On Mon, May 9, 2016 at 11:37 PM, Peter R <pete...@gmx.com> wrote: > It is a standard result that there are > m! / [n! (m-n)!] > ways of picking n numbers from a set of m numbers, so there are > > (2^32)! / [2! (2^32 - 2)!] ~ 2^63 > possible pairs in a set of 2^32 transactions. So wouldn’t you have to > perform approximately 2^63 comparisons in order to identify which pair of > transactions are the two that collide? > > Perhaps I made an error or there is a faster way to scan your set to find the > collision. Happy to be corrected…
$ echo -n Perhaps. 00000000f2736d91 |sha256sum 359dfa6d4c2eb2ac81535392d68af4b5e1cb6d9c6321e8f111d3244329b6a4d8 $ echo -n Perhaps. 0000000011ac0388 |sha256sum 359dfa6d4c2eb2ac44d54d0ceeb2212500cb34617b9360695432f6c0fde9b006 Try search term "collision", or there may be an undergrad Data structures and algorithms coarse online-- you want something covering "cycle finding". (Though even ignoring efficient cycle finding, your factorial argument doesn't hold... you can simply sort the data... Search term "quicksort" for a relevant algorithm). _______________________________________________ bitcoin-dev mailing list bitcoin-dev@lists.linuxfoundation.org https://lists.linuxfoundation.org/mailman/listinfo/bitcoin-dev