Trash Chain
Reverse engineering a hash function

# Problem

It seems that my problems with hashing just keep multiplying...
nc umbccd.io 3100
Author: RJ

# Solution

The challenge is as follows:
Welcome to TrashChain! In this challenge, you will enter two sequences of integers which are used to compute two hashes. If the two hashes match, you get the flag! Restrictions:
• Integers must be greater than 1.
• Chain 2 must be at least 3 integers longer than chain 1
• All integers in chain 1 must be less than the smallest element in chain 2
Type "done" when you are finished inputting numbers for each chain.
Hash function:
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def H(val, prev_hash, hash_num):
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return (prev_hash * pow(val + hash_num, B, A) % A)
Copied!
Computing hashes:
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hashes = []
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for chain_num in range(len(chains)):
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cur_hash = 1
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for i, val in enumerate(chains[chain_num]):
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cur_hash = H(val, cur_hash, i+1)
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hashes.append(cur_hash)
Copied!
Leverage the fact that
$A^B\mod{A}=0$
. If we simply pass in A as the input, the hash will be 0, and we can make the two hashes collide by forcing them both to be 0.
Since chain 2 must be 3 numbers longer than chain 1, we can simply use
$A^{nB}\mod{A}=0$
, where n is any positive integer.
Testing our theory:
Of course, this works on the actual server as well.