# (Homework Solution): How many bits are necessary to store the key for an ideal block cipher with B-bit

How many bits are necessary to store the key for an ideal block cipher with B-bit sized block? Is this cipher practical for large values of B? Why or why not? Please answer all aspects of this question neatly so I can easily read it.

-> In ideal block cipher we break the message into some blocks of given length and encrypt that message by encrypting each block with given key one by one and produce the output of the same block length. Encryption key is also of the same length. This block length may vary from 4 to 64 or some multipes of these. In this way we encrypt the whole message into a ciphertext using ideal block cipher. When a N bit size block is considered then message will be encrypted using all the possible combinations of N bit size block and for every integer we will give an output of N size block.

Since we are using B bit size block so total combinations will be 2^B (pow(2,B)).
And for B bits there will be a different output and every output will be of B bits.
So this will give the total size of the key that will be equal to :

=   B * 2^B

-> No, this cipher is not Practical for Larger values of B.

Reason: This cipher is impractical for larger values of B because if we increase value of B it will increase the size of the key. For B=32 the size would be equal to around 10^11 (pow(10,11)) which is very large size to implement, transmit, store and process the key for decryption. So there will be some logistic problems while dealing with such larger size keys. So this is not practical for large value of B.

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