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Re^2: The 10**21 Problem (Part 2)

by oiskuu (Friar)
on May 18, 2014 at 09:21 UTC ( #1086495=note: print w/ replies, xml ) Need Help??

in reply to Re: The 10**21 Problem (Part 2)
in thread The 10**21 Problem (Part 2)

$ perl -lwe '$/ = \128; ++$hist[tr/\0//c] while <>; print "$_: $hist[$ +_]" for 0..$#hist;' lookzpM.byte-* 0: 12003438 1: 148200 2: 147874 ... 42: 330481 43: 33028

This means you have plenty of room to pack the lines as: {N} ({i}{x})* {pad}.
Compute q8 = (i ^ m7) & 127; q9 = (x ^ HASH_LEN);

Average 15.984 elements per line. Lots of looping and branches can be avoided this way.

Update. One more thing occurs to me. Indeed, those files compress rather well: lzma packs them 1:2500!

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