Question: Consider inserting the keys 10, 22, 31, 4, 15, 28, 17, 88, 59 into a hash table of length m = 11 using open addressing

Consider inserting the keys 10, 22, 31, 4, 15, 28, 17, 88, 59 into a hash table of length m = 11 using open addressing with the auxiliary hash function h′(k) = k. Illustrate the result of inserting these keys using linear probing, using quadratic probing with c₁ = 1 and c₂ = 3, and using double hashing with h₁(k) = k and h₂(k) = 1 + (k mod (m – 1)).

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a Linear probing hk hk i mod m b Quadratic probing hk hk c1i c2i 2 mod m c Double hashing hk h1k ih2k mod m Inserting the keys 10 22 31415 28 17 8859 ... View full answer

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