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What I understand so far: For nbit 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 nbit block size, there are 2^{n} plaintexts (P)
 there are P! possible keys
 The bits necessary to represent each of the possible mappings is log_{2}(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 8bit 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 32bit 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.
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