Mutex.new
end
#Send message out to everyone but sender
def broadcast message, sender = nil
#Need to use \r\n for our Windows friends
message = message.strip << "\r\n"
#Mutex for safety - GServer uses threads
@mutex.synchronize do
@chatters.each do |chatter|
begin
chatter.print message unless chatter == sender
rescue
@chatters.delete chatter
end
end
end
end
#Handle each connection
def serve io
io.print 'Name: '
name = io.gets
#They might disconnect
return if name.nil?
name.strip!
broadcast "--+ #{name} has joined +--"
#Add to our list of connections
@mutex.synchronize do
@chatters << io
end
#Get and broadcast input until connection returns nil
loop do
message = io.gets
if message
broadcast "#{name}> #{message}", io
else
break
end
end
broadcast "--+ #{name} has left +--"
end
end
#Start up the server on port 7000
#Accept connections for any IP address
#Allow up to 100 connections
#Send information to stderr
#Turn on informational messages
ChatServer.new(7000, '0.0.0.0', 100, $stderr, true).start.join
|
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.
| #F.23 | F# | let c = 'A'
let n = 65
printfn "%d" (int c)
printfn "%c" (char n) |
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.
| #Factor | Factor | CHAR: katakana-letter-a .
"ア" first .
12450 1string print |
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. | #Python | Python | from __future__ import print_function
from pprint import pprint
from math import sqrt
def cholesky(A):
L = [[0.0] * len(A) for _ in xrange(len(A))]
for i in xrange(len(A)):
for j in xrange(i+1):
s = sum(L[i][k] * L[j][k] for k in xrange(j))
L[i][j] = sqrt(A[i][i] - s) if (i == j) else \
(1.0 / L[j][j] * (A[i][j] - s))
return L
if __name__ == "__main__":
m1 = [[25, 15, -5],
[15, 18, 0],
[-5, 0, 11]]
pprint(cholesky(m1))
print()
m2 = [[18, 22, 54, 42],
[22, 70, 86, 62],
[54, 86, 174, 134],
[42, 62, 134, 106]]
pprint(cholesky(m2), width=120) |
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
| #ooRexx | ooRexx |
a = .array~new(4) -- creates an array of 4 items, with all slots empty
say a~size a~items -- size is 4, but there are 0 items
a[1] = "Fred" -- assigns a value to the first item
a[5] = "Mike" -- assigns a value to the fifth slot, expanding the size
say a~size a~items -- size is now 5, with 2 items
|
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
| #Raku | Raku | .say for combinations(5,3); |
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.
| #TI-83_BASIC | TI-83 BASIC | If condition
statement |
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}}}
.
| #Python | Python | # Python 2.7
def chinese_remainder(n, a):
sum = 0
prod = reduce(lambda a, b: a*b, n)
for n_i, a_i in zip(n, a):
p = prod / n_i
sum += a_i * mul_inv(p, n_i) * p
return sum % prod
def mul_inv(a, b):
b0 = b
x0, x1 = 0, 1
if b == 1: return 1
while a > 1:
q = a / b
a, b = b, a%b
x0, x1 = x1 - q * x0, x0
if x1 < 0: x1 += b0
return x1
if __name__ == '__main__':
n = [3, 5, 7]
a = [2, 3, 2]
print chinese_remainder(n, a) |
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.
| #PHL | PHL | module classes;
extern printf;
class @MyClass {
field @Integer myField { get:get_myField, set:set_myField };
new [
this.set_myField(2);
]
@Void method [
this.set_myField(this::get_myField + 1);
]
};
@Integer main [
var obj = new @MyClass;
printf("obj.myField: %i\n", obj::get_myField);
obj::method;
printf("obj.myField: %i\n", obj::get_myField);
return 0;
] |
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.
| #PHP | PHP | class MyClass {
public static $classVar;
public $instanceVar; // can also initialize it here
function __construct() {
$this->instanceVar = 0;
}
function someMethod() {
$this->instanceVar = 1;
self::$classVar = 3;
}
}
$myObj = new MyClass(); |
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)
| #Ruby | Ruby | Point = Struct.new(:x, :y)
def distance(p1, p2)
Math.hypot(p1.x - p2.x, p1.y - p2.y)
end
def closest_bruteforce(points)
mindist, minpts = Float::MAX, []
points.combination(2) do |pi,pj|
dist = distance(pi, pj)
if dist < mindist
mindist = dist
minpts = [pi, pj]
end
end
[mindist, minpts]
end
def closest_recursive(points)
return closest_bruteforce(points) if points.length <= 3
xP = points.sort_by(&:x)
mid = points.length / 2
xm = xP[mid].x
dL, pairL = closest_recursive(xP[0,mid])
dR, pairR = closest_recursive(xP[mid..-1])
dmin, dpair = dL<dR ? [dL, pairL] : [dR, pairR]
yP = xP.find_all {|p| (xm - p.x).abs < dmin}.sort_by(&:y)
closest, closestPair = dmin, dpair
0.upto(yP.length - 2) do |i|
(i+1).upto(yP.length - 1) do |k|
break if (yP[k].y - yP[i].y) >= dmin
dist = distance(yP[i], yP[k])
if dist < closest
closest = dist
closestPair = [yP[i], yP[k]]
end
end
end
[closest, closestPair]
end
points = Array.new(100) {Point.new(rand, rand)}
p ans1 = closest_bruteforce(points)
p ans2 = closest_recursive(points)
fail "bogus!" if ans1[0] != ans2[0]
require 'benchmark'
points = Array.new(10000) {Point.new(rand, rand)}
Benchmark.bm(12) do |x|
x.report("bruteforce") {ans1 = closest_bruteforce(points)}
x.report("recursive") {ans2 = closest_recursive(points)}
end |
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
| #PARI.2FGP | PARI/GP | circ(a, b, r)={
if(a==b, return("impossible"));
my(h=(b-a)/2,t=sqrt(r^2-abs(h)^2)/abs(h)*h);
[a+h+t*I,a+h-t*I]
};
circ(0.1234 + 0.9876*I, 0.8765 + 0.2345*I, 2)
circ(0.0000 + 2.0000*I, 0.0000 + 0.0000*I, 1)
circ(0.1234 + 0.9876*I, 0.1234 + 0.9876*I, 2)
circ(0.1234 + 0.9876*I, 0.8765 + 0.2345*I, .5)
circ(0.1234 + 0.9876*I, 0.1234 + 0.9876*I, 0) |
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).
| #Racket | Racket | #lang racket
(require racket/date)
; Any CE Year that was the first of a 60-year cycle
(define base-year 1984)
(define celestial-stems '("甲" "乙" "丙" "丁" "戊" "己" "庚" "辛" "壬" "癸"))
(define terrestrial-branches '("子" "丑" "寅" "卯" "辰" "巳" "午" "未" "申" "酉" "戌" "亥"))
(define zodiac-animals
'("Rat" "Ox" "Tiger" "Rabbit" "Dragon" "Snake" "Horse" "Goat" "Monkey" "Rooster" "Dog" "Pig"))
(define elements '("Wood" "Fire" "Earth" "Metal" "Water"))
(define aspects '("yang" "yin"))
(define pinyin
(map cons
(append celestial-stems terrestrial-branches)
(list "jiă" "yĭ" "bĭng" "dīng" "wù" "jĭ" "gēng" "xīn" "rén" "gŭi"
"zĭ" "chŏu" "yín" "măo" "chén" "sì" "wŭ" "wèi" "shēn" "yŏu" "xū" "hài")))
(define (this-year) (date-year (current-date)))
(define (pinyin-for han) (cdr (assoc han pinyin)))
(define (han/pinyin-nth n hans) (let ((han (list-ref hans n))) (values han (pinyin-for han))))
(define (chinese-zodiac ce-year)
(let* ((cycle-year (- ce-year base-year))
(stem-number (modulo cycle-year (length celestial-stems)))
(element-number (quotient stem-number 2))
(aspect-number (modulo cycle-year (length aspects)))
(branch-number (modulo cycle-year (length terrestrial-branches)))
(element (list-ref elements element-number))
(zodiac-animal (list-ref zodiac-animals branch-number))
(aspect (list-ref aspects aspect-number)))
(let-values (([stem-han stem-pinyin] (han/pinyin-nth stem-number celestial-stems))
([branch-han branch-pinyin] (han/pinyin-nth branch-number terrestrial-branches)))
(list ce-year stem-han branch-han stem-pinyin branch-pinyin element zodiac-animal aspect))))
(module+ test
(for ((ce-year (in-list '(1935 1938 1941 1947 1968 1972 1976))))
(apply printf "~a: ~a~a (~a-~a, ~a ~a; ~a)~%" (chinese-zodiac ce-year)))) |
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).
| #Raku | Raku | sub Chinese-zodiac ( Int $year ) {
my @heaven = <甲 jiă 乙 yĭ 丙 bĭng 丁 dīng 戊 wù 己 jĭ 庚 gēng 辛 xīn 壬 rén 癸 gŭi>.pairup;
my @earth = <子 zĭ 丑 chŏu 寅 yín 卯 măo 辰 chén 巳 sì 午 wŭ 未 wèi 申 shēn 酉 yŏu 戌 xū 亥 hài>.pairup;
my @animal = <Rat Ox Tiger Rabbit Dragon Snake Horse Goat Monkey Rooster Dog Pig>;
my @element = <Wood Fire Earth Metal Water>;
my @aspect = <yang yin>;
my $cycle_year = ($year - 4) % 60;
my $i2 = $cycle_year % 2;
my $i10 = $cycle_year % 10;
my $i12 = $cycle_year % 12;
%(
'Han' => @heaven[$i10].key ~ @earth[$i12].key,
'pinyin' => @heaven[$i10].value ~ @earth[$i12].value,
'heaven' => @heaven[$i10],
'earth' => @earth[$i12],
'element' => @element[$i10 div 2],
'animal' => @animal[$i12],
'aspect' => @aspect[$i2],
'cycle' => $cycle_year + 1
)
}
# TESTING
printf "%d: %s (%s, %s %s; %s - year %d in the cycle)\n",
$_, .&Chinese-zodiac<Han pinyin element animal aspect cycle>
for 1935, 1938, 1968, 1972, 1976, 1984, Date.today.year; |
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
| #Icon_and_Unicon | Icon and Unicon | every dir := !["./","/"] do {
write("file ", f := dir || "input.txt", if stat(f) then " exists." else " doesn't exist.")
write("directory ", f := dir || "docs", if stat(f) then " exists." else " doesn't exist.")
} |
http://rosettacode.org/wiki/Chaos_game | Chaos game | The Chaos Game is a method of generating the attractor of an iterated function system (IFS).
One of the best-known and simplest examples creates a fractal, using a polygon and an initial point selected at random.
Task
Play the Chaos Game using the corners of an equilateral triangle as the reference points. Add a starting point at random (preferably inside the triangle). Then add the next point halfway between the starting point and one of the reference points. This reference point is chosen at random.
After a sufficient number of iterations, the image of a Sierpinski Triangle should emerge.
See also
The Game of Chaos
| #Phix | Phix | --
-- demo\rosetta\Chaos_game.exw
-- ===========================
--
with javascript_semantics
include pGUI.e
Ihandle dlg, canvas
cdCanvas cddbuffer, cdcanvas
enum TRI,SQ1,SQ2,SQ3,PENT
sequence descs = {"Sierpinsky Triangle",
"Square 1",
"Square 2",
"Square 3",
"Pentagon"}
integer mode = TRI
function redraw_cb(Ihandle /*ih*/, integer /*posx*/, /*posy*/)
atom {w,h} = IupGetIntInt(canvas, "DRAWSIZE")
atom {x,y} = {w*0.05,h*0.05}
{w,h} = {w*0.9,h*0.9}
sequence points = iff(mode<SQ1?{{x,y},{x+w/2,y+h},{x+w,y}}:
iff(mode<PENT?{{x,y},{x,y+h},{x+w,y+h},{x+w,y}}
:{{x+w/6,y},{x,y+h*2/3},{x+w/2,y+h},{x+w,y+h*2/3},{x+w*5/6,y}}))
cdCanvasActivate(cddbuffer)
integer last = 0
for i=1 to 1000 do
integer r = rand(length(points))
if mode=TRI or r!=last then
atom {nx,ny} = points[r]
{x,y} = {(x+nx)/2,(y+ny)/2}
cdCanvasPixel(cddbuffer, x, y, CD_GREY)
if mode=SQ2
or mode=SQ3 then
r = mod(r,length(points))+1
if mode=SQ3 then
r = mod(r,length(points))+1
end if
end if
last = r
end if
end for
cdCanvasFlush(cddbuffer)
IupSetStrAttribute(dlg, "TITLE", "Chaos Game (%s)", {descs[mode]})
return IUP_DEFAULT
end function
function timer_cb(Ihandle /*ih*/)
IupUpdate(canvas)
return IUP_IGNORE
end function
function map_cb(Ihandle ih)
cdcanvas = cdCreateCanvas(CD_IUP, ih)
cddbuffer = cdCreateCanvas(CD_DBUFFER, cdcanvas)
cdCanvasSetBackground(cddbuffer, CD_WHITE)
cdCanvasSetForeground(cddbuffer, CD_GRAY)
return IUP_DEFAULT
end function
function key_cb(Ihandle /*ih*/, atom c)
if c=K_ESC then return IUP_CLOSE end if
if c=' ' then
mode += 1
if mode>PENT then
mode = TRI
end if
cdCanvasClear(cddbuffer)
IupRedraw(canvas)
end if
return IUP_CONTINUE
end function
procedure main()
IupOpen()
canvas = IupCanvas("RASTERSIZE=640x640")
IupSetCallbacks(canvas, {"MAP_CB", Icallback("map_cb"),
"ACTION", Icallback("redraw_cb")})
dlg = IupDialog(canvas, `TITLE="Chaos Game"`)
IupSetCallback(dlg, "KEY_CB", Icallback("key_cb"))
IupShow(dlg)
IupSetAttribute(canvas, "RASTERSIZE", NULL)
Ihandle timer = IupTimer(Icallback("timer_cb"), 40)
if platform()!=JS then
IupMainLoop()
IupClose()
end if
end procedure
main()
|
http://rosettacode.org/wiki/Chat_server | Chat server | Task
Write a server for a minimal text based chat.
People should be able to connect via ‘telnet’, sign on with a nickname, and type messages which will then be seen by all other connected users. Arrivals and departures of chat members should generate appropriate notification messages.
| #Rust | Rust |
use std::collections::HashMap;
use std::io;
use std::io::prelude::*;
use std::io::BufReader;
use std::net::{TcpListener, TcpStream};
use std::sync::{Arc, RwLock};
use std::thread;
type Username = String;
/// Sends a message to all clients except the sending client.
fn broadcast_message(
user: &str,
clients: &mut HashMap<String, TcpStream>,
message: &str,
) -> io::Result<()> {
for (client, stream) in clients.iter_mut() {
if client != user {
writeln!(stream, "{}", message)?;
}
}
Ok(())
}
fn chat_loop(listener: &TcpListener) -> io::Result<()> {
let local_clients: Arc<RwLock<HashMap<Username, TcpStream>>> =
Arc::new(RwLock::new(HashMap::new()));
println!("Accepting connections on {}", listener.local_addr()?.port());
for stream in listener.incoming() {
match stream {
Ok(stream) => {
let client_clients = Arc::clone(&local_clients);
thread::spawn(move || -> io::Result<()> {
let mut reader = BufReader::new(stream.try_clone()?);
let mut writer = stream;
let mut name = String::new();
loop {
write!(writer, "Please enter a username: ")?;
reader.read_line(&mut name)?;
name = name.trim().to_owned();
let clients = client_clients.read().unwrap();
if !clients.contains_key(&name) {
writeln!(writer, "Welcome, {}!", &name)?;
break;
}
writeln!(writer, "That username is taken.")?;
name.clear();
}
{
let mut clients = client_clients.write().unwrap();
clients.insert(name.clone(), writer);
broadcast_message(
&name,
&mut *clients,
&format!("{} has joined the chat room.", &name),
)?;
}
for line in reader.lines() {
let mut clients = client_clients.write().unwrap();
broadcast_message(&name, &mut *clients, &format!("{}: {}", &name, line?))?;
}
{
let mut clients = client_clients.write().unwrap();
clients.remove(&name);
broadcast_message(
&name,
&mut *clients,
&format!("{} has left the chat room.", &name),
)?;
}
Ok(())
});
}
Err(e) => {
println!("Connection failed: {}", e);
}
}
}
Ok(())
}
fn main() {
let listener = TcpListener::bind(("localhost", 7000)).unwrap();
chat_loop(&listener).unwrap();
}
|
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.
| #FALSE | FALSE | 'A."
"65, |
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.
| #Fantom | Fantom | fansh> 97.toChar
a
fansh> 'a'.toInt
97 |
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. | #q | q | solve:{[A;B] $[0h>type A;B%A;inv[A] mmu B]}
ak:{[m;k] (),/:m[;k]til k:k-1}
akk:{[m;k] m[k;k:k-1]}
transpose:{$[0h=type x;flip x;enlist each x]}
mult:{[A;B]$[0h=type A;A mmu B;A*B]}
cholesky:{[A]
{[A;L;n]
l_k:solve[L;ak[A;n]];
l_kk:first over sqrt[akk[A;n] - mult[transpose l_k;l_k]];
({$[0h<type x;enlist x;x]}L,'0f),enlist raze transpose[l_k],l_kk
}[A]/[sqrt A[0;0];1_1+til count first A]
}
show cholesky (25 15 -5f;15 18 0f;-5 0 11f)
-1"";
show cholesky (18 22 54 42f;22 70 86 62f;54 86 174 134f;42 62 134 106f) |
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
| #Oz | Oz | declare
%% Lists (immutable, recursive)
Xs = [1 2 3 4]
%% Add element at the front (cons)
Xs0 = 0|Xs
{Show {Length Xs}} %% output: 4
%% Records (immutable maps with a label)
Rec = label(1:2 symbol:3)
{Show Rec} %% output: label(2 symbol:3)
{Show Rec.1} %% output: 2
%% create a new record with an added field
Rec2 = {AdjoinAt Rec 2 value}
{Show Rec2} %% output: label(2 value symbol:3)
%% Dictionaries (mutable maps)
Dict = {Dictionary.new}
Dict.1 := 1
Dict.symbol := 3
{Show Dict.1} %% output: 1
%% Arrays (mutable with integer keys)
Arr = {Array.new 1 10 initValue}
Arr.1 := 3
{Show Arr.1} %% output: 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
| #REXX | REXX | /*REXX program displays combination sets for X things taken Y at a time. */
parse arg x y $ . /*get optional arguments from the C.L. */
if x=='' | x=="," then x= 5 /*No X specified? Then use default.*/
if y=='' | y=="," then y= 3; oy= y; y= abs(y) /* " Y " " " " */
if $=='' | $=="," then $='123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ',
"~!@#$%^&*()_+`{}|[]\:;<>?,./█┌┐└┘±≥≤≈∙" /*some extended chars*/
/* [↑] No $ specified? Use default.*/
if y>x then do; say y " can't be greater than " x; exit 1; end
say "────────────" x ' things taken ' y " at a time:"
say "────────────" combN(x,y) ' combinations.'
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
combN: procedure expose $ oy; parse arg x,y; xp= x+1; xm= xp-y; !.= 0
if x=0 | y=0 then return 'no'
do i=1 for y; !.i= i
end /*i*/
do j=1; L=
do d=1 for y; L= L substr($, !.d, 1)
end /*d*/
if oy>0 then say L; !.y= !.y + 1 /*don't show if OY<0 */
if !.y==xp then if .combN(y-1) then leave
end /*j*/
return j
/*──────────────────────────────────────────────────────────────────────────────────────*/
.combN: procedure expose !. y xm; parse arg d; if d==0 then return 1; p= !.d
do u=d to y; !.u= p+1; if !.u==xm+u then return .combN(u-1); p= !.u
end /*u*/ /* ↑ */
return 0 /*recursive call──►──────┘ */ |
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.
| #Toka | Toka | 100 100 = [ ." True\n" ] ifTrue
100 200 = [ ." True\n" ] ifTrue |
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}}}
.
| #R | R | mul_inv <- function(a, b)
{
b0 <- b
x0 <- 0L
x1 <- 1L
if (b == 1) return(1L)
while(a > 1){
q <- as.integer(a/b)
t <- b
b <- a %% b
a <- t
t <- x0
x0 <- x1 - q*x0
x1 <- t
}
if (x1 < 0) x1 <- x1 + b0
return(x1)
}
chinese_remainder <- function(n, a)
{
len <- length(n)
prod <- 1L
sum <- 0L
for (i in 1:len) prod <- prod * n[i]
for (i in 1:len){
p <- as.integer(prod / n[i])
sum <- sum + a[i] * mul_inv(p, n[i]) * p
}
return(sum %% prod)
}
n <- c(3L, 5L, 7L)
a <- c(2L, 3L, 2L)
chinese_remainder(n, a) |
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.
| #PicoLisp | PicoLisp | (class +Rectangle)
# dx dy
(dm area> () # Define a a method that calculates the rectangle's area
(* (: dx) (: dy)) )
(println # Create a rectangle, and print its area
(area> (new '(+Rectangle) 'dx 3 'dy 4)) ) |
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.
| #Pop11 | Pop11 | uses objectclass;
define :class MyClass;
slot value = 1;
enddefine; |
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)
| #Run_BASIC | Run BASIC | n =10 ' 10 data points input
dim x(n)
dim y(n)
pt1 = 0 ' 1st point
pt2 = 0 ' 2nd point
for i =1 to n ' read in data
read x(i)
read y(i)
next i
minDist = 1000000
for i =1 to n -1
for j =i +1 to n
distXsq =(x(i) -x(j))^2
disYsq =(y(i) -y(j))^2
d =abs((dxSq +disYsq)^0.5)
if d <minDist then
minDist =d
pt1 =i
pt2 =j
end if
next j
next i
print "Distance ="; minDist; " between ("; x(pt1); ", "; y(pt1); ") and ("; x(pt2); ", "; y(pt2); ")"
end
data 0.654682, 0.925557
data 0.409382, 0.619391
data 0.891663, 0.888594
data 0.716629, 0.996200
data 0.477721, 0.946355
data 0.925092, 0.818220
data 0.624291, 0.142924
data 0.211332, 0.221507
data 0.293786, 0.691701
data 0.839186, 0.72826 |
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
| #Perl | Perl | use strict;
sub circles {
my ($x1, $y1, $x2, $y2, $r) = @_;
return "Radius is zero" if $r == 0;
return "Coincident points gives infinite number of circles" if $x1 == $x2 and $y1 == $y2;
# delta x, delta y between points
my ($dx, $dy) = ($x2 - $x1, $y2 - $y1);
my $q = sqrt($dx**2 + $dy**2);
return "Separation of points greater than diameter" if $q > 2*$r;
# halfway point
my ($x3, $y3) = (($x1 + $x2) / 2, ($y1 + $y2) / 2);
# distance along the mirror line
my $d = sqrt($r**2-($q/2)**2);
# pair of solutions
sprintf '(%.4f, %.4f) and (%.4f, %.4f)',
$x3 - $d*$dy/$q, $y3 + $d*$dx/$q,
$x3 + $d*$dy/$q, $y3 - $d*$dx/$q;
}
my @arr = (
[0.1234, 0.9876, 0.8765, 0.2345, 2.0],
[0.0000, 2.0000, 0.0000, 0.0000, 1.0],
[0.1234, 0.9876, 0.1234, 0.9876, 2.0],
[0.1234, 0.9876, 0.8765, 0.2345, 0.5],
[0.1234, 0.9876, 0.1234, 0.9876, 0.0]
);
printf "(%.4f, %.4f) and (%.4f, %.4f) with radius %.1f: %s\n", @$_[0..4], circles @$_ for @arr; |
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).
| #Ring | Ring |
yinyang = ["yang", "yin"]
elements = ["Wood", "Fire", "Earth", "Metal", "Water"]
animals = ["Rat", "Ox", "Tiger", "Rabbit", "Dragon", "Snake",
"Horse", "Goat", "Monkey", "Rooster", "Dog", "Pig"]
years = [1801, 1861, 1984, 2020, 2186, 76543]
output = ""
for year in years
yy = year % 2 + 1
element = (year - 4) % 5 + 1
animal = (year - 4) % 12 + 1
output = string(year) + " is the year of the "
output += elements[element] + " " + animals[animal] + " (" + yinyang[yy] + ")."
? output
next
|
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).
| #Ruby | Ruby | # encoding: utf-8
pinyin = {
'甲' => 'jiă',
'乙' => 'yĭ',
'丙' => 'bĭng',
'丁' => 'dīng',
'戊' => 'wù',
'己' => 'jĭ',
'庚' => 'gēng',
'辛' => 'xīn',
'壬' => 'rén',
'癸' => 'gŭi',
'子' => 'zĭ',
'丑' => 'chŏu',
'寅' => 'yín',
'卯' => 'măo',
'辰' => 'chén',
'巳' => 'sì',
'午' => 'wŭ',
'未' => 'wèi',
'申' => 'shēn',
'酉' => 'yŏu',
'戌' => 'xū',
'亥' => 'hài'
}
celestial = %w(甲 乙 丙 丁 戊 己 庚 辛 壬 癸)
terrestrial = %w(子 丑 寅 卯 辰 巳 午 未 申 酉 戌 亥)
animals = %w(Rat Ox Tiger Rabbit Dragon Snake
Horse Goat Monkey Rooster Dog Pig)
elements = %w(Wood Fire Earth Metal Water)
aspects = %w(yang yin)
BASE = 4
args = if !ARGV.empty?
ARGV
else
[Time.new.year]
end
args.each do |arg|
ce_year = Integer(arg)
print "#{ce_year}: " if ARGV.length > 1
cycle_year = ce_year - BASE
stem_number = cycle_year % 10
stem_han = celestial[stem_number]
stem_pinyin = pinyin[stem_han]
element_number = stem_number / 2
element = elements[element_number]
branch_number = cycle_year % 12
branch_han = terrestrial[branch_number]
branch_pinyin = pinyin[branch_han]
animal = animals[branch_number]
aspect_number = cycle_year % 2
aspect = aspects[aspect_number]
index = cycle_year % 60 + 1
print stem_han, branch_han
puts " (#{stem_pinyin}-#{branch_pinyin}, #{element} #{animal}; #{aspect} - year #{index} of the cycle)"
end |
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
| #IDL | IDL |
print, FILE_TEST('input.txt')
print, FILE_TEST(PATH_SEP()+'input.txt')
print, FILE_TEST('docs', /DIRECTORY)
print, FILE_TEST(PATH_SEP()+'docs', /DIRECTORY)
|
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
| #J | J | require 'files'
fexist 'input.txt'
fexist '/input.txt'
direxist=: 2 = ftype
direxist 'docs'
direxist '/docs' |
http://rosettacode.org/wiki/Chaos_game | Chaos game | The Chaos Game is a method of generating the attractor of an iterated function system (IFS).
One of the best-known and simplest examples creates a fractal, using a polygon and an initial point selected at random.
Task
Play the Chaos Game using the corners of an equilateral triangle as the reference points. Add a starting point at random (preferably inside the triangle). Then add the next point halfway between the starting point and one of the reference points. This reference point is chosen at random.
After a sufficient number of iterations, the image of a Sierpinski Triangle should emerge.
See also
The Game of Chaos
| #Plain_English | Plain English | To run:
Start up.
Initialize our reference points.
Clear the screen to the lightest gray color.
Play the chaos game.
Refresh the screen.
Wait for the escape key.
Shut down.
To play the chaos game:
Pick a spot within 2 inches of the screen's center.
Loop.
Draw the spot.
If a counter is past 20000, exit.
Pick a reference spot.
Find a middle spot of the spot and the reference spot.
Put the middle spot into the spot.
Repeat.
To find a middle spot of a spot and another spot:
Put the spot's x coord plus the other spot's x coord divided by 2 into the middle spot's x coord.
Put the spot's y coord plus the other spot's y coord divided by 2 into the middle spot's y coord.
The top spot is a spot.
The left spot is a spot.
The right spot is a spot.
To initialize our reference points:
Move up 2-1/2 inches.
Put the context's spot into the top spot.
Turn right. Turn 1/6 of the way around.
Move 5 inches.
Put the context's spot into the right spot.
Turn 1/3 of the way around.
Move 5 inches.
Put the context's spot into the left spot.
To pick a reference spot:
Pick a number between 1 and 3.
If the number is 1, put the top spot into the reference spot.
If the number is 2, put the right spot into the reference spot.
If the number is 3, put the left spot into the reference spot. |
http://rosettacode.org/wiki/Chaos_game | Chaos game | The Chaos Game is a method of generating the attractor of an iterated function system (IFS).
One of the best-known and simplest examples creates a fractal, using a polygon and an initial point selected at random.
Task
Play the Chaos Game using the corners of an equilateral triangle as the reference points. Add a starting point at random (preferably inside the triangle). Then add the next point halfway between the starting point and one of the reference points. This reference point is chosen at random.
After a sufficient number of iterations, the image of a Sierpinski Triangle should emerge.
See also
The Game of Chaos
| #Processing | Processing | size(300, 260);
background(#ffffff); // white
int x = floor(random(width));
int y = floor(random(height));
int colour = #ffffff;
for (int i=0; i<30000; i++) {
int v = floor(random(3));
switch (v) {
case 0:
x = x / 2;
y = y / 2;
colour = #00ff00; // green
break;
case 1:
x = width/2 + (width/2 - x)/2;
y = height - (height - y)/2;
colour = #ff0000; // red
break;
case 2:
x = width - (width - x)/2;
y = y / 2;
colour = #0000ff; // blue
}
set(x, height-y, colour);
} |
http://rosettacode.org/wiki/Chat_server | Chat server | Task
Write a server for a minimal text based chat.
People should be able to connect via ‘telnet’, sign on with a nickname, and type messages which will then be seen by all other connected users. Arrivals and departures of chat members should generate appropriate notification messages.
| #Tcl | Tcl | package require Tcl 8.6
# Write a message to everyone except the sender of the message
proc writeEveryoneElse {sender message} {
dict for {who ch} $::cmap {
if {$who ne $sender} {
puts $ch $message
}
}
}
# How to read a line (up to 256 chars long) in a coroutine
proc cgets {ch var} {
upvar 1 $var v
while {[gets $ch v] < 0} {
if {[eof $ch] || [chan pending input $ch] > 256} {
return false
}
yield
}
return true
}
# The chatting, as seen by one user
proc chat {ch addr port} {
### CONNECTION CODE ###
#Log "connection from ${addr}:${port} on channel $ch"
fconfigure $ch -buffering none -blocking 0 -encoding utf-8
fileevent $ch readable [info coroutine]
global cmap
try {
### GET THE NICKNAME OF THE USER ###
puts -nonewline $ch "Please enter your name: "
if {![cgets $ch name]} {
return
}
#Log "Mapping ${addr}:${port} to ${name} on channel $ch"
dict set cmap $name $ch
writeEveryoneElse $name "+++ $name arrived +++"
### MAIN CHAT LOOP ###
while {[cgets $ch line]} {
writeEveryoneElse $name "$name> $line"
}
} finally {
### DISCONNECTION CODE ###
if {[info exists name]} {
writeEveryoneElse $name "--- $name left ---"
dict unset cmap $name
}
close $ch
#Log "disconnection from ${addr}:${port} on channel $ch"
}
}
# Service the socket by making corouines running [chat]
socket -server {coroutine c[incr count] chat} 4004
set ::cmap {}; # Dictionary mapping nicks to channels
vwait forever; # Run event loop |
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.
| #Forth | Forth | char a
dup . \ 97
emit \ a |
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.
| #Fortran | Fortran | WRITE(*,*) ACHAR(97), IACHAR("a")
WRITE(*,*) CHAR(97), ICHAR("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. | #R | R | t(chol(matrix(c(25, 15, -5, 15, 18, 0, -5, 0, 11), nrow=3, ncol=3)))
# [,1] [,2] [,3]
# [1,] 5 0 0
# [2,] 3 3 0
# [3,] -1 1 3
t(chol(matrix(c(18, 22, 54, 42, 22, 70, 86, 62, 54, 86, 174, 134, 42, 62, 134, 106), nrow=4, ncol=4)))
# [,1] [,2] [,3] [,4]
# [1,] 4.242641 0.000000 0.000000 0.000000
# [2,] 5.185450 6.565905 0.000000 0.000000
# [3,] 12.727922 3.046038 1.649742 0.000000
# [4,] 9.899495 1.624554 1.849711 1.392621 |
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. | #Racket | Racket |
#lang racket
(require math)
(define (cholesky A)
(define mref matrix-ref)
(define n (matrix-num-rows A))
(define L (for/vector ([_ n]) (for/vector ([_ n]) 0)))
(define (set L i j x) (vector-set! (vector-ref L i) j x))
(define (ref L i j) (vector-ref (vector-ref L i) j))
(for* ([i n] [k n])
(set L i k
(cond
[(= i k)
(sqrt (- (mref A i i) (for/sum ([j k]) (sqr (ref L k j)))))]
[(> i k)
(/ (- (mref A i k) (for/sum ([j k]) (* (ref L i j) (ref L k j))))
(ref L k k))]
[else 0])))
L)
(cholesky (matrix [[25 15 -5]
[15 18 0]
[-5 0 11]]))
(cholesky (matrix [[18 22 54 42]
[22 70 86 62]
[54 86 174 134]
[42 62 134 106]]))
|
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
| #PARI.2FGP | PARI/GP | v = vector(0);
v = [];
cv = vectorv(0);
cv = []~;
m = matrix(1,1);
s = Set(v);
l = List(v);
vs = vectorsmall(0);
M = Map() |
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
| #Ring | Ring |
# Project : Combinations
n = 5
k = 3
temp = []
comb = []
num = com(n, k)
while true
temp = []
for n = 1 to 3
tm = random(4) + 1
add(temp, tm)
next
bool1 = (temp[1] = temp[2]) and (temp[1] = temp[3]) and (temp[2] = temp[3])
bool2 = (temp[1] < temp[2]) and (temp[2] < temp[3])
if not bool1 and bool2
add(comb, temp)
ok
for p = 1 to len(comb) - 1
for q = p + 1 to len(comb)
if (comb[p][1] = comb[q][1]) and (comb[p][2] = comb[q][2]) and (comb[p][3] = comb[q][3])
del(comb, p)
ok
next
next
if len(comb) = num
exit
ok
end
comb = sortfirst(comb, 1)
see showarray(comb) + nl
func com(n, k)
res1 = 1
for n1 = n - k + 1 to n
res1 = res1 * n1
next
res2 = 1
for n2 = 1 to k
res2 = res2 * n2
next
res3 = res1/res2
return res3
func showarray(vect)
svect = ""
for nrs = 1 to len(vect)
svect = "[" + vect[nrs][1] + " " + vect[nrs][2] + " " + vect[nrs][3] + "]" + nl
see svect
next
Func sortfirst(alist, ind)
aList = sort(aList,ind)
for n = 1 to len(alist)-1
for m= n + 1 to len(aList)
if alist[n][1] = alist[m][1] and alist[m][2] < alist[n][2]
temp = alist[n]
alist[n] = alist[m]
alist[m] = temp
ok
next
next
for n = 1 to len(alist)-1
for m= n + 1 to len(aList)
if alist[n][1] = alist[m][1] and alist[n][2] = alist[m][2] and alist[m][3] < alist[n][3]
temp = alist[n]
alist[n] = alist[m]
alist[m] = temp
ok
next
next
return aList
|
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.
| #TorqueScript | TorqueScript | // numbers and objects
if(%num == 1)
{
foo();
}
else if(%obj == MyObject.getID())
{
bar();
}
else
{
deusEx();
}
// strings
if(%str $= "Hello World")
{
foo();
}
else if(%str $= "Bye World")
{
bar();
}
else
{
deusEx();
} |
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}}}
.
| #Racket | Racket | #lang racket
(require (only-in math/number-theory solve-chinese))
(define as '(2 3 2))
(define ns '(3 5 7))
(solve-chinese as ns) |
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}}}
.
| #Raku | Raku | # returns x where (a * x) % b == 1
sub mul-inv($a is copy, $b is copy) {
return 1 if $b == 1;
my ($b0, @x) = $b, 0, 1;
($a, $b, @x) = (
$b,
$a % $b,
@x[1] - ($a div $b)*@x[0],
@x[0]
) while $a > 1;
@x[1] += $b0 if @x[1] < 0;
return @x[1];
}
sub chinese-remainder(*@n) {
my \N = [*] @n;
-> *@a {
N R% [+] map {
my \p = N div @n[$_];
@a[$_] * mul-inv(p, @n[$_]) * p
}, ^@n
}
}
say chinese-remainder(3, 5, 7)(2, 3, 2); |
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.
| #Portugol | Portugol |
programa {
inclua biblioteca Objetos --> obj
// "constructor" returns address of object
funcao inteiro my_class_new(inteiro value) {
inteiro this = obj.criar_objeto()
obj.atribuir_propriedade(this, "variable", value) // add property to object
retorne this
}
// "method" takes the address returned by criar_objeto
funcao my_class_some_method(inteiro this) {
my_class_set_variable(this, 1)
}
// "setter"
funcao my_class_set_variable(inteiro this, inteiro value) {
obj.atribuir_propriedade(this, "variable", value)
}
// "getter"
funcao inteiro my_class_get_variable(inteiro this) {
retorne obj.obter_propriedade_tipo_inteiro(this, "variable")
}
funcao inicio() {
inteiro this = my_class_new(0)
escreva("variable = ", my_class_get_variable(this), "\n")
my_class_some_method(this)
escreva("variable = ", my_class_get_variable(this), "\n")
my_class_set_variable(this, 99)
escreva("variable = ", my_class_get_variable(this), "\n")
obj.liberar_objeto(this)
}
}
|
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.
| #PowerShell | PowerShell |
Add-Type -Language CSharp -TypeDefinition @'
public class MyClass
{
public MyClass()
{
}
public void SomeMethod()
{
}
private int _variable;
public int Variable
{
get { return _variable; }
set { _variable = value; }
}
public static void Main()
{
// instantiate it
MyClass instance = new MyClass();
// invoke the method
instance.SomeMethod();
// set the variable
instance.Variable = 99;
// get the variable
System.Console.WriteLine( "Variable=" + instance.Variable.ToString() );
}
}
'@
|
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)
| #Rust | Rust |
//! We interpret complex numbers as points in the Cartesian plane, here. We also use the
//! [sweepline/plane sweep closest pairs algorithm][algorithm] instead of the divide-and-conquer
//! algorithm, since it's (arguably) easier to implement, and an efficient implementation does not
//! require use of unsafe.
//!
//! [algorithm]: http://www.cs.mcgill.ca/~cs251/ClosestPair/ClosestPairPS.html
extern crate num;
use num::complex::Complex;
use std::cmp::{Ordering, PartialOrd};
use std::collections::BTreeSet;
type Point = Complex<f32>;
/// Wrapper around `Point` (i.e. `Complex<f32>`) so that we can use a `TreeSet`
#[derive(PartialEq)]
struct YSortedPoint {
point: Point,
}
impl PartialOrd for YSortedPoint {
fn partial_cmp(&self, other: &YSortedPoint) -> Option<Ordering> {
(self.point.im, self.point.re).partial_cmp(&(other.point.im, other.point.re))
}
}
impl Ord for YSortedPoint {
fn cmp(&self, other: &YSortedPoint) -> Ordering {
self.partial_cmp(other).unwrap()
}
}
impl Eq for YSortedPoint {}
fn closest_pair(points: &mut [Point]) -> Option<(Point, Point)> {
if points.len() < 2 {
return None;
}
points.sort_by(|a, b| (a.re, a.im).partial_cmp(&(b.re, b.im)).unwrap());
let mut closest_pair = (points[0], points[1]);
let mut closest_distance_sqr = (points[0] - points[1]).norm_sqr();
let mut closest_distance = closest_distance_sqr.sqrt();
// the strip that we inspect for closest pairs as we sweep right
let mut strip: BTreeSet<YSortedPoint> = BTreeSet::new();
strip.insert(YSortedPoint { point: points[0] });
strip.insert(YSortedPoint { point: points[1] });
// index of the leftmost point on the strip (on points)
let mut leftmost_idx = 0;
// Start the sweep!
for (idx, point) in points.iter().enumerate().skip(2) {
// Remove all points farther than `closest_distance` away from `point`
// along the x-axis
while leftmost_idx < idx {
let leftmost_point = &points[leftmost_idx];
if (leftmost_point.re - point.re).powi(2) < closest_distance_sqr {
break;
}
strip.remove(&YSortedPoint {
point: *leftmost_point,
});
leftmost_idx += 1;
}
// Compare to points in bounding box
{
let low_bound = YSortedPoint {
point: Point {
re: ::std::f32::INFINITY,
im: point.im - closest_distance,
},
};
let mut strip_iter = strip.iter().skip_while(|&p| p < &low_bound);
loop {
let point2 = match strip_iter.next() {
None => break,
Some(p) => p.point,
};
if point2.im - point.im >= closest_distance {
// we've reached the end of the box
break;
}
let dist_sqr = (*point - point2).norm_sqr();
if dist_sqr < closest_distance_sqr {
closest_pair = (point2, *point);
closest_distance_sqr = dist_sqr;
closest_distance = dist_sqr.sqrt();
}
}
}
// Insert point into strip
strip.insert(YSortedPoint { point: *point });
}
Some(closest_pair)
}
pub fn main() {
let mut test_data = [
Complex::new(0.654682, 0.925557),
Complex::new(0.409382, 0.619391),
Complex::new(0.891663, 0.888594),
Complex::new(0.716629, 0.996200),
Complex::new(0.477721, 0.946355),
Complex::new(0.925092, 0.818220),
Complex::new(0.624291, 0.142924),
Complex::new(0.211332, 0.221507),
Complex::new(0.293786, 0.691701),
Complex::new(0.839186, 0.728260),
];
let (p1, p2) = closest_pair(&mut test_data[..]).unwrap();
println!("Closest pair: {} and {}", p1, p2);
println!("Distance: {}", (p1 - p2).norm_sqr().sqrt());
}
|
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
| #Phix | Phix | with javascript_semantics
constant tests = {{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}}
for i=1 to length(tests) do
atom {x1,y1,x2,y2,r} = tests[i],
xd = x2-x1, yd = y1-y2,
s2 = xd*xd+yd*yd, sep = sqrt(s2),
xh = (x1+x2)/2, yh = (y1+y2)/2
string txt
if sep=0 then
txt = "same points/"&iff(r=0?"radius is zero":"infinite solutions")
elsif sep=2*r then
txt = sprintf("opposite ends of diameter with centre {%.4f,%.4f}",{xh,yh})
elsif sep>2*r then
txt = sprintf("too far apart (%.4f > %.4f)",{sep,2*r})
else
atom md = sqrt(r*r-s2/4), xs = md*xd/sep, ys = md*yd/sep
txt = sprintf("{%.4f,%.4f} and {%.4f,%.4f}",{xh+ys,yh+xs,xh-ys,yh-xs})
end if
printf(1,"points {%.4f,%.4f}, {%.4f,%.4f} with radius %.1f ==> %s\n",{x1,y1,x2,y2,r,txt})
end for
|
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).
| #Rust | Rust | fn chinese_zodiac(year: usize) -> String {
static ANIMALS: [&str; 12] = [
"Rat", "Ox", "Tiger", "Rabbit", "Dragon", "Snake",
"Horse", "Goat", "Monkey", "Rooster", "Dog", "Pig",
];
static ASPECTS: [&str; 2] = ["Yang", "Yin"];
static ELEMENTS: [&str; 5] = ["Wood", "Fire", "Earth", "Metal", "Water"];
static STEMS: [char; 10] = [
'甲', '乙', '丙', '丁', '戊', '己', '庚', '辛', '壬', '癸',
];
static BRANCHES: [char; 12] = [
'子', '丑', '寅', '卯', '辰', '巳', '午', '未', '申', '酉', '戌', '亥',
];
static S_NAMES: [&str; 10] = [
"jiă", "yĭ", "bĭng", "dīng", "wù", "jĭ", "gēng", "xīn", "rén", "gŭi",
];
static B_NAMES: [&str; 12] = [
"zĭ", "chŏu", "yín", "măo", "chén", "sì",
"wŭ", "wèi", "shēn", "yŏu", "xū", "hài",
];
let y = year - 4;
let s = y % 10;
let b = y % 12;
let stem = STEMS[s];
let branch = BRANCHES[b];
let s_name = S_NAMES[s];
let b_name = B_NAMES[b];
let element = ELEMENTS[s / 2];
let animal = ANIMALS[b];
let aspect = ASPECTS[s % 2];
let cycle = y % 60 + 1;
format!(
"{} {}{} {:9} {:7} {:7} {:6} {:02}/60",
year,
stem,
branch,
format!("{}-{}", s_name, b_name),
element,
animal,
aspect,
cycle
)
}
fn main() {
let years = [1935, 1938, 1968, 1972, 1976, 1984, 2017];
println!("Year Chinese Pinyin Element Animal Aspect Cycle");
println!("---- ------- --------- ------- ------- ------ -----");
for &year in &years {
println!("{}", chinese_zodiac(year));
}
} |
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).
| #Scala | Scala | object Zodiac extends App {
val years = Seq(1935, 1938, 1968, 1972, 1976, 1984, 1985, 2017, 2018)
private def animals =
Seq("Rat",
"Ox",
"Tiger",
"Rabbit",
"Dragon",
"Snake",
"Horse",
"Goat",
"Monkey",
"Rooster",
"Dog",
"Pig")
private def animalChars =
Seq("子", "丑", "寅", "卯", "辰", "巳", "午", "未", "申", "酉", "戌", "亥")
private def elements = Seq("Wood", "Fire", "Earth", "Metal", "Water")
private def elementChars =
Seq(Array("甲", "丙", "戊", "庚", "壬"), Array("乙", "丁", "己", "辛", "癸"))
private def getYY(year: Int) = if (year % 2 == 0) "yang" else "yin"
for (year <- years) {
println(year
+ " is the year of the " + elements(math.floor((year - 4) % 10 / 2).toInt) + " "
+ animals((year - 4) % 12)
+ " (" + getYY(year) + "). "
+ elementChars(year % 2)(math.floor((year - 4) % 10 / 2).toInt)
+ animalChars((year - 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
| #Java | Java | import java.io.File;
public class FileExistsTest {
public static boolean isFileExists(String filename) {
boolean exists = new File(filename).exists();
return exists;
}
public static void test(String type, String filename) {
System.out.println("The following " + type + " called " + filename +
(isFileExists(filename) ? " exists." : " not exists.")
);
}
public static void main(String args[]) {
test("file", "input.txt");
test("file", File.separator + "input.txt");
test("directory", "docs");
test("directory", File.separator + "docs" + File.separator);
}
} |
http://rosettacode.org/wiki/Chaos_game | Chaos game | The Chaos Game is a method of generating the attractor of an iterated function system (IFS).
One of the best-known and simplest examples creates a fractal, using a polygon and an initial point selected at random.
Task
Play the Chaos Game using the corners of an equilateral triangle as the reference points. Add a starting point at random (preferably inside the triangle). Then add the next point halfway between the starting point and one of the reference points. This reference point is chosen at random.
After a sufficient number of iterations, the image of a Sierpinski Triangle should emerge.
See also
The Game of Chaos
| #Python | Python |
import argparse
import random
import shapely.geometry as geometry
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.animation as animation
def main(args):
# Styles
plt.style.use("ggplot")
# Creating figure
fig = plt.figure()
line, = plt.plot([], [], ".")
# Limit axes
plt.xlim(0, 1)
plt.ylim(0, 1)
# Titles
title = "Chaos Game"
plt.title(title)
fig.canvas.set_window_title(title)
# Getting data
data = get_data(args.frames)
# Creating animation
line_ani = animation.FuncAnimation(
fig=fig,
func=update_line,
frames=args.frames,
fargs=(data, line),
interval=args.interval,
repeat=False
)
# To save the animation install ffmpeg and uncomment
# line_ani.save("chaos_game.gif")
plt.show()
def get_data(n):
"""
Get data to plot
"""
leg = 1
triangle = get_triangle(leg)
cur_point = gen_point_within_poly(triangle)
data = []
for _ in range(n):
data.append((cur_point.x, cur_point.y))
cur_point = next_point(triangle, cur_point)
return data
def get_triangle(n):
"""
Create right triangle
"""
ax = ay = 0.0
a = ax, ay
bx = 0.5 * n
by = 0.75 * (n ** 2)
b = bx, by
cx = n
cy = 0.0
c = cx, cy
triangle = geometry.Polygon([a, b, c])
return triangle
def gen_point_within_poly(poly):
"""
Generate random point inside given polygon
"""
minx, miny, maxx, maxy = poly.bounds
while True:
x = random.uniform(minx, maxx)
y = random.uniform(miny, maxy)
point = geometry.Point(x, y)
if point.within(poly):
return point
def next_point(poly, point):
"""
Generate next point according to chaos game rules
"""
vertices = poly.boundary.coords[:-1] # Last point is the same as the first one
random_vertex = geometry.Point(random.choice(vertices))
line = geometry.linestring.LineString([point, random_vertex])
return line.centroid
def update_line(num, data, line):
"""
Update line with new points
"""
new_data = zip(*data[:num]) or [(), ()]
line.set_data(new_data)
return line,
if __name__ == "__main__":
arg_parser = argparse.ArgumentParser(description="Chaos Game by Suenweek (c) 2017")
arg_parser.add_argument("-f", dest="frames", type=int, default=1000)
arg_parser.add_argument("-i", dest="interval", type=int, default=10)
main(arg_parser.parse_args())
|
http://rosettacode.org/wiki/Chat_server | Chat server | Task
Write a server for a minimal text based chat.
People should be able to connect via ‘telnet’, sign on with a nickname, and type messages which will then be seen by all other connected users. Arrivals and departures of chat members should generate appropriate notification messages.
| #Visual_Basic_.NET | Visual Basic .NET | Imports System.Net.Sockets
Imports System.Text
Imports System.Threading
Module Module1
Class State
Private ReadOnly client As TcpClient
Private ReadOnly sb As New StringBuilder
Public Sub New(name As String, client As TcpClient)
Me.Name = name
Me.client = client
End Sub
Public ReadOnly Property Name As String
Public Sub Send(text As String)
Dim bytes = Encoding.ASCII.GetBytes(String.Format("{0}" & vbCrLf, text))
client.GetStream().Write(bytes, 0, bytes.Length)
End Sub
End Class
ReadOnly connections As New Dictionary(Of Integer, State)
Dim listen As TcpListener
Dim serverThread As Thread
Sub Main()
listen = New TcpListener(Net.IPAddress.Parse("127.0.0.1"), 4004)
serverThread = New Thread(New ThreadStart(AddressOf DoListen))
serverThread.Start()
End Sub
Private Sub DoListen()
listen.Start()
Console.WriteLine("Server: Started server")
Do
Console.Write("Server: Waiting...")
Dim client = listen.AcceptTcpClient()
Console.WriteLine(" Connected")
' New thread with client
Dim clientThread As New Thread(New ParameterizedThreadStart(AddressOf DoClient))
clientThread.Start(client)
Loop
End Sub
Private Sub DoClient(client As TcpClient)
Console.WriteLine("Client (Thread: {0}): Connected!", Thread.CurrentThread.ManagedThreadId)
Dim bytes = Encoding.ASCII.GetBytes("Enter name: ")
client.GetStream().Write(bytes, 0, bytes.Length)
Dim done As Boolean
Dim name As String
Do
If Not client.Connected Then
Console.WriteLine("Client (Thread: {0}): Terminated!", Thread.CurrentThread.ManagedThreadId)
client.Close()
Thread.CurrentThread.Abort() ' Kill thread
End If
name = Receive(client)
done = True
For Each cl In connections
Dim state = cl.Value
If state.Name = name Then
bytes = Encoding.ASCII.GetBytes("Name already registered. Please enter your name: ")
client.GetStream().Write(bytes, 0, bytes.Length)
done = False
End If
Next
Loop While Not done
connections.Add(Thread.CurrentThread.ManagedThreadId, New State(name, client))
Console.WriteLine(vbTab & "Total connections: {0}", connections.Count)
Broadcast(String.Format("+++ {0} arrived +++", name))
Do
Dim text = Receive(client)
If text = "/quit" Then
Broadcast(String.Format("Connection from {0} closed.", name))
connections.Remove(Thread.CurrentThread.ManagedThreadId)
Console.WriteLine(vbTab & "Total connections: {0}", connections.Count)
Exit Do
End If
If Not client.Connected Then
Exit Do
End If
Broadcast(String.Format("{0}> {1}", name, text))
Loop
Console.WriteLine("Client (Thread: {0}): Terminated!", Thread.CurrentThread.ManagedThreadId)
client.Close()
Thread.CurrentThread.Abort()
End Sub
Private Function Receive(client As TcpClient) As String
Dim sb As New StringBuilder
Do
If client.Available > 0 Then
While client.Available > 0
Dim ch = Chr(client.GetStream.ReadByte())
If ch = vbCr Then
' ignore
Continue While
End If
If ch = vbLf Then
Return sb.ToString()
End If
sb.Append(ch)
End While
' pause
Thread.Sleep(100)
End If
Loop
End Function
Private Sub Broadcast(text As String)
Console.WriteLine(text)
For Each client In connections
If client.Key <> Thread.CurrentThread.ManagedThreadId Then
Dim state = client.Value
state.Send(text)
End If
Next
End Sub
End Module |
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.
| #Free_Pascal | Free Pascal |
' FreeBASIC v1.05.0 win64
Print "a - > "; Asc("a")
Print "98 -> "; Chr(98)
Print
Print "Press any key to exit the program"
Sleep
End
|
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.
| #FreeBASIC | FreeBASIC |
' FreeBASIC v1.05.0 win64
Print "a - > "; Asc("a")
Print "98 -> "; Chr(98)
Print
Print "Press any key to exit the program"
Sleep
End
|
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. | #Raku | Raku | sub cholesky(@A) {
my @L = @A »*» 0;
for ^@A -> $i {
for 0..$i -> $j {
@L[$i][$j] = ($i == $j ?? &sqrt !! 1/@L[$j][$j] * * )(
@A[$i][$j] - [+] (@L[$i;*] Z* @L[$j;*])[^$j]
);
}
}
return @L;
}
.say for cholesky [
[25],
[15, 18],
[-5, 0, 11],
];
.say for cholesky [
[18, 22, 54, 42],
[22, 70, 86, 62],
[54, 86, 174, 134],
[42, 62, 134, 106],
]; |
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
| #Pascal | Pascal | var
MyArray: array[1..5] of real;
begin
MyArray[1] := 4.35;
end; |
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
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n
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1
)
…
(
n
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1
)
k
(
k
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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
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k
)
=
n
+
k
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1
C
k
=
(
n
+
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!
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!
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
| #Ruby | Ruby | def comb(m, n)
(0...n).to_a.combination(m).to_a
end
comb(3, 5) # => [[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]] |
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