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As long as the hash stored securely is over a canonical representation of the data (i.e. the variables and their encoding cannot overlap) then storing the hash is secure. It would be similar security as having a signature over the data, but those are larger. One thing that I would make sure of is that you store the hash using a transaction; otherwise the ...

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Late important addtion: I now realize the code attempts to find collisions for a 64-bit $\operatorname{hash}$ accepting 64-bit messages. If that $\operatorname{hash}$ was a bijection, it would not collide. There is a continuum between a bijection and a random function, and no insurance that $\operatorname{hash}$ behaves mostly like the later. On the contrary,...

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Too unreputable to comment... I expect it is an implementation issue - the high-level description of the method seems reasonable. Can it find a collision if you instead use the prefixes 0x01000099 and 0xDEADBD5C? Spoiler: e.g. 0x010000992287FF50 vs. 0xDEADBD5C05F19159 The method used for finding this collision is essentially the same as the one you describe, ...

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This seems like home work so I will stop short of a full solution. Yes you are allowed to query the functions $H_1$ and $H_2$ it's almost the only thing you can do. So you can collect a pool of input output pairs for each. And then what can you do with two such collections of input output pairs? You may want to index one one of them for efficient lookup.

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They so that by taking an authenticated message, and applying a carefully crafted difference to the message, you can ensure half the bits of the authentication tag will be preserved. You can repeat the attack on different authenticated cipher texts you captured(or perhaps caused) or (more relevant) different solutions for the linear problem as it is under-...

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TEA was used in the XBox and was so weak, that it allowed the XBox hackers to alter the code and still get the same hash: https://web.archive.org/web/20090416175601/http://www.xbox-linux.org/wiki/17_Mistakes_Microsoft_Made_in_the_Xbox_Security_System If you flip both bit 16 and 31 of a 32 bit word, the hash will be the same. We could easily patch a jump in ...

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