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Aug 13, 2017 at 13:14 comment added Paul Uszak I think that they're implied (otherwise the question becomes moot), but we'll have to wait for the horse's mouth.
Aug 13, 2017 at 12:45 comment added deviantfan About bit operations, the data could very well be in binary format in our Turing machine. Setting a bit to 0, with these 4 operations, could be a multiplication of 2 (including the usual overflow rules, and/or thinking as algebraic structure, or whatever), setting to 1 is setting to 0 and then adding 1, and flipping is just adding 1. ... Sure, it's a bit cumbersome per hand, but there are no hard time limits here.
Aug 13, 2017 at 12:42 comment added deviantfan Well, I wouldn't call 100 data blocks (instead of one) part of the algorithm. Same for multiple repetitions of the SHA, for me they are not part of SHA. ... and SHA isn't even an "encryption" scheme.
Aug 13, 2017 at 12:25 comment added Paul Uszak I would disagreed with all known... I believe that the spirit of the question doesn't allow for hand/calculator processing of 14 AES rounds x 100 blocks of data + 1000 SHA rounds of a KDF. I'm often wrong though...
Aug 13, 2017 at 12:22 comment added Paul Uszak If we restrict ourselves to the exact 4 simple operations of the question, bit manipulation becomes hard. So XOR, left and right shifts /rolls are pretty cumbersome. Not impossible, but tricky.
Aug 13, 2017 at 8:22 history answered deviantfan CC BY-SA 3.0