mirror of
https://github.com/openvswitch/ovs
synced 2025-10-13 14:07:02 +00:00
156 lines
5.2 KiB
C
156 lines
5.2 KiB
C
/*
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* Copyright (c) 2009 Nicira Networks.
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*
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* Licensed under the Apache License, Version 2.0 (the "License");
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* you may not use this file except in compliance with the License.
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* You may obtain a copy of the License at:
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*
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* http://www.apache.org/licenses/LICENSE-2.0
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*
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* Unless required by applicable law or agreed to in writing, software
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* distributed under the License is distributed on an "AS IS" BASIS,
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* WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
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* See the License for the specific language governing permissions and
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* limitations under the License.
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*/
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#include <config.h>
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#include <inttypes.h>
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#include <stdio.h>
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#include <stdlib.h>
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#include <string.h>
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#include "hash.h"
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#undef NDEBUG
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#include <assert.h>
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static void
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set_bit(uint32_t array[3], int bit)
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{
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assert(bit >= 0 && bit <= 96);
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memset(array, 0, sizeof(uint32_t) * 3);
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if (bit < 96) {
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array[bit / 32] = UINT32_C(1) << (bit % 32);
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}
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}
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static uint32_t
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hash_words_cb(uint32_t input)
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{
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return hash_words(&input, 1, 0);
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}
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static uint32_t
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hash_int_cb(uint32_t input)
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{
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return hash_int(input, 0);
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}
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static void
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check_word_hash(uint32_t (*hash)(uint32_t), const char *name,
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int min_unique)
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{
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int i, j;
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for (i = 0; i <= 32; i++) {
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uint32_t in1 = i < 32 ? UINT32_C(1) << i : 0;
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for (j = i + 1; j <= 32; j++) {
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uint32_t in2 = j < 32 ? UINT32_C(1) << j : 0;
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uint32_t out1 = hash(in1);
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uint32_t out2 = hash(in2);
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const uint32_t unique_mask = (UINT32_C(1) << min_unique) - 1;
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int ofs;
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for (ofs = 0; ofs < 32 - min_unique; ofs++) {
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uint32_t bits1 = (out1 >> ofs) & unique_mask;
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uint32_t bits2 = (out2 >> ofs) & unique_mask;
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if (bits1 == bits2) {
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printf("Partial collision for '%s':\n", name);
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printf("%s(%08"PRIx32") = %08"PRIx32"\n", name, in1, out1);
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printf("%s(%08"PRIx32") = %08"PRIx32"\n", name, in2, out2);
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printf("%d bits of output starting at bit %d "
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"are both 0x%"PRIx32"\n", min_unique, ofs, bits1);
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exit(1);
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}
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}
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}
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}
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}
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int
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main(void)
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{
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int i, j;
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/* Check that all hashes computed with hash_words with one 1-bit (or no
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* 1-bits) set within a single 32-bit word have different values in all
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* 11-bit consecutive runs.
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*
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* Given a random distribution, the probability of at least one collision
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* in any set of 11 bits is approximately
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*
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* 1 - ((2**11 - 1)/2**11)**C(33,2)
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* == 1 - (2047/2048)**528
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* =~ 0.22
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*
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* There are 21 ways to pick 11 consecutive bits in a 32-bit word, so if we
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* assumed independence then the chance of having no collisions in any of
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* those 11-bit runs would be (1-0.22)**21 =~ .0044. Obviously
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* independence must be a bad assumption :-)
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*/
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check_word_hash(hash_words_cb, "hash_words", 11);
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/* Check that all hash functions of with one 1-bit (or no 1-bits) set
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* within three 32-bit words have different values in their lowest 12
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* bits.
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*
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* Given a random distribution, the probability of at least one collision
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* in 12 bits is approximately
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*
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* 1 - ((2**12 - 1)/2**12)**C(97,2)
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* == 1 - (4095/4096)**4656
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* =~ 0.68
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*
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* so we are doing pretty well to not have any collisions in 12 bits.
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*/
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for (i = 0; i <= 96; i++) {
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for (j = i + 1; j <= 96; j++) {
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uint32_t in1[3], in2[3];
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uint32_t out1, out2;
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const int min_unique = 12;
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const uint32_t unique_mask = (UINT32_C(1) << min_unique) - 1;
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set_bit(in1, i);
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set_bit(in2, j);
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out1 = hash_words(in1, 3, 0);
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out2 = hash_words(in2, 3, 0);
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if ((out1 & unique_mask) == (out2 & unique_mask)) {
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printf("Partial collision:\n");
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printf("hash(1 << %d) == %08"PRIx32"\n", i, out1);
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printf("hash(1 << %d) == %08"PRIx32"\n", j, out2);
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printf("The low-order %d bits of output are both "
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"0x%"PRIx32"\n", min_unique, out1 & unique_mask);
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exit(1);
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}
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}
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}
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/* Check that all hashes computed with hash_int with one 1-bit (or no
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* 1-bits) set within a single 32-bit word have different values in all
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* 14-bit consecutive runs.
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*
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* Given a random distribution, the probability of at least one collision
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* in any set of 14 bits is approximately
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*
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* 1 - ((2**14 - 1)/2**14)**C(33,2)
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* == 1 - (16,383/16,834)**528
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* =~ 0.031
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*
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* There are 18 ways to pick 14 consecutive bits in a 32-bit word, so if we
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* assumed independence then the chance of having no collisions in any of
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* those 14-bit runs would be (1-0.03)**18 =~ 0.56. This seems reasonable.
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*/
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check_word_hash(hash_int_cb, "hash_int", 14);
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return 0;
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}
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