New answers tagged sha-2
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Cryptographic hash functions by design cannot be collision-free since they operate on arbitrary-sized input to fixed-sized outputs like 256 for SHA-256 and 512 for SHA-512. $$H:\{0,1\}^* \to \{0,1\}^b$$ where $b$ is the $H$'s output size.
By the pigeonhole principle, collisions are inevitable. Simply consider 100 holes and 101 pigeons. With this condition ...
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