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

Linear probing: The hash function for the linear probing is: h (k, i) = (h’ (k) + i) mod m where h’k…

Answered: 3. (10%) Consider inserting the keys…

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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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)).

3. (10%) 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 cl = 1 and c2 = 3, and
using double hashing with h1(k) =k and h2(k) = 1+ (k mod(m – 1)).

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