87b60bf7e8
we can use very simple function which is monotonic with respect to re-hashing: n ^= n >> 16; n ^= n << 10; h = (n >> (16 - o)) & ((1 << o) - 1); where o is table order. Statistical analysis for both backbone routing table and local OSPF routing tables gives values near theoretical optimum for uniform distribution (see ips.c for formulae). The trick is very simple: We always calculate a 16-bit hash value n and use o most significant bits (this gives us monotonity wrt. rehashing if we sort the chains by the value of n). The first shift/xor pair reduces the IP address to a 16-bit one, the second pair makes higher bits of the 16-bit value uniformly distributed even for tables containing lots of long prefixes (typical interior routing case with 24-bit or even longer prefixes).
94 lines
2.1 KiB
C
94 lines
2.1 KiB
C
#include <stdio.h>
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#include <math.h>
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#include <unistd.h>
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#include <stdlib.h>
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int h[65536];
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/*
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* Probability analysis of hashing function:
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*
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* Let n be number of items and k number of boxes. For uniform distribution
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* we get:
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*
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* Expected value of "item i is in given box": Xi = 1/k
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* Expected number of items in given box: a = EX = E(sum Xi) = sum E(Xi) = n/k
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* Expected square value: E(X^2) = E((sum Xi)^2) = E((sum_i Xi^2) + (sum_i,j i<>j XiXj)) =
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* = sum_i E(Xi^2) + sum_i,j i<>j E(XiXj) =
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* = sum_i E(Xi) [Xi is binary] + sum_i,j i<>j E(XiXj) [those are independent] =
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* = n/k + n*(n-1)/k^2
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* Variance: var X = E(X^2) - (EX)^2 = n/k + n*(n-1)/k^2 - n^2/k^2 =
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* = n/k - n/k^2 = a * (1-1/k)
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* Probability of fixed box being zero: Pz = ((k-1)/k)^n = (1-1/k)^n = (1-1/k)^(ak) =
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* = ((1-1/k)^k)^a which we can approximate by e^-a.
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*/
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unsigned int hf(unsigned int n)
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{
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#if 0
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n = (n ^ (n >> 16)) & 0xffff;
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n = (n ^ (n << 8)) & 0xffff;
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#elif 1
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n = (n >> 16) ^ n;
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n = (n ^ (n << 10)) & 0xffff;
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#elif 0
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n = (n >> 16) ^ n;
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n *= 259309;
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#elif 0
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n ^= (n >> 20);
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n ^= (n >> 10);
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n ^= (n >> 5);
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#elif 0
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n = (n * 259309) + ((n >> 16) * 123479);
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#else
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return random();
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#endif
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return n;
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}
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int
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main(int argc, char **argv)
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{
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int cnt=0;
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int i;
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int bits = atol(argv[1]);
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int z = 1 << bits;
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int max = atol(argv[2]);
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while (max--)
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{
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unsigned int i, e;
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if (scanf("%x/%d", &i, &e) != 2)
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if (feof(stdin))
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break;
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else
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fprintf(stderr, "BUGGG\n");
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// i >>= (32-e);
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// i |= (i >> e);
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cnt++;
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h[(hf(i) >> 1*(16 - bits)) & (z-1)]++;
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}
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// printf(">>> %d addresses\n", cnt);
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#if 0
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for(i=0; i<z; i++)
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printf("%d\t%d\n", i, h[i]);
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#else
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{
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int min=cnt, max=0, zer=0;
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double delta=0;
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double avg = (double) cnt / z;
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double exdelta = avg*(1-1/z);
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double exzer = exp(-avg);
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for(i=0; i<z; i++) {
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if (h[i] < min) min=h[i];
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if (h[i] > max) max=h[i];
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delta += (h[i] - avg) * (h[i] - avg);
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if (!h[i]) zer++;
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}
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printf("size=%5d, min=%d, max=%2d, delta=%-7.6g (%-7.6g), avg=%-5.3g, zero=%g%% (%g%%)\n", z, min, max, delta/z, exdelta, avg, zer/(double)z*100, exzer*100);
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}
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#endif
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return 0;
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}
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