Chinese remainder theorem

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

Suppose $n_{1}$, $n_{2}$, $\ldots$, $n_{k}$ are positive integers that are pairwise coprime. Then, for any given sequence of integers $a_{1}$, $a_{2}$, $\dots$, $a_{k}$, there exists an integer $x$ solving the following system of simultaneous congruences.

${\begin{aligned}x&\equiv a_{1}{\pmod {n_{1}}}\\x&\equiv a_{2}{\pmod {n_{2}}}\\&{}\ \ \vdots \\x&\equiv a_{k}{\pmod {n_{k}}}\end{aligned}}$

Furthermore, all solutions $x$ of this system are congruent modulo the product, $N=n_{1}n_{2}\ldots n_{k}$.

Your task is to 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 $s$ where $0\leq s\leq n_{1}n_{2}\ldots n_{k}$.

Show the functionality of this program by printing the result such that the $n$'s are $[3,5,7]$ and the $a$'s are $[2,3,2]$.

Algorithm: The following algorithm only applies if the $n_{i}$'s are pairwise coprime.

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

$x\equiv a_{i}{\pmod {n_{i}}}\quad \mathrm {for} \;i=1,\ldots ,k$

Again, to begin, the product $N=n_{1}n_{2}\ldots n_{k}$ is defined. Then a solution $x$ can be found as follows.

For each $i$, the integers $n_{i}$ and $N/n_{i}$ are coprime. Using the Extended Euclidean algorithm we can find integers $r_{i}$ and $s_{i}$ such that $r_{i}n_{i}+s_{i}N/n_{i}=1$. Then, one solution to the system of simultaneous congruences is:

$x=\sum _{i=1}^{k}a_{i}s_{i}N/n_{i}$

and the minimal solution,

$x{\pmod {N}}$.

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 pariwise 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>

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

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<<< .

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>chvec(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

Perl 6

Translation of: C

<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

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 uses the Chinese Remainder Theorem (Sun Tzu). */ parse arg Ns As . /*get optional arguments from CL.*/ if Ns== then Ns = '3,5,7' /*Ns not specified? Use default.*/ if As== then As = '2,3,2' /*As " " " " */ Ns=space(translate(Ns,,',')); #=words(Ns) /*elide superfluous blanks*/ As=space(translate(As,,',')); _=words(As) /* " " " */ 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, Ns.*/
     n.j=word(Ns,j);  N=N*n.j         /*get an  N.j  and calculate prod*/
     a.j=word(As,j)                   /* "   "  A.j  from the  As.     */
     end   /*j*/

 do x=1  for N                        /*use a simple algebraic method. */
    do i=1  for #                     /*process each   A.i   number.   */
    if x//n.i\==a.i  then iterate x   /*is the modulus correct for F ? */
    end   /*i*/                       /* [↑]  limit solution to product*/
 say 'found a solution with x=' x     /*announce a possible solution.  */
 exit                                 /*stick a fork in it, we're done.*/
 end   /*x*/

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

                                      /*stick a fork in it, we're done.*/</lang>

output

found a solution with x= 23

congruences sets

<lang rexx>/*REXX program uses the Chinese Remainder Theorem (Sun Tzu). */ parse arg Ns As . /*get optional arguments from CL.*/ if Ns== then Ns = '3,5,7' /*Ns not specified? Use default.*/ if As== then As = '2,3,2' /*As " " " " */ Ns=space(translate(Ns,,',')); #=words(Ns) /*elide superfluous blanks*/ As=space(translate(As,,',')); _=words(As) /* " " " */ 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, Ns.*/
     n.j=word(Ns,j);  N=N*n.j         /*get an  N.j  and calculate prod*/
     a.j=word(As,j)                   /* "   "  A.j  from the  As.     */
     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 list of modulo values. */
   end   /*i*/
                                      /* [↓] find common number in sets*/
 do x=1  for N;   if @.1.x==  then iterate           /*find a number.*/
   do v=2  to #;  if @.v.x==  then iterate x;  end   /*In all sets ? */
 say 'found a solution with X=' x     /*we found the lowest solution.  */
 exit                                 /*stick a fork in it, we're done.*/
 end   /*x*/

say 'no solution found.' /*oops, there's not a solution. */

                                      /*stick a fork in it, we're done.*/</lang>

output

found a solution with X= 23

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>

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