O algoritmu za generiranje


SUBMITTED BY: Guest

DATE: Aug. 25, 2012, 9:39 p.m.

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  1. 12 down vote accepted
  2. I don't believe that there's any way to generate the vanity hashes without iterating. In base 58, there's log2(58)≈5.858 bits per letter, so fixing 8 letters would need in average 588/2=2log2(58)⋅8/2≈246 iterations. Note that Bitcoin addresses always start with a 1 by convention (this comes from the version field), and the next character in the base 58 representation is usually among "23456789ABCDEFGHJKLMNPQR" (not the full 58 character set), which halves the number of tries once more (if your wished starting character is in this set, otherwise it gets impossible). If you allow your vanity string to appear anywhere in the hash, you can divide this number again by 25, leading to 240.5 iterations (on average) (which also are quite good parallelizable).
  3. Also, the actual example he gave had an "s" rather than an "i" in the vanity "ByteCoins" part, so the odds of that are a bit greater (i.e. the needed time is even smaller).
  4. However, you can do it securely (i.e. without the service gaining access to your private key):
  5. Let the user generate a private key, a, and submit the corresponding public key, a⋅B (where B is the base point of the group). The service can then generate public keys as a⋅B+x⋅B, where x is incremented to generate different public keys.
  6. Actually, if you have a⋅B+x⋅B, then a⋅B+(x+1)⋅B=(a⋅B+x⋅B)+B, i.e. you need only one EC point addition (of the base point) and the necessary hashes for each try.
  7. Then, by telling the user x, the user has private key a+x with public key (a+x)⋅B and H((a+x)⋅B) has the required property, if the service did its job.

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