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Topic: bruteforcing sha256 (ed: not really it turns out, ty for not raping me too much) (Read 1208 times)

newbie
Activity: 28
Merit: 0
Also google "Rainbow Tables". It's been done for some algorithms, but their real-life usage is limited by the spread of multiple hashing, which would require exponential storage.
ahh, wpa cracking Smiley my friend successfully cracked his own wifi (testing obv.) only took 4hrs
sr. member
Activity: 252
Merit: 250
Also google "Rainbow Tables". It's been done for some algorithms, but their real-life usage is limited by the spread of multiple hashing, which would require exponential storage.
newbie
Activity: 28
Merit: 0
thanks for the input guys, a bit more detailed than the "yeah your probably right" answers i was getting from various chatrooms

more reading *sigh* Roll Eyes
donator
Activity: 1218
Merit: 1079
Gerald Davis
donator
Activity: 1218
Merit: 1079
Gerald Davis
essentially, isn't this what mining does?
we do all the hard work, save the results - eventually every single hash will be decrypted & linked to its digest...

sha256 security? getting less by the hash

First of all a couple points.

1) We don't save all the work.  We discard about 99.9999999999999999999999999999999999999999% of the work.

2) Even if we saved some work the hashes only provide a lookup for the source (plaintext).  Have you looked at what we hash?  Not very useful outside bitcoin

3) There are 2^256 potential hashes.  The network is ~9TH.  Lets pretend it has been 9TH since the begining.  Since the start of bitcoin we have hashed (and thrown away but lets just pretend we didn't thrown all those hashes away) 5.67648E+20 hashes.

That is 1 in 1/203,985,725,726,711,000,000,000,000,000,000,000,000,000,000,000,000,000,000 of the potential hashes in SHA-256.

If the hashrate was 1000x higher (9 petahashes) and we hashed for the next milenium and saved all those hashes and constructed a high efficiency lookup table we would have roughly 1 in 1/ 407,971,451,453,423,000,000,000,000,000,000,000,000,000,000,000,000 of the SHA-256 space cataloged.

Smiley

2^256 is a much (by magnitudes) larger number than you think it is.  However large conceptually you think it is take that and square it and you likely still aren't close.

 
legendary
Activity: 1148
Merit: 1008
If you want to walk on water, get out of the boat
essentially, isn't this what mining does?
we do all the hard work, save the results - eventually every single hash will be decrypted & linked to its digest...

sha256 security? getting less by the hash
No.
hero member
Activity: 504
Merit: 502
we do all the hard work, save the results - eventually every single hash will be decrypted & linked to its digest...

I have no idea what you mean.  A hash is a digest.  A hash can't be "decrypted".

The attack against a hashing algorithm is to, say, have a target hash, and be able to find an input text that produces that hash without brute forcing.

Mining, on the other hand, is brute forcing the hash.  Or rather a range of hashes; we have a set of allowable target hashes and the input text is adjusted (via the nonce) until the input produces one of the target hashes.  Each potential input has to be tried though, there is no short cut.

The next text is a brand new block; in what way does knowledge of the previous brute forcings help you short cut your search for the next one?

I suppose the blocks do record particular outputs, of the 115792089237316195423570985008687907853269984665640564039457584007913129639936 possible sha256 values, 156 thousand ish blocks permanently record outputs for particular inputs.  How does that help with an attack?  156khashes can be calculated in less than a second.
newbie
Activity: 28
Merit: 0
essentially, isn't this what mining does?
we do all the hard work, save the results - eventually every single hash will be decrypted & linked to its digest...

sha256 security? getting less by the hash
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