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/*
* Copyright (c) 2009, 2012, 2014 Nicira, Inc.
*
* Licensed under the Apache License, Version 2.0 (the "License");
* you may not use this file except in compliance with the License.
* You may obtain a copy of the License at:
*
* http://www.apache.org/licenses/LICENSE-2.0
*
* Unless required by applicable law or agreed to in writing, software
* distributed under the License is distributed on an "AS IS" BASIS,
* WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
* See the License for the specific language governing permissions and
* limitations under the License.
*/
#include <config.h>
#include <inttypes.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include "hash.h"
#include "jhash.h"
#include "ovstest.h"
#undef NDEBUG
#include <assert.h>
static void
set_bit(uint32_t array[3], int bit)
{
assert(bit >= 0 && bit <= 96);
memset(array, 0, sizeof(uint32_t) * 3);
if (bit < 96) {
array[bit / 32] = UINT32_C(1) << (bit % 32);
}
}
static uint32_t
hash_words_cb(uint32_t input)
{
return hash_words(&input, 1, 0);
}
static uint32_t
jhash_words_cb(uint32_t input)
{
return jhash_words(&input, 1, 0);
}
static uint32_t
hash_int_cb(uint32_t input)
{
return hash_int(input, 0);
}
static void
check_word_hash(uint32_t (*hash)(uint32_t), const char *name,
int min_unique)
{
int i, j;
for (i = 0; i <= 32; i++) {
uint32_t in1 = i < 32 ? UINT32_C(1) << i : 0;
for (j = i + 1; j <= 32; j++) {
uint32_t in2 = j < 32 ? UINT32_C(1) << j : 0;
uint32_t out1 = hash(in1);
uint32_t out2 = hash(in2);
const uint32_t unique_mask = (UINT32_C(1) << min_unique) - 1;
int ofs;
for (ofs = 0; ofs < 32 - min_unique; ofs++) {
uint32_t bits1 = (out1 >> ofs) & unique_mask;
uint32_t bits2 = (out2 >> ofs) & unique_mask;
if (bits1 == bits2) {
printf("Partial collision for '%s':\n", name);
printf("%s(%08"PRIx32") = %08"PRIx32"\n", name, in1, out1);
printf("%s(%08"PRIx32") = %08"PRIx32"\n", name, in2, out2);
printf("%d bits of output starting at bit %d "
"are both 0x%"PRIx32"\n", min_unique, ofs, bits1);
exit(1);
}
}
}
}
}
static void
check_3word_hash(uint32_t (*hash)(const uint32_t[], size_t, uint32_t),
const char *name)
{
int i, j;
for (i = 0; i <= 96; i++) {
for (j = i + 1; j <= 96; j++) {
uint32_t in1[3], in2[3];
uint32_t out1, out2;
const int min_unique = 12;
const uint32_t unique_mask = (UINT32_C(1) << min_unique) - 1;
set_bit(in1, i);
set_bit(in2, j);
out1 = hash(in1, 3, 0);
out2 = hash(in2, 3, 0);
if ((out1 & unique_mask) == (out2 & unique_mask)) {
printf("%s has a partial collision:\n", name);
printf("hash(1 << %d) == %08"PRIx32"\n", i, out1);
printf("hash(1 << %d) == %08"PRIx32"\n", j, out2);
printf("The low-order %d bits of output are both "
"0x%"PRIx32"\n", min_unique, out1 & unique_mask);
}
}
}
}
static void
test_hash_main(int argc OVS_UNUSED, char *argv[] OVS_UNUSED)
{
/* Check that all hashes computed with hash_words with one 1-bit (or no
* 1-bits) set within a single 32-bit word have different values in all
* 11-bit consecutive runs.
*
* Given a random distribution, the probability of at least one collision
* in any set of 11 bits is approximately
*
* 1 - ((2**11 - 1)/2**11)**C(33,2)
* == 1 - (2047/2048)**528
* =~ 0.22
*
* There are 21 ways to pick 11 consecutive bits in a 32-bit word, so if we
* assumed independence then the chance of having no collisions in any of
* those 11-bit runs would be (1-0.22)**21 =~ .0044. Obviously
* independence must be a bad assumption :-)
*/
check_word_hash(hash_words_cb, "hash_words", 11);
check_word_hash(jhash_words_cb, "jhash_words", 11);
/* Check that all hash functions of with one 1-bit (or no 1-bits) set
* within three 32-bit words have different values in their lowest 12
* bits.
*
* Given a random distribution, the probability of at least one collision
* in 12 bits is approximately
*
* 1 - ((2**12 - 1)/2**12)**C(97,2)
* == 1 - (4095/4096)**4656
* =~ 0.68
*
* so we are doing pretty well to not have any collisions in 12 bits.
*/
check_3word_hash(hash_words, "hash_words");
check_3word_hash(jhash_words, "jhash_words");
/* Check that all hashes computed with hash_int with one 1-bit (or no
* 1-bits) set within a single 32-bit word have different values in all
* 12-bit consecutive runs.
*
* Given a random distribution, the probability of at least one collision
* in any set of 12 bits is approximately
*
* 1 - ((2**12 - 1)/2**12)**C(33,2)
* == 1 - (4,095/4,096)**528
* =~ 0.12
*
* There are 20 ways to pick 12 consecutive bits in a 32-bit word, so if we
* assumed independence then the chance of having no collisions in any of
* those 12-bit runs would be (1-0.12)**20 =~ 0.078. This refutes our
* assumption of independence, which makes it seem like a good hash
* function.
*/
check_word_hash(hash_int_cb, "hash_int", 12);
}
OVSTEST_REGISTER("test-hash", test_hash_main);
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