Introduction to Java Programming and Data Structures, Comprehensive Version (11th Edition)
11th Edition
ISBN: 9780134670942
Author: Y. Daniel Liang
Publisher: PEARSON
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Chapter 27.4, Problem 27.4.5CP
Program Plan Intro
Open Addressing:
- Open Addressing is a method of finding an open location in the hash table at the time of collision.
- There are several variations for open addressing such as linear probing, quadratic probing, and double hashing.
Quadratic Probing:
- Quadratic probing is one other variation of open addressing.
- It is introduced to avoid the clustering problem in linear probing.
- Quadratic probing will look at the cells at indices (k + j2) % n, for j ≥ 0, i.e., k, (k + 1) % n, (k + 4) % n, (k + 9) % n, ..., and so on.
- Starting from the initial index, quadratic probing will add an increment of 2 to k to define a search sequence.
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Using quadratic probing, create a hash table HT[0..8, 0..2] for the collection of keys 17, 9, 34, 56, 11, 4, 71, 86, 55, 10, 39, 49, 52, 82, 31, 13, 22, 35, 44, 20, 60, and 28.Use the hashing formula H(X) = X mod 9. What conclusions have you drawn?
Using double hasing, insert the following keys: {333, 335, 123, 617, 93, 63, 17, 37} into a hash table of size 17 using k mod m as h1(k), and h2(k) = 8 - k mod 8 as the offset.
Suppose we have a hash table of size 11 and use the hash function h(key) = (key + i2) % 11, where i = 0, 12, 22, ..., 102. After inserting entries with keys 35, 29, 54, 43, 121, 33, 44, and 187. What is the index of key 187?
index = {0, 1, 2, ..., 10}
Chapter 27 Solutions
Introduction to Java Programming and Data Structures, Comprehensive Version (11th Edition)
Ch. 27.2 - Prob. 27.2.1CPCh. 27.3 - Prob. 27.3.1CPCh. 27.3 - Prob. 27.3.2CPCh. 27.3 - Prob. 27.3.3CPCh. 27.3 - Prob. 27.3.4CPCh. 27.3 - Prob. 27.3.5CPCh. 27.3 - Prob. 27.3.6CPCh. 27.3 - If N is an integer power of the power of 2, is N /...Ch. 27.3 - Prob. 27.3.8CPCh. 27.3 - Prob. 27.3.9CP
Ch. 27.4 - Prob. 27.4.1CPCh. 27.4 - Prob. 27.4.2CPCh. 27.4 - Prob. 27.4.3CPCh. 27.4 - Prob. 27.4.4CPCh. 27.4 - Prob. 27.4.5CPCh. 27.4 - Prob. 27.4.6CPCh. 27.5 - Prob. 27.5.1CPCh. 27.6 - Prob. 27.6.1CPCh. 27.6 - Prob. 27.6.2CPCh. 27.6 - Prob. 27.6.3CPCh. 27.7 - Prob. 27.7.1CPCh. 27.7 - What are the integers resulted from 32 1, 32 2,...Ch. 27.7 - Prob. 27.7.3CPCh. 27.7 - Describe how the put(key, value) method is...Ch. 27.7 - Prob. 27.7.5CPCh. 27.7 - Show the output of the following code:...Ch. 27.7 - If x is a negative int value, will x (N 1) be...Ch. 27.8 - Prob. 27.8.1CPCh. 27.8 - Prob. 27.8.2CPCh. 27.8 - Can lines 100103 in Listing 27.4 be removed?Ch. 27.8 - Prob. 27.8.4CPCh. 27 - Prob. 27.1PECh. 27 - Prob. 27.2PECh. 27 - (Modify MyHashMap with duplicate keys) Modify...Ch. 27 - Prob. 27.6PECh. 27 - Prob. 27.7PECh. 27 - Prob. 27.8PECh. 27 - Prob. 27.10PECh. 27 - Prob. 27.11PECh. 27 - (setToList) Write the following method that...Ch. 27 - (The Date class) Design a class named Date that...Ch. 27 - (The Point class) Design a class named Point that...
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- For the set of keys {17, 9, 34, 56, 11, 4, 71, 86, 55, 10, 39, 49, 52, 82, 31, 13, 22,35, 44, 20, 60, 28} obtain a hash table HT[0..8, 0..2] following quadratic probing.Make use of the hash function H(X) = X mod 9. What are your observations?arrow_forwardSuppose that 50 keys are to be inserted into an initially empty hash table using quadratic probing. What should be the size of the hash table to guarantee that all the collisions are resolved?arrow_forwardLet us consider an empty hash table with 10 positions indexed from 0 to 9. Please illustrate the content of the hash table after inserting the elements 79, 8, 39, 48, 3, and 60 using(1) Quadratic Probing (2) Double hashing with h(k) = k%10 and h’(k)= 7-k%7arrow_forward
- Does a hash table of size m contain the same number of linked lists at all times? No matter how hard I attempt, I cannot identify the purpose of a hash function. Provide an example to illustrate your point.arrow_forwardShow what occurs when the keys 5, 28, 19, 15, 20, 33, 12, 17, and 10 are inserted into a hash table with collisions addressed by chaining. If the table has nine slots, then the hash function should be h.k/D k mod nine.arrow_forwardInsert the keys <13, 19, 35, 71, 31, 6, 23, 4> into hash table of size m=11 using linear probing hashing. Here, h(k, i) =((k mod m) + i) mod m, i=0,1,2,…. How many times you have to increment i to resolve collisions?arrow_forward
- For a n-bit hash function and m-bit messages, there are 2m-n (I,e; Two to the power m-n) messages per hash value. For example, m=1024 and n=128 there are 2 896 (two to the power 896) messages per hash value. Still. Why is it difficult to find hash collision?arrow_forwardConsider a Hash Table with 6 bins. Show the content of the Hash Table (that uses chaining to handle collisions) after the following numbers are inserted into the table in order: 0 1 6 7arrow_forwardIn simple uniform hashing, each key is assumed to have equal probability to map to any of the hashes in a given table of size m. Given an open-address table of size 100 and 2 random keys, what is the probability that they hash to the same value? What is the probability that they hash to different values? PLEASE give me a written paragraph answerarrow_forward
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