Hash Table Analysis. e(k) = bp(k). kp(k) = n b.

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1 Septemer 1, 2016 Hash Tale Analysis Assume that we have a hash tale structured as a vector of lists, resolving collisions y sequential search of uckets, as in the generic hash tale template HashTale<K,D,H> distriuted in the file hashtl.h. Let n denote the numer of items in the tale [tale size] and denote the numer of uckets (or slots) in the tale implementation. At this point we make no assumptions aout the hash function. Proposition 1. Denote y e(k) the expected numer of uckets of size k and y p(k) the proaility a given ucket has exactly k items. Then e(k) = p(k). Proof. e(k) = expected value = (frequency) (proaility) = p(k). Proposition 2. Let p(k) denote the proaility a given ucket has exactly k items. Then the expected ucket size is n kp(k) = n. k=0 Proof. The expected value of ucket size is the sum of (the expected value of size of uckets of size k) (proaility a ucket has size k), where expected value of size of uckets of size k = k proaility a ucket has size k = p(k) duh definition of p(k) Sustituting, we get the formula on the left for expected value of ucket size. On the other hand, there are n items in uckets, so the average ucket size is n/, the formula on the right. Proposition 3. Let a[k] denote the actual numer of uckets of size k. Then the size of the tale is n = k 0 ka[k]. Proof. The numer of items in uckets of size k is k a[k]. Since each item is in at most one ucket, the sum on the right does not overcount the numer of items. Since each item is in at least one ucket, the sum on the right does not undercount either. So the sum must e the numer n of items. 1

2 Remark. The smallest safe upper index for the sum in the formula aove is the maximum ucket size, ecause there are no items in uckets with size greater than the maximum. We could also say that n is the smallest safe upper index for the sum, if we want a data independent answer. n is attained in the most dismal case where all items end up in the same ucket, which would make a[i] = 0 for i < n and a[n] = 1, reducing the sum to n = n a[n]. Assume now that we have assigned keys to uckets randomly, so that each key has a uniformly likely proaility of eing assigned to a ucket. (This is called simple uniform hashing in the textook - see page 259.) Proposition 4. The proaility that a given ucket is empty is ( ) n 1 p(0) =. Proof. The proaility that one particular item is not in a given ucket is the proaility it is in some other ucket. There are uckets in total, so there are 1 other uckets. Thus the proaility that one particular item is not in a given ucket is q = ( 1)/. The proaility that two particular items are oth not in that ucket is the product q q = q 2. Reasoning y induction, it follows that the proaility that k particular items are not in a given ucket is q k 1 q = q k. Finally, then, the proaility that none of the n items is in the given ucket is q n. Proposition 5. Assume simple uniform hashing and 0 k n. The proaility that a given ucket has size k is ( )( ) k ( ) n k n 1 1 p(k) =. k Proof. This is a refinement of the argument aove. Note that the proaility that an item is in a given ucket is p = 1/ and the proaility it is not in that ucket is q = 1 p = ( 1)/. If we have a particular choice of k items, the proaility that exactly these items are in a given ucket is p k q n k, ecause those k items are each in the ucket, accounting for the p k factor, and the remaining n k items are not in the ucket, accounting for the q n k factor. There are ( n k) independent choices of exactly k items to put in the ucket, so the proaility that the given ucket has size k is the sum of p k q n k over all these independent ways of selecting k items: ( n k) p k q n k = i=1 ( ) n p k q n k k 2

3 Proposition 6. Assume simple uniform hashing. The ucket size distriution e(k) satisfies the iterative calculation (recurrence): ( ) n 1 e(0) = initial condition ( ) ( ) n k e(k) = e(k 1) for k = 1... n. k 1 e(k) = 0 for k > n. 3 Proof. We use the notation estalished aove: p = 1/, q = ( 1)/. Note that p + q = 1 and p/q = 1/( 1). First the initializer: e(0) = p(0) Proposition 2 ( ) n 1 = Proposition 4 Then the recurrence: e(k) e(k 1) = p(k) p(k 1) Proposition 2 = p(k) cancel p(k 1) ( n ) k p k q n k = ( n ) Proposition 5 k 1 p k 1 q n k+1 (n k + 1)p = applying definitions and simplifying kq ( ) ( ) n k = sustituting p/q = 1/( 1) k 1 Multiplying y e(k 1) completes the proof for the iterative formula. Finally, note that no ucket can have more than n items, so e(k) = 0 when k > n.

4 Hash Tale Analysis Algorithms. Assumptions: n = tale size, = numer of uckets Analysis of a specific hash tale, with a specified hash function, numer of uckets, and loaded data set, can e an important feature of hash tale implementations, simply due to the variaility of performance of hashing, particularly on non-random data where a ias in the hashing specified may exist. The results aove provide ackground needed to devise algorithms required for such an analysis. Algorithm 1: Maximum Bucket Size Runtime : Θ() + cost of calls to size = Θ( + n) typical unsigned long MaxBucketSize () unsigned long max = 0 for ( size_t i = 0; i < ; ++ i) s = size of ucket i if ( max < s) max = s return max ; Algorithm 2: Actual Bucket Size Distriution Runtime : Θ( + n) void ActualBucketSizeDistriution ( Vector < unsigned long >& sizecount ) // calculate and set size of distriution m = MaxBucketSize (); sizecount. SetSize (1+m,0); 4 // calculate distriution for ( size_t i = 0; i < ; ++ i) ++ sizecount [ size of ucket i]; Algorithm 3: Numer of Non - Empty Buckets Runtime : 1 + cost of uck size distriution = Θ( + n) typical Numer of non - empty uckets = - sizecount [0];

5 Algorithm 4: Simple Uniform Hashing Bucket Size Distriution Runtime : Θ(n) void ExpectedBucketSizeDistriution ( Vector < float >& expdcount ) expdcount. SetSize (1+ n,0.0); // set size of distriution expdcount [0] = ( -1)/ ; expdcount [0] = pow ( expdcount [0],n -1); expdcount [0] *= ( -1); for ( size_t k = 1; k < 1+n; ++k) expdcount [k] = ((n-k +1)/(( -1)* k)) * expdcount [k -1]; 5 Remark. The distriution calculated in Algorithm 4 is a inomial distriution, so named ecause of the relationship with inomial choice and the expansion of the inomial (A + B) n [with A = and B = ( 1)/)]: (A + B) n = A n + na n 1 B ( n k ) A k B n k B n In this context, ecause the 1 1 factors are much smaller than the factors, it is very closely approximated y the Poisson distriution. 1 Algorithm 4 calculates many more terms than would normally e useful in practice - most of them are so small they equate to zero in float representation. It is therefore usually a waste of space and time to allocate the entire vector of size 1 + n and calculate all of its elements. In the case here, it is more prudent to allocate space of size 1 + m and, if a few more values are needed, calculate them on the fly (continuing the same iterative process, starting at the (1 + m)th value expdcount[m]) without storing them. 1 See entries in Wikipedia and NIST Handook.