Transaction

3e0c06d0f7f538c2aaa754ccfa9849506e208ff3045bcedee1d85ec99cd33077
Timestamp (utc)
2024-03-22 12:50:11
Fee Paid
0.00000018 BSV
(
0.00687662 BSV
-
0.00687644 BSV
)
Fee Rate
10.04 sat/KB
Version
1
Confirmations
94,326
Size Stats
1,792 B

2 Outputs

Total Output:
0.00687644 BSV
  • j"1LAnZuoQdcKCkpDBKQMCgziGMoPC4VQUckM<div class="post"><div class="quoteheader"><a href="https://bitcointalk.org/index.php?topic=360.msg3044#msg3044">Quote from: theymos on July 15, 2010, 01:45:03 AM</a></div><div class="quote">You're trying to solve:<br/>SHA256(SHA256(d'))==256-bit number<br/><br/>This still has 256 bits of security, and you have to do two hashes per attempt to brute-force it.<br/></div>I agree, basically I get what he is saying at the top, but it isn't as bad as everyone thinks.<br/><br/>To brute force your 256bit hash, you either guess the starting string or the second hash string. Which means you have to guess, convert to 256, convert to 256 again and see if that matches. If you attack the second hash, you can't start a guess at 0000000.....1 for example, you know it's already another 256bit number that have to start with. That's like trying to brute force a password and knowing ahead of time the guy's password was 100 digits long. Does it make it any easier? Well no because the first 99 digits could be 0 and the last 1 but either way you still have to try all in between.<br/><br/>Mathematically, I see no advantage. Either you guess the first string to generate the hash or guess the first hash that generates the second hash. Either way you are going to have to brute force the entire key-space to do this. The only advantage is that instead of having "one" correct answer, you have "two" correct answers out of the billion trillion million guesses possible. &nbsp;<img alt="Grin" border="0" src="/static/img/emoticons/grin.gif"/></div> text/html
    https://whatsonchain.com/tx/3e0c06d0f7f538c2aaa754ccfa9849506e208ff3045bcedee1d85ec99cd33077