Modular multiplicative inverse

Question

Your task is to given two integer numbers, a and b calculate the modular multiplicative inverse of a modulo b, if it exists.

The modular inverse of a modulo b is a number c such that ac ≡ 1 (mod b). This number is unique modulo b for any pair of a and b. It exists only if the greatest common divisor of a and b is 1.

The Wikipedia page for modular multiplicative inverse can be consulted if you require more information about the topic.

Input and Output

Input is given as either two integers or a list of two integers. Your program should output either a single number, the modular multiplicative inverse that is in the interval 0 < c < b, or a value indicating there is no inverse. The value can be anything, except a number in the range (0,b), and may also be an exception. The value should however be the same for cases in which there is no inverse.

0 < a < b can be assumed

Rules

• The program should finish at some point, and should solve each test case in less than 60 seconds
• Standard loopholes apply

Test cases

Test cases below are given in the format, a, b -> output

1, 2 -> 1
3, 6 -> Does not exist
7, 87 -> 25
25, 87 -> 7
2, 91 -> 46
13, 91 -> Does not exist
19, 1212393831 -> 701912218
31, 73714876143 -> 45180085378
3, 73714876143 -> Does not exist

Scoring

This is code golf, so the shortest code for each language wins.

This and this are similar questions, but both ask for specific situations.

Answer 1

Mathematica, 14 bytes

Obligatory Mathematica builtin:

ModularInverse

It's a function that takes two arguments (a and b), and returns the inverse of a mod b if it exists. If not, it returns the error ModularInverse: a is not invertible modulo b..

Answer 2

JavaScript (ES6), 79 73 62 61 bytes

Returns false if the inverse does not exist.

It uses the extended Euclidean algorithm and solves all test cases almost instantly.

f=(a,b,c=!(n=b),d=1)=>a?f(b%a,a,d,c-(b-b%a)/a*d):b<2&&(c+n)%n

Test cases

f=(a,b,c=!(n=b),d=1)=>a?f(b%a,a,d,c-(b-b%a)/a*d):b<2&&(c+n)%n

console.log(f(1, 2)) // -> 1
console.log(f(3, 6)) // -> Does not exist
console.log(f(7, 87)) // -> 25
console.log(f(25, 87)) // -> 7
console.log(f(2, 91)) // -> 46
console.log(f(13, 91)) // -> Does not exist
console.log(f(19, 1212393831)) // -> 701912218
console.log(f(31, 73714876143)) // -> 45180085378
console.log(f(3, 73714876143)) // -> Does not exist

Answer 3

Python 2, 34 bytes

f=lambda a,b:a==1or-~b*f(-b%a,a)/a

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Recursive function that gives True for print f(1,2), which I believe to be acceptable, and errors for invalid inputs.

We are trying to find \$x\$ in \$a\cdot x\equiv 1\pmod{b}\$.

This can be written as \$a\cdot x-1=k\cdot b\$ where \$k\$ is an integer.

Taking \$\mod{a}\$ of this gives \$-1\equiv k\cdot b\pmod{a}\$. Moving the minus gives \$-k\cdot b\equiv1\pmod{a}\$, where we have to solve for \$k\$.

Seeing how it resembles the initial scenario, allow us to recurse to solve for \$k\$ by calling the function with \$f(-b\%a,a)\$ (works because Python gives positive values for modulo with a negative argument).

The program recurses for until \$a\$ becomes 1, which only happens if the original \$a\$ and \$b\$ are coprime to each other (ie there exists a multiplicative inverse), or ends in an error caused by division by 0.

This value of \$k\$ can be substituted in the equation \$a\cdot x-1=k\cdot b\$ to give \$x\$ as \$\frac{k\cdot b+1}{a}\$.

Answer 4

Jelly, 2 bytes

æi

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This uses a builtin for modular inverse, and returns 0 for no modular inverse.

Jelly, 7 bytes

R×%⁸’¬T

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Outputs empty set (represented as empty string) on no modular inverse. Runs out of memory on TIO for the largest test-cases, but should work given enough memory.

How it Works

R×%⁸’¬T  
R        Generate range of b
 ×       Multiply each by a
  %⁸     Mod each by b
    ’    Decrement (Map 1 to 0 and all else to truthy)
     ¬   Logical NOT
      T  Get the index of the truthy element.

If you want to work for larger test-cases, try this (relatively ungolfed) version, which requires much time rather than memory:

Jelly, 9 bytes

×⁴%³’¬ø1#

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How it Works

×⁴%³’¬ø1#
        #   Get the first
      ø1      one integer
            which meets:
×⁴            When multiplied by a
  %³          And modulo-d by b
    ’         Decrement
     ¬        Is falsy

Answer 5

Mathematica, 18 bytes

PowerMod[#,-1,#2]&

input

[31, 73714876143]

Answer 6

R + numbers, 15 bytes

numbers::modinv

returns NA for those a without inverses mod b.

R-Fiddle to try it!

R, 33 bytes (non-competing)

This will fail on very large b since it actually creates a vector of size 32*b bits.

function(a,b)which((1:b*a)%%b==1)

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Returns integer(0) (an empty list) for those a without inverses mod b.

Answer 7

Japt, 9 8 bytes

Takes the inputs in reverse order. Outputs -1 for no match. Craps out as the bigger integer gets larger.

Ç*V%UÃb1

Test it

• Saved 1 byte thanks to ETH pointing out an errant, and very obvious, space.

Answer 8

Python 3 + gmpy, 23 bytes

I don't think it can get any shorter in Python.

gmpy.invert
import gmpy

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Answer 9

Python 3, 49 bytes

lambda a,b:[c for c in range(b)if-~c*a%b==1][0]+1

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Python 3, 50 bytes

lambda a,b:[c for c in range(1,b+1)if c*a%b==1][0]

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This throws IndexError: list index out of range in case there is no modular multiplicative inverse, as it is allowed by the rules.

Answer 10

Python 2, 51 49 54 53 51 49 bytes

^{-1 byte thanks to officialaimm
-1 byte thanks to Shaggy}

a,b=input()
i=a<2
while(a*i%b-1)*b%a:i+=1
print+i

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Prints 0 when there is no solution.

Answer 11

8th, 6 bytes

Code

invmod

Explanation

invmod is a 8th word that calculates the value of the inverse of a, modulo b. It returns null on overflow or other errors.

Usage and test cases

ok> 1 2 invmod .
1
ok> 3 6 invmod .
null
ok> 7 87 invmod .
25
ok> 25 87 invmod .
7
ok> 2 91 invmod .
46
ok> 13 91 invmod .
null
ok> 19 1212393831 invmod .
701912218
ok> 31 73714876143 invmod .
45180085378
ok> 3 73714876143 invmod .
null

Answer 12

J, 28 bytes

4 :'(1=x+.y)*x y&|@^<:5 p:y'

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Uses Euler's theorem. Returns 0 if the inverse does not exist.

Explanation

4 :'(1=x+.y)*x y&|@^<:5 p:y'  Input: a (LHS), b (RHS)
4 :'                       '  Define an explicit dyad - this is to use the special
                              form `m&|@^` to perform modular exponentiation
                          y   Get b
                      5 p:    Euler totient
                    <:        Decrement
             x                Get a
                   ^          Exponentiate
               y&|@             Modulo b
       x+.y                   GCD of a and b
     1=                       Equals 1
            *                 Multiply

Answer 13

Pyth, 10 bytes

3 bytes saved thanks to @Jakube.

xm%*szdQQ1

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Returns -1 for no multiplicative inverse.

Code Breakdown

xm%*szdQQ1      Let Q be the first input.
 m      Q       This maps over [0 ... Q) with a variable d.
   *szd         Now d is multiplied by the evaluated second input.
  %    Q        Now the remained modulo Q is retrieved.
x        1      Then, the first index of 1 is retrieved from that mapping.

Pyth, 15 13 bytes

KEhfq1%*QTKSK

Throws an exception in case no multiplicative inverse exists.

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Pyth, 15 bytes

Iq1iQKEfq1%*QTK

This adds lots of bytes for handling the case where no such number exists. The program can be shortened significantly if that case would not need to be handled:

fq1%*QTK

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Answer 14

Pari/GP, 11 bytes

a->b->1/a%b

Throws an error when there is no inverse.

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Answer 15

C (gcc), 115 bytes

#define L long long
L g(L a,L b,L c,L d){return a?g(b%a,a,d-b/a*c,c):b-1?0:d;}L f(L a,L b){return(g(a,b,1,0)+b)%b;}

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Extended Euclidean algorithm, recursive version

C (gcc), 119 bytes

long long f(a,b,c,d,t,n)long long a,b,c,d,t,n;{for(c=1,d=0,n=b;a;a=t)t=d-b/a*c,d=c,c=t,t=b%a,b=a;return b-1?0:(d+n)%n;}

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Extended Euclidean algorithm, iterative version

Answer 16

C (gcc), 48 110 104 bytes

#define f(a,b)g(a,b,!b,1,b)
long g(a,b,c,d,n)long a,b,c,d,n;{a=a?g(b%a,a,d,c-(b-b%a)/a*d):!--b*(c+n)%n;}

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This should work with all inputs (that fit within a long) within 60 seconds.

Edit. I'm already abusing the n variable so I might as well assume that gcc puts the first assignment in %rax.

Answer 17

APL (Dyalog Unicode), 36 38 bytes

{⌈/i×1=⍵|(i←⍳⍵)×⍵|⍺}

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Explanation:

                   ⍵|⍺} ⍝ Get ⍺ mod ⍵
             (i←⍳⍵)×     ⍝ Multiply the result by all numbers up to ⍵
          ⍵|            ⍝ Take result mod ⍵
     i×1=                ⍝ Find all numbers (1,⍵) where the mod is 1
{⌈/                      ⍝ And take the largest

Much thanks to Adam in the APL Orchard chatroom for the help with this one!

Formula obtained from this site

First iteration:

{((⍵|⍺),⍵){+/⌈/⍵×1=(¯1↑⍺)|⍵×⊃⍺}⍳⍵}

Answer 18

Python 3.8, 22 bytes

lambda a,b:pow(a,-1,b)

Obligatory Python 3.8+ builtin. Outputs the result, or gives the error ValueError: base is not invertible for the given modulus if there is no result.

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Explanation:

To quote the Python 3.8 release notes:

For integers, the three-argument form of the pow() function now permits the exponent to be negative in the case where the base is relatively prime to the modulus. It then computes a modular inverse to the base when the exponent is -1, and a suitable power of that inverse for other negative exponents. For example, to compute the modular multiplicative inverse of 38 modulo 137, write:

>>> pow(38, -1, 137)
119
>>> 119 * 38 % 137
1

Modular inverses arise in the solution of linear Diophantine equations. For example, to find integer solutions for 4258𝑥 + 147𝑦 = 369, first rewrite as 4258𝑥 ≡ 369 (mod 147) then solve:

>>> x = 369 * pow(4258, -1, 147) % 147
y = (4258 * x - 369) // -147
4258 * x + 147 * y
369

(Contributed by Mark Dickinson in bpo-36027.)

Answer 19

[Pyt], 1 byte

ɯ

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Takes inputs in the order b a (on separate lines). Outputs None if there is no such multiplicate inverse. Gotta love built-ins.

Answer 20

Python 2 + sympy, 74 bytes

from sympy import*
def f(a,m):i,_,g=numbers.igcdex(a,m);return g==1and i%m

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Taken from Jelly source code.

Answer 21

Axiom, 45 bytes

f(x:PI,y:PI):NNI==(gcd(x,y)=1=>invmod(x,y);0)

0 for error else return z with x*z Mod y =1

Answer 22

Python 2, 52 bytes

-3 bytes thanks to Mr. Xcoder.

f=lambda a,b,i=1:i*a%b==1and i or i<b and f(a,b,i+1)

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Outputs False on no solution and errors out as b gets larger.

Embedded TIO

I'm just testing out iframes in Stack Snippets and they work absolutely fantastic.

html,body{height:100%;}iframe{height:100%;width:100%;border:none;}

<iframe src="https://tio.run/##PY3BCsIwEETv/Yq5CEndy6ZotbhfIh4SanBB09Lm4tdHU8HTvIHhzfzOjym5UqI8/SuMHp4CqfCgrd8FEfZphGJaoJeAWqLZJnu2Jd/XvEJwNUxwlmA6wrFmTzj1FdzhT4QzV@DuR7emidULTdhMA@ZFU/4@tGrLBw"></iframe>

Answer 23

JavaScript (ES6), 42 41 39 38 bytes

Outputs false for no match. Will throw a overflow error as the second number gets to be too large.

x=>y=>(g=z=>x*z%y==1?z:z<y&&g(++z))(1)

Answer 24

Jelly, 27 bytes

²%³
⁴Ç⁹Сx⁸
ÆṪ’BṚçL$P%³×gỊ¥

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Uses Euler's theorem with modular exponentiation. Since Jelly doesn't have a builtin to perform modular exponentiation, it had to be implemented, and it took most of the bytes.

Answer 25

Axiom, 99 bytes

w(a,b,x,u)==(a=0=>(b*b=1=>b*x;0);w(b rem a,a,u,x-u*(b quo a)));h(a,b)==(b=0=>0;(b+w(a,b,0,1))rem b)

it use the function h(); h(a,b) return 0 if error [not exist inverse] otherwise it return the z such that a*z mod b = 1 This would be ok even if arguments are negative...

this would be the general egcd() function that retunr a list of int (so they can be negative too)

egcd(aa:INT,bb:INT):List INT==
   x:=u:=-1   -- because the type is INT
   (a,b,x,u):=(aa,bb,0,1)
   repeat
      a=0=>break
      (q,r):=(b quo a, b rem a)
      (b,a,x,u):=(a,r,u,x-u*q)
   [b,x, (b-x*aa)quo bb]

this is how use it

(7) -> h(31,73714876143)
   (7)  45180085378
                                                    Type: PositiveInteger

i find the base algo in internet from https://pastebin.com/A13ybryc