Comments (4)
c are bytes with popcount==4
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It is guaranteed... because wyrand visit all 2^64 status. if some seed works, one day you will visit these working seed. ;-)
However, 2^64 is a bit number, take it easy, usually it works quickly.
We just need to find a number that popount(Ni^Nj)==32 which is easy.
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Now I looked at the code again (didn't do any analysis though) and somehow I still have a subtle feeling, that it would be possible to construct the seed in a way which would make secret[i]^secret[j]
always yield popcount of 32 (it depends on popcount of all pair-wise combinations of numbers in the c[]
array and on the 0-63 previous iterations of the j
inner for loop as well as on the 0-4 previous iterations of the outer/main for loop).
How did you generate the numbers in c[]
? If it's guaranteed that for each constellation of these numbers there exists at least one combination yielding different popcount than 32, then all is good đ.
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Oh man, my bad - I misread ok=0
as ok=1
. Thanks for the explanation!
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