1. Hashing and Linear Probing for Collision 9. Perform a search for key 1040237. How many (hash table) elements will have tobe compared with this search key before it can concluded that 1040237 is found inthe array?10. Perform a search for key 1150246. How many (hash table) elements will have tobe compared with this search key before it can concluded that 1150246 is found inthe array? Consider the followingscenario/assumptions:• A hash table can store at most 11 elements. Indices are numbered from 0 to 10. Itis currently empty (no elements yet).• The hashing function is defined as H(k) = k % 11• The collision resolution technique is Linear Probing.Insert the following keys in the hash table in the sequence that they appear below:• 160237• 270238• 380237• 490240• 600237• 710242• 820237• 930244• 1040237• 1150246Write on a piece of paper the contents of the resulting has table after inserting thesekeys. The table should contain 10 elements, with only one address that is left open,i.e., it is not occupied by any key. Answer the questions below by filling up the blanks.

Given hash funcion H(K) = K% 11 resolved using Lienar Probing where we find next empty slot in case…

9. Perform a search for key 1040237. How many (hash table) elements will have to be compared with this search key before it can concluded that 1040237 is found in the array? 10. Perform a search for key 1150246. How many (hash table) elements will have to be compared with this search key before it can concluded that 1150246 is found in the array?

9. Perform a search for key 1040237. How many (hash table) elements will have to be compared with this search key before it can concluded that 1040237 is found in the array? 10. Perform a search for key 1150246. How many (hash table) elements will have to be compared with this search key before it can concluded that 1150246 is found in the array?

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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Assume a hash table utilizes an array of 13 elements and that collisions are handled by separate chaining. Considering the hash function is defined as: h(k)=k mod 13. i) Draw the contents of the table after inserting elements with the following keys: 36, 243, 261, 180, 217, 180, 21, 16, 182, 202, 91, 97, 166, 78, 33, 70, 51, 58.
Course: Data Structure and Algorithms Language: Java Kindly something and Answer in 1 hour. Read Carefully and give answer with all necesary details. See the image for askii codes. Question6: In this Problem, you are required to insert some keys into a hash table, using given hash functions. You have to Draw a Hash table with the inserted keys. Write total number of collisions encountered when a particular collision resolution technique is used. size= 13, H(X) = sum of Ascii codes of key % HTSIZE Keys: Mia, Bea, Zoe, Jan, Ada, Sam, Leo, Meo, Ben, Tim, Ted, Zod Use Linear probing: H’(X) = [H(X) +f(i))] % HTSIZE where f(i)=i where i=0,1,2,….
3. Given the following keys: 68, 39, 89, 28, 79 Take alpha = 0.6 Compute m (size of hash table), and select it to be a prime number. Now illustrate how you would construct the hash table. Use double hashing as collision resolving technique. The original hash function, is h(k)= k mod m The second hash function for double hashing, h₂(k)=8-(k mod 8)
Draw the contents of the hash table given the following conditions: The size of the hash table is 25. • Linear Probing is used to resolve collisions. • The hash function H(k) should be calculated in the following way where k is the element to be hashed: R(k) = k mod (summation of last two digits of your student id + 7) If R(k)<10 H(k)=R(k) + 8 else H(k)=R(k) - 4 What values will be in the hash table after the following sequence of insertions? 55, 65, 8, 18, 20, 11, 56, 27, 15, 167 [Note: Draw the values using a hash table and show your work for partial credit.]
Create a hash table of size 8 using modular division method. Use hash function h(k) = 2k + 3 where k is the key value. Apply quadratic probing method in resolving collision. Insert the following keys in the hash table: 3, 2, 9, 6, 11, 13. Show the final hash table.
Linear hash index:Consider the Extendible Hashing index shown in figure below.A bucket is split each time an overflow page gets created Show the state of the index after each operation:a) Insertion of entry 2b) Insertion of entry 40c) Insertion of entry 39
Draw the contents of the hash table given the following conditions: • The size of the hash table is 18. • Linear Probing is used to resolve collisions. • The hash function H(k) should be calculated in the following way where k is the element to be hashed: R(k) = k mod (summation of last two digits of your student id) If R(k)<10 H(k)=R(k)+8 else H(k)=R(k)-2 What values will be in the hash table after the following sequence of insertions? 35, 18, 8, 13, 11, 49, 16, 27 [Note: Draw the values using a hash table and show your work for partial credit.]
The hash table array has capacity of 10. Capacity is the number of slots present in the array that is used to build the hashtable. The hash function returns the absolute value of the key mod the capacity of the hash table. a) Insert these keys in the hash table: 3,23,11,21,1,7,77,8 where the hash table uses quadratic probing to resolve collisions. b) Search and Delete 3 and 11 from the table. Be careful about changing the status of the table slot to “deleted” after deleting each item. c)Search 23 and 21 from the table and print their position.
2-A linear probing hash table of length 10 uses the hash function h(x) = x mod 10 + 1. mod is the reminder opertatro. For example the h(42) = 42 mod 10 + 1 = 2 + 1 = 3 so we insert 42 in t he position 3 in the array. After inserting 6 integer keys into an initially empty hash table, the array of keys is 1 2 3 4 5 7 8 9. 42 23 34 52 46 33 a- insert the key 35 b- insert the key 12 c- insert the key 10
Trace Linear Probing with My Data A The hash table is size 7 and uses open addressing and linear probing. The hash function is hashCode(phone number) = last four digits % table size 555-1432 VoterA 555-6135 VoterB 555-2747 VoterC 555-6118 VoterD 555-5733 VoterE 555-4722 VoterF 555-7523 VoterG Show the hashCode of each Voter. Show the table after inserting the above voters in the listed order. Assume you want to retrieve each voter. How many seeks does it take to find each voter? How many seeks would it take to know that 555-5357 VoterH is not in the table? What does the table look like if you remove VoterF? Now how many seeks does it take to know that 555-5357 VoterH is not in the table? How many seeks does it take to retrieve voter G?
2. Suppose you have the following hash table, implemented using linear probing. The hash function we are using is the identity function, h(x) = x. 0 9 1 A B с D E 18 2 3 4 S 12 3 14 7 8 a) In which order could the elements have been added to the hash table? There are several correct answers, and you should give all of them. Assume that the hash table has never been resized, and no elements have been deleted yet. 12, 14, 3, 9, 4, 18, 21 12, 9, 18, 3, 14, 21,4 12, 3, 14, 18, 4, 9, 21 9, 12, 14, 3, 4, 21, 18 9, 14, 4, 18, 12, 3, 21 6 4 21
Draw a hash table given these numbers and hashing function, build each of the hashing options, and decide if it is efficient, and what function or size might be a better match. Explain why it is efficient or what changes you would make for it to be efficient. a. Scenario 1 a. Hashing function: k mod 4 b. Linear Probing c. Array from 0 to 3 d. Values: 19, 7, 12, 11 b. Scenario 2 a. Hashing function: k mod 10 b. Quadratic Probing c. Array from 0 to 9 d. Values: 20 39 23 56 34 29 55 13 c. Scenario 3 a. Hashing function: first three digits of a phone number b. Overflow chaining c. File with base address of 0 and ends at 999 d. Values: 3135552314, 7343455523, 3134445555, 3134441234, 7342346555, e. 5555342232, 4072984555, 2692185552 d. Scenario 4 a. Hashing function: k mod 100 b. Internal Chaining c. Array 0 to 99 d. Values: 314 325 623 2234 425 1234 2132 2361 1245 123 436 742