# Gavin Andresen
# 2010-11-20 14:57:17
# https://bitcointalk.org/index.php?topic=1854.msg23068#msg23068

Everybody is looking for: @p{par}

h = f(x), where h @p{lt} TARGET @p{par}

I think you're asking if there is some weakness in SHA256 (or hashing in general) that can make solving that problem trivial. @p{par}

It seems unlikely to me, but I'm not a professional cryptographer.  If TARGET were '4', then the problem is equivalent to: @p{brk}
h = f(x) where h = 0 OR h=1 OR h=2 OR h=3 @p{par}

Maybe there's some tricky method that reuses work and makes solving the TARGET=4 case more than 4 times easier than solving for TARGET=1 (which is "find this specific hash")... but I just don't see that doing anything more than what has already been pointed out in previous posts:  it is just a quicker way of hashing, so difficulty would go up. @p{brk}