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I read in Wikipedia about the Diffie–Hellman key exchange. But i can't imagine numbers which are hard to guess.

Can anybody give me a real-world example for $p$, $g$ and the two random secrets?

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up vote 2 down vote accepted

I am Bob and I am communicating with Alice. You are Eve, the eavesdropper and you see the following information being passed between us.

$p$ = a 396 digit prime 315791951375393537137595555337555955191395351995195751755791151795317131135377351919777977373317997317733397199751739199735799971153399111973979977771537137371797357935195531355957399953977139577337393111951779135151171355371173379337573915193973715113971779315731713793579595533511197399993313719939759551175175337795317333957313779755351991151933337157555517575773115995775199513553337335137111

$g$ = 5

$g^a \mod p$ = 27489633325073800398775665555817886420250962668110351939566023888497836125868152406573384812160988816636135839671600830972123154326698347543908237984222952075526254924036242003399678149299821208547691422947419851811974028751045560581548575054043995970503234125889791605774889110477621403114012112713190957314561511096280698589221374817602716179951101187316281123669357781523803509664317497207729

$g^b \mod p$ = 1871328666565825995546798366128332313007950715331716749713204590388482544710311757501721471367857473733174282285058381076423826262539615001751155796241435075482428300917592530410990583546909080238556317378367406354698458062808098314556896127603125072486660646729701592117222647853144199702370081278435240604316086143749700366105405028431580266698489313733272288229202699113269909480645581599706251

and now you tell me what is the secret that Alice and I share $g^{ab} \mod p$.

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thanks thats exactly what i've looked for.. i'll give it a try =) – wutzebaer Sep 8 '13 at 12:12
when using this algorhitm, is it a security problem if i use always the same p? or does it not matter? – wutzebaer Sep 8 '13 at 12:23
can you tell me how much digits "a" has? – wutzebaer Sep 8 '13 at 16:35
@wutzebaer Well you are supposed to use this exchange to establish a common key for a symmetric cipher like AES and then use AES for your actual communication. Symmetric ciphers are way faster than asymmetric ciphers. And then at the beginning of the next cipher I would use another prime p and have a new common key. As for $a$, nope I can't tell you how many digits does it have. That's part of the fun. – Fixed Point Sep 9 '13 at 5:09

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