2 ms·

You're missing something in your definition of a second-preimage attack, which is that Y != X.

• Gankro 10y ago
  Ah good point. It could be that the first-preimage attack gets you exactly X, in which case it fails to handle second-preimage. I would expect this to be incredibly unlikely (especially since they seem to generally involve generating some random garbage). I would also expect most first-preimage attacks have a way to "continue" to a third value, eventually.
• shawabawa3 10y ago
  That's kind of implicit. If X = Y, it's not an attack, it's the primary purpose of hashing