So in 3DES is applied one encryption block followed by a decryption block, followed by an encryption block, to not be vulnerable to meet in the middle attacks?
Here forward 2^56 possibilities are compared with backward 2^56 possibilities. So doesn't it result in 2^56 * 2 ^56 = 2 ^112 possiblities ? So time should be O(2^112) right ? How is it O(2^56) ? Can anyone clarify?
My understanding is it's similar to how you can have two for-loops that goes for n times, one after the other, and the complexity is still O(n), not O(n^2). O(2n) is still O(n).
Thank you. It was very useful tutorial for me.
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So in 3DES is applied one encryption block followed by a decryption block, followed by an encryption block, to not be vulnerable to meet in the middle attacks?
good explanation!
Thank you so muchhhh!!
Heya 😂
in case the Key length is n bit for both keys and pain text is of n bits what is the Approx computational load ?
Thanks!
Here forward 2^56 possibilities are compared with backward 2^56 possibilities. So doesn't it result in 2^56 * 2 ^56 = 2 ^112 possiblities ? So time should be O(2^112) right ? How is it O(2^56) ? Can anyone clarify?
My understanding is it's similar to how you can have two for-loops that goes for n times, one after the other, and the complexity is still O(n), not O(n^2). O(2n) is still O(n).
Thank You For This Beautfiul Explanation.
Excellent, thank you
THANKS!!!
how does attacker know P' ?
This is an KPA (Known Plaintext Attack), thats why the adversary knows P'.
still not able to get easier