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Title: SHA256 1 round equivance Post by: Balthazar2012 on August 22, 2013, 07:04:21 PM My question is the following:
I have calculated the average number of operations in 1 round of SHA256 as follows: Additions: 9.25 Bitwise Rotations: 9 Bitwise Shifts: 1.5 Bitwise AND: 5 Bitwise EXOR: 10 I have certain proposals that save the following number of operations in the SHA256 algorithm for Bitcoin mining: Additions: 24 Bitwise Rotations: 12 Bitwise Shifts: 6 Bitwise AND: 0 Bitwise EXOR: 12 My question is, How can I represent the above in terms of a SHA256 round? Do I just consider the additions? So that the saved operations above are equivalent to 24/9.25 = 2.5946 SHA256 rounds? Any help is greatly appreciated. |