Brute Forcing Encryption

[Jared Russell]

Originally Posted by jtdowney

This would leave us with only the option of brute forcing the algorithm. Now he did indicate that the input was 12 characters which narrows the field somewhat and that the input consists of A-Z, a-z, 0-9, and space. This leaves us with 12 (r) spots for one of 63 (n) values. Using (n+r-1)!/r!(n-1)! we arrive at 21,944,067,106,416 possible values.

Now I for one don't have enough time on the University supercomputer to brute force that but if someone has a few graphics cards laying around you could probably build a computer that could crack it in a weeks time.

1) Andrew defined his alphabet as A-Z, a-z, space. He did not say that 0-9 were in the input. The alphabet size is therefore 53, not 63.

2) You used (n+r-1)! / r!(n-1)! to compute the number of possible inputs. This is the formula for the number of COMBINATIONS with repetition. A combination is an un-ordered collection of elements. Order is important for the input string (as "123" hashes to something different from "321" with all of the schemes discussed in this thread); the correct formula to use is therefore the number of PERMUTATIONS with repetition. This is simply n^r as I had done above.

n^r = 53^12 = 4.91258904 × 10^20

Originally Posted by jtdowney

Let me first say for the most part all previous posts in this thread are confusing cryptographic primitives. Please take no offense from this correction as cryptography is a difficult subject.

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