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#include <stdio.h>
#include <math.h>
#include <unistd.h>
#include <stdlib.h>
int h[65536];
/*
* Probability analysis of hashing function:
*
* Let n be number of items and k number of boxes. For uniform distribution
* we get:
*
* Expected value of "item i is in given box": Xi = 1/k
* Expected number of items in given box: a = EX = E(sum Xi) = sum E(Xi) = n/k
* Expected square value: E(X^2) = E((sum Xi)^2) = E((sum_i Xi^2) + (sum_i,j i<>j XiXj)) =
* = sum_i E(Xi^2) + sum_i,j i<>j E(XiXj) =
* = sum_i E(Xi) [Xi is binary] + sum_i,j i<>j E(XiXj) [those are independent] =
* = n/k + n*(n-1)/k^2
* Variance: var X = E(X^2) - (EX)^2 = n/k + n*(n-1)/k^2 - n^2/k^2 =
* = n/k - n/k^2 = a * (1-1/k)
* Probability of fixed box being zero: Pz = ((k-1)/k)^n = (1-1/k)^n = (1-1/k)^(ak) =
* = ((1-1/k)^k)^a which we can approximate by e^-a.
*/
uint hf(uint n)
{
#if 0
n = (n ^ (n >> 16)) & 0xffff;
n = (n ^ (n << 8)) & 0xffff;
#elif 1
n = (n >> 16) ^ n;
n = (n ^ (n << 10)) & 0xffff;
#elif 0
n = (n >> 16) ^ n;
n *= 259309;
#elif 0
n ^= (n >> 20);
n ^= (n >> 10);
n ^= (n >> 5);
#elif 0
n = (n * 259309) + ((n >> 16) * 123479);
#else
return random();
#endif
return n;
}
int
main(int argc, char **argv)
{
int cnt=0;
int i;
int bits = atol(argv[1]);
int z = 1 << bits;
int max = atol(argv[2]);
while (max--)
{
uint i, e;
if (scanf("%x/%d", &i, &e) != 2)
if (feof(stdin))
break;
else
fprintf(stderr, "BUGGG\n");
// i >>= (32-e);
// i |= (i >> e);
cnt++;
h[(hf(i) >> 1*(16 - bits)) & (z-1)]++;
}
// printf(">>> %d addresses\n", cnt);
#if 0
for(i=0; i<z; i++)
printf("%d\t%d\n", i, h[i]);
#else
{
int min=cnt, max=0, zer=0;
double delta=0;
double avg = (double) cnt / z;
double exdelta = avg*(1-1/(double)z);
double exzer = exp(-avg);
for(i=0; i<z; i++) {
if (h[i] < min) min=h[i];
if (h[i] > max) max=h[i];
delta += (h[i] - avg) * (h[i] - avg);
if (!h[i]) zer++;
}
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);
}
#endif
return 0;
}
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