chap11-solutions - Selected Solutions for Chapter 11 Hash...

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Selected Solutions for Chapter 11: Hash Tables Solution to Exercise 11.2-1 For each pair of keys k; l , where k ¤ l , define the indicator random variable X kl D I f h.k/ D h.l/ g . Since we assume simple uniform hashing, Pr f X kl D 1 g D Pr f h.k/ D h.l/ g D 1=m , and so E ŒX kl ± D 1=m . Now define the random variable Y to be the total number of collisions, so that Y D P k ¤ l X kl . The expected number of collisions is E ŒY ± D E ±X k ¤ l X kl ² D X k ¤ l E ŒX kl ± (linearity of expectation) D n 2 ! 1 m D n.n N 1/ 2 ± 1 m D n.n N 1/ 2m : Solution to Exercise 11.2-4 The flag in each slot will indicate whether the slot is free. ± A free slot is in the free list, a doubly linked list of all free slots in the table. The slot thus contains two pointers. ± A used slot contains an element and a pointer (possibly NIL) to the next element that hashes to this slot. (Of course, that pointer points to another slot in the table.)
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11-2 Selected Solutions for Chapter 11: Hash Tables Operations ± Insertion: ± If the element hashes to a free slot, just remove the slot from the free list and store the element there (with a NIL pointer). The free list must be doubly linked in order for this deletion to run in O.1/ time. ± If the element hashes to a used slot j , check whether the element x already there “belongs” there (its key also hashes to slot j ). ± If so, add the new element to the chain of elements in this slot. To do so, allocate a free slot (e.g., take the head of the free list) for the new element and put this new slot at the head of the list pointed to by the hashed-to slot ( j ).
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This note was uploaded on 04/20/2010 for the course IE ie200 taught by Professor . during the Spring '10 term at 카이스트, 한국과학기술원.

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chap11-solutions - Selected Solutions for Chapter 11 Hash...

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