text: /*
   * 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>
  #undef NDEBUG
  #include "hash.h"
  #include <assert.h>
  #include <inttypes.h>
  #include <stdio.h>
  #include <stdlib.h>
  #include <string.h>
  #include "jhash.h"
  #include "ovstest.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 in0[3], in1[3], in2[3];
              uint32_t out0,out1, out2;
              const int min_unique = 12;
              const uint32_t unique_mask = (UINT32_C(1) << min_unique) - 1;
  
              set_bit(in0, i);
              set_bit(in1, i);
              set_bit(in2, j);
              out0 = hash(in0, 3, 0);
              out1 = hash(in1, 3, 0);
              out2 = hash(in2, 3, 0);
  
              if (out0 != out1) {
                  printf("%s hash not the same for non-64 aligned data "
                         "%08"PRIx32" != %08"PRIx32"\n", name, out0, out1);
              }
              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 - (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
       *
       * 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);
  