D-H is a key exchange mechanism , a way to exchange cryptographic keys over public channels, that is , I want to agree a number with my buddy Simon but i don’t want to tell him what the number is.
Agreeing in Prime and Base Numbers
So there is a few requirements for this to work , and also a bunch of rules that have to be applied to the values we gonna communicate over an open space:
So all these numbers are communicated to each other over public transport , and this is the requirement: Each user will generate a “secret exponent” that can’t have any common factors between them. So here’s some code to make this up
def giveMePrimes():primeList = list()for num in range(1000,10001):if all(num%i!=0 for i in range(2,int(math.sqrt(num))+1)):primeList.append(num)return primeList
This appends the the prime numbers in the range() to a list.
For the Base and Secret Exponents we pick numbers just randomzingin a range:
secure_random.choice = random.SystemRandom()secure_random.choice(range(1,10001))
Let’s assume we have these numbers:
P (prime) = 2833N (base) = 3667J(Jerry's Secret exponent) = 6531S(Simon's Secret exponent) = 8249
Only the secret exponent are the ones we want to keep private , Jerry and Simon know only their “secret exponent” ,P and N , in short only P and N have been communicated over public channels.
Now each party will compute a number with the values they have (P , N and each individual “secret exponent”)
Jerry will compute: (N(base) to the J(secret exponent) modulus P (prime))
JC = N ** J % P (2642)
and Simon:
SC = N ** S % P (1037)
And they will pass each other these values (Jerry passes JC to Simon , and Simon passes SC to Jerry).
Good ! so now to Recap , let’s see what values each individual knows:
Jerry: Prime , Base , Jerry’s secret exponent , JC and SC
Simon: Prime , Base , Simon’s secret exponent , JC and SC
Random: Prime , Base , JC and SC
So now is when the magic happens , both of them apply the following function: ( peer’s computed number to the “secret exponent” modulus Prime)
Jerry:
SC ** J % P
Simon
JC ** S % P
And if you do the math this translates to:
Jerry
1037 ** 6531 % 2833 = 1747L
Simon
2642 ** 8249 % 2833 = 1747L
Great (L == Python’s literal ) , now we have have a number(1747L) that Simon and Jerry know that derived of a bunch of numbers that were transmitted publicly.
With this number as a “key” Simon and Jerry can start communicating privately using a cipher.
I’ve put an example of the code that i’ve used for this , it kind of works: