A commercial file encryption program from the early 1990s used standard DES with

Question

A commercial file encryption program from the early 1990s used standard DES with 56 key bits. In those days, performing an exhaustive key search was considerably harder than nowadays, and thus the key length was sufficient for some applications. Unfortunately, the implementation of the key generation was flawed, which we are going to analyze. Assume that you can test 106 keys per second on a conventional PC.
The key is generated from a password consisting of 8 characters. The key is a simple concatenation of the 8 ASCII characters, yielding 64 = 8 x 8 key bits. With the permutation PC1 in the key schedule, the least significant bit (LSB) of each 8-bit character is ignored, yielding 56 key bits.
1. What is the size of the key space if all 8 characters are randomly chosen 8-bit ASCII characters? How long does an average key search take with a single PC?
2. How many key bits are used, if the 8 characters are randomly chosen 7-bit ASCII characters (i.e., the most significant bit is always zero)? How long does an average key search take with a single PC?

Explanation / Answer

1)
size of the key space is can be calcualted as: 8x8 = 64 bits
with permuation, the LSB of each bit is ignored so yield 56 key bits.

To search for a single pc maximum it takes 2^56 attempts to find key.
2^56 = 72,057,594,037,927,936

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2)
since for DES the key size must be 64 bit. It wont accept less than <64 bits.
So it would take 7x8 = 56 key bits...no LSB ignored since bit is always zero (already
mentioned in question)

To search for a single pc maximum it takes 2^56 attempts to find key.
2^56 = 72,057,594,037,927,936

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