2 * Copyright (c) 2009, 2012, 2014 Nicira, Inc.
4 * Licensed under the Apache License, Version 2.0 (the "License");
5 * you may not use this file except in compliance with the License.
6 * You may obtain a copy of the License at:
8 * http://www.apache.org/licenses/LICENSE-2.0
10 * Unless required by applicable law or agreed to in writing, software
11 * distributed under the License is distributed on an "AS IS" BASIS,
12 * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
13 * See the License for the specific language governing permissions and
14 * limitations under the License.
30 set_bit(uint32_t array[3], int bit)
32 assert(bit >= 0 && bit <= 96);
33 memset(array, 0, sizeof(uint32_t) * 3);
35 array[bit / 32] = UINT32_C(1) << (bit % 32);
40 hash_words_cb(uint32_t input)
42 return hash_words(&input, 1, 0);
46 jhash_words_cb(uint32_t input)
48 return jhash_words(&input, 1, 0);
52 hash_int_cb(uint32_t input)
54 return hash_int(input, 0);
58 check_word_hash(uint32_t (*hash)(uint32_t), const char *name,
63 for (i = 0; i <= 32; i++) {
64 uint32_t in1 = i < 32 ? UINT32_C(1) << i : 0;
65 for (j = i + 1; j <= 32; j++) {
66 uint32_t in2 = j < 32 ? UINT32_C(1) << j : 0;
67 uint32_t out1 = hash(in1);
68 uint32_t out2 = hash(in2);
69 const uint32_t unique_mask = (UINT32_C(1) << min_unique) - 1;
71 for (ofs = 0; ofs < 32 - min_unique; ofs++) {
72 uint32_t bits1 = (out1 >> ofs) & unique_mask;
73 uint32_t bits2 = (out2 >> ofs) & unique_mask;
75 printf("Partial collision for '%s':\n", name);
76 printf("%s(%08"PRIx32") = %08"PRIx32"\n", name, in1, out1);
77 printf("%s(%08"PRIx32") = %08"PRIx32"\n", name, in2, out2);
78 printf("%d bits of output starting at bit %d "
79 "are both 0x%"PRIx32"\n", min_unique, ofs, bits1);
88 check_3word_hash(uint32_t (*hash)(const uint32_t[], size_t, uint32_t),
93 for (i = 0; i <= 96; i++) {
94 for (j = i + 1; j <= 96; j++) {
95 uint32_t in0[3], in1[3], in2[3];
96 uint32_t out0,out1, out2;
97 const int min_unique = 12;
98 const uint32_t unique_mask = (UINT32_C(1) << min_unique) - 1;
103 out0 = hash(in0, 3, 0);
104 out1 = hash(in1, 3, 0);
105 out2 = hash(in2, 3, 0);
108 printf("%s hash not the same for non-64 aligned data "
109 "%08"PRIx32" != %08"PRIx32"\n", name, out0, out1);
111 if ((out1 & unique_mask) == (out2 & unique_mask)) {
112 printf("%s has a partial collision:\n", name);
113 printf("hash(1 << %d) == %08"PRIx32"\n", i, out1);
114 printf("hash(1 << %d) == %08"PRIx32"\n", j, out2);
115 printf("The low-order %d bits of output are both "
116 "0x%"PRIx32"\n", min_unique, out1 & unique_mask);
123 test_hash_main(int argc OVS_UNUSED, char *argv[] OVS_UNUSED)
125 /* Check that all hashes computed with hash_words with one 1-bit (or no
126 * 1-bits) set within a single 32-bit word have different values in all
127 * 11-bit consecutive runs.
129 * Given a random distribution, the probability of at least one collision
130 * in any set of 11 bits is approximately
132 * 1 - ((2**11 - 1)/2**11)**C(33,2)
133 * == 1 - (2047/2048)**528
136 * There are 21 ways to pick 11 consecutive bits in a 32-bit word, so if we
137 * assumed independence then the chance of having no collisions in any of
138 * those 11-bit runs would be (1-0.22)**21 =~ .0044. Obviously
139 * independence must be a bad assumption :-)
141 check_word_hash(hash_words_cb, "hash_words", 11);
142 check_word_hash(jhash_words_cb, "jhash_words", 11);
144 /* Check that all hash functions of with one 1-bit (or no 1-bits) set
145 * within three 32-bit words have different values in their lowest 12
148 * Given a random distribution, the probability of at least one collision
149 * in 12 bits is approximately
151 * 1 - ((2**12 - 1)/2**12)**C(97,2)
152 * == 1 - (4095/4096)**4656
155 * so we are doing pretty well to not have any collisions in 12 bits.
157 check_3word_hash(hash_words, "hash_words");
158 check_3word_hash(jhash_words, "jhash_words");
160 /* Check that all hashes computed with hash_int with one 1-bit (or no
161 * 1-bits) set within a single 32-bit word have different values in all
162 * 12-bit consecutive runs.
164 * Given a random distribution, the probability of at least one collision
165 * in any set of 12 bits is approximately
167 * 1 - ((2**12 - 1)/2**12)**C(33,2)
168 * == 1 - (4,095/4,096)**528
171 * There are 20 ways to pick 12 consecutive bits in a 32-bit word, so if we
172 * assumed independence then the chance of having no collisions in any of
173 * those 12-bit runs would be (1-0.12)**20 =~ 0.078. This refutes our
174 * assumption of independence, which makes it seem like a good hash
177 check_word_hash(hash_int_cb, "hash_int", 12);
180 OVSTEST_REGISTER("test-hash", test_hash_main);