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test-hash: Improve comments.
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This commit tries to simplify and further clarify the test cases
in test-hash.

Signed-off-by: Alex Wang <[email protected]>
Acked-by: Ben Pfaff <[email protected]>
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@yew011
yew011 committed Mar 16, 2015
1 parent 5f38375 commit 6cc5985
Showing 1 changed file with 51 additions and 68 deletions.
119 changes: 51 additions & 68 deletions tests/test-hash.c
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Original file line number Diff line number Diff line change
Expand Up @@ -242,94 +242,77 @@ check_256byte_hash(void (*hash)(const void *, size_t, uint32_t, ovs_u128 *),
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.
/*
* The following tests check that all hashes computed with hash_function
* with one 1-bit (or no 1-bits) set within a X-bit word have different
* values in all N-bit consecutive comparisons.
*
* test_function(hash_function, test_name, N)
*
* Given a random distribution, the probability of at least one collision
* in any set of 11 bits is approximately
* in any set of N bits is approximately
*
* 1 - (proportion of same_bits)
* **(binomial_coefficient(n_bits_in_data + 1, 2))
* == 1 - ((2**11 - 1)/2**11)**C(33,2)
* == 1 - (2047/2048)**528
* =~ 0.22
* 1 - (prob of no collisions)
* **(combination of all possible comparisons)
* == 1 - ((2**N - 1)/2**N)**C(X+1,2)
* == p
*
* There are 21 ways to pick 11 consecutive bits in a 32-bit word, so if we
* There are (X-N) ways to pick N consecutive bits in a X-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
* those X-bit runs would be (1-p)**(X-N) == q. If this q is very small
* and we can also find a relatively small 'magic number' N such that there
* is no collision in any comparison, then it means we have a pretty good
* hash function.
*
* 1 - ((2**12 - 1)/2**12)**C(97,2)
* == 1 - (4095/4096)**4656
* =~ 0.68
* The values of each parameters mentioned above for the tested hash
* functions are summarized as follow:
*
* 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
* hash_function X N p q
* ------------- --- --- ------- -------
*
* 1 - ((2**12 - 1)/2**12)**C(33,2)
* == 1 - (4,095/4,096)**528
* =~ 0.12
* hash_words_cb 32 11 0.22 0.0044
* jhash_words_cb 32 11 0.22 0.0044
* hash_int_cb 32 12 0.12 0.0078
* hash_bytes128 128 19 0.0156 0.174
*
* 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_words_cb, "hash_words", 11);
check_word_hash(jhash_words_cb, "jhash_words", 11);
check_word_hash(hash_int_cb, "hash_int", 12);
check_hash_bytes128(hash_bytes128, "hash_bytes128", 19);

/* Check that all hashes computed with hash_bytes128 with one 1-bit (or no
* 1-bits) set within a single 128-bit word have different values in all
* 19-bit consecutive runs.
/*
* The following tests check that all hashes computed with hash_function
* with one 1-bit (or no 1-bits) set within Y X-bit word have different
* values in their lowest N bits.
*
* test_function(hash_function, test_name, N)
*
* Given a random distribution, the probability of at least one collision
* in any set of 19 bits is approximately
* in any set of N bits is approximately
*
* 1 - ((2**19 - 1)/2**19)**C(129,2)
* == 1 - (524,287/524,288)**8256
* =~ 0.0156
* 1 - (prob of no collisions)
* **(combination of all possible comparisons)
* == 1 - ((2**N - 1)/2**N)**C(Y*X+1,2)
* == p
*
* There are 111 ways to pick 19 consecutive bits in a 128-bit word, so if
* we assumed independence then the chance of having no collisions in any of
* those 19-bit runs would be (1-0.0156)**111 =~ 0.174. This refutes our
* assumption of independence, which makes it seem like a good hash
* function.
*/
check_hash_bytes128(hash_bytes128, "hash_bytes128", 19);

/* Check that all hashes computed with hash_bytes128 with 1-bit (or no
* 1-bits) set within 16 128-bit words have different values in their
* lowest 23 bits.
* If this p is not very small and we can also find a relatively small
* 'magic number' N such that there is no collision in any comparison,
* then it means we have a pretty good hash function.
*
* Given a random distribution, the probability of at least one collision
* in any set of 23 bits is approximately
* The values of each parameters mentioned above for the tested hash
* functions are summarized as follow:
*
* 1 - ((2**23 - 1)/2**23)**C(2049,2)
* == 1 - (8,388,607/8,388,608)**2,098,176
* =~ 0.22
* hash_function Y X N p
* ------------- --- --- --- -------
*
* hash_words 3 32 12 0.68
* jhash_words 3 32 12 0.68
* hash_bytes128 16 128 23 0.22
*
* so we are doing pretty well to not have any collisions in 23 bits.
*/
check_3word_hash(hash_words, "hash_words");
check_3word_hash(jhash_words, "jhash_words");
check_256byte_hash(hash_bytes128, "hash_bytes128", 23);
}

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