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Re: The 10**21 Problem (Part 3)by oiskuu (Hermit) 
on May 16, 2014 at 12:35 UTC ( #1086299=note: print w/ replies, xml )  Need Help?? 
Some further analysis.Consider the last hashing step. Xor with q8 was a ±127 adjustment at most. Multiplied by hp=1000003, the difference is less than 2**27. It is therefore unlikely that a change in q8 would carry into sign of d8. Now, hp % 1001 == 4, which is like a shift. In a sense, all the magic bits have a weight of a power of two. What this means is q8 can be stepped without altering 2 lowest result bits, provided d8 does not change sign and d >= 4*step && d < 10014*step.
Note that if this heuristic is enabled, it may leave no time for prefetching. Try without lookups. Perhaps some other modulus might have better karma, a modulus that reduces hp to odd value (e.g. 1000003 % 1039 == 485)? Update: Actually, scratch that. I forgot q8 selects q9, which of course toggles the low bits. There might be a way to set further restrictions on allowed q8's, but this is getting pretty convoluted.
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