—

The first half of collision resolution is to visit table indices via this

recurrence:

j = ((5*j) + 1) mod 2**i

For any initial j in range(2**i), repeating that 2**i times generates each

int in range(2**i) exactly once (see any text on random-number generation for

proof).

—

and

—

Note that because perturb is unsigned, if the recurrence

is executed often enough perturb eventually becomes and remains 0. At that

point (very rarely reached) the recurrence is on (just) 5*j+1 again, and

that’s certain to find an empty slot eventually (since it generates every int

in range(2**i), and we make sure there’s always at least one empty slot).

—

]]>j = (5*j) + 1 + perturb;

perturb >>= PERTURB_SHIFT;

use j % 2**i as the next table index

J’ai beaucoup apprécié ! ]]>

I have just one question. What’s the call that gets the accuracy of 0.8. I found the method, the first parameter is the classifier, but which is the second parameter? ]]>

I have been looking for something like this and dind’t find anything for a long.

I’m a systems engineering student and never understood this trees stuff like i just did with your explanation, thank you so much!

]]>