# Diffie-Hellman Key Exchange

First published at:

http://www.adamsinfo.com/diffie-hellman-key-exchange/

Diffie-Hellman Key Exchange is a popular mathematical key exchange algorithm. It allows two parties to establish a ‘key’ over an insecure medium such as the internet. As you will see, it doesn’t matter whether the intercepting party captures each piece of transmitted information, they will not be able to break the key in any way, other than the usual brute force method.

Diffie-Hellman Key Exchange is not an encryption method, it is generall but not always used pre encryption to decide on a shared encryption key.

We will call the communicating parties Bill and Ben. Let Roger be the intercepting party. You can work out these calculations on a calculator:

Bill and Ben transmit and agree on a public prime number (p) and a ‘generator’ (g) which is an integer less than ‘p’. Bill now decides on a random private number (a) which he does not transmit, Ben also agrees on a random private number (b) which he does not transmit either.

In this example, Bill and Ben decide that:

p=137

g=13

Roger catches p(137) and g(13)

Bill decides privately that a=31

Ben decides privately that b=23

In actual fact these numbers will be much larger to hinder brute force. We’re going to use small numbers in our example though.

Bill now computes:

j = (ga)modp [ programatically j=(g^a)%p]

j = (1331)mod137

j = 20

Ben now computes:

k = (gb)modp [ programatically k=(g^b)%p ]

k = (1323)mod137

k = 24

Ben now transmits k to Bill, and Bill transmits j to Ben

Roger captures (j)20 and (k)24

Bill now computes:

x = kamodp [programatically =(k^a)%p

x = 2431mod137

x = 91

Bill now knows that the shared encryption key is 72. He does not [need to] transmit it.

Ben calculates:

x = jbmodp [programatically =(j^b)%p

x = 2023mod137

x = 92

Ben also now knows that the shared encryption key is 72. He does not [need to] transmit it.

At this point, expand outwards:

x = 92

x = jbmodp = kamodp

((ga)modp)bmodp == ((gb)modp)amodp

Despite the fact that Roger has caught each individual transmission, j, k, g, and p, he can not work out x.

Now that both sides know the key, we can now agree that they’re going to encrypt using AAE - Adam’s Amazing Encryption.

Bob takes his phrase to encrypt - “password” and adds ‘72′ to each character using the ASCII alphabet:

Bob can now turn his phrase to a hex string:

\x70\x61\x73\x73\x77\x6f\x72\x64

And then add 72 to each character, making sure it wraps around 255:

\xb8\xa9\xbb\xbb\xbf\xb7\xba\xac

Ben can now decrypt using the opposite method.

In actual fact, not only would the key be substantially longer than ‘72′, but our encryption algorithm of choice ‘AAE’ would also be replaced with something more cryptographically sound :-) - Possibly RC4

Enjoy!

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