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 generally 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 = (**g ^{a}**)mod

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

j = (

**13**)mod

^{31}**137**

j = 20

Ben now computes:

k = (**g ^{b}**)mod

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

k = (

**13**)mod

^{23}**137**

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** = **k ^{a}**mod

**p**[programatically =(k^a)%p

**= 24**

x

x

^{31}mod137

**x**= 92

Bill now knows that the shared encryption key is 92. He does not [need to] transmit it.

Ben calculates:

**x** = **j**^{b}mod**p** [programatically =(j^b)%p**
x** = 20

^{23}mod137

**x**= 92

Ben also now knows that the shared encryption key is 92. He does not [need to] transmit it.

At this point, expand outwards:

x = 92

x = **j**^{b}mod**p = ****k ^{a}**mod

**p**

((

**g**)mod

^{a}**p**)

^{b}mod

**p ==**((

**g**)mod

^{b}**p**)

^{a}mod

**p**

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 ’92′ 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 92 to each character, making sure it wraps around 255:

\xcc\xbd\xcf\xcf\xd3\xcb\xce\xc0

Ben can now decrypt using the opposite method.

In actual fact, not only would the key be substantially longer than ’92′, but our encryption algorithm of choice ‘AAE’ would also be replaced with something more cryptographically sound – Possibly RC4

Might I point out that ‘x’ (and the encryption key) should be 92 in all of its references.

Thank you very much Airdrik. I have adjusted this

Would you mind giving me an explanation of what the mod function/operator is? I tried to follow along with my calculator but someone forgot to put the “mod” key on my calculator. Does it go by some other name?

Sure cymerej. You may see ‘Modulo’ or ‘%’. Modulo does actually have multiple meanings/operations, however in this case and most common cases, it will result in the remainder after integer division.

i.e.

5mod3 would equal 2, as 5/3 is 1 remainder 2.

6mod3 would equal 0, as 6/3 is 2 remainder 0.

8mod7 would equal 1, as 8/7 is 1 remainder 1.

You can find a ‘Mod’ button on the Windows calc.exe in scientific mode. Hope this helps.