Chinese remainder theorem
You are encouraged to solve this task according to the task description, using any language you may know.
Suppose Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n_1} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n_2} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ldots} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n_k} are positive integers that are pairwise coprime. Then, for any given sequence of integers Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_1} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_2} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \dots} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_k} , there exists an integer Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} solving the following system of simultaneous congruences.
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} x &\equiv a_1 \pmod{n_1} \\ x &\equiv a_2 \pmod{n_2} \\ &{}\ \ \vdots \\ x &\equiv a_k \pmod{n_k} \end{align}}
Furthermore, all solutions Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} of this system are congruent modulo the product, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle N=n_1n_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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s} where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0 \leq s \leq n_1n_2\ldots n_k} .
Show the functionality of this program by printing the result such that the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} 's are Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [3,5,7]} and the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} 's are Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [2,3,2]} .
Algorithm: The following algorithm only applies if the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n_i} 's are pairwise coprime.
Suppose, as above, that a solution is required for the system of congruences:
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \equiv a_i \pmod{n_i} \quad\mathrm{for}\; i = 1, \ldots, k}
Again, to begin, the product Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle N = n_1n_2 \ldots n_k} is defined. Then a solution Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} can be found as follows.
For each Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i} , the integers Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n_i} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle N/n_i} are coprime. Using the Extended Euclidean algorithm we can find integers Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r_i} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s_i} such that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r_i n_i + s_i N/n_i = 1} . Then, one solution to the system of simultaneous congruences is:
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x = \sum_{i=1}^k a_i s_i N/n_i}
and the minimal solution,
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \pmod{N}} .
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
<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
<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<<< .
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
<lang haskell>module CRT ( chineseRemainder ) where
egcd :: Integral a => a -> a -> (a, a) egcd _ 0 = (1, 0) egcd a b = (t, s - q * t)
where (s, t) = egcd b r (q, r) = a `quotRem` b
modInv :: Integral a => a -> a -> Maybe a modInv a b = case egcd a b of
(x, y) | a * x + b * y == 1 -> Just x
| otherwise -> Nothing
chineseRemainder :: Integral a => [a] -> [a] -> Maybe a chineseRemainder residues modulii = do
inverses <- sequence $ zipWith modInv crtModulii modulii return . (`mod` modPI) . sum $ zipWith (*) crtModulii (zipWith (*) residues inverses) where modPI = product modulii crtModulii = map (modPI `div`) modulii
</lang>
- Output:
λ> chineseRemainder [10, 4, 12] [11, 12, 13] Just 1000 λ> chineseRemainder [10, 4, 9] [11, 22, 19] Nothing λ> chineseRemainder [2, 3, 2] [3, 5, 7] Just 23
Icon and Unicon
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.
Mathematica
Very easy, because it is a built-in function: <lang Mathematica >ChineseRemainder[{2, 3, 2}, {3, 5, 7}] 23</lang>
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 6
<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
Python
<lang python>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
def chinese_remainder(n, a, lena):
p = i = prod = 1; sm = 0
for i in range(lena): prod *= n[i]
for i in range(lena):
p = prod / n[i]
sm += a[i] * mul_inv(p, n[i]) * p
return sm % prod
if __name__ == '__main__':
n = [3, 5, 7] a = [2, 3, 2] print(chinese_remainder(n, a, len(n)))</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
<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
zkl
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>