What would happen if SHA256's pre-image or collision resistance would be broken? I'm experimenting with some self study cryptography and I was curious about the two following scenarios (simple answers and/or in relation to bitcoin)

If SHA256 was found to lack pre-image resistance, would it make solving the puzzle easier?

Alternatively, if the algorithm was found to be not collision resistant - would the puzzle again be easier to solve?

Thanks
Any collision resistant hashing function Hk : {0,1}* → {0,1}k is pre-image resistant regarding the uniform distribution on {0,1}2k.


A ⇒ B ⇔ ¬B ⇒ ¬A: I.e. if a function isn't pre-image resistant it is not collision resistant either.

SHA256 is a function that maps a potentially unlimited set of numbers to a smaller set of numbers.


  Hk : {0,1}* → {0,1}k


Collision resistance is a property that loosely says, it is difficult to find two inverse images X ≠ X' that have the same image H(X) = H(X').

More to the point:
A function H is collision resistant, if any algorithm can only find a collision with a negligible probability in probabilistic polynomial time.

I.e. if SHA256 would turn out to not be collision resistant, one could try to pick hashes that would succeed at the current difficulty, use the above predicted inverse function of SHA256 to calculate inverse image candidates, and finally check whether one can satisfy them with the currently available block input.

My gut feeling would be though, that it would be pretty difficult to find inverse images that also satisfy the required structure of the block input, especially getting right the hash of the parent block, and matching an address that one controls for the coinbase transaction.

Unfortunately, I have no idea whether the complexity of that would be greater or smaller than bruteforce mining.