Diffie-Hellman in JavaScript

Question

Diffie-Hellman is a key exchange that allows 2 people to share a symmetric key without interaction beforehand. First, a person shares an equation; in this case, we use:

$$3^x \mod{17}$$

Next, each person generates a random, usually prime, number. Then, they plug it in the equation. Let's use 5 and 7:

• Alice: \$3^5 \mod{17} \equiv 9 \$
• Bob: \$3^7 \mod{17} \equiv 6 \$

Finally, they share their generated numbers and calculate a "Shared Secret", this is done by (\$U\$ represents your private number):

\$\text{Public}^U \mod{17}\$

In this case:

• Alice: \$6^5 \mod{17} \equiv 2\$
• Bob: \$9^7 \mod{17} \equiv 2\$

Resulting in both parties having the same number. Confused? See this video.

I've implemented a function to generate that key. Thoughts?

function KeyGen(p1, p2, n){
  // Get a binary string, reverse it
  var bin = String((n).toString(2)).split("").reverse().join("");
  // Base for growth
  var grow = n;
  // Holds values for totals
  var tota = [];
  var total = 1;
  // The main loop
  for(var i = 0; i < bin.length; i++){
    tota[i] = 1;
    if(bin.substring(i, i + 1) === "1"){
      for (var l = grow; l > 0;l--){
        tota[i] *= p1;
        tota[i] %= p2;
      }
    }
    total *= tota[i];
    total %= p2;
    grow *= n;
  }
  return total;
}

// Example of gen
var gen = KeyGen(3, 17, 5);

Answer 1

Variable Naming

It took me a while to figure out what are p1, p2 and n (and you don't mention them in your example).

String operation

You could use math operation to do the computation:

function KeyGen(base, modulo, exponent) {
  var result = 1;
  while(exponent > 0){
    if(exponent % 2 == 1){
        result = (result * base) % modulo;
    }
    base = (base * base) % modulo;
    exponent = exponent >>> 1;
  }
  return result;
}

The memory footprint is then very small (no array required).

Big numbers

Issues appear when you try to use big integers (more than 54 bits - like in your comment). To handle them correctly, I would recommand an external library like BigInt:

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e=this.value;return this.sign?new o(m(e,1),!0):b(e,1,this.sign)},u.prototype.prev=function(){var e=this.value;return e-1>-r?new u(e-1):new o(i,!0)};var _=[1];while(_[_.length-1]<=t)_.push(2*_[_.length-1]);var D=_.length,P=_[D-1];o.prototype.shiftLeft=function(e){if(!H(e))return e.isNegative()?this.shiftRight(e.abs()):this.times(G[2].pow(e));e=+e;if(e<0)return this.shiftRight(-e);var t=this;while(e>=D)t=t.multiply(P),e-=D-1;return t.multiply(_[e])},u.prototype.shiftLeft=o.prototype.shiftLeft,o.prototype.shiftRight=function(e){var t;if(!H(e))return e.isNegative()?this.shiftLeft(e.abs()):(t=this.divmod(G[2].pow(e)),t.remainder.isNegative()?t.quotient.prev():t.quotient);e=+e;if(e<0)return this.shiftLeft(-e);var n=this;while(e>=D){if(n.isZero())return n;t=A(n,P),n=t[1].isNegative()?t[0].prev():t[0],e-=D-1}return t=A(n,_[e]),t[1].isNegative()?t[0].prev():t[0]},u.prototype.shiftRight=o.prototype.shiftRight,o.prototype.not=function(){return this.negate().prev()},u.prototype.not=o.prototype.not,o.prototype.and=function(e){return B(this,e,function(e,t){return e&t})},u.prototype.and=o.prototype.and,o.prototype.or=function(e){return B(this,e,function(e,t){return e|t})},u.prototype.or=o.prototype.or,o.prototype.xor=function(e){return B(this,e,function(e,t){return e^t})},u.prototype.xor=o.prototype.xor;var j=1<<30,F=(t&-t)*(t&-t)|j,X=function(e,t){var n=G[0],r=G[1],i=e.length;if(2<=t&&t<=36&&i<=s/Math.log(t))return new u(parseInt(e,t));t=Q(t);var o=[],a,f=e[0]==="-";for(a=f?1:0;a<e.length;a++){var l=e[a].toLowerCase(),c=l.charCodeAt(0);if(48<=c&&c<=57)o.push(Q(l));else if(97<=c&&c<=122)o.push(Q(l.charCodeAt(0)-87));else{if(l!=="<")throw new Error(l+" is not a valid character");var h=a;do a++;while(e[a]!==">");o.push(Q(e.slice(h+1,a)))}}o.reverse();for(a=0;a<o.length;a++)n=n.add(o[a].times(r)),r=r.times(t);return f?n.negate():n};o.prototype.toString=function(t){t===e&&(t=10);if(t!==10)return $(this,t);var n=this.value,r=n.length,i=String(n[--r]),s="0000000",o;while(--r>=0)o=String(n[r]),i+=s.slice(o.length)+o;var u=this.sign?"-":"";return u+i},u.prototype.toString=function(t){return t===e&&(t=10),t!=10?$(this,t):String(this.value)},o.prototype.valueOf=function(){return+this.toString()},o.prototype.toJSNumber=o.prototype.valueOf,u.prototype.valueOf=function(){return this.value},u.prototype.toJSNumber=u.prototype.valueOf;var G=function(e,t){return typeof e=="undefined"?G[0]:typeof t!="undefined"?+t===10?Q(e):X(e,t):Q(e)};for(var Y=0;Y<1e3;Y++)G[Y]=new u(Y),Y>0&&(G[-Y]=new u(-Y));return G.one=G[1],G.zero=G[0],G.minusOne=G[-1],G.max=q,G.min=R,G.gcd=U,G.lcm=z,G.isInstance=function(e){return e instanceof o||e instanceof u},G.randBetween=W,G}();typeof module!="undefined"&&module.hasOwnProperty("exports")&&(module.exports=bigInt);


function compute(){
  var base = bigInt(document.getElementById('base').value);
  var exponent = bigInt(document.getElementById('exponent').value);
  var modulo = bigInt(document.getElementById('modulo').value);

  var res = base.modPow(exponent, modulo);
  document.getElementById('result').value = res;
}

<label>Base:<input id="base" value="3"></label><br>
<label>Exponent:<input id="exponent" value="23984081230374123749712934791249172394719237497219347129374498"></label><br>
<label>Modulo:<input id="modulo" value="112457129983317064494133258034491756790943511028023366901014968560410379195027"></label><br>
<input type="button" value="Compute" onclick="compute()"><br>
Result: <input id="result">

The result seems to be exact.

PS: The modPow implementation is actually quite readable.

Answer 2

I found a really efficient way to solve this problem! It requires calculating the values before hand, my code goes as follows:

function KeyGen(p1, p2, n) {
    var inc = 0;
    var bin = [];
    bin[0] = p1;
    for (var i = 1; i <= n.toString(2).length; i++){
      bin[i] = (bin[i-1] * bin[i-1]) % p2;
      console.log(bin[i] + " and " + bin[i-1]);
    }
    console.log("loop");
    return n.toString(2).split('').reverse().reduce(function(total, char) {
        if(char === '1') {
              total = (total * bin[inc]) % p2;
              console.log("current "+ total);
            }
        inc++;
        return total % p2;
    }, 1);
}