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Re: The 10**21 Problem (Part I)

by oiskuu (Pilgrim)
on May 05, 2014 at 20:08 UTC ( #1085114=note: print w/ replies, xml ) Need Help??


in reply to The 10**21 Problem (Part I)

Regarding the hash function and the modulus.

Firstly, the if (m9 == -1) continue; clause could just as well be omitted—this may only result in false positives, not a false negative. Unlikely to matter though. I would also test if skipping the 10/13 cmps in the innermost loop(s) might improve performance!

Apart from the oddball -1 case, the py_hash() function does not feed high order bits back into the register. Bits propagate higher; low bits of the hash value depend on low bits of input only. The python modulus is trickier. It ((1001+v%1001) % 1001) can be evaluated via unsigned modulus: ((v-(v<0)*620U) % 1001). The sign bit therefore mixes into the 3rd lowest bit.

Compilers know how to replace a divide-by-constant with multiplication by its reciprocal. (Div operation typically has a very high latency.) A division by 10, for example, may be substituted with mul 0xcccccccd; shift right by 3(+32). The shift size varies:

udiv by 9993: magic=2746836385 (0xa3b965a1); shift=14; udiv by 8089: magic=2174828291 (0x81a13f03); shift=12 udiv by 1001: magic=98685563 (0x05e1d27b); shift=10; udiv by 1419: magic=24214051 (0x01717a23); shift=3 udiv by 6019: magic=2854273 (0x002b8d81); shift=2 udiv by 4765: magic=3605429 (0x003703b5); shift=2 udiv by 6147: magic=1397419 (0x001552ab); shift=1 udiv by 2049: magic=4192257 (0x003ff801); shift=1
Evidently, the choice of modulus can strengthen the hash function. Mod=2049 is particularly weak: (-1U%2049 +1) == 1024. Sign mixes into 10th lowest.

Update: it is easily shown that modulus must be odd. Hash function either preserves or inverts the lowest bit of r. Even modulus would also preserve it, making (r^v)&1 constant. But that does not fit the problem.


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