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copied!<p>Somebody complained that the "best" comment was bringing up the worst comments. IMHO, they're funnier, and so "better", but here's the honest best comment I've ever <a href="http://svn.python.org/view/python/trunk/Objects/dictobject.c?rev=53656&view=markup" rel="nofollow noreferrer">read</a>:</p> <pre><code>/* Major subtleties ahead: Most hash schemes depend on having a "good" hash function, in the sense of simulating randomness. Python doesn't: its most important hash functions (for strings and ints) are very regular in common cases: >>> map(hash, (0, 1, 2, 3)) [0, 1, 2, 3] >>> map(hash, ("namea", "nameb", "namec", "named")) [-1658398457, -1658398460, -1658398459, -1658398462] >>> This isn't necessarily bad! To the contrary, in a table of size 2**i, taking the low-order i bits as the initial table index is extremely fast, and there are no collisions at all for dicts indexed by a contiguous range of ints. The same is approximately true when keys are "consecutive" strings. So this gives better-than-random behavior in common cases, and that's very desirable. OTOH, when collisions occur, the tendency to fill contiguous slices of the hash table makes a good collision resolution strategy crucial. Taking only the last i bits of the hash code is also vulnerable: for example, consider [i << 16 for i in range(20000)] as a set of keys. Since ints are their own hash codes, and this fits in a dict of size 2**15, the last 15 bits of every hash code are all 0: they *all* map to the same table index. But catering to unusual cases should not slow the usual ones, so we just take the last i bits anyway. It's up to collision resolution to do the rest. If we *usually* find the key we're looking for on the first try (and, it turns out, we usually do -- the table load factor is kept under 2/3, so the odds are solidly in our favor), then it makes best sense to keep the initial index computation dirt cheap. 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). By itself, this doesn't help much: like linear probing (setting j += 1, or j -= 1, on each loop trip), it scans the table entries in a fixed order. This would be bad, except that's not the only thing we do, and it's actually *good* in the common cases where hash keys are consecutive. In an example that's really too small to make this entirely clear, for a table of size 2**3 the order of indices is: 0 -> 1 -> 6 -> 7 -> 4 -> 5 -> 2 -> 3 -> 0 [and here it's repeating] If two things come in at index 5, the first place we look after is index 2, not 6, so if another comes in at index 6 the collision at 5 didn't hurt it. Linear probing is deadly in this case because there the fixed probe order is the *same* as the order consecutive keys are likely to arrive. But it's extremely unlikely hash codes will follow a 5*j+1 recurrence by accident, and certain that consecutive hash codes do not. The other half of the strategy is to get the other bits of the hash code into play. This is done by initializing a (unsigned) vrbl "perturb" to the full hash code, and changing the recurrence to: j = (5*j) + 1 + perturb; perturb >>= PERTURB_SHIFT; use j % 2**i as the next table index; Now the probe sequence depends (eventually) on every bit in the hash code, and the pseudo-scrambling property of recurring on 5*j+1 is more valuable, because it quickly magnifies small differences in the bits that didn't affect the initial index. 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). Selecting a good value for PERTURB_SHIFT is a balancing act. You want it small so that the high bits of the hash code continue to affect the probe sequence across iterations; but you want it large so that in really bad cases the high-order hash bits have an effect on early iterations. 5 was "the best" in minimizing total collisions across experiments Tim Peters ran (on both normal and pathological cases), but 4 and 6 weren't significantly worse. Historical: Reimer Behrends contributed the idea of using a polynomial-based approach, using repeated multiplication by x in GF(2**n) where an irreducible polynomial for each table size was chosen such that x was a primitive root. Christian Tismer later extended that to use division by x instead, as an efficient way to get the high bits of the hash code into play. This scheme also gave excellent collision statistics, but was more expensive: two if-tests were required inside the loop; computing "the next" index took about the same number of operations but without as much potential parallelism (e.g., computing 5*j can go on at the same time as computing 1+perturb in the above, and then shifting perturb can be done while the table index is being masked); and the dictobject struct required a member to hold the table's polynomial. In Tim's experiments the current scheme ran faster, produced equally good collision statistics, needed less code & used less memory. Theoretical Python 2.5 headache: hash codes are only C "long", but sizeof(Py_ssize_t) > sizeof(long) may be possible. In that case, and if a dict is genuinely huge, then only the slots directly reachable via indexing by a C long can be the first slot in a probe sequence. The probe sequence will still eventually reach every slot in the table, but the collision rate on initial probes may be much higher than this scheme was designed for. Getting a hash code as fat as Py_ssize_t is the only real cure. But in practice, this probably won't make a lick of difference for many years (at which point everyone will have terabytes of RAM on 64-bit boxes). */ </code></pre>

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