CODEWITHSUNDEEP
javascriptalgorithmrsainversemodular-arithmetic
Lizziej
Lizziej

Reputation: 95

Calculating the Modular Inverse in JavaScript

I am trying to take ed = 1 mod((p-1)(q-1)) and solve for d, just like the RSA algorithm.

e = 5, (p-1)*(q-1) = 249996

I've tried a lot of code in javascript such as:

function modInverse(){
var e = 5;
var p = 499;
var q = 503;
var d = e.modInverse((p-1) * (q-1));
DisplayResult(d, "privateKeyResultLabel")
}

or

function modInverse(){ 
System.out.println(BigInteger.valueOf(5).modInverse(BigInteger.valueOf(249996)));
}

I just can't figure out the correct way to solve for d, the modular inverse, in javascript.

Upvotes: 9

Views: 9569

Answers (2)

Vivek Pradhan
Vivek Pradhan

Reputation: 4847

I was just going through the definition of modular multiplicative inverse and from what I understand:

ax = 1 (mod m)
=> m is a divisor of ax -1 and x is the inverse we are looking for
=> ax - 1 = q*m (where q is some integer)
And the most important thing is gcd(a, m) = 1
i.e. a and m are co-primes

In your case:

ed = 1 mod((p-1)(q-1)) //p, q and e are given 
=> ed - 1 = z*((p-1)(q-1)) //where z is some integer and we need to find d

Again from the wikipedia entry, one can compute the modular inverse using the extended Euclidean GCD Algorithm which does the following:

ax + by = g //where g = gcd(a,b) i.e. a and b are co-primes
//The extended gcd algorithm gives us the value of x and y as well.

In your case the equation would be something like this:

ed - z*((p-1)(q-1)) = 1; //Compare it with the structure given above

a -> e
x -> d
b -> (p-1)(q-1)
y -> z

So if we just apply that algorithm to this case, we will get the values of d and z.

For ax + by = gcd(a,b), the extended gcd algorithm could look something like (source):

function xgcd(a, b) { 

  if (b == 0) {
    return [1, 0, a];
  }

  temp = xgcd(b, a % b);
  x = temp[0];
  y = temp[1];
  d = temp[2];
  return [y, x-y*Math.floor(a/b), d];
}

This algorithm runs in time O(log(m)^2), assuming |a| < m, and is generally more efficient than exponentiation.

I don't know if there is an inbuilt function for this in javascript. I doubt if there is, and I am a fan of algorithms, so I thought you might want to give this approach a try. You can fiddle with it and change it to handle your range of values and I hope it gets you started in the right direction.

Upvotes: 10

Dipu
Dipu

Reputation: 8279

This implementation of modular inverse can accept any type of inputs. If input types are not supported, NaN is returned. Also, it does not use recursion.

function modInverse(a, m) {
  // validate inputs
  [a, m] = [Number(a), Number(m)]
  if (Number.isNaN(a) || Number.isNaN(m)) {
    return NaN // invalid input
  }
  a = (a % m + m) % m
  if (!a || m < 2) {
    return NaN // invalid input
  }
  // find the gcd
  const s = []
  let b = m
  while(b) {
    [a, b] = [b, a % b]
    s.push({a, b})
  }
  if (a !== 1) {
    return NaN // inverse does not exists
  }
  // find the inverse
  let x = 1
  let y = 0
  for(let i = s.length - 2; i >= 0; --i) {
    [x, y] = [y,  x - y * Math.floor(s[i].a / s[i].b)]
  }
  return (y % m + m) % m
}

// Tests
console.log(modInverse(1, 2))       // = 1
console.log(modInverse(3, 6))       // = NaN
console.log(modInverse(25, 87))     // = 7
console.log(modInverse(7, 87))      // = 25
console.log(modInverse(19, 1212393831))     // = 701912218
console.log(modInverse(31, 73714876143))    // = 45180085378
console.log(modInverse(3, 73714876143))     // = NaN
console.log(modInverse(-7, 87))     // = 62
console.log(modInverse(-25, 87))    // = 80
console.log(modInverse(0, 3))       // = NaN
console.log(modInverse(0, 0))       // = NaN

Upvotes: 8

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