Bitcoin Forum
May 05, 2024, 06:31:13 PM *
News: Latest Bitcoin Core release: 27.0 [Torrent]
 
   Home   Help Search Login Register More  
Pages: [1]
  Print  
Author Topic: SHA256 and RIPEMD160 collisions  (Read 477 times)
coder0x15 (OP)
Jr. Member
*
Offline Offline

Activity: 36
Merit: 3


View Profile
August 21, 2017, 07:23:52 AM
 #1

H = ripemd160(sha256(ECMultiply(A,G)))

A - 32-byte number,

H - 20-byte number,

G - generator point,

ECMultiply - elliptic curve point multiplication.

Is it true that there is many A that have the same H?
It is a common myth that Bitcoin is ruled by a majority of miners. This is not true. Bitcoin miners "vote" on the ordering of transactions, but that's all they do. They can't vote to change the network rules.
Advertised sites are not endorsed by the Bitcoin Forum. They may be unsafe, untrustworthy, or illegal in your jurisdiction.
1714933873
Hero Member
*
Offline Offline

Posts: 1714933873

View Profile Personal Message (Offline)

Ignore
1714933873
Reply with quote  #2

1714933873
Report to moderator
amaclin1
Sr. Member
****
Offline Offline

Activity: 770
Merit: 305


View Profile
August 21, 2017, 09:01:57 AM
 #2

yes
https://en.wikipedia.org/wiki/Pigeonhole_principle

Bitcoin SV GUI client for Windows and Linux
https://github.com/AlisterMaclin/bitcoin-sv/releases
efeaydin
Full Member
***
Offline Offline

Activity: 181
Merit: 102


View Profile
August 21, 2017, 09:14:58 AM
 #3

H = ripemd160(sha256(ECMultiply(A,G)))

A - 32-byte number,

H - 20-byte number,

G - generator point,

ECMultiply - elliptic curve point multiplication.

Is it true that there is many A that have the same H?

Yes, it is possible but the possibility is too low.

SHA256 hashes are almost unique. I don't think that anyone has found a SHA256.
Pages: [1]
  Print  
 
Jump to:  

Powered by MySQL Powered by PHP Powered by SMF 1.1.19 | SMF © 2006-2009, Simple Machines Valid XHTML 1.0! Valid CSS!