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135 lines
6.7 KiB
Go
135 lines
6.7 KiB
Go
/*
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Package jump implements the "jump consistent hash" algorithm.
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Example
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h := jump.Hash(256, 1024) // h = 520
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Reference C++ implementation[1]
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int32_t JumpConsistentHash(uint64_t key, int32_t num_buckets) {
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int64_t b = -1, j = 0;
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while (j < num_buckets) {
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b = j;
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key = key * 2862933555777941757ULL + 1;
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j = (b + 1) * (double(1LL << 31) / double((key >> 33) + 1));
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}
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return b;
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}
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Explanation of the algorithm
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Jump consistent hash works by computing when its output changes as the
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number of buckets increases. Let ch(key, num_buckets) be the consistent hash
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for the key when there are num_buckets buckets. Clearly, for any key, k,
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ch(k, 1) is 0, since there is only the one bucket. In order for the
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consistent hash function to balanced, ch(k, 2) will have to stay at 0 for
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half the keys, k, while it will have to jump to 1 for the other half. In
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general, ch(k, n+1) has to stay the same as ch(k, n) for n/(n+1) of the
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keys, and jump to n for the other 1/(n+1) of the keys.
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Here are examples of the consistent hash values for three keys, k1, k2, and
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k3, as num_buckets goes up:
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│ 1 │ 2 │ 3 │ 4 │ 5 │ 6 │ 7 │ 8 │ 9 │ 10 │ 11 │ 12 │ 13 │ 14
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───┼───┼───┼───┼───┼───┼───┼───┼───┼───┼────┼────┼────┼────┼────
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k1 │ 0 │ 0 │ 2 │ 2 │ 4 │ 4 │ 4 │ 4 │ 4 │ 4 │ 4 │ 4 │ 4 │ 4
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───┼───┼───┼───┼───┼───┼───┼───┼───┼───┼────┼────┼────┼────┼────
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k2 │ 0 │ 1 │ 1 │ 1 │ 1 │ 1 │ 1 │ 7 │ 7 │ 7 │ 7 │ 7 │ 7 │ 7
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───┼───┼───┼───┼───┼───┼───┼───┼───┼───┼────┼────┼────┼────┼────
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k3 │ 0 │ 1 │ 1 │ 1 │ 1 │ 5 │ 5 │ 7 │ 7 │ 7 │ 10 │ 10 │ 10 │ 10
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A linear time algorithm can be defined by using the formula for the
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probability of ch(key, j) jumping when j increases. It essentially walks
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across a row of this table. Given a key and number of buckets, the algorithm
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considers each successive bucket, j, from 1 to num_buckets1, and uses
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ch(key, j) to compute ch(key, j+1). At each bucket, j, it decides whether to
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keep ch(k, j+1) the same as ch(k, j), or to jump its value to j. In order to
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jump for the right fraction of keys, it uses a pseudorandom number
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generator with the key as its seed. To jump for 1/(j+1) of keys, it
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generates a uniform random number between 0.0 and 1.0, and jumps if the
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value is less than 1/(j+1). At the end of the loop, it has computed
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ch(k, num_buckets), which is the desired answer. In code:
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int ch(int key, int num_buckets) {
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random.seed(key);
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int b = 0; // This will track ch(key,j+1).
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for (int j = 1; j < num_buckets; j++) {
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if (random.next() < 1.0 / (j + 1)) b = j;
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}
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return b;
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}
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We can convert this to a logarithmic time algorithm by exploiting that
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ch(key, j+1) is usually unchanged as j increases, only jumping occasionally.
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The algorithm will only compute the destinations of jumps the j’s for
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which ch(key, j+1) ≠ ch(key, j). Also notice that for these j’s, ch(key,
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j+1) = j. To develop the algorithm, we will treat ch(key, j) as a random
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variable, so that we can use the notation for random variables to analyze
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the fractions of keys for which various propositions are true. That will
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lead us to a closed form expression for a pseudorandom variable whose value
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gives the destination of the next jump.
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Suppose that the algorithm is tracking the bucket numbers of the jumps for a
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particular key, k. And suppose that b was the destination of the last jump,
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that is, ch(k, b) ≠ ch(k, b+1), and ch(k, b+1) = b. Now, we want to find the
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next jump, the smallest j such that ch(k, j+1) ≠ ch(k, b+1), or
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equivalently, the largest j such that ch(k, j) = ch(k, b+1). We will make a
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pseudorandom variable whose value is that j. To get a probabilistic
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constraint on j, note that for any bucket number, i, we have j ≥ i if and
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only if the consistent hash hasn’t changed by i, that is, if and only if
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ch(k, i) = ch(k, b+1). Hence, the distribution of j must satisfy
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P(j ≥ i) = P( ch(k, i) = ch(k, b+1) )
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Fortunately, it is easy to compute that probability. Notice that since P(
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ch(k, 10) = ch(k, 11) ) is 10/11, and P( ch(k, 11) = ch(k, 12) ) is 11/12,
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then P( ch(k, 10) = ch(k, 12) ) is 10/11 * 11/12 = 10/12. In general, if n ≥
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m, P( ch(k, n) = ch(k, m) ) = m / n. Thus for any i > b,
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P(j ≥ i) = P( ch(k, i) = ch(k, b+1) ) = (b+1) / i .
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Now, we generate a pseudorandom variable, r, (depending on k and j) that is
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uniformly distributed between 0 and 1. Since we want P(j ≥ i) = (b+1) / i,
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we set P(j ≥ i) iff r ≤ (b+1) / i. Solving the inequality for i yields P(j ≥
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i) iff i ≤ (b+1) / r. Since i is a lower bound on j, j will equal the
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largest i for which P(j ≥ i), thus the largest i satisfying i ≤ (b+1) / r.
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Thus, by the definition of the floor function, j = floor((b+1) / r).
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Using this formula, jump consistent hash finds ch(key, num_buckets) by
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choosing successive jump destinations until it finds a position at or past
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num_buckets. It then knows that the previous jump destination is the answer.
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int ch(int key, int num_buckets) {
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random.seed(key);
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int b = -1; // bucket number before the previous jump
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int j = 0; // bucket number before the current jump
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while (j < num_buckets) {
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b = j;
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r = random.next();
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j = floor((b + 1) / r);
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}
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return = b;
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}
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To turn this into the actual code of figure 1, we need to implement random.
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We want it to be fast, and yet to also to have well distributed successive
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values. We use a 64bit linear congruential generator; the particular
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multiplier we use produces random numbers that are especially well
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distributed in higher dimensions (i.e., when successive random values are
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used to form tuples). We use the key as the seed. (For keys that don’t fit
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into 64 bits, a 64 bit hash of the key should be used.) The congruential
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generator updates the seed on each iteration, and the code derives a double
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from the current seed. Tests show that this generator has good speed and
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distribution.
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It is worth noting that unlike the algorithm of Karger et al., jump
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consistent hash does not require the key to be hashed if it is already an
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integer. This is because jump consistent hash has an embedded pseudorandom
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number generator that essentially rehashes the key on every iteration. The
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hash is not especially good (i.e., linear congruential), but since it is
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applied repeatedly, additional hashing of the input key is not necessary.
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[1] http://arxiv.org/pdf/1406.2294v1.pdf
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*/
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package jump
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