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How are keys mapped to cipher texts in block ciphers with large block sizes?

What I understand so far: For n-bit block size, there are 2n plaintexts (P) there are P! possible keys The bits necessary to represent each of the possible mappings is log2(P!) rounded up Basically, my calculator craps out at a block size of 8 bits because the key length is so massive to map all::Listen

What I understand so far:

  • For n-bit block size, there are 2n plaintexts (P)
  • there are P! possible keys
  • The bits necessary to represent each of the possible mappings is log2(P!) rounded up

Basically, my calculator craps out at a block size of 8 bits because the key length is so massive to map all possible keys for 8-bit block size.

What I am led to understand is that there is some way to use much smaller key sizes (like 128 bits), but from what I have read, there are encryptions that use larger block sizes like 32 bits, 64 bits, etc where there is no feasible way to map every possible ciphertext to a key.

It seems to me that if there was a 32-bit block size being used, for example, then with 128 bits, you can only map a very small chunk of the possible ciphertexts. I’m not even sure how to ask what I’m trying to understand…how does this work? I’m just confused by this. I guess what I’m trying to ask is, if it’s impossible to map a specific key to a specific, unique ciphertext value, what does it actually map to if I just pick 32 random bits? It seems that each possible key would have to map to multiple ciphertexts.

Shanghai Man: Blockchain Week with Vitalik still happening, ‘Bitcoin’ searches on WeChat hit 26M in a day

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