Difference between revisions of "Encryption"

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Line 72: Line 72:
     s.push({a, b})
     s.push({a, b})
   }
   }
   if (a !== 1n) {
   if (a!==1n) {
     return NaN // inverse does not exists
     return NaN // inverse does not exists
   }
   }
Line 78: Line 78:
   let x = 1n
   let x = 1n
   let y = 0n
   let y = 0n
   for(let i = s.length - 2; i >= 0; --i) {
   for(let i=s.length-2; i>=0;--i) {
     [x, y] = [y, x - y * (s[i].a / s[i].b)]
     [x,y] = [y,x-y*(s[i].a/s[i].b)]
   }
   }
   return (y%m+m)%m
   return (y%m+m)%m

Revision as of 22:28, 19 September 2021


Fermat's Little Theorem

Fermat's little theorem tells us that

xp mod p = x

if p is prime for x<p

Verify this by trying prime and non-prime values for p. You can can generate prime numbers from https://bigprimes.org/

document.body.append(
  addInput('p','101'),
  addInput('m','3'),
  $m('button',{onclick:function(){
    document.getElementById('result').value =
      pow(getbig('m'),getbig('p'),getbig('p'));
  }},`m<sup>p</sup> mod p`),
  $m('input',{id:'result'})
);

//Utility functions
function pow(n,e,m){
  if (e<=0) return 1n;
  let r = pow(n,e/2n,m);
  return (r*r*(e%2n===1n?n:1n))%m;
}

function $m(tag,prop,children){
  let ret = document.createElement(tag);
  for(let k in prop)
    ret[k] = prop[k];
  if (typeof(children)==='string')
    ret.innerHTML = children;
  if (Array.isArray(children))
    for(let c of children)
      ret.append(c);
  return ret;
}
function getbig(id){return BigInt(document.getElementById(id).value)}
function addInput(id,value){
  return $m('div',{},[$m('label',{},id),$m('input',{id,value})]);  
}

Generate public/private key pairs


let addInput = (id,value) => {
  return $m('div',{},[$m('label',{},id),$m('input',{id,value})]);  
}
document.body.append(
  addInput('p','101'),
  addInput('q','103'),
  addInput('e','3'),
  $m('button',{},'Generate public/private key'),
  addInput('d',''),
  addInput('n','')
);


function modInverse(a, m){
  a = (a%m+m)%m;
  if (!a||m<2n) {
    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!==1n) {
    return NaN // inverse does not exists
  }
  // find the inverse
  let x = 1n
  let y = 0n
  for(let i=s.length-2; i>=0;--i) {
    [x,y] = [y,x-y*(s[i].a/s[i].b)]
  }
  return (y%m+m)%m
}
//Utility functions
function pow(n,e,m){
  if (e<=0) return 1n;
  let r = pow(n,e/2n,m);
  return (r*r*(e%2n===1n?n:1n))%m;
}

function $m(tag,prop,children){
  let ret = document.createElement(tag);
  for(let k in prop)
    ret[k] = prop[k];
  if (typeof(children)==='string')
    ret.innerHTML = children;
  if (Array.isArray(children))
    for(let c of children)
      ret.append(c);
  return ret;
}
function getbig(id){return BigInt(document.getElementById(id).value)}
function addInput(id,value){
  return $m('div',{},[$m('label',{},`${id} `),$m('input',{id,value})]);  
}