Chinese remainder theorem: Difference between revisions

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ReadChar
ReadChar
END CRT.</lang>
END CRT.</lang>
{{out}}
<pre>23</pre>


=={{header|Nim}}==
=={{header|Nim}}==

Revision as of 23:34, 28 August 2018

Task
Chinese remainder theorem
You are encouraged to solve this task according to the task description, using any language you may know.

Suppose   ,   ,   ,     are positive integers that are pairwise co-prime.  

Then, for any given sequence of integers   ,   ,   ,   ,   there exists an integer     solving the following system of simultaneous congruences:

Furthermore, all solutions     of this system are congruent modulo the product,   .


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     where   .


Show the functionality of this program by printing the result such that the   's   are     and the   's   are   .


Algorithm:   The following algorithm only applies if the   's   are pairwise co-prime.

Suppose, as above, that a solution is required for the system of congruences:

Again, to begin, the product     is defined.

Then a solution     can be found as follows:

For each   ,   the integers     and     are co-prime.

Using the   Extended Euclidean algorithm,   we can find integers     and     such that   .

Then, one solution to the system of simultaneous congruences is:

and the minimal solution,

.



360 Assembly

Translation of: REXX

<lang 360asm>* Chinese remainder theorem 06/09/2015 CHINESE CSECT

        USING  CHINESE,R12        base addr
        LR     R12,R15

BEGIN LA R9,1 m=1

        LA     R6,1               j=1

LOOPJ C R6,NN do j=1 to nn

        BH     ELOOPJ
        LR     R1,R6              j
        SLA    R1,2               j*4
        M      R8,N-4(R1)         m=m*n(j)
        LA     R6,1(R6)           j=j+1
        B      LOOPJ

ELOOPJ LA R6,1 x=1 LOOPX CR R6,R9 do x=1 to m

        BH     ELOOPX
        LA     R7,1               i=1

LOOPI C R7,NN do i=1 to nn

        BH     ELOOPI
        LR     R1,R7              i
        SLA    R1,2               i*4
        LR     R5,R6              x
        LA     R4,0
        D      R4,N-4(R1)         x//n(i)
        C      R4,A-4(R1)         if x//n(i)^=a(i)
        BNE    ITERX              then iterate x
        LA     R7,1(R7)           i=i+1
        B      LOOPI

ELOOPI MVC PG(2),=C'x='

        XDECO  R6,PG+2            edit x
        XPRNT  PG,14              print buffer
        B      RETURN

ITERX LA R6,1(R6) x=x+1

        B      LOOPX

ELOOPX XPRNT NOSOL,17 print RETURN XR R15,R15 rc=0

        BR     R14

NN DC F'3' N DC F'3',F'5',F'7' A DC F'2',F'3',F'2' PG DS CL80 NOSOL DC CL17'no solution found'

        YREGS
        END    CHINESE</lang>
Output:
x=          23

Ada

Using the package Mod_Inv from [[1]].

<lang Ada>with Ada.Text_IO, Mod_Inv;

procedure Chin_Rema is

  N: array(Positive range <>) of Positive := (3, 5, 7);
  A: array(Positive range <>) of Positive := (2, 3, 2);   
  Tmp: Positive;
  Prod: Positive := 1;
  Sum: Natural := 0;

begin

  for I in N'Range loop
     Prod := Prod * N(I);
  end loop;
  
  for I in A'Range loop
     Tmp := Prod / N(I);
     Sum := Sum + A(I) * Mod_Inv.Inverse(Tmp, N(I)) * Tmp;
  end loop;
  Ada.Text_IO.Put_Line(Integer'Image(Sum mod Prod));

end Chin_Rema;</lang>

Bracmat

Translation of: C

<lang bracmat>( ( mul-inv

 =   a b b0 q x0 x1
   .   !arg:(?a.?b:?b0)
     & ( !b:1
       |   0:?x0
         & 1:?x1
         &   whl
           ' ( !a:>1
             &   (!b.mod$(!a.!b):?q.!x1+-1*!q*!x0.!x0)
               : (?a.?b.?x0.?x1)
             )
         & ( !x1:<0&!b0+!x1
           | !x1
           )
       )
 )

& ( chinese-remainder

 =   n a as p ns ni prod sum
   .   !arg:(?n.?a)
     & 1:?prod
     & 0:?sum
     & !n:?ns
     & whl'(!ns:%?ni ?ns&!prod*!ni:?prod)
     & !n:?ns
     & !a:?as
     &   whl
       ' ( !ns:%?ni ?ns
         & !as:%?ai ?as
         & div$(!prod.!ni):?p
         & !sum+!ai*mul-inv$(!p.!ni)*!p:?sum
         )
     & mod$(!sum.!prod):?arg
     & !arg
 )

& 3 5 7:?n & 2 3 2:?a & put$(str$(chinese-remainder$(!n.!a) \n)) );</lang> Output:

23

C

When n are not pairwise coprime, the program crashes due to division by zero, which is one way to convey error. <lang c>#include <stdio.h>

// returns x where (a * x) % b == 1 int mul_inv(int a, int b) { int b0 = b, t, q; int x0 = 0, x1 = 1; if (b == 1) return 1; while (a > 1) { q = a / b; t = b, b = a % b, a = t; t = x0, x0 = x1 - q * x0, x1 = t; } if (x1 < 0) x1 += b0; return x1; }

int chinese_remainder(int *n, int *a, int len) { int p, i, prod = 1, sum = 0;

for (i = 0; i < len; i++) prod *= n[i];

for (i = 0; i < len; i++) { p = prod / n[i]; sum += a[i] * mul_inv(p, n[i]) * p; }

return sum % prod; }

int main(void) { int n[] = { 3, 5, 7 }; int a[] = { 2, 3, 2 };

printf("%d\n", chinese_remainder(n, a, sizeof(n)/sizeof(n[0]))); return 0; }</lang>

C#

<lang csharp>using System; using System.Linq;

namespace ChineseRemainderTheorem {

   class Program
   {
       static void Main(string[] args)
       {
           int[] n = { 3, 5, 7 };
           int[] a = { 2, 3, 2 };
           int result = ChineseRemainderTheorem.Solve(n, a);
           int counter = 0;
           int maxCount = n.Length - 1;
           while (counter <= maxCount)
           {
               Console.WriteLine($"{result} ≡ {a[counter]} (mod {n[counter]})");
               counter++;
           }
       }
   }
   public static class ChineseRemainderTheorem
   {
       public static int Solve(int[] n, int[] a)
       {
           int prod = n.Aggregate(1, (i, j) => i * j);
           int p;
           int sm = 0;
           for (int i = 0; i < n.Length; i++)
           {
               p = prod / n[i];
               sm += a[i] * ModularMultiplicativeInverse(p, n[i]) * p;
           }
           return sm % prod;
       }
       private static int ModularMultiplicativeInverse(int a, int mod)
       {
           int b = a % mod;
           for (int x = 1; x < mod; x++)
           {
               if ((b * x) % mod == 1)
               {
                   return x;
               }
           }
           return 1;
       }
   }

}</lang>

Clojure

Modeled after the Python version http://rosettacode.org/wiki/Category:Python <lang lisp>(ns test-p.core

 (:require [clojure.math.numeric-tower :as math]))

(defn extended-gcd

 "The extended Euclidean algorithm
 Returns a list containing the GCD and the Bézout coefficients
 corresponding to the inputs. "
 [a b]
 (cond (zero? a) [(math/abs b) 0 1]
       (zero? b) [(math/abs a) 1 0]
       :else (loop [s 0
                    s0 1
                    t 1
                    t0 0
                    r (math/abs b)
                    r0 (math/abs a)]
               (if (zero? r)
                 [r0 s0 t0]
                 (let [q (quot r0 r)]
                   (recur (- s0 (* q s)) s
                          (- t0 (* q t)) t
                          (- r0 (* q r)) r))))))

(defn chinese_remainder

 " Main routine to return the chinese remainder "
 [n a]
 (let [prod (apply * n)
       reducer (fn [sum [n_i a_i]]
                 (let [p (quot prod n_i)           ; p = prod / n_i
                       egcd (extended-gcd p n_i)   ; Extended gcd
                       inv_p (second egcd)]        ; Second item is the inverse
                   (+ sum (* a_i inv_p p))))
       sum-prod (reduce reducer 0 (map vector n a))] ; Replaces the Python for loop to sum
                                                     ; (map vector n a) is same as
       ;                                             ; Python's version Zip (n, a)
   (mod sum-prod prod)))                             ; Result line

(def n [3 5 7]) (def a [2 3 2])

(println (chinese_remainder n a)) </lang>

Output:

23

Coffeescript

<lang coffeescript>crt = (n,a) -> sum = 0 prod = n.reduce (a,c) -> a*c for [ni,ai] in _.zip n,a p = prod // ni sum += ai * p * mulInv p,ni sum % prod

mulInv = (a,b) -> b0 = b [x0,x1] = [0,1] if b==1 then return 1 while a > 1 q = a // b [a,b] = [b, a % b] [x0,x1] = [x1-q*x0, x0] if x1 < 0 then x1 += b0 x1

print crt [3,5,7], [2,3,2]</lang> Output:

23

Common Lisp

Using function invmod from [[2]]. <lang lisp> (defun chinese-remainder (am) "Calculates the Chinese Remainder for the given set of integer modulo pairs.

Note: All the ni and the N must be coprimes."
 (loop :for (a . m) :in am
       :with mtot = (reduce #'* (mapcar #'(lambda(X) (cdr X)) am))
       :with sum  = 0
       :finally (return (mod sum mtot))
       :do
  (incf sum (* a (invmod (/ mtot m) m) (/ mtot m)))))

</lang>

Output:

* (chinese-remainder '((2 . 3) (3 . 5) (2 . 7)))

23
* (chinese-remainder '((10 . 11) (4 . 12) (12 . 13)))

1000
* (chinese-remainder '((19 . 100) (0 . 23)))

1219
* (chinese-remainder '((10 . 11) (4 . 22) (9 . 19)))

debugger invoked on a SIMPLE-ERROR in thread
#<THREAD "main thread" RUNNING {1002A8B1B3}>:
  invmod: Values 418 and 11 are not coprimes.

Type HELP for debugger help, or (SB-EXT:EXIT) to exit from SBCL.

restarts (invokable by number or by possibly-abbreviated name):
  0: [ABORT] Exit debugger, returning to top level.

(INVMOD 418 11)
0] 

D

Translation of: Python

<lang d>import std.stdio, std.algorithm;

T chineseRemainder(T)(in T[] n, in T[] a) pure nothrow @safe @nogc in {

   assert(n.length == a.length);

} body {

   static T mulInv(T)(T a, T b) pure nothrow @safe @nogc {
       auto b0 = b;
       T x0 = 0, x1 = 1;
       if (b == 1)
           return T(1);
       while (a > 1) {
           immutable q = a / b;
           immutable amb = a % b;
           a = b;
           b = amb;
           immutable xqx = x1 - q * x0;
           x1 = x0;
           x0 = xqx;
       }
       if (x1 < 0)
           x1 += b0;
       return x1;
   }
   immutable prod = reduce!q{a * b}(T(1), n);
   T p = 1, sm = 0;
   foreach (immutable i, immutable ni; n) {
       p = prod / ni;
       sm += a[i] * mulInv(p, ni) * p;
   }
   return sm % prod;

}

void main() {

   immutable n = [3, 5, 7],
             a = [2, 3, 2];
   chineseRemainder(n, a).writeln;

}</lang>

Output:
23

EchoLisp

egcd - extended gcd - and crt-solve - chinese remainder theorem solve - are included in math.lib. <lang scheme> (lib 'math) math.lib v1.10 ® EchoLisp Lib: math.lib loaded.

(crt-solve '(2 3 2) '(3 5 7))

  → 23

(crt-solve '(2 3 2) '(7 1005 15)) 💥 error: mod[i] must be co-primes : assertion failed : 1005 </lang>

Elixir

Translation of: Ruby
Works with: Elixir version 1.2

Brute-force: <lang elixir>defmodule Chinese do

 def remainder(mods, remainders) do
   max = Enum.reduce(mods, fn x,acc -> x*acc end)
   Enum.zip(mods, remainders)
   |> Enum.map(fn {m,r} -> Enum.take_every(r..max, m) |> MapSet.new end)
   |> Enum.reduce(fn set,acc -> MapSet.intersection(set, acc) end)
   |> MapSet.to_list
 end

end

IO.inspect Chinese.remainder([3,5,7], [2,3,2]) IO.inspect Chinese.remainder([10,4,9], [11,22,19]) IO.inspect Chinese.remainder([11,12,13], [10,4,12])</lang>

Output:
[23]
[]
[1000]

Erlang

Translation of: OCaml

<lang erlang>-module(crt). -import(lists, [zip/2, unzip/1, foldl/3, sum/1]). -export([egcd/2, mod/2, mod_inv/2, chinese_remainder/1]).

egcd(_, 0) -> {1, 0}; egcd(A, B) ->

   {S, T} = egcd(B, A rem B),
   {T, S - (A div B)*T}.

mod_inv(A, B) ->

   {X, Y} = egcd(A, B),
   if
       A*X + B*Y =:= 1 -> X;
       true -> undefined
   end.

mod(A, M) ->

   X = A rem M,
   if
       X < 0 -> X + M;
       true -> X
   end.

calc_inverses([], []) -> []; calc_inverses([N | Ns], [M | Ms]) ->

   case mod_inv(N, M) of
       undefined -> undefined;
       Inv -> [Inv | calc_inverses(Ns, Ms)]
   end.

chinese_remainder(Congruences) ->

   {Residues, Modulii} = unzip(Congruences),
   ModPI = foldl(fun(A, B) -> A*B end, 1, Modulii),
   CRT_Modulii = [ModPI div M || M <- Modulii],
   case calc_inverses(CRT_Modulii, Modulii) of
       undefined -> undefined;
       Inverses ->
           Solution = sum([A*B || {A,B} <- zip(CRT_Modulii,
                                   [A*B || {A,B} <- zip(Residues, Inverses)])]),
           mod(Solution, ModPI)
   end.</lang>
Output:
16> crt:chinese_remainder([{10,11}, {4,12}, {12,13}]).
1000
17> crt:chinese_remainder([{10,11}, {4,22}, {9,19}]).
undefined
18> crt:chinese_remainder([{2,3}, {3,5}, {2,7}]).
23

F#

sieving

<lang fsharp>let rec sieve cs x N =

   match cs with
   | [] -> Some(x)
   | (a,n)::rest ->
       let arrProgress = Seq.unfold (fun x -> Some(x, x+N)) x
       let firstXmodNequalA = Seq.tryFind (fun x -> a = x % n)
       match firstXmodNequalA (Seq.take n arrProgress) with
       | None -> None
       | Some(x) -> sieve rest x (N*n)

[ [(2,3);(3,5);(2,7)];

 [(10,11); (4,22); (9,19)];
 [(10,11); (4,12); (12,13)] ]

|> List.iter (fun congruences ->

   let cs =
       congruences
       |> List.map (fun (a,n) -> (a % n, n))
       |> List.sortBy (snd>>(~-)) 
   let an = List.head cs
   match sieve (List.tail cs) (fst an) (snd an) with
   | None    -> printfn "no solution"
   | Some(x) -> printfn "result = %i" x

)</lang>

Output:
result = 23
no solution
result = 1000

Or for those who prefer unsieved

This uses Greatest_common_divisor#F.23 to verify valid input, can be simplified if you know input has a solution.
This uses Modular_inverse#F.23 <lang fsharp> //Chinese Division Theorem: Nigel Galloway: April 3rd., 2017 let CD n g =

 match Seq.fold(fun n g->if (gcd n g)=1 then n*g else 0) 1 g with
 |0 -> None
 |fN-> Some ((Seq.fold2(fun n i g -> n+i*(fN/g)*(MI g ((fN/g)%g))) 0 n g)%fN)

</lang>

Output:
CD [10;4;12] [11;12;13] -> Some 1000
CD [10;4;9] [11;22;19]  -> None
CD [2;3;2] [3;5;7]      -> Some 23

Factor

<lang factor>USING: math.algebra prettyprint ; { 2 3 2 } { 3 5 7 } chinese-remainder .</lang>

Output:
23

Forth

Tested with GNU FORTH <lang forth>: egcd ( a b -- a b )

   dup 0= IF
       2drop 1 0
   ELSE
       dup -rot /mod               \ -- b r=a%b q=a/b
       -rot recurse                \ -- q (s,t) = egcd(b, r)
       >r swap r@ * - r> swap      \ -- t (s - q*t)
   THEN ;
egcd>gcd ( a b x y -- n ) \ calculate gcd from egcd
   rot * -rot * + ;
mod-inv ( a m -- a' ) \ modular inverse with coprime check
   2dup egcd over >r egcd>gcd r> swap 1 <> -24 and throw ;
array-product ( adr count -- n )
   1 -rot  cells bounds ?DO  i @ *  cell +LOOP ;
crt-from-array ( adr1 adr2 count -- n )
   2dup array-product   locals| M count m[] a[] |
   0  \ result
   count 0 DO
       m[] i cells + @
       dup M swap /
       dup rot mod-inv *
       a[] i cells + @ * +
   LOOP  M mod ;

create crt-residues[] 10 cells allot create crt-moduli[] 10 cells allot

crt ( .... n -- n ) \ takes pairs of "n (mod m)" from stack.
   10 min  locals| n |
   n 0 DO
       crt-moduli[] i cells + !
       crt-residues[] i cells + !
   LOOP
   crt-residues[] crt-moduli[] n crt-from-array ;

</lang>

Output:
Gforth 0.7.2, Copyright (C) 1995-2008 Free Software Foundation, Inc.
Gforth comes with ABSOLUTELY NO WARRANTY; for details type `license'
Type `bye' to exit
10 11  4 12  12 13  3 crt . 1000  ok
10 11  4 22   9 19  3 crt . 
:2: Invalid numeric argument
10 11  4 22   9 19  3 >>>crt<<< .

FunL

<lang funl>import integers.modinv

def crt( congruences ) =

   N = product( n | (_, n) <- congruences )
   sum( a*modinv(N/n, n)*N/n | (a, n) <- congruences ) mod N

println( crt([(2, 3), (3, 5), (2, 7)]) )</lang>

Output:
23

Go

Go has the Extended Euclidean algorithm in the GCD function for big integers in the standard library. GCD will return 1 only if numbers are coprime, so a result != 1 indicates the error condition. <lang go>package main

import (

   "fmt"
   "math/big"

)

var one = big.NewInt(1)

func crt(a, n []*big.Int) (*big.Int, error) {

   p := new(big.Int).Set(n[0])
   for _, n1 := range n[1:] {
       p.Mul(p, n1)
   }
   var x, q, s, z big.Int
   for i, n1 := range n {
       q.Div(p, n1)
       z.GCD(nil, &s, n1, &q)
       if z.Cmp(one) != 0 {
           return nil, fmt.Errorf("%d not coprime", n1)
       }
       x.Add(&x, s.Mul(a[i], s.Mul(&s, &q)))
   }
   return x.Mod(&x, p), nil

}

func main() {

   n := []*big.Int{
       big.NewInt(3),
       big.NewInt(5),
       big.NewInt(7),
   }
   a := []*big.Int{
       big.NewInt(2),
       big.NewInt(3),
       big.NewInt(2),
   }
   fmt.Println(crt(a, n))

}</lang>

Output:

Two values, the solution x and an error value.

23 <nil>

Haskell

Translation of: Erlang

<lang haskell>import Control.Monad (zipWithM)

egcd :: Int -> Int -> (Int, Int) egcd _ 0 = (1, 0) egcd a b = (t, s - q * t)

 where
   (s, t) = egcd b r
   (q, r) = a `quotRem` b

modInv :: Int -> Int -> Maybe Int modInv a b =

 case egcd a b of
   (x, y)
     | a * x + b * y == 1 -> Just x
     | otherwise -> Nothing

chineseRemainder :: [Int] -> [Int] -> Maybe Int chineseRemainder residues modulii = do

 inverses <- zipWithM modInv crtModulii modulii
 return . (`mod` modPI) . sum $
   zipWith (*) crtModulii (zipWith (*) residues inverses)
 where
   modPI = product modulii
   crtModulii = (modPI `div`) <$> modulii

main :: IO () main =

 mapM_ print $
 uncurry chineseRemainder <$>
 [ ([10, 4, 12], [11, 12, 13])
 , ([10, 4, 9], [11, 22, 19])
 , ([2, 3, 2], [3, 5, 7])
 ]</lang>
Output:
Just 1000
Nothing
Just 23

Icon and Unicon

Translation of: Python

with error check added.

Works in both languages: <lang unicon>link numbers # for gcd()

procedure main()

   write(cr([3,5,7],[2,3,2]) | "No solution!")
   write(cr([10,4,9],[11,22,19]) | "No solution!")

end

procedure cr(n,a)

   if 1 ~= gcd(n[i := !*n],a[i]) then fail  # Not pairwise coprime
   (prod := 1, sm := 0)
   every prod *:= !n
   every p := prod/(ni := n[i := !*n]) do sm +:= a[i] * mul_inv(p,ni) * p
   return sm%prod

end

procedure mul_inv(a,b)

   if b = 1 then return 1
   (b0 := b, x0 := 0, x1 := 1)
   while q := (1 < a)/b do {
       (t := a, a := b, b := t%b)
       (t := x0, x0 := x1-q*t, x1 := t)
       }
   return if x1 < 0 then x1+b0 else x1

end</lang>

Output:

->crt
23
No solution!
->

J

Solution (brute force):<lang j> crt =: (1 + ] - {:@:[ -: {.@:[ | ])^:_&0@:,:</lang> Example: <lang j> 3 5 7 crt 2 3 2 23

  11 12 13 crt 10 4 12

1000</lang> Notes: This is a brute force approach and does not meet the requirement for explicit notification of an an unsolvable set of equations (it just spins forever). A much more thorough and educational approach can be found on the J wiki's Essay on the Chinese Remainder Thereom.

Java

Translation of Python via D

Works with: Java version 8

<lang java>import static java.util.Arrays.stream;

public class ChineseRemainderTheorem {

   public static int chineseRemainder(int[] n, int[] a) {
       int prod = stream(n).reduce(1, (i, j) -> i * j);
       int p, sm = 0;
       for (int i = 0; i < n.length; i++) {
           p = prod / n[i];
           sm += a[i] * mulInv(p, n[i]) * p;
       }
       return sm % prod;
   }
   private static int mulInv(int a, int b) {
       int b0 = b;
       int x0 = 0;
       int x1 = 1;
       if (b == 1)
           return 1;
       while (a > 1) {
           int q = a / b;
           int amb = a % b;
           a = b;
           b = amb;
           int xqx = x1 - q * x0;
           x1 = x0;
           x0 = xqx;
       }
       if (x1 < 0)
           x1 += b0;
       return x1;
   }
   public static void main(String[] args) {
       int[] n = {3, 5, 7};
       int[] a = {2, 3, 2};
       System.out.println(chineseRemainder(n, a));
   }

}</lang>

23

jq

This implementation is similar to the one in C, but raises an error if there is no solution, as illustrated in the last example. <lang jq># mul_inv(a;b) returns x where (a * x) % b == 1, or else null def mul_inv(a; b):

 # state: [a, b, x0, x1]
 def iterate:
   .[0] as $a | .[1] as $b
   | if $a > 1 then
       if $b == 0 then null
       else ($a / $b | floor) as $q
          | [$b, ($a % $b), (.[3] - ($q * .[2])), .[2]] | iterate
       end
     else .
     end ;
 if (b == 1) then 1
 else [a,b,0,1] | iterate
      | if . == null then .
        else  .[3] | if . <  0 then . + b else . end
        end
 end;

def chinese_remainder(mods; remainders):

 (reduce mods[] as $i (1; . * $i)) as $prod
 | reduce range(0; mods|length) as $i
     (0;
      ($prod/mods[$i]) as $p
      | mul_inv($p; mods[$i]) as $mi
      | if $mi == null then error("nogo: p=\($p) mods[\($i)]=\(mods[$i])")
        else . + (remainders[$i] * $mi * $p)
        end )
 | . % $prod ;</lang>

Examples:

chinese_remainder([3,5,7]; [2,3,2])   
# => 23
chinese_remainder([100,23]; [19,0])
# => 1219
chinese_remainder([10,4,9]; [11,22,19])
# jq: error: nogo: p=36 mods[0]=10

Julia

Works with: Julia version 0.6
Translation of: Python

<lang julia>function chineseremainder(n::Array{Int}, a::Array{Int})

   sum = 0
   prd = prod(n)
   for (ni, ai) in zip(n, a)
       p   = prd ÷ ni
       sum += ai * mulinv(p, ni) * p
   end
   return sum % prd

end

function mulinv(a::Int, b::Int)

   @assert(a % b != 0, "$a is multiple of $b")
   @assert(b % a != 0, "$b is multiple of $a")
   b0 = b
   x0, x1 = 0, 1
   if b == 1 return 1 end
   while a > 1
       q = a ÷ b
       a, b = b, a % b
       x0, x1 = x1 - q * x0, x0
   end
   if x1 < 0 x1 += b0 end
   return x1

end

@show chineseremainder([3, 5, 7], [2, 3, 2])</lang>

Output:
chineseremainder([3, 5, 7], [2, 3, 2]) = 23

Kotlin

Translation of: C

<lang scala>// version 1.1.2

/* returns x where (a * x) % b == 1 */ fun multInv(a: Int, b: Int): Int {

   if (b == 1) return 1
   var aa = a
   var bb = b
   var x0 = 0
   var x1 = 1
   while (aa > 1) {
       val q = aa / bb
       var t = bb
       bb = aa % bb
       aa = t
       t = x0
       x0 = x1 - q * x0
       x1 = t
   }
   if (x1 < 0) x1 += b
   return x1

}

fun chineseRemainder(n: IntArray, a: IntArray): Int {

   val prod = n.fold(1) { acc, i -> acc * i }
   var sum = 0
   for (i in 0 until n.size) {
       val p = prod / n[i]
       sum += a[i] * multInv(p, n[i]) * p
   }
   return sum % prod

}

fun main(args: Array<String>) {

   val n = intArrayOf(3, 5, 7)
   val a = intArrayOf(2, 3, 2)
   println(chineseRemainder(n, a))

}</lang>

Output:
23

Maple

This is a Maple built-in procedure, so it is trivial: <lang Maple>> chrem( [2, 3, 2], [3, 5, 7] );

                                          23

</lang>


Mathematica / Wolfram Language

Very easy, because it is a built-in function: <lang Mathematica >ChineseRemainder[{2, 3, 2}, {3, 5, 7}] 23</lang>

MATLAB / Octave

<lang MATLAB>function f = chineseRemainder(r, m)

 s = prod(m) ./ m;
 [~, t] = gcd(s, m);
 f = s .* t * r';</lang>
Output:

<lang MATLAB>>> chineseRemainder([2 3 2], [3 5 7])

ans = 23</lang>

Modula-2

<lang modula2>MODULE CRT; FROM FormatString IMPORT FormatString; FROM Terminal IMPORT WriteString,WriteLn,ReadChar;

PROCEDURE WriteInt(n : INTEGER); VAR buf : ARRAY[0..15] OF CHAR; BEGIN

   FormatString("%i", buf, n);
   WriteString(buf)

END WriteInt;

PROCEDURE MulInv(a,b : INTEGER) : INTEGER; VAR

   b0,x0,x1,q,amb,xqx : INTEGER;

BEGIN

   b0 := b;
   x0 := 0;
   x1 := 1;
   IF b=1 THEN
       RETURN 1
   END;
   WHILE a>1 DO
       q := a DIV b;
       amb := a MOD b;
       a := b;
       b := amb;
       xqx := x1 - q * x0;
       x1 := x0;
       x0 := xqx
   END;
   IF x1<0 THEN
       x1 := x1 + b0
   END;
   RETURN x1

END MulInv;

PROCEDURE ChineseRemainder(n,a : ARRAY OF INTEGER) : INTEGER; VAR

   i : CARDINAL;
   prod,p,sm : INTEGER;

BEGIN

   prod := n[0];
   FOR i:=1 TO HIGH(n) DO
       prod := prod * n[i]
   END;
   sm := 0;
   FOR i:=0 TO HIGH(n) DO
       p := prod DIV n[i];
       sm := sm + a[i] * MulInv(p, n[i]) * p
   END;
   RETURN sm MOD prod

END ChineseRemainder;

TYPE TA = ARRAY[0..2] OF INTEGER; VAR n,a : TA; BEGIN

   n := TA{3, 5, 7};
   a := TA{2, 3, 2};
   WriteInt(ChineseRemainder(n, a));
   WriteLn;
   ReadChar

END CRT.</lang>

Output:
23

Nim

Translation of: C

<lang nim>proc mulInv(a0, b0): int =

 var (a, b, x0) = (a0, b0, 0)
 result = 1
 if b == 1: return
 while a > 1:
   let q = a div b
   a = a mod b
   swap a, b
   result = result - q * x0
   swap x0, result
 if result < 0: result += b0

proc chineseRemainder[T](n, a: T): int =

 var prod = 1
 var sum = 0
 for x in n: prod *= x
 for i in 0 .. <n.len:
   let p = prod div n[i]
   sum += a[i] * mulInv(p, n[i]) * p
 sum mod prod

echo chineseRemainder([3,5,7], [2,3,2])</lang> Output:

23

OCaml

This is using the Jane Street Ocaml Core library. <lang ocaml>open Core.Std open Option.Monad_infix

let rec egcd a b =

  if b = 0 then (1, 0)
  else
     let q = a/b and r = a mod b in
     let (s, t) = egcd b r in
        (t, s - q*t)


let mod_inv a b =

  let (x, y) = egcd a b in
     if a*x + b*y = 1 then Some x else None


let calc_inverses ns ms =

  let rec list_inverses ns ms l =
     match (ns, ms) with
        | ([], []) -> Some l
        | ([], _)
        | (_, []) -> assert false
        | (n::ns, m::ms) ->
           let inv = mod_inv n m in
              match inv with
                 | None -> None
                 | Some v -> list_inverses ns ms (v::l)
  in
     list_inverses ns ms [] >>= fun l -> Some (List.rev l)


let chinese_remainder congruences =

  let (residues, modulii) = List.unzip congruences in
  let mod_pi = List.reduce_exn modulii ~f:( * ) in
  let crt_modulii = List.map modulii ~f:(fun m -> mod_pi / m) in
  calc_inverses crt_modulii modulii >>=
     fun inverses ->
        Some (List.map3_exn residues inverses crt_modulii ~f:(fun a b c -> a*b*c)
              |> List.reduce_exn ~f:(+)
              |> fun n -> let n' = n mod mod_pi in if n' < 0 then n' + mod_pi else n')

</lang>

Output:
utop # chinese_remainder [(10, 11); (4, 12); (12, 13)];;
- : int option = Some 1000 
                                                                                                        
utop # chinese_remainder [(10, 11); (4, 22); (9, 19)];;
- : int option = None  

PARI/GP

<lang parigp>chivec(residues, moduli)={

 my(m=Mod(0,1));
 for(i=1,#residues,
   m=chinese(Mod(residues[i],moduli[i]),m)
 );
 lift(m)

}; chivec([2,3,2], [3,5,7])</lang>

Output:
23

Pari's chinese function takes a vector in the form [Mod(a1,n1), Mod(a2, n2), ...], so we can do this directly: <lang parigp>lift( chinese([Mod(2,3),Mod(3,5),Mod(2,7)]) )</lang> or to take the residue/moduli array as above: <lang parigp>chivec(residues,moduli)={

 lift(chinese(vector(#residues,i,Mod(residues[i],moduli[i]))))

}</lang>

Perl

There are at least three CPAN modules for this: ntheory (Math::Prime::Util), Math::ModInt, and Math::Pari. All three handle bigints.

Library: ntheory

<lang perl>use ntheory qw/chinese/; say chinese([2,3], [3,5], [2,7]);</lang>

Output:
23

The function returns undef if no common residue class exists. The combined modulus can be obtained using the lcm function applied to the moduli (e.g. lcm(3,5,7) = 105 in the example above).

<lang perl>use Math::ModInt qw(mod); use Math::ModInt::ChineseRemainder qw(cr_combine); say cr_combine(mod(2,3),mod(3,5),mod(2,7));</lang>

Output:
mod(23, 105)

This returns a Math::ModInt object, which if no common residue class exists will be a special undefined object. The modulus and residue methods may be used to extract the integer components.

Non-pairwise-coprime

All three modules will also handle cases where the moduli are not pairwise co-prime but a solution exists, e.g.: <lang perl>use ntheory qw/chinese lcm/; say chinese( [2328,16256], [410,5418] ), " mod ", lcm(16256,5418);</lang>

Output:
28450328 mod 44037504

Perl 6

Translation of: C
Works with: Rakudo version 2015.12

<lang perl6># 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);</lang>

Output:
23

PicoLisp

<lang 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)</lang>

PureBasic

<lang 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()</lang>

Output:
[( 2 . 3 )( 3 . 5 )( 2 . 7 )]
x = 23
[( 10 . 11 )( 4 . 12 )( 12 . 13 )]
x = 1000
[( 11 . 10 )( 22 . 4 )( 19 . 9 )]
x = Division by zero

Python

<lang 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)</lang>
Output:
23

R

Translation of: C

<lang rsplus>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)</lang>

Output:
23


Racket

This is more of a demonstration of the built-in function "solve-chinese", than anything. A bit cheeky, I know... but if you've got a dog, why bark yourself?

Take a look in the "math/number-theory" package it's full of goodies! URL removed -- I can't be doing the Dutch recaptchas I'm getting. <lang 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)</lang>

Output:
23

REXX

algebraic

<lang rexx>/*REXX program demonstrates Sun Tzu's (or Sunzi's) Chinese Remainder Theorem. */ parse arg Ns As . /*get optional arguments from the C.L. */ if Ns== | Ns=="," then Ns = '3,5,7' /*Ns not specified? Then use default.*/ if As== | As=="," then As = '2,3,2' /*As " " " " " */

      say 'Ns: ' Ns
      say 'As: ' As;                   say

Ns=space(translate(Ns, , ',')); #=words(Ns) /*elide any superfluous blanks from N's*/ As=space(translate(As, , ',')); _=words(As) /* " " " " " A's*/ if #\==_ then do; say "size of number sets don't match."; exit 131; end if #==0 then do; say "size of the N set isn't valid."; exit 132; end if _==0 then do; say "size of the A set isn't valid."; exit 133; end N=1 /*the product─to─be for prod(n.j). */

     do j=1  for #                              /*process each number for  As  and Ns. */
     n.j=word(Ns,j);  N=N*n.j                   /*get an  N.j  and calculate product.  */
     a.j=word(As,j)                             /* "   "  A.j  from the  As  list.     */
     end   /*j*/
     do    x=1  for N                           /*use a simple algebraic method.       */
        do i=1  for #                           /*process each   N.i  and  A.i  number.*/
        if x//n.i\==a.i  then iterate x         /*is modulus correct for the number X ?*/
        end   /*i*/                             /* [↑]  limit solution to the product. */
     say 'found a solution with X='   x         /*display one possible solution.       */
     exit                                       /*stick a fork in it,  we're all done. */
     end      /*x*/

say 'no solution found.' /*oops, announce that solution ¬ found.*/</lang>

output   when using the default inputs:
Ns:  3,5,7
As:  2,3,2

found a solution with X= 23

congruences sets

<lang rexx>/*REXX program demonstrates Sun Tzu's (or Sunzi's) Chinese Remainder Theorem. */ parse arg Ns As . /*get optional arguments from the C.L. */ if Ns== | Ns=="," then Ns = '3,5,7' /*Ns not specified? Then use default.*/ if As== | As=="," then As = '2,3,2' /*As " " " " " */

      say 'Ns: ' Ns
      say 'As: ' As;                   say

Ns=space(translate(Ns, , ',')); #=words(Ns) /*elide any superfluous blanks from N's*/ As=space(translate(As, , ',')); _=words(As) /* " " " " " A's*/ if #\==_ then do; say "size of number sets don't match."; exit 131; end if #==0 then do; say "size of the N set isn't valid."; exit 132; end if _==0 then do; say "size of the A set isn't valid."; exit 133; end N=1 /*the product─to─be for prod(n.j). */

     do j=1  for #                              /*process each number for  As  and Ns. */
     n.j=word(Ns,j);  N=N*n.j                   /*get an  N.j  and calculate product.  */
     a.j=word(As,j)                             /* "   "  A.j  from the  As  list.     */
     end   /*j*/

@.= /* [↓] converts congruences ───► sets.*/

     do i=1  for #;  _=a.i;  @.i._=a.i;  p=a.i
       do N; p=p+n.i;  @.i.p=p;  end            /*build a (array) list of modulo values*/
     end   /*i*/
                                                /* [↓]  find common number in the sets.*/
 do   x=1  for N;  if @.1.x==  then iterate                       /*locate a number. */
   do v=2  to #;   if @.v.x==  then iterate x;  end               /*Is in all sets ? */
 say 'found a solution with X='    x            /*display one possible solution.       */
 exit                                           /*stick a fork in it,  we're all done. */
 end   /*x*/

say 'no solution found.' /*oops, announce that solution ¬ found.*/</lang>

output   is identical to the 1st REXX version.



Ruby

Brute-force. <lang ruby> def chinese_remainder(mods, remainders)

 max = mods.inject( :* )                            
 series = remainders.zip( mods ).map{|r,m| r.step( max, m ).to_a } 
 series.inject( :& ).first #returns nil when empty

end

p chinese_remainder([3,5,7], [2,3,2]) #=> 23 p chinese_remainder([10,4,9], [11,22,19]) #=> nil </lang>

Similar to above, but working with large(r) numbers. <lang ruby> def extended_gcd(a, b)

 last_remainder, remainder = a.abs, b.abs
 x, last_x, y, last_y = 0, 1, 1, 0
 while remainder != 0
   last_remainder, (quotient, remainder) = remainder, last_remainder.divmod(remainder)
   x, last_x = last_x - quotient*x, x
   y, last_y = last_y - quotient*y, y
 end
 return last_remainder, last_x * (a < 0 ? -1 : 1)

end

def invmod(e, et)

 g, x = extended_gcd(e, et)
 if g != 1
   raise 'Multiplicative inverse modulo does not exist!'
 end
 x % et

end

def chinese_remainder(mods, remainders)

 max = mods.inject( :* )  # product of all moduli
 series = remainders.zip(mods).map{ |r,m| (r * max * invmod(max/m, m) / m) }
 series.inject( :+ ) % max 

end

p chinese_remainder([3,5,7], [2,3,2]) #=> 23 p chinese_remainder([17353461355013928499, 3882485124428619605195281, 13563122655762143587], [7631415079307304117, 1248561880341424820456626, 2756437267211517231]) #=> 937307771161836294247413550632295202816 p chinese_remainder([10,4,9], [11,22,19]) #=> nil </lang>

Rust

This example does not show the output mentioned in the task description on this page (or a page linked to from here). Please ensure that it meets all task requirements and remove this message.
Note that phrases in task descriptions such as "print and display" and "print and show" for example, indicate that (reasonable length) output be a part of a language's solution.


<lang rust>fn mul_inv(mut a: i32,mut b: i32)-> i32 { let b0=b;let mut t;let mut q; let mut x0=0;let mut x1=1; if b==1 {return 1; } while a>1 { q=a/b; t=b; b=a%b; a=t; t=x0; x0=x1-q*x0; x1=t; } if x1<0 {x1+=b0; } x1 }

fn chinese_remainder(n: &mut[i32],a: &mut[i32],len: usize)->i32 { let mut p=0;let mut prod=1;let mut sum=0; for i in 0..len { prod*=n[i]; } for i in 0..len { p=prod/n[i]; sum += a[i]*mul_inv(p, n[i])*p; } sum%prod }


fn main() {

   let mut n = [3,5,7];

let mut a = [2,3,2]; let s = a.len();

   println!("{}",chinese_remainder(&mut n,&mut a,s));
  

}</lang>

Scala

Output:

Best seen running in your browser either by ScalaFiddle (ES aka JavaScript, non JVM) or Scastie (remote JVM).

<lang Scala>import scala.util.{Success, Try}

object ChineseRemainderTheorem extends App {

 def chineseRemainder(n: List[Int], a: List[Int]): Option[Int] = {
   require(n.size == a.size)
   val prod = n.product
   def iter(n: List[Int], a: List[Int], sm: Int): Int = {
     def mulInv(a: Int, b: Int): Int = {
       def loop(a: Int, b: Int, x0: Int, x1: Int): Int = {
         if (a > 1) loop(b, a % b, x1 - (a / b) * x0, x0) else x1
       }
       if (b == 1) 1
       else {
         val x1 = loop(a, b, 0, 1)
         if (x1 < 0) x1 + b else x1
       }
     }
     if (n.nonEmpty) {
       val p = prod / n.head
       iter(n.tail, a.tail, sm + a.head * mulInv(p, n.head) * p)
     } else sm
   }
   Try {
     iter(n, a, 0) % prod
   } match {
     case Success(v) => Some(v)
     case _          => None
   }
 }
 println(chineseRemainder(List(3, 5, 7), List(2, 3, 2)))
 println(chineseRemainder(List(11, 12, 13), List(10, 4, 12)))
 println(chineseRemainder(List(11, 22, 19), List(10, 4, 9)))

}</lang>

Seed7

<lang seed7>$ include "seed7_05.s7i";

 include "bigint.s7i";

const func integer: modInverse (in integer: a, in integer: b) is

 return ord(modInverse(bigInteger conv a, bigInteger conv b));

const proc: main is func

 local
   const array integer: n is [] (3, 5, 7);
   const array integer: a is [] (2, 3, 2);
   var integer: num is 0;
   var integer: prod is 1;
   var integer: sum is 0;
   var integer: index is 0;
 begin
   for num range n do
     prod *:= num;
   end for;
   for key index range a do
     num := prod div n[index];
     sum +:= a[index] * modInverse(num, n[index]) * num;
   end for;
   writeln(sum mod prod);
 end func;</lang>
Output:
23

Sidef

Translation of: Perl 6

<lang ruby>func chinese_remainder(*n) {

   var N = n.prod
   func (*a) {
       n.range.sum { |i|
           var p = (N / n[i])
           a[i] * p.invmod(n[i]) * p
       } % N
   }

}

say chinese_remainder(3, 5, 7)(2, 3, 2)</lang>

Output:
23

Swift

<lang swift>import Darwin

/*

* Function: euclid
* Usage: (r,s) = euclid(m,n)
* --------------------------
* The extended Euclidean algorithm subsequently performs
* Euclidean divisions till the remainder is zero and then
* returns the Bézout coefficients r and s.
*/

func euclid(_ m:Int, _ n:Int) -> (Int,Int) {

   if m % n == 0 {
       return (0,1)
   } else {
       let rs = euclid(n % m, m)
       let r = rs.1 - rs.0 * (n / m)
       let s = rs.0
       return (r,s)
   }

}

/*

* Function: gcd
* Usage: x = gcd(m,n)
* -------------------
* The greatest common divisor of two numbers a and b
* is expressed by ax + by = gcd(a,b) where x and y are
* the Bézout coefficients as determined by the extended
* euclidean algorithm.
*/

func gcd(_ m:Int, _ n:Int) -> Int {

   let rs = euclid(m, n)
   return m * rs.0 + n * rs.1

}

/*

* Function: coprime
* Usage: truth = coprime(m,n)
* ---------------------------
* If two values are coprime, their greatest common
* divisor is 1.
*/

func coprime(_ m:Int, _ n:Int) -> Bool {

   return gcd(m,n) == 1 ? true : false

}

coprime(14,26) //coprime(2,4)

/*

* Function: crt
* Usage: x = crt(a,n)
* -------------------
* The Chinese Remainder Theorem supposes that given the
* integers n_1...n_k that are pairwise co-prime, then for
* any sequence of integers a_1...a_k there exists an integer
* x that solves the system of linear congruences:
*
*   x === a_1 (mod n_1)
*   ...
*   x === a_k (mod n_k)
*/

func crt(_ a_i:[Int], _ n_i:[Int]) -> Int {

   // There is no identity operator for elements of [Int].
   // The offset of the elements of an enumerated sequence
   // can be used instead, to determine if two elements of the same
   // array are the same.
   let divs = n_i.enumerated()
   
   // Check if elements of n_i are pairwise coprime divs.filter{ $0.0 < n.0 }
   divs.forEach{
       n in divs.filter{ $0.0 < n.0 }.forEach{
           assert(coprime(n.1, $0.1))
       }
   }
   
   // Calculate factor N
   let N = n_i.map{$0}.reduce(1, *)
   
   // Euclidean algorithm determines s_i (and r_i)
   var s:[Int] = []
   
   // Using euclidean algorithm to calculate r_i, s_i
   n_i.forEach{ s += [euclid($0, N / $0).1] }
   
   // Solve for x
   var x = 0
   a_i.enumerated().forEach{
       x += $0.1 * s[$0.0] * N / n_i[$0.0]
   }
   // Return minimal solution
   return x % N

}

let a = [2,3,2] let n = [3,5,7]

let x = crt(a,n)

print(x)</lang>

Output:
23

Tcl

Translation of: C

<lang tcl>proc ::tcl::mathfunc::mulinv {a b} {

   if {$b == 1} {return 1}
   set b0 $b; set x0 0; set x1 1
   while {$a > 1} {

set x0 [expr {$x1 - ($a / $b) * [set x1 $x0]}] set b [expr {$a % [set a $b]}]

   }
   incr x1 [expr {($x1 < 0) * $b0}]

} proc chineseRemainder {nList aList} {

   set sum 0; set prod [::tcl::mathop::* {*}$nList]
   foreach n $nList a $aList {

set p [expr {$prod / $n}] incr sum [expr {$a * mulinv($p, $n) * $p}]

   }
   expr {$sum % $prod}

} puts [chineseRemainder {3 5 7} {2 3 2}]</lang>

Output:
23

uBasic/4tH

Translation of: C

<lang>@(000) = 3 : @(001) = 5 : @(002) = 7 @(100) = 2 : @(101) = 3 : @(102) = 2

Print Func (_Chinese_Remainder (3))

' -------------------------------------

@(000) = 11 : @(001) = 12 : @(002) = 13 @(100) = 10 : @(101) = 04 : @(102) = 12

Print Func (_Chinese_Remainder (3))

' -------------------------------------

End

                                      ' returns x where (a * x) % b == 1

_Mul_Inv Param (2) ' ( a b -- n)

 Local (4)
 c@ = b@
 d@ = 0
 e@ = 1
 If b@ = 1 Then Return (1)
 Do While a@ > 1
    f@ = a@ / b@
    Push b@ : b@ = a@ % b@ : a@ = Pop()
    Push d@ : d@ = e@ - f@ * d@ : e@ = Pop()
 Loop
 If e@ < 0 Then e@ = e@ + c@

Return (e@)


_Chinese_Remainder Param (1) ' ( len -- n)

 Local (5)
 b@ = 1
 c@ = 0
 For d@ = 0 Step 1 While d@ < a@
   b@ = b@ * @(d@)
 Next
 For d@ = 0 Step 1 While d@ < a@
   e@ = b@ / @(d@)
   c@ = c@ + (@(100 + d@) * Func (_Mul_Inv (e@, @(d@))) * e@)
 Next

Return (c@ % b@) </lang>

Output:
23
1000

0 OK, 0:1034

zkl

Translation of: Go

Using the GMP library, gcdExt is the Extended Euclidean algorithm. <lang zkl>var BN=Import("zklBigNum"), one=BN(1);

fcn crt(xs,ys){

  p:=xs.reduce('*,BN(1));
  X:=BN(0);
  foreach x,y in (xs.zip(ys)){
     q:=p/x;
     z,s,_:=q.gcdExt(x);
     if(z!=one) throw(Exception.ValueError("%d not coprime".fmt(x)));
     X+=y*s*q;
  }
  return(X % p);

}</lang> <lang zkl>println(crt(T(3,5,7), T(2,3,2))); //-->23 println(crt(T(11,12,13),T(10,4,12))); //-->1000 println(crt(T(11,22,19), T(10,4,9))); //-->ValueError: 11 not coprime</lang>

ZX Spectrum Basic

Translation of: C

<lang zxbasic>10 DIM n(3): DIM a(3) 20 FOR i=1 TO 3 30 READ n(i),a(i) 40 NEXT i 50 DATA 3,2,5,3,7,2 100 LET prod=1: LET sum=0 110 FOR i=1 TO 3: LET prod=prod*n(i): NEXT i 120 FOR i=1 TO 3 130 LET p=INT (prod/n(i)): LET a=p: LET b=n(i) 140 GO SUB 1000 150 LET sum=sum+a(i)*x1*p 160 NEXT i 170 PRINT FN m(sum,prod) 180 STOP 200 DEF FN m(a,b)=a-INT (a/b)*b: REM Modulus function 1000 LET b0=b: LET x0=0: LET x1=1 1010 IF b=1 THEN RETURN 1020 IF a<=1 THEN GO TO 1100 1030 LET q=INT (a/b) 1040 LET t=b: LET b=FN m(a,b): LET a=t 1050 LET t=x0: LET x0=x1-q*x0: LET x1=t 1060 GO TO 1020 1100 IF x1<0 THEN LET x1=x1+b0 1110 RETURN </lang>

23