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Mar 16, 2020 at 0:07 vote accept Biology nerd
Mar 15, 2020 at 16:40 history edited kelalaka CC BY-SA 4.0
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Mar 15, 2020 at 16:34 comment added kelalaka If you consider how the SHA256 operates, that will be more clear. Since the new inputs are different the hash value will be different.
Mar 15, 2020 at 15:58 comment added Biology nerd So if x1 and x2 collide in the first 80 bits, and I perform a length extension attack to produce H(s∥x1) and H(s∥x2) - then H(s∥x1) and H(s∥x2) won't collide in the first 80 bits either? I'm wondering why this would be the case........
Mar 15, 2020 at 15:20 history edited kelalaka CC BY-SA 4.0
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Mar 15, 2020 at 15:09 history answered kelalaka CC BY-SA 4.0