What is the theoretical minimum number of logical operations an ASIC needs to perform to compute double iterated SHA256, i.e., sha(sha(•))?
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SHA256D, which is what Bitcoin uses, is 128 rounds, comprising
768 additions,
640 ORs
896 XORs
And a bunch of bit shifts but bit shifts are free on an ASIC.
(source)
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Those numbers don't seem correct. I count 522 additions across one iteration of SHA256, for example. (1044 across two) http://en.wikipedia.org/wiki/SHA-2#Pseudocode – Nick ODell Apr 19 '15 at 08:32
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1...and I'm counting 600 additions for one SHA256 block: 48*3 + 64*7 + 8. Honestly I think this should be closed as off-topic and re-posted to http://crypto.stackexchange.com/ where you could attract the attention of a crypto expert. – Christopher Gurnee Apr 19 '15 at 15:03