# Modular inverse

Modular inverse
You are encouraged to solve this task according to the task description, using any language you may know.

From Wikipedia:

In modular arithmetic,   the modular multiplicative inverse of an integer   a   modulo   m   is an integer   x   such that

${\displaystyle a\,x\equiv 1{\pmod {m}}.}$

Or in other words, such that:

${\displaystyle \exists k\in \mathbb {Z} ,\qquad a\,x=1+k\,m}$

It can be shown that such an inverse exists   if and only if   a   and   m   are coprime,   but we will ignore this for this task.

Either by implementing the algorithm, by using a dedicated library or by using a built-in function in your language,   compute the modular inverse of   42 modulo 2017.

## 11l

Translation of: C

<lang 11l>F mul_inv(=a, =b)

  V b0 = b
V x0 = 0
V x1 = 1
I b == 1 {R 1}

  L a > 1
V q = a I/ b
(a, b) = (b, a % b)
(x0, x1) = (x1 - q * x0, x0)

  I x1 < 0 {x1 += b0}
R x1


print(mul_inv(42, 2017))</lang>

Output:
1969


## 8th

<lang Forth> \ return "extended gcd" of a and b; The result satisfies the equation: \ a*x + b*y = gcd(a,b)

n:xgcd \ a b -- gcd x y
 dup 0 n:= if
1 swap            \ -- a 1 0
else
tuck n:/mod
-rot recurse
tuck 4 roll
n:* n:neg n:+
then ;


\ Return modular inverse of n modulo mod, or null if it doesn't exist (n and mod \ not coprime):

n:invmod \ n mod -- invmod
 dup >r
n:xgcd rot 1 n:= not if
2drop null
else
drop dup 0 n:< if r@ n:+ then
then
rdrop ;


42 2017 n:invmod . cr bye </lang>

Output:
1969


## Action!

<lang Action!>INT FUNC ModInverse(INT a,b)

 INT t,nt,r,nr,q,tmp

 IF b<0 THEN b=-b FI
IF a<0 THEN a=b-(-a MOD b) FI
t=0 nt=1
r=b nr=a MOD b
WHILE nr#0
DO
q=r/nr
tmp=nt nt=t-q*nt t=tmp
tmp=nr nr=r-q*nr r=tmp
OD
IF r>1 THEN
RETURN (-1)
FI
IF t<0 THEN
t==+b
FI


RETURN (t)

PROC Test(INT a,b)

 INT res

 res=ModInverse(a,b)
IF res>=0 THEN
PrintF("%I MODINV %I=%I%E",a,b,res)
ELSE
PrintF("%I MODINV %I has no result%E",a,b)
FI


RETURN

PROC Main()

 Test(42,2017)
Test(40,1)
Test(52,-217)
Test(-486,217)
Test(40,2018)


RETURN</lang>

Output:
42 MODINV 2017=1969
40 MODINV 1=0
52 MODINV -217=96
-486 MODINV 217=121
40 MODINV 2018 has no result


 -- inv_mod calculates the inverse of a mod n. We should have n>0 and, at the end, the contract is a*Result=1 mod n
-- If this is false then we raise an exception (don't forget the -gnata option when you compile
function inv_mod (a : Integer; n : Positive) return Integer with post=> (a * inv_mod'Result) mod n = 1 is
-- To calculate the inverse we do as if we would calculate the GCD with the Euclid extended algorithm
-- (but we just keep the coefficient on a)
function inverse (a, b, u, v : Integer) return Integer is
(if b=0 then u else inverse (b, a mod b, v, u-(v*a)/b));
begin
return inverse (a, n, 1, 0);
end inv_mod;


begin

 -- This will output -48 (which is correct)
Put_Line (inv_mod (42,2017)'img);
-- The further line will raise an exception since the GCD will not be 1
Put_Line (inv_mod (42,77)'img);
exception when others => Put_Line ("The inverse doesn't exist.");


end modular_inverse; </lang>

## ALGOL 68

<lang algol68> BEGIN

  PROC modular inverse = (INT a, m) INT :
BEGIN
PROC extended gcd = (INT x, y) []INT :


CO

  Algol 68 allows us to return three INTs in several ways.  A [3]INT
is used here but it could just as well be a STRUCT.


CO

     BEGIN


INT v := 1, a := 1, u := 0, b := 0, g := x, w := y; WHILE w>0 DO INT q := g % w, t := a - q * u; a := u; u := t; t := b - q * v; b := v; v := t; t := g - q * w; g := w; w := t OD; a PLUSAB (a < 0 | u | 0); (a, b, g)

     END;
[] INT egcd = extended gcd (a, m);
(egcd[3] > 1 | 0 | egcd[1] MOD m)
END;
printf (($"42 ^ -1 (mod 2017) = ", g(0)$, modular inverse (42, 2017)))


CO

  Note that if ϕ(m) is known, then a^-1 = a^(ϕ(m)-1) mod m which
allows an alternative implementation in terms of modular
exponentiation but, in general, this requires the factorization of
m.  If m is prime the factorization is trivial and ϕ(m) = m-1.
2017 is prime which may, or may not, be ironic within the context
of the Rosetta Code conditions.


CO END </lang>

Output:
42 ^ -1 (mod 2017) = 1969


## Arturo

Translation of: D

<lang rebol>modInverse: function [a,b][

   if b = 1 -> return 1

   b0: b   x0: 0   x1: 1
z: a

   while [z > 1][
q: z / b        t: b
b: z % b        z: t
t: x0           x0: x1 - q * x0
x1: t
]
(x1 < 0) ? -> x1 + b0
-> x1


]

print modInverse 42 2017</lang>

Output:
1969

## AutoHotkey

Translation of C. <lang AutoHotkey>MsgBox, % ModInv(42, 2017)

ModInv(a, b) { if (b = 1) return 1 b0 := b, x0 := 0, x1 :=1 while (a > 1) { q := a // b , t  := b , b  := Mod(a, b) , a  := t , t  := x0 , x0 := x1 - q * x0 , x1 := t } if (x1 < 0) x1 += b0 return x1 }</lang>

Output:
1969

## AWK

<lang AWK>

1. syntax: GAWK -f MODULAR_INVERSE.AWK
2. converted from C

BEGIN {

   printf("%s\n",mod_inv(42,2017))
exit(0)


} function mod_inv(a,b, b0,t,q,x0,x1) {

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


} </lang>

Output:
1969


## BASIC

### ASIC

Translation of: Pascal

<lang basic> REM Modular inverse E = 42 T = 2017 GOSUB CalcModInv: PRINT ModInv END

CalcModInv: REM Increments E Step times until Bal is greater than T REM Repeats until Bal = 1 (MOD = 1) and Count REM Bal will not be greater than T + E D = 0 IF E < T THEN

 Bal = E
Count = 1
Loop:
Step = T - Bal
Step = Step / E
Step = Step + 1
REM So ... Step = (T - Bal) / E + 1
StepTimesE = Step * E
Bal = Bal + StepTimesE
Count = Count + Step
Bal = Bal - T
IF Bal <> 1 THEN Loop:
D = Count


ENDIF ModInv = D RETURN </lang>

Output:
1969


## Batch File

Based from C's second implementation

Translation of: Perl

<lang dos>@echo off setlocal enabledelayedexpansion %== Calls the "function" ==% call :ModInv 42 2017 result echo !result! call :ModInv 40 1 result echo !result! call :ModInv 52 -217 result echo !result! call :ModInv -486 217 result echo !result! call :ModInv 40 2018 result echo !result! pause>nul exit /b 0

%== The "function" ==%

ModInv

set a=%1 set b=%2

if !b! lss 0 (set /a b=-b) if !a! lss 0 (set /a a=b - ^(-a %% b^))

set t=0&set nt=1&set r=!b!&set /a nr=a%%b

:while_loop if !nr! neq 0 ( set /a q=r/nr set /a tmp=nt set /a nt=t - ^(q*nt^) set /a t=tmp

set /a tmp=nr set /a nr=r - ^(q*nr^) set /a r=tmp goto while_loop )

if !r! gtr 1 (set %3=-1&goto :EOF) if !t! lss 0 set /a t+=b set %3=!t! goto :EOF</lang>

Output:
1969
0
96
121
-1

## BCPL

<lang bcpl>get "libhdr"

let mulinv(a, b) =

   b<0 -> mulinv(a, -b),
a<0 -> mulinv(b - (-a rem b), b),
valof
$( let t, nt, r, nr = 0, 1, b, a rem b until nr = 0$(  let tmp, q = ?, r / nr
tmp := nt ; nt := t - q*nt ; t := tmp
tmp := nr ; nr := r - q*nr ; r := tmp
$) resultis r>1 -> -1, t<0 -> t + b, t$)



let show(a, b) be $( let mi = mulinv(a, b)  test mi>=0 do writef("%N, %N -> %N*N", a, b, mi) or writef("%N, %N -> no inverse*N", a, b) $)

let start() be $( show(42, 2017)  show(40, 1) show(52, -217) show(-486, 217) show(40, 2018) $)</lang>

Output:
42, 2017 -> 1969
40, 1 -> 0
52, -217 -> 96
-486, 217 -> 121
40, 2018 -> no inverse

## Bracmat

Translation of: Julia

<lang bracmat>( ( mod-inv

 =   a b b0 x0 x1 q
.   !arg:(?a.?b)
& ( !b:1
|   (!b.0.1):(?b0.?x0.?x1)
&   whl
' ( !a:>1
& div$(!a.!b):?q & (!b.mod$(!a.!b)):(?a.?b)
& (!x1+-1*!q*!x0.!x0):(?x0.?x1)
)
& (!x:>0|!x1+!b0)
)
)


& out$(mod-inv$(42.2017)) };</lang> Output

1969

## C

<lang c>#include <stdio.h>

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 main(void) { printf("%d\n", mul_inv(42, 2017)); return 0; }</lang>

The above method has some problems. Most importantly, when given a pair (a,b) with no solution, it generates an FP exception. When given b=1, it returns 1 which is not a valid result mod 1. When given negative a or b the results are incorrect. The following generates results that should match Pari/GP for numbers in the int range.

Translation of: Perl

<lang c>#include <stdio.h>

int mul_inv(int a, int b) {

       int t, nt, r, nr, q, tmp;
if (b < 0) b = -b;
if (a < 0) a = b - (-a % b);
t = 0;  nt = 1;  r = b;  nr = a % b;
while (nr != 0) {
q = r/nr;
tmp = nt;  nt = t - q*nt;  t = tmp;
tmp = nr;  nr = r - q*nr;  r = tmp;
}
if (r > 1) return -1;  /* No inverse */
if (t < 0) t += b;
return t;


} int main(void) {

       printf("%d\n", mul_inv(42, 2017));
printf("%d\n", mul_inv(40, 1));
printf("%d\n", mul_inv(52, -217));  /* Pari semantics for negative modulus */
printf("%d\n", mul_inv(-486, 217));
printf("%d\n", mul_inv(40, 2018));
return 0;


}</lang>

Output:
1969
0
96
121
-1


## C#

<lang csharp>public class Program {

   static void Main()
{
System.Console.WriteLine(42.ModInverse(2017));
}


}

public static class IntExtensions {

   public static int ModInverse(this int a, int m)
{
if (m == 1) return 0;
int m0 = m;
(int x, int y) = (1, 0);

       while (a > 1) {
int q = a / m;
(a, m) = (m, a % m);
(x, y) = (y, x - q * y);
}
return x < 0 ? x + m0 : x;
}


}</lang>

## C++

### Iterative implementation

Translation of: C

<lang cpp>#include <iostream>

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 main(void) { std::cout << mul_inv(42, 2017) << std::endl; return 0; }</lang>

### Recursive implementation

<lang cpp>#include <iostream>

short ObtainMultiplicativeInverse(int a, int b, int s0 = 1, int s1 = 0) {

   return b==0? s0: ObtainMultiplicativeInverse(b, a%b, s1, s0 - s1*(a/b));


}

int main(int argc, char* argv[]) {

   std::cout << ObtainMultiplicativeInverse(42, 2017) << std::endl;
return 0;


}</lang>

## Clojure

<lang lisp>(ns test-p.core

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


(defn extended-gcd

 "The extended Euclidean algorithm--using Clojure code from RosettaCode for Extended Eucliean
(see http://en.wikipedia.orwiki/Extended_Euclidean_algorithm)
Returns a list containing the GCD and the Bézout coefficients
corresponding to the inputs with the result: gcd followed by bezout coefficients "
[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 mul_inv

 " Get inverse using extended gcd.  Extended GCD returns
gcd followed by bezout coefficients. We want the 1st coefficients
(i.e. second of extend-gcd result).  We compute mod base so result
is between 0..(base-1) "
[a b]
(let [b (if (neg? b) (- b) b)
a (if (neg? a) (- b (mod (- a) b)) a)
egcd (extended-gcd a b)]
(if (= (first egcd) 1)
(mod (second egcd) b)
(str "No inverse since gcd is: " (first egcd)))))


(println (mul_inv 42 2017)) (println (mul_inv 40 1)) (println (mul_inv 52 -217)) (println (mul_inv -486 217)) (println (mul_inv 40 2018))

</lang>

Output:

1969
0
96
121
No inverse since gcd is: 2


## CLU

Translation of: Perl

<lang clu>mul_inv = proc (a, b: int) returns (int) signals (no_inverse)

   if b<0 then b := -b end
if a<0 then a := b - (-a // b) end
t: int := 0
nt: int := 1
r: int := b
nr: int := a // b
while nr ~= 0 do
q: int := r / nr
t, nt := nt, t - q*nt
r, nr := nr, r - q*nr
end
if r>1 then signal no_inverse end
if t<0 then t := t+b end
return(t)


end mul_inv

start_up = proc ()

   pair = struct[a, b: int]
tests: sequence[pair] := sequence[pair]$[pair${a: 42, b: 2017},
pair${a: 40, b: 1}, pair${a: 52, b: -217},
pair${a: -486, b: 217}, pair${a: 40, b: 2018}]

po: stream := stream$primary_output() for test: pair in sequence[pair]$elements(tests) do
stream$puts(po, int$unparse(test.a) || ", "
|| int$unparse(test.b) || " -> ") stream$putl(po, int$unparse(mul_inv(test.a, test.b))) except when no_inverse: stream$putl(po, "no modular inverse")
end
end


end start_up</lang>

Output:
42, 2017 -> 1969
40, 1 -> 0
52, -217 -> 96
-486, 217 -> 121
40, 2018 -> no modular inverse

## Common Lisp

<lang lisp>

Calculates the GCD of a and b based on the Extended Euclidean Algorithm. The function also returns
the Bézout coefficients s and t, such that gcd(a, b) = as + bt.
The algorithm is described on page http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm#Iterative_method_2

(defun egcd (a b)

 (do ((r (cons b a) (cons (- (cdr r) (* (car r) q)) (car r))) ; (r+1 r) i.e. the latest is first.
(s (cons 0 1) (cons (- (cdr s) (* (car s) q)) (car s))) ; (s+1 s)
(u (cons 1 0) (cons (- (cdr u) (* (car u) q)) (car u))) ; (t+1 t)
(q nil))
((zerop (car r)) (values (cdr r) (cdr s) (cdr u)))       ; exit when r+1 = 0 and return r s t
(setq q (floor (/ (cdr r) (car r))))))                     ; inside loop; calculate the q

Calculates the inverse module for a = 1 (mod m).
Note
The inverse is only defined when a and m are coprimes, i.e. gcd(a, m) = 1.”

(defun invmod (a m)

 (multiple-value-bind (r s k) (egcd a m)
(unless (= 1 r) (error "invmod: Values ~a and ~a are not coprimes." a m))
s))


</lang>

Output:
* (invmod 42 2017)

-48
* (mod -48 2017)

1969


## Crystal

Translation of: Ruby

<lang ruby>def modinv(a0, m0)

 return 1 if m0 == 1
a, m = a0, m0
x0, inv = 0, 1
while a > 1
inv -= (a // m) * x0
a, m = m, a % m
x0, inv = inv, x0
end
inv += m0 if inv < 0
inv


end</lang>

Output:
> modinv(42,2017)
=> 1969

## D

Translation of: C

<lang d>T modInverse(T)(T a, T b) pure nothrow {

   if (b == 1)
return 1;
T b0 = b,
x0 = 0,
x1 = 1;

   while (a > 1) {
immutable q = a / b;
auto t = b;
b = a % b;
a = t;
t = x0;
x0 = x1 - q * x0;
x1 = t;
}
return (x1 < 0) ? (x1 + b0) : x1;


}

void main() {

   import std.stdio;
writeln(modInverse(42, 2017));


}</lang>

Output:
1969

## dc

Translation of: C

This solution prints the inverse u only if it exists (a*u = 1 mod m). <lang dc> dc -e "[m=]P?dsm[a=]P?dsa1sv[dsb~rsqlbrldlqlv*-lvsdsvd0<x]dsxxldd[dlmr+]sx0>xdla*lm%[p]sx1=x" </lang> If ~ is not implemented, it can be replaced by SdSnlnld/LnLd%.

Replace [p]sx1=x at the end by [pq]sx1=x16i6E6F7420636F7072696D65P if an error message "not coprime" is desired.

Output:
m=2 800^1+
a=37
342411551958695219479776173037037562556082184118925013641969995739234\
344644689214483533004909620355470582887300743869103978073598454778206\
829469635119691272637318902731800747596752668736012071540136041369140\
1228044652005748974399041408477572

m=2017
a=42
1969

m=42
a=7


See #Pascal.

## Draco

<lang draco>proc mulinv(int a, b) int:

   int t, nt, r, nr, q, tmp;
if b<0 then b := -b fi;
if a<0 then a := b - (-a % b) fi;
t := 0; nt := 1; r := b; nr := a % b;
while nr /= 0 do
q := r / nr;
tmp := nt; nt := t - q*nt; t := tmp;
tmp := nr; nr := r - q*nr; r := tmp
od;
if r>1 then -1
elif t<0 then t+b
else t
fi


corp

proc show(int a, b) void:

   int mi;
mi := mulinv(a, b);
if mi>=0
then writeln(a:5, ", ", b:5, " -> ", mi:5)
else writeln(a:5, ", ", b:5, " -> no inverse")
fi


corp

proc main() void:

   show(42, 2017);
show(40, 1);
show(52, -217);
show(-486, 217);
show(40, 2018)


corp</lang>

Output:
   42,  2017 ->  1969
40,     1 ->     0
52,  -217 ->    96
-486,   217 ->   121
40,  2018 -> no inverse

## EchoLisp

<lang scheme> (lib 'math) ;; for egcd = extended gcd

(define (mod-inv x m)

   (define-values (g inv q) (egcd x m))
(unless (= 1 g) (error 'not-coprimes (list x m) ))
(if (< inv 0) (+ m inv) inv))


(mod-inv 42 2017) → 1969 (mod-inv 42 666) 🔴 error: not-coprimes (42 666) </lang>

## Elixir

Translation of: Ruby

<lang elixir>defmodule Modular do

 def extended_gcd(a, b) do
{last_remainder, last_x} = extended_gcd(abs(a), abs(b), 1, 0, 0, 1)
{last_remainder, last_x * (if a < 0, do: -1, else: 1)}
end

defp extended_gcd(last_remainder, 0, last_x, _, _, _), do: {last_remainder, last_x}
defp extended_gcd(last_remainder, remainder, last_x, x, last_y, y) do
quotient   = div(last_remainder, remainder)
remainder2 = rem(last_remainder, remainder)
extended_gcd(remainder, remainder2, x, last_x - quotient*x, y, last_y - quotient*y)
end

def inverse(e, et) do
{g, x} = extended_gcd(e, et)
if g != 1, do: raise "The maths are broken!"
rem(x+et, et)
end
end


IO.puts Modular.inverse(42,2017)</lang>

Output:
1969


## ERRE

<lang ERRE>PROGRAM MOD_INV

!$INTEGER PROCEDURE MUL_INV(A,B->T)  LOCAL NT,R,NR,Q,TMP IF B<0 THEN B=-B IF A<0 THEN A=B-(-A MOD B) T=0 NT=1 R=B NR=A MOD B WHILE NR<>0 DO Q=R DIV NR TMP=NT NT=T-Q*NT T=TMP TMP=NR NR=R-Q*NR R=TMP END WHILE IF (R>1) THEN T=-1 EXIT PROCEDURE ! NO INVERSE IF (T<0) THEN T+=B  END PROCEDURE BEGIN  MUL_INV(42,2017->T) PRINT(T) MUL_INV(40,1->T) PRINT(T) MUL_INV(52,-217->T) PRINT(T) ! pari semantics for negative modulus MUL_INV(-486,217->T) PRINT(T) MUL_INV(40,2018->T) PRINT(T)  END PROGRAM </lang> Output:  1969 0 96 121 -1  ## F# <lang fsharp> // Calculate the inverse of a (mod m) // See here for eea specs: // https://en.wikipedia.org/wiki/Extended_Euclidean_algorithm let modInv m a =  let rec eea t t' r r' = match r' with | 0 -> t | _ -> let div = r/r' eea t' (t - div * t') r' (r - div * r') (m + eea 0 1 m a) % m  </lang> Input: // Inverse of 347 (mod 29) modInv 29 347  Output: 28  ## Factor <lang>USE: math.functions 42 2017 mod-inv</lang> Output: 1969  ## Forth ANS Forth with double-number word set <lang forth> invmod { a m | v b c -- inv }  m to v 1 to c 0 to b begin a while v a / >r c b s>d c s>d r@ 1 m*/ d- d>s to c to b a v s>d a s>d r> 1 m*/ d- d>s to a to v repeat b m mod dup to b 0< if m b + else b then ;  </lang> ANS Forth version without locals <lang forth> modinv ( a m - inv)  dup 1- \ a m (m != 1)? if \ a m tuck 1 0 \ m0 a m 1 0 begin \ m0 a m inv x0 2>r over 1 > \ m0 a m (a > 1)? R: inv x0 while \ m0 a m R: inv x0 tuck /mod \ m0 m (a mod m) (a/m) R: inv x0 r> tuck * \ m0 a' m' x0 (a/m)*x0 R: inv r> swap - \ m0 a' m' x0 (inv-q) R: repeat \ m0 a' m' inv' x0' 2drop \ m0 R: inv x0 2r> drop \ m0 inv R: dup 0< \ m0 inv (inv < 0)? if over + then \ m0 (inv + m0) then \ x inv' nip \ inv  </lang> 42 2017 invmod . 1969 42 2017 modinv . 1969  ## FreeBASIC <lang freebasic>' version 10-07-2018 ' compile with: fbc -s console Type ext_euclid  Dim As Integer a, b  End Type ' "Table method" aka "The Magic Box" Function magic_box(x As Integer, y As Integer) As ext_euclid  Dim As Integer a(1 To 128), b(1 To 128), d(1 To 128), k(1 To 128)   a(1) = 1 : b(1) = 0 : d(1) = x a(2) = 0 : b(2) = 1 : d(2) = y : k(2) = x \ y   Dim As Integer i = 2   While Abs(d(i)) <> 1 i += 1 a(i) = a(i -2) - k(i -1) * a(i -1) b(i) = b(i -2) - k(i -1) * b(i -1) d(i) = d(i -2) Mod d(i -1) k(i) = d(i -1) \ d(i) 'Print a(i),b(i),d(i),k(i) If d(i -1) Mod d(i) = 0 Then Exit While Wend If d(i) = -1 Then ' -1 * (ab + by) = -1 * -1 ==> -ab -by = 1 a(i) = -a(i) b(i) = -b(i) End If   Function = Type( a(i), b(i) )  End Function ' ------=< MAIN >=------ Dim As Integer x, y, gcd Dim As ext_euclid result Do  Read x, y If x = 0 AndAlso y = 0 Then Exit Do result = magic_box(x, y) With result gcd = .a * x + .b * y Print "a * "; Str(x); " + b * "; Str(y); Print " = GCD("; Str(x); ", "; Str(y); ") ="; gcd If gcd > 1 Then Print "No solution, numbers are not coprime" Else Print "a = "; .a; ", b = ";.b Print "The Modular inverse of "; x; " modulo "; y; " = "; While .a < 0 : .a += IIf(y > 0, y, -y) : Wend Print .a 'Print "The Modular inverse of "; y; " modulo "; x; " = "; 'While .b < 0 : .b += IIf(x > 0, x, -x) : Wend 'Print .b End if End With Print  Loop Data 42, 2017 Data 40, 1 Data 52, -217 Data -486, 217 Data 40, 2018 Data 0, 0 ' empty keyboard buffer While Inkey <> "" : Wend Print : Print "hit any key to end program" Sleep End</lang> Output: a * 42 + b * 2017 = GCD(42, 2017) = 1 a = -48, b = 1 The Modular inverse of 42 modulo 2017 = 1969 a * 40 + b * 1 = GCD(40, 1) = 1 a = 0, b = 1 The Modular inverse of 40 modulo 1 = 0 a * 52 + b * -217 = GCD(52, -217) = 1 a = 96, b = 23 The Modular inverse of 52 modulo -217 = 96 a * -486 + b * 217 = GCD(-486, 217) = 1 a = -96, b = -215 The Modular inverse of -486 modulo 217 = 121 a * 40 + b * 2018 = GCD(40, 2018) = 2 No solution, numbers are not coprime ## Frink <lang frink>println[modInverse[42, 2017]]</lang> Output: 1969  ## FunL <lang funl>import integers.egcd def modinv( a, m ) =  val (g, x, _) = egcd( a, m )   if g != 1 then error( a + ' and ' + m + ' not coprime' ) val res = x % m   if res < 0 then res + m else res  println( modinv(42, 2017) )</lang> Output: 1969  ## Go The standard library function uses the extended Euclidean algorithm internally. <lang go>package main import ( "fmt" "math/big" ) func main() { a := big.NewInt(42) m := big.NewInt(2017) k := new(big.Int).ModInverse(a, m) fmt.Println(k) }</lang> Output: 1969  ## GW-BASIC Translation of: Pascal Works with: PC-BASIC version any <lang qbasic> 10 ' Modular inverse 20 LET E% = 42 30 LET T% = 2017 40 GOSUB 1000 50 PRINT MODINV% 60 END 990 ' increments e stp (step) times until bal is greater than t 992 ' repeats until bal = 1 (mod = 1) and returns count 994 ' bal will not be greater than t + e 1000 LET D% = 0 1010 IF E% >= T% THEN GOTO 1140 1020 LET BAL% = E% 1025 ' At least one iteration is necessary 1030 LET STP% = ((T% - BAL%) \ E%) + 1 1040 LET BAL% = BAL% + STP% * E% 1050 LET COUNT% = 1 + STP% 1060 LET BAL% = BAL% - T% 1070 WHILE BAL% <> 1 1080 LET STP% = ((T% - BAL%) \ E%) + 1 1090 LET BAL% = BAL% + STP% * E% 1100 LET COUNT% = COUNT% + STP% 1110 LET BAL% = BAL% - T% 1120 WEND 1130 LET D% = COUNT% 1140 LET MODINV% = D% 1150 RETURN </lang> Output:  1969  ## Haskell <lang haskell>-- Given a and m, return Just x such that ax = 1 mod m. -- If there is no such x return Nothing. modInv :: Int -> Int -> Maybe Int modInv a m  | 1 == g = Just (mkPos i) | otherwise = Nothing where (i, _, g) = gcdExt a m mkPos x | x < 0 = x + m | otherwise = x  -- Extended Euclidean algorithm. -- Given non-negative a and b, return x, y and g -- such that ax + by = g, where g = gcd(a,b). -- Note that x or y may be negative. gcdExt :: Int -> Int -> (Int, Int, Int) gcdExt a 0 = (1, 0, a) gcdExt a b =  let (q, r) = a quotRem b (s, t, g) = gcdExt b r in (t, s - q * t, g)  main :: IO () main = mapM_ print [2 modInv 4, 42 modInv 2017]</lang> Output: Nothing Just 1969 ## Icon and Unicon Translation of: C <lang unicon>procedure main(args)  a := integer(args[1]) | 42 b := integer(args[2]) | 2017 write(mul_inv(a,b))  end procedure mul_inv(a,b)  if b == 1 then return 1 (b0 := b, x0 := 0, x1 := 1) while a > 1 do { q := a/b (t := b, b := a%b, a := t) (t := x0, x0 := x1-q*x0, x1 := t) } return if (x1 > 0) then x1 else x1+b0  end</lang> Output: ->mi 1969 ->  Adding a coprime test: <lang unicon>link numbers procedure main(args)  a := integer(args[1]) | 42 b := integer(args[2]) | 2017 write(mul_inv(a,b))  end procedure mul_inv(a,b)  if b == 1 then return 1 if gcd(a,b) ~= 1 then return "not coprime" (b0 := b, x0 := 0, x1 := 1) while a > 1 do { q := a/b (t := b, b := a%b, a := t) (t := x0, x0 := x1-q*x0, x1 := t) } return if (x1 > 0) then x1 else x1+b0  end</lang> ## IS-BASIC <lang IS-BASIC>100 PRINT MODINV(42,2017) 120 DEF MODINV(A,B) 130 LET B=ABS(B) 140 IF A<0 THEN LET A=B-MOD(-A,B) 150 LET T=0:LET NT=1:LET R=B:LET NR=MOD(A,B) 160 DO WHILE NR<>0 170 LET Q=INT(R/NR) 180 LET TMP=NT:LET NT=T-Q*NT:LET T=TMP 190 LET TMP=NR:LET NR=R-Q*NR:LET R=TMP 200 LOOP 210 IF R>1 THEN 220 LET MODINV=-1 230 ELSE IF T<0 THEN 240 LET MODINV=T+B 250 ELSE 260 LET MODINV=T 270 END IF 280 END DEF</lang> ## J Solution:<lang j> modInv =: dyad def 'x y&|@^ <: 5 p: y'"0</lang> Example:<lang j> 42 modInv 2017 1969</lang> Notes: • Calculates the modular inverse as a^( totient(m) - 1 ) mod m. • 5 p: y is Euler's totient function of y. • J has a fast implementation of modular exponentiation (which avoids the exponentiation altogether), invoked with the form m&|@^ (hence, we use explicitly-named arguments for this entry, as opposed to the "variable free" tacit style: the m&| construct must freeze the value before it can be used but we want to use different values in that expression at different times...). ## Java The BigInteger library has a method for this: <lang java>System.out.println(BigInteger.valueOf(42).modInverse(BigInteger.valueOf(2017)));</lang> Output: 1969 ## JavaScript Using brute force. <lang javascript>var modInverse = function(a, b) {  a %= b; for (var x = 1; x < b; x++) { if ((a*x)%b == 1) { return x; } }  }</lang> ## jq Works with: jq Works with gojq, the Go implementation of jq <lang jq># Integer division: 1. If$j is 0, then an error condition is raised;
2. otherwise, assuming infinite-precision integer arithmetic,
3. if the input and $j are integers, then the result will be an integer. def idivide($j):

 . as $i | ($i % $j) as$mod
| ($i -$mod) / $j ;  1. the multiplicative inverse of . modulo$n

def modInv($n):  if$n == 1 then 1
else . as $this | { r :$n,
t   : 0,
newR: length, # abs
newT: 1}
| until(.newR == 0;
.newR as $newR | (.r | idivide($newR)) as $q | {r :$newR,
t   : .newT,
newT: (.t - $q * .newT), newR: (.r -$q * $newR) } )   | if (.r|length) != 1 then "\($this) and \($n) are not co-prime." | error else .t | if . < 0 then . +$n
elif $this < 0 then - . else . end end end ;  1. Example: 42 | modInv(2017) </lang> Output: 1969  ## Julia Works with: Julia version 1.2 ### Built-in Julia includes a built-in function for this: <lang julia>invmod(a, b)</lang> ### C translation Translation of: C The following code works on any integer type. To maximize performance, we ensure (via a promotion rule) that the operands are the same type (and use built-ins zero(T) and one(T) for initialization of temporary variables to ensure that they remain of the same type throughout execution). <lang julia>function modinv{T<:Integer}(a::T, b::T)  b0 = b x0, x1 = zero(T), one(T) while a > 1 q = div(a, b) a, b = b, a % b x0, x1 = x1 - q * x0, x0 end x1 < 0 ? x1 + b0 : x1  end modinv(a::Integer, b::Integer) = modinv(promote(a,b)...)</lang> Output: julia> invmod(42, 2017) 1969 julia> modinv(42, 2017) 1969  ## Kotlin <lang scala>// version 1.0.6 import java.math.BigInteger fun main(args: Array<String>) {  val a = BigInteger.valueOf(42) val m = BigInteger.valueOf(2017) println(a.modInverse(m))  }</lang> Output: 1969  ## Maple <lang Maple> 1/42 mod 2017; </lang> Output:  1969  ## Mathematica/Wolfram Language The built-in function FindInstance works well for this <lang Mathematica>modInv[a_, m_] := Block[{x,k}, x /. FindInstance[a x == 1 + k m, {x, k}, Integers]]</lang>  Another way by using the built-in function PowerMod : <lang Mathematica>PowerMod[a,-1,m]</lang> For example : modInv[42, 2017] {1969} PowerMod[42, -1, 2017] 1969 ## МК-61/52 <lang>П1 П2 <-> П0 0 П5 1 П6 ИП1 1 - x=0 14 С/П ИП0 1 - /-/ x<0 50 ИП0 ИП1 / [x] П4 ИП1 П3 ИП0 ^ ИП1 / [x] ИП1 * - П1 ИП3 П0 ИП5 П3 ИП6 ИП4 ИП5 * - П5 ИП3 П6 БП 14 ИП6 x<0 55 ИП2 + С/П</lang> ## Modula-2 Translation of: C <lang Modula-2>MODULE ModularInverse;  FROM InOut IMPORT WriteString, WriteInt, WriteLn;   TYPE Data = RECORD x : INTEGER; y : INTEGER END;   VAR c : INTEGER; ab : ARRAY [1..5] OF Data;  PROCEDURE mi(VAR a, b : INTEGER): INTEGER;  VAR t, nt, r, nr, q, tmp : INTEGER;  BEGIN  b := ABS(b); IF a < 0 THEN a := b - (-a MOD b) END; t := 0; nt := 1; r := b; nr := a MOD b; WHILE (nr # 0) DO q := r / nr; tmp := nt; nt := t - q * nt; t := tmp; tmp := nr; nr := r - q * nr; r := tmp; END; IF (r > 1) THEN RETURN -1 END; IF (t < 0) THEN RETURN t + b END; RETURN t;  END mi; BEGIN  ab[1].x := 42; ab[1].y := 2017; ab[2].x := 40; ab[2].y := 1; ab[3].x := 52; ab[3].y := -217; ab[4].x := -486; ab[4].y := 217; ab[5].x := 40; ab[5].y := 2018; WriteLn; WriteString("Modular inverse"); WriteLn; FOR c := 1 TO 5 DO WriteInt(ab[c].x, 6); WriteString(", "); WriteInt(ab[c].y, 6); WriteString(" = "); WriteInt(mi(ab[c].x, ab[c].y),6); WriteLn; END;  END ModularInverse.</lang> Output: Modular inverse 42, 2017 = 1969 40, 1 = 0 52, -217 = 96 -486, 217 = 121 40, 2018 = -1 ## newLISP <lang NewLisp> (define (modular-multiplicative-inverse a n)  (if (< n 0) (setf n (abs n))) (if (< a 0) (setf a (- n (% (- 0 a) n)))) (setf t 0) (setf nt 1) (setf r n) (setf nr (mod a n)) (while (not (zero? nr)) (setf q (int (div r nr))) (setf tmp nt) (setf nt (sub t (mul q nt))) (setf t tmp) (setf tmp nr) (setf nr (sub r (mul q nr))) (setf r tmp)) (if (> r 1) (setf retvalue nil)) (if (< t 0) (setf retvalue (add t n)) (setf retvalue t)) retvalue)  (println (modular-multiplicative-inverse 42 2017)) </lang> Output: 1969  ## Nim Translation of: C <lang nim>proc modInv(a0, b0: int): int =  var (a, b, x0) = (a0, b0, 0) result = 1 if b == 1: return while a > 1: result = result - (a div b) * x0 a = a mod b swap a, b swap x0, result if result < 0: result += b0  echo modInv(42, 2017)</lang> Output: 1969 ## OCaml ### Translation of: C <lang ocaml>let mul_inv a = function 1 -> 1 | b ->  let rec aux a b x0 x1 = if a <= 1 then x1 else if b = 0 then failwith "mul_inv" else aux b (a mod b) (x1 - (a / b) * x0) x0 in let x = aux a b 0 1 in if x < 0 then x + b else x</lang>  Testing: # mul_inv 42 2017 ;; - : int = 1969  ### Translation of: Haskell <lang ocaml>let rec gcd_ext a = function  | 0 -> (1, 0, a) | b -> let s, t, g = gcd_ext b (a mod b) in (t, s - (a / b) * t, g)  let mod_inv a m =  let mk_pos x = if x < 0 then x + m else x in match gcd_ext a m with | i, _, 1 -> mk_pos i | _ -> failwith "mod_inv"</lang>  Testing: # mod_inv 42 2017 ;; - : int = 1969  ## Oforth Usage : a modulus invmod <lang Oforth>// euclid ( a b -- u v r ) // Return r = gcd(a, b) and (u, v) / r = au + bv euclid(a, b) | q u u1 v v1 |  b 0 < ifTrue: [ b neg ->b ] a 0 < ifTrue: [ b a neg b mod - ->a ]   1 dup ->u ->v1 0 dup ->v ->u1   while(b) [ b a b /mod ->q ->b ->a u1 u u1 q * - ->u1 ->u v1 v v1 q * - ->v1 ->v ] u v a ;  invmod(a, modulus)  a modulus euclid 1 == ifFalse: [ drop drop null return ] drop dup 0 < ifTrue: [ modulus + ] ;</lang>  Output: 42 2017 invmod println 1969  ## PARI/GP <lang parigp>Mod(1/42,2017)</lang> ## Pascal <lang Pascal> // increments e step times until bal is greater than t // repeats until bal = 1 (mod = 1) and returns count // bal will not be greater than t + e function modInv(e, t : integer) : integer;  var d : integer; bal, count, step : integer; begin d := 0; if e < t then begin count := 1; bal := e; repeat step := ((t-bal) DIV e)+1; bal := bal + step * e; count := count + step; bal := bal - t; until bal = 1; d := count; end; modInv := d; end;</lang>  Testing:  Writeln(modInv(42,2017));  Output: 1969 ## Perl Various CPAN modules can do this, such as: <lang perl>use bigint; say 42->bmodinv(2017); 1. or use Math::ModInt qw/mod/; say mod(42, 2017)->inverse->residue; 1. or use Math::Pari qw/PARI lift/; say lift PARI "Mod(1/42,2017)"; 1. or use Math::GMP qw/:constant/; say 42->bmodinv(2017); 1. or use ntheory qw/invmod/; say invmod(42, 2017);</lang> or we can write our own: <lang perl>sub invmod {  my($a,$n) = @_; my($t,$nt,$r,$nr) = (0, 1,$n, $a %$n);
while ($nr != 0) { # Use this instead of int($r/$nr) to get exact unsigned integer answers my$quot = int( ($r - ($r % $nr)) /$nr );
($nt,$t) = ($t-$quot*$nt,$nt);
($nr,$r) = ($r-$quot*$nr,$nr);
}
return if $r > 1;$t += $n if$t < 0;
if ($a < 0)$a = $n - (-$a % $n); $t = 0; $nt = 1;$r = $n;$nr = $a %$n; while ($nr != 0) {$quot= intval($r/$nr); $tmp =$nt; $nt =$t - $quot*$nt; $t =$tmp; $tmp =$nr; $nr =$r - $quot*$nr; $r =$tmp; } if ($r > 1) return -1; if ($t < 0) $t +=$n; return $t; } printf("%d\n", invmod(42, 2017)); ?></lang> Output: 1969 ## PicoLisp Translation of: C <lang PicoLisp>(de modinv (A B)  (let (B0 B X0 0 X1 1 Q 0 T1 0) (while (< 1 A) (setq Q (/ A B) T1 B B (% A B) A T1 T1 X0 X0 (- X1 (* Q X0)) X1 T1 ) ) (if (lt0 X1) (+ X1 B0) X1) ) )  (println  (modinv 42 2017) )  (bye)</lang> ## PL/I Translation of: REXX <lang pli>*process source attributes xref or(!); /*-------------------------------------------------------------------- * 13.07.2015 Walter Pachl *-------------------------------------------------------------------*/ minv: Proc Options(main); Dcl (x,y) Bin Fixed(31); x=42; y=2017; Put Edit('modular inverse of',x,' by ',y,' ---> ',modinv(x,y)) (Skip,3(a,f(4))); modinv: Proc(a,b) Returns(Bin Fixed(31)); Dcl (a,b,ob,ox,d,t) Bin Fixed(31); ob=b; ox=0; d=1;  If b=1 Then; Else Do; Do While(a>1); q=a/b; r=mod(a,b); a=b; b=r; t=ox; ox=d-q*ox; d=t; End; End; If d<0 Then d=d+ob; Return(d); End; End;</lang>  Output: modular inverse of 42 by 2017 ---> 1969 ## PowerShell <lang powershell>function invmod($a,$n){  if ([int]$n -lt 0) {$n = -$n}
if ([int]$a -lt 0) {$a = $n - ((-$a) % $n)} $t = 0 $nt = 1$r = $n$nr = $a %$n while ($nr -ne 0) {$q = [Math]::truncate($r/$nr) $tmp =$nt $nt =$t - $q*$nt $t =$tmp $tmp =$nr $nr =$r - $q*$nr $r =$tmp } if ($r -gt 1) {return -1} if ($t -lt 0) {$t +=$n} return $t } invmod 42 2017</lang> Output: PS> .\INVMOD.PS1 1969 PS>  ## Prolog <lang Prolog> egcd(_, 0, 1, 0) :- !. egcd(A, B, X, Y) :-  divmod(A, B, Q, R), egcd(B, R, S, X), Y is S - Q*X.  modinv(A, B, N) :-  egcd(A, B, X, Y), A*X + B*Y =:= 1, N is X mod B.  </lang> Output: ?- modinv(42, 2017, N). N = 1969. ?- modinv(42, 64, X). false.  ## PureBasic Using brute force. <lang PureBasic>EnableExplicit Declare main() Declare.i mi(a.i, b.i) If OpenConsole("MODULAR-INVERSE")  main() : Input() : End  EndIf Macro ModularInverse(a, b)  PrintN(~"\tMODULAR-INVERSE(" + RSet(Str(a),5) + "," + RSet(Str(b),5)+") = " + RSet(Str(mi(a, b)),5))  EndMacro Procedure main()  ModularInverse(42, 2017) ; = 1969 ModularInverse(40, 1) ; = 0 ModularInverse(52, -217) ; = 96 ModularInverse(-486, 217) ; = 121 ModularInverse(40, 2018) ; = -1  EndProcedure Procedure.i mi(a.i, b.i)  Define x.i = 1, y.i = Int(Abs(b)), r.i = 0 If y = 1 : ProcedureReturn 0 : EndIf While x < y r = (a * x) % b If r = 1 Or (y + r) = 1 Break EndIf x + 1 Wend If x > y - 1 : x = -1 : EndIf ProcedureReturn x  EndProcedure</lang> Output:  MODULAR-INVERSE( 42, 2017) = 1969 MODULAR-INVERSE( 40, 1) = 0 MODULAR-INVERSE( 52, -217) = 96 MODULAR-INVERSE( -486, 217) = 121 MODULAR-INVERSE( 40, 2018) = -1 ## Python ### Iteration and error-handling Implementation of this pseudocode with this. <lang python>>>> def extended_gcd(aa, bb):  lastremainder, remainder = abs(aa), abs(bb) x, lastx, y, lasty = 0, 1, 1, 0 while remainder: lastremainder, (quotient, remainder) = remainder, divmod(lastremainder, remainder) x, lastx = lastx - quotient*x, x y, lasty = lasty - quotient*y, y return lastremainder, lastx * (-1 if aa < 0 else 1), lasty * (-1 if bb < 0 else 1)  >>> def modinv(a, m): g, x, y = extended_gcd(a, m) if g != 1: raise ValueError return x % m >>> modinv(42, 2017) 1969 >>> </lang> ### Recursion and an option type Or, using functional composition as an alternative to iterative mutation, and wrapping the resulting value in an option type, to allow for the expression of computations which establish the absence of a modular inverse: <lang python>from functools import (reduce) from itertools import (chain) 1. modInv :: Int -> Int -> Maybe Int def modInv(a):  return lambda m: ( lambda ig=gcdExt(a)(m): ( lambda i=ig[0]: ( Just(i + m if 0 > i else i) if 1 == ig[2] else ( Nothing() ) ) )() )()  1. gcdExt :: Int -> Int -> (Int, Int, Int) def gcdExt(x):  def go(a, b): if 0 == b: return (1, 0, a) else: (q, r) = divmod(a, b) (s, t, g) = go(b, r) return (t, s - q * t, g) return lambda y: go(x, y)  1. TEST --------------------------------------------------- 1. Numbers between 2010 and 2015 which do yield modular inverses for 42: 1. main :: IO () def main():  print ( mapMaybe( lambda y: bindMay(modInv(42)(y))( lambda mInv: Just((y, mInv)) ) )( enumFromTo(2010)(2025) ) )  1. -> [(2011, 814), (2015, 48), (2017, 1969), (2021, 1203)] 1. GENERIC ABSTRACTIONS ------------------------------------ 1. enumFromTo :: Int -> Int -> [Int] def enumFromTo(m):  return lambda n: list(range(m, 1 + n))  1. bindMay (>>=) :: Maybe a -> (a -> Maybe b) -> Maybe b def bindMay(m):  return lambda mf: ( m if m.get('Nothing') else mf(m.get('Just')) )  1. Just :: a -> Maybe a def Just(x):  return {'type': 'Maybe', 'Nothing': False, 'Just': x}  1. mapMaybe :: (a -> Maybe b) -> [a] -> [b] def mapMaybe(mf):  return lambda xs: reduce( lambda a, x: maybe(a)(lambda j: a + [j])(mf(x)), xs, [] )  1. maybe :: b -> (a -> b) -> Maybe a -> b def maybe(v):  return lambda f: lambda m: v if m.get('Nothing') else ( f(m.get('Just')) )  1. Nothing :: Maybe a def Nothing():  return {'type': 'Maybe', 'Nothing': True}  1. MAIN --- main()</lang> Output: [(2011, 814), (2015, 48), (2017, 1969), (2021, 1203)] ## Quackery Translation of: Forth <lang Quackery> [ dup 1 != if  [ tuck 1 0 [ swap temp put temp put over 1 > while tuck /mod swap temp take tuck * temp take swap - again ] 2drop temp release temp take dup 0 < if [ over + ] ] nip ] is modinv ( n n --> n )   42 2017 modinv echo</lang>  Output: 1969 ## Racket <lang racket> (require math) (modular-inverse 42 2017) </lang> Output: <lang racket> 1969 </lang> ## Raku (formerly Perl 6) <lang perl6>sub inverse($n, :$modulo) {  my ($c, $d,$uc, $vc,$ud, $vd) = ($n % $modulo,$modulo, 1, 0, 0, 1);
my $q; while$c != 0 {
($q,$c, $d) = ($d div $c,$d % $c,$c);
($uc,$vc, $ud,$vd) = ($ud -$q*$uc,$vd - $q*$vc, $uc,$vc);
}
return $ud %$modulo;


}

say inverse 42, :modulo(2017)</lang>

## REXX

<lang rexx>/*REXX program calculates and displays the modular inverse of an integer X modulo Y.*/ parse arg x y . /*obtain two integers from the C.L. */ if x== | x=="," then x= 42 /*Not specified? Then use the default.*/ if y== | y=="," then y= 2017 /* " " " " " " */ say 'modular inverse of ' x " by " y ' ───► ' modInv(x,y) exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ modInv: parse arg a,b 1 ob; z= 0 /*B & OB are obtained from the 2nd arg.*/

       $= 1 /*initialize modular inverse to unity. */ if b\=1 then do while a>1 parse value a/b a//b b z with q b a t z=$  -  q * z
$= trunc(t) end /*while*/   if$<0  then $=$ + ob                   /*Negative?  Then add the original  B. */
return $</lang>  output when using the default inputs of: 42 2017 modular inverse of 42 by 2017 ───► 1969  ## Ring <lang ring> see "42 %! 2017 = " + multInv(42, 2017) + nl func multInv a,b  b0 = b x0 = 0 multInv = 1 if b = 1 return 0 ok while a > 1 q = floor(a / b) t = b b = a % b a = t t = x0 x0 = multInv - q * x0 multInv = t end if multInv < 0 multInv = multInv + b0 ok return multInv  </lang> Output: 42 %! 2017 = 1969  ## Ruby <lang ruby>#based on pseudo code from http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm#Iterative_method_2 and from translating the python implementation. 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 'The maths are broken!' end x % et  end</lang>  > invmod(42,2017) => 1969 Simplified equivalent implementation <lang ruby> def modinv(a, m) # compute a^-1 mod m if possible  raise "NO INVERSE - #{a} and #{m} not coprime" unless a.gcd(m) == 1 return m if m == 1 m0, inv, x0 = m, 1, 0 while a > 1 inv -= (a / m) * x0 a, m = m, a % m inv, x0 = x0, inv end inv += m0 if inv < 0 inv  end </lang> > modinv(42,2017) => 1969  ## Run BASIC <lang runbasic>print multInv(42, 2017) end function multInv(a,b) b0 = b multInv = 1 if b = 1 then goto [endFun] while a > 1 q = a / b t = b b = a mod b a = t t = x0 x0 = multInv - q * x0 multInv = int(t) wend if multInv < 0 then multInv = multInv + b0 [endFun] end function</lang> Output: 1969  ## Rust <lang rust>fn mod_inv(a: isize, module: isize) -> isize {  let mut mn = (module, a); let mut xy = (0, 1); while mn.1 != 0 { xy = (xy.1, xy.0 - (mn.0 / mn.1) * xy.1); mn = (mn.1, mn.0 % mn.1); } while xy.0 < 0 { xy.0 += module; } xy.0  } fn main() {  println!("{}", mod_inv(42, 2017))  }</lang> Output: 1969  Alternative implementation <lang rust> fn modinv(a0: isize, m0: isize) -> isize {  if m0 == 1 { return 1 }   let (mut a, mut m, mut x0, mut inv) = (a0, m0, 0, 1);   while a > 1 { inv -= (a / m) * x0; a = a % m; std::mem::swap(&mut a, &mut m); std::mem::swap(&mut x0, &mut inv); } if inv < 0 { inv += m0 } inv  } fn main() {  println!("{}", modinv(42, 2017))  }</lang> Output: 1969  ## Scala Based on the Handbook of Applied Cryptography, Chapter 2. See http://cacr.uwaterloo.ca/hac/ . <lang scala> def gcdExt(u: Int, v: Int): (Int, Int, Int) = {  @tailrec def aux(a: Int, b: Int, x: Int, y: Int, x1: Int, x2: Int, y1: Int, y2: Int): (Int, Int, Int) = { if(b == 0) (x, y, a) else { val (q, r) = (a / b, a % b) aux(b, r, x2 - q * x1, y2 - q * y1, x, x1, y, y1) } } aux(u, v, 1, 0, 0, 1, 1, 0)  } def modInv(a: Int, m: Int): Option[Int] = {  val (i, j, g) = gcdExt(a, m) if (g == 1) Option(if (i < 0) i + m else i) else Option.empty  }</lang> Translated from C++ (on this page) <lang scala> def modInv(a: Int, m: Int, x:Int = 1, y:Int = 0) : Int = if (m == 0) x else modInv(m, a%m, y, x - y*(a/m)) </lang> Output: scala> modInv(2,4) res1: Option[Int] = None scala> modInv(42, 2017) res2: Option[Int] = Some(1976)  ## Seed7 The library bigint.s7i defines the bigInteger function modInverse. It returns the modular multiplicative inverse of a modulo b when a and b are coprime (gcd(a, b) = 1). If a and b are not coprime (gcd(a, b) <> 1) the exception RANGE_ERROR is raised. <lang seed7>const func bigInteger: modInverse (in var bigInteger: a,  in var bigInteger: b) is func result var bigInteger: modularInverse is 0_; local var bigInteger: b_bak is 0_; var bigInteger: x is 0_; var bigInteger: y is 1_; var bigInteger: lastx is 1_; var bigInteger: lasty is 0_; var bigInteger: temp is 0_; var bigInteger: quotient is 0_; begin if b < 0_ then raise RANGE_ERROR; end if; if a < 0_ and b <> 0_ then a := a mod b; end if; b_bak := b; while b <> 0_ do temp := b; quotient := a div b; b := a rem b; a := temp;   temp := x; x := lastx - quotient * x; lastx := temp;   temp := y; y := lasty - quotient * y; lasty := temp; end while; if a = 1_ then modularInverse := lastx; if modularInverse < 0_ then modularInverse +:= b_bak; end if; else raise RANGE_ERROR; end if; end func;</lang>  Original source: [1] ## Sidef Built-in: <lang ruby>say 42.modinv(2017)</lang> Algorithm implementation: <lang ruby>func invmod(a, n) {  var (t, nt, r, nr) = (0, 1, n, a % n) while (nr != 0) { var quot = int((r - (r % nr)) / nr); (nt, t) = (t - quot*nt, nt); (nr, r) = (r - quot*nr, nr); } r > 1 && return() t < 0 && (t += n) t  } say invmod(42, 2017)</lang> Output: 1969  ## Swift <lang swift>extension BinaryInteger {  @inlinable public func modInv(_ mod: Self) -> Self { var (m, n) = (mod, self) var (x, y) = (Self(0), Self(1))   while n != 0 { (x, y) = (y, x - (m / n) * y) (m, n) = (n, m % n) }   while x < 0 { x += mod }   return x }  } print(42.modInv(2017))</lang> Output: 1969 ## Tcl Translation of: Haskell <lang tcl>proc gcdExt {a b} {  if {$b == 0} {


return [list 1 0 $a]  } set q [expr {$a / $b}] set r [expr {$a % $b}] lassign [gcdExt$b $r] s t g return [list$t [expr {$s -$q*$t}]$g]


} proc modInv {a m} {

   lassign [gcdExt $a$m] i -> g
if {$g != 1} {  return -code error "no inverse exists of$a %! $m"  } while {$i < 0} {incr i $m} return$i


}</lang> Demonstrating <lang tcl>puts "42 %! 2017 = [modInv 42 2017]" catch {

   puts "2 %! 4 = [modInv 2 4]"


} msg; puts $msg</lang> Output: 42 %! 2017 = 1969 no inverse exists of 2 %! 4  ## Tiny BASIC 2017 causes integer overflow, so I'll do the inverse of 42 modulo 331 instead. <lang tinybasic>  PRINT "Modular inverse." PRINT "What is the modulus?" INPUT M PRINT "What number is to be inverted?" INPUT X PRINT "Solution is:"  10 LET A = A + 1  GOTO 20  15 IF B = 1 THEN PRINT A  IF B = 1 THEN END IF A = M-1 THEN PRINT "nonexistent" IF A = M-1 THEN END GOTO 10  20 LET B = A*X 30 IF B < M THEN GOTO 15  LET B = B - M GOTO 30  </lang> Output: Modular inverse. What is the modulus? 331 What number is to be inverted? 42 Solution is: 134 Another version: Translation of: GW-BASIC <lang tinybasic>  REM Modular inverse LET E=42 LET T=2017 GOSUB 100 PRINT M END   REM Increments E S (step) times until B is greater than T REM Repeats until B = 1 (MOD = 1) and C (count) REM B will not be greater than T + E  100 LET D=0  IF E>=T THEN GOTO 130 LET B=E REM At least one iteration is necessary LET S=((T-B)/E)+1 LET B=B+S*E LET C=1+S LET B=B-T  110 IF B=1 THEN GOTO 120  LET S=((T-B)/E)+1 LET B=B+S*E LET C=C+S LET B=B-T GOTO 110  120 LET D=C 130 LET M=D  RETURN  </lang> Output: 1969  ## tsql <lang tsql>;WITH Iterate(N,A,B,X0,X1) AS ( SELECT 1 ,CASE WHEN @a < 0 THEN @b-(-@a % @b) ELSE @a END ,CASE WHEN @b < 0 THEN -@b ELSE @b END ,0 ,1 UNION ALL SELECT N+1 ,B ,A%B ,X1-((A/B)*X0) ,X0 FROM Iterate WHERE A != 1 AND B != 0 ), ModularInverse(Result) AS ( SELECT -1 FROM Iterate WHERE A != 1 AND B = 0 UNION ALL SELECT TOP(1) CASE WHEN X1 < 0 THEN X1+@b ELSE X1 END AS Result FROM Iterate WHERE (SELECT COUNT(*) FROM Iterate WHERE A != 1 AND B = 0) = 0 ORDER BY N DESC ) SELECT * FROM ModularInverse</lang> ## uBasic/4tH Translation of: C <lang>Print FUNC(_MulInv(42, 2017)) End _MulInv Param(2)  Local(5)   c@ = b@ f@ = 0 g@ = 1   If b@ = 1 Then Return   Do While a@ > 1 e@ = a@ / b@ d@ = b@ b@ = a@ % b@ a@ = d@   d@ = f@ f@ = g@ - e@ * f@ g@ = d@ Loop   If g@ < 0 Then g@ = g@ + c@  Return (g@)</lang> Translation of: Perl <lang>Print FUNC(_mul_inv(42, 2017)) Print FUNC(_mul_inv(40, 1)) Print FUNC(_mul_inv(52, -217)) Print FUNC(_mul_inv(-486, 217)) Print FUNC(_mul_inv(40, 2018)) End _mul_inv Param(2)  Local(6)   If (b@ < 0) b@ = -b@ If (a@ < 0) a@ = b@ - (-a@ % b@) c@ = 0 : d@ = 1 : e@ = b@ : f@ = a@ % b@   Do Until (f@ = 0) g@ = e@/f@ h@ = d@ : d@ = c@ - g@*d@ : c@ = h@ h@ = f@ : f@ = e@ - g@*f@ : e@ = h@ Loop   If (e@ > 1) Return (-1) ' No inverse' If (c@ < 0) c@ = c@ + b@  Return (c@)</lang> Output: 1969 0 96 121 -1 0 OK, 0:156 ## UNIX Shell Works with: Bourne Again SHell Works with: Korn Shell Works with: Zsh Translation of: PowerShell <lang sh>function invmod {  typeset -i a=$1 n=$2 if (( n < 0 )); then (( n = -n )); fi if (( a < 0 )); then (( a = n - (-a) % n )); fi   typeset -i t=0 nt=1 r=n nr q tmp (( nr = a % n )) while (( nr )); do (( q = r/nr )) (( tmp = nt )) (( nt = t - q*nt )) (( t = tmp )) (( tmp = nr )) (( nr = r - q*nr )) (( r = tmp )) done if (( r > 1 )); then return 1 fi while (( t < 0 )); do (( t += n )); done printf '%s\n' "$t"


}

invmod 42 2017</lang>

Output:
1969

## VBA

Translation of: Phix
<lang vb>

Private Function mul_inv(a As Long, n As Long) As Variant

   If n < 0 Then n = -n
If a < 0 Then a = n - ((-a) Mod n)
Dim t As Long: t = 0
Dim nt As Long: nt = 1
Dim r As Long: r = n
Dim nr As Long: nr = a
Dim q As Long
Do While nr <> 0
q = r \ nr
tmp = t
t = nt
nt = tmp - q * nt
tmp = r
r = nr
nr = tmp - q * nr
Loop
If r > 1 Then
mul_inv = "a is not invertible"
Else
If t < 0 Then t = t + n
mul_inv = t
End If


End Function Public Sub mi()

   Debug.Print mul_inv(42, 2017)
Debug.Print mul_inv(40, 1)
Debug.Print mul_inv(52, -217) '/* Pari semantics for negative modulus */
Debug.Print mul_inv(-486, 217)
Debug.Print mul_inv(40, 2018)

End Sub</lang>
Output:
 1969
0
96
121
a is not invertible

## Wren

Library: Wren-big

<lang ecmascript>import "/big" for BigInt

var a = BigInt.new(42) var b = BigInt.new(2017) System.print(a.modInv(b))</lang>

Output:
1969


## XPL0

<lang XPL0>code IntOut=11, Text=12; int X; def A=42, M=2017; [for X:= 2 to M-1 do

   if rem(A*X/M) = 1 then [IntOut(0, X);  exit];


Text(0, "Does not exist"); ]</lang>

Output:
1969


## zkl

  if(b==0) return(1,0,a);

modInv(2,4)  //-->Void