If that holds true for all six bytes.
6 bytes 5 candidates per byte. 5^6 = 15,625 possible pins.
15,625 is a short list compared to 100^6.
VTL: How long would it take to test 15,625 possible pins generated from six sets of five candidates. ?
If the correct byte value always ends up a short list. Reduction by shortlisting might be the best compromise between speed and dependability.
But it does not. For the cmp+jmp instruction clusters on the same flash page/cache line/pipeline fetch only the first one yields a larger latency, the following has no latency spike, so it would never make it to the short list.
Then how do we ever get bytes B1 & B2?
I figured that. I have too many pokers in the fire myself.