8. Let's look at the actions of HashSet. Assume the following: • the hash table array has an initial capacity of 10 • the hash table uses chaining for resolving collisions • the hash function returns the integer key's value, mod the size of the table, i.e., h(1) = i % 10 • rehashing occurs at the end of an add where the load factor is 2 0.5 and doubles the capacity of the hash table Draw an array diagram to show the final state of the hash table after the following operations are performed. Leave a box empty if an array element is null or is unused. You do not have to redraw an entirely new hash table after each element is added or removed, but since the final answer depends on every add/remove being done correctly, you may wish to redraw the table at various important stages. YOU MUST SHOW ALL WORK FOR FULL CREDIT. Set set = new HashSet( ); set.add(47); set.add(97); set.add(52); set.add(-34); set.remove(4); set.remove(52); set.add(44); set.add(77); set.add(-232); set.add(-4); set.add(444); Size = Capacity = Load factor =.

8. Let's look at the actions of HashSet. Assume the following: • the hash table array has an initial capacity of 10 • the hash table uses chaining for resolving collisions • the hash function returns the integer key's value, mod the size of the table, i.e., h(1) = i % 10 • rehashing occurs at the end of an add where the load factor is 2 0.5 and doubles the capacity of the hash table Draw an array diagram to show the final state of the hash table after the following operations are performed. Leave a box empty if an array element is null or is unused. You do not have to redraw an entirely new hash table after each element is added or removed, but since the final answer depends on every add/remove being done correctly, you may wish to redraw the table at various important stages. YOU MUST SHOW ALL WORK FOR FULL CREDIT. Set set = new HashSet( ); set.add(47); set.add(97); set.add(52); set.add(-34); set.remove(4); set.remove(52); set.add(44); set.add(77); set.add(-232); set.add(-4); set.add(444); Size = Capacity = Load factor =.

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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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,….
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.
3. Double hashing is one of the methods to resolve collision. Write a function to implement this method. The equations used in this method are given below. Note: implement everything within the double hash function. P = (P + INCREMENT(Key)) mod TABLE_SIZE INCREMENT(Key) = 1 + (Key mod INCR)
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.]
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.]
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
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
1. Given the hash function: h(i) = i % 13 and double hash function: d(i) = 1 + i % 12.Show the array after inserting the following keys: 18, 26, 35, 9, 64, 47, 96, 36, 70 , in this order. Use double hashing to resolve collisions. 2. Given the hash function: h(i) = i % 10Show the array after inserting the following keys: 24, 42, 34, 62, 73, 90, 65, 55 in this order. Use quadratic probing to resolve collisions.
81. The process of finding some other position when hash address is occupied is classified as a. collision resolution b. address space resolution c. multiple hashing resolution d. chaining resolution
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)
Outline the steps in inserting a data M into a hash table. Assume that the collisionresolution technique used is double hashing.
Consider the following scenario/assumptions: . A hash table can store at most 11 elements. Indices are numbered from 0 to 10. It is 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: • 172458 • 282459 • 392458 • 502461 • 612458 • 722463 832458 • 942465 1052458 • 1162467 Write on a piece of paper the contents of the resulting has table after inserting these keys. 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, 1. What is the address (array index) of the memory space that is still open? 2. What is H(722463)? 3. In which address (array index) was key 832458 stored? 4. Which key value was stored in address (array index) 8? 5. The value of H(k) is called k's home address. How many keys were not…