If there's more to it, then please expand the question.
I just wonder is there possible collision in hash sha256 or in hash RMD160?
Oh I see, then the answer is there's little to no chance of collision.
There's a possibility, yes because 2^160 (
the "pubKeyhash") is a finite number but still too large to have a reasonable chance of collision happening in our lifetime. (
at least with the current hardware)
Not to mention, SHA256 hash which is 2^256.
Take note those hashes are not exclusive to Bitcoin or Cryptocurrencies so if ever there's a legitimate collision happened in the past,
It would've been in the news by now.
Thank for clarification, so I can conclusion hash RMD160 is possible to collision the reason;
1. The input have 2**256 combination (SHA256)
2. The output only have 2**160 combination (RMD160)
Lets break with this key;
- pvkey
00000000000000000000000000000000000000000000000300000e4692d2217b
00000000000000000000000000000000000000000000000300005a15da2a15b1
- pubkey (comp)
0358ec223b4c813a10033d5519da9ca50f0c8754b914fbd5500c6cbe9ea290cff5
0355a3cc6b5399da908a19fac59abf1382c7dd599198cb323a5fcbde7fad39e8cf
- SHA256
9ec24e06cacf3033c8afaf1ae266c9aec885523144992c82dc1e736b8219dda2
53e94b4ccadab2d3a3ad3323b7a07bc9752e7fcdcb25ccc1384b332b558630ef
- RMD160
20d45a6a7627 d06f3ad6e0f34707d19a2f9f12f1 (12 prefixes collision)
20d45a6a7627 38faae370bae7cc4517b0ccf01b2 (12 prefixes collision)
but this just paper calculation and until full collision found this is only hypothesis, maybe ratio 1:2 between hash sha256 with hash RMD160 is like coordinate in curve secp256k1 which one X coordinate have two Y coordinate
**
Thank for all response