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Feb 11, 2016 at 4:43 vote accept winsticknova
Feb 10, 2016 at 5:22 comment added winsticknova So x1 would just be 0 as well. So this would be similar if there were two x's, x1 and x2, and they were the same because then that would XOR to 0. Therefore, this function isn't collision resistant. But does preimage resistance and second preimage resistance imply collision resistance? Also, from what you said, I think that the function is second preimage resistant, but like I said earlier, if you XOR the same number it XORs to 0 which means that the function is not preimage resistant?
Feb 10, 2016 at 1:00 history answered poncho CC BY-SA 3.0