Chinese remainder theorem: Difference between revisions
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var (b0, x0, x1) = (0, 0, 1); |
var (b0, x0, x1) = (0, 0, 1); |
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while (a > 1) { |
while (a > 1) { |
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(a, b, x0, x1) = (b, a % b, x1 - x0*int(a / b), x0); |
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}; |
}; |
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x1 < 0 ? x1+b0 : x1; |
x1 < 0 ? x1+b0 : x1; |
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func chinese_remainder(*n) { |
func chinese_remainder(*n) { |
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var N = n«*»; |
var N = n«*»; |
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func (*a) { |
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n.range.map { |i| |
n.range.map { |i| |
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var p = int(N / n[i]); |
var p = int(N / n[i]); |
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Revision as of 20:03, 21 October 2015
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 coprime. 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, .
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 where .
Show the functionality of this program by printing the result such that the 's are and the 's are .
![{\displaystyle [3,5,7]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8d9ca0ab1bf25a3bcb4ebd62e32b7803bcb6de90)
![{\displaystyle [2,3,2]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bff761c73efae3b48ef8af23c52077a39828b98a)
Algorithm: The following algorithm only applies if the 's are pairwise coprime.
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 coprime. 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
<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
<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>
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
<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
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) |> Enum.into(HashSet.new) end)
|> Enum.reduce(fn set,acc -> Set.intersection(set, acc) end)
|> Set.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
<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<<< .
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
<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.
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
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>
Nim
<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.
<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/; say chinese( [2328,16256], [410,5418] ), " mod ", lcm(16256,5418);</lang>
- Output:
28450328 mod 44037504
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
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>
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
<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== then Ns = '3,5,7' /*Ns not specified? Then use default.*/ if As== then As = '2,3,2' /*As " " " " " */
say 'Ns: ' Ns
say 'As: ' As; say
Ns=space(translate(Ns,,',')); #=words(Ns) /*elide any 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 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
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== then Ns = '3,5,7' /*Ns not specified? Then use default.*/ if As== then As = '2,3,2' /*As " " " " " */
say 'Ns: ' Ns
say 'As: ' As; say
Ns=space(translate(Ns,,',')); #=words(Ns) /*elide any 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 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 the same as 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>
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
<lang ruby>func mul_inv(a, b) {
b == 1 && return 1;
var (b0, x0, x1) = (0, 0, 1);
while (a > 1) {
(a, b, x0, x1) = (b, a % b, x1 - x0*int(a / b), x0);
};
x1 < 0 ? x1+b0 : x1;
}
func chinese_remainder(*n) {
var N = n«*»;
func (*a) {
n.range.map { |i|
var p = int(N / n[i]);
a[i] * mul_inv(p, n[i]) * p;
}.sum
}
}
say chinese_remainder(3, 5, 7)(2, 3, 2);</lang>
- Output:
23
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
uBasic/4tH
<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
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>