You are given an empty hash table of size 7 that uses open addressing (no chaining). Insert the following keys 15 17 8 23 35 into the table using each of the following approaches. If overflow occurs mark where it occurs and indicate the element that causes the overflow. a. h(x) = x % 7; linear probing b. h(x) = x % 7; quadratic probing 1 1 3 4 4 5 61 6. c. h(x)= x%7; double hashing with g(x) = x/7 +1 (using integer division) LO 3,

You are given an empty hash table of size 7 that uses open addressing (no chaining). Insert the following keys 15 17 8 23 35 into the table using each of the following approaches. If overflow occurs mark where it occurs and indicate the element that causes the overflow. a. h(x) = x % 7; linear probing b. h(x) = x % 7; quadratic probing 1 1 3 4 4 5 61 6. c. h(x)= x%7; double hashing with g(x) = x/7 +1 (using integer division) LO 3,

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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Suppose you have a hash table with seven entries (indexed 0 through 6). This table uses open addressing with the hash function that maps each letter to its alphabet code (a = A = 0, etc.) modulo 7. Rehashing is accomplished using linear-probing with a jump of 1. Describe the state of the table after each of the letters D, a, d, H, a, and h are added to the table.
Question 5 Assume you have a hash table of size 13, insert the following keys using open addressing with linear probing, assuming that the H(Key) = key % size and the bucket size is equal to 1. Keys: 25, 66, 710, 150, 82, 111, 70,17 a. index key 0 1 66 2 3 4 82 5 70 6 17 7 150 8 710 9 111 10 11 25 12 b. index key 1 2 66 3 4 5 82 6 70 7 17 8 150 9 710 10 111 11 12 13 25 c. index key 0 1 66 2 3 4 82 5 70 6 17 7 150 8 111 9 710 10 11 12 25 d. index key 0 1 66 2 3 4 82 5 70 6 17 7 150 8 710 9 111 10 11 12 25
Using Hopscotch hashing with a max hop of 4, hash the following keys. Use a table of size 13 (indexes 0-12). A: 6 B: 7 C: 9 D: 7 E: 6 F: 7 G: 8
Consider a OBHTin which the keys are student identifiers (strings of 6 digits). Assume the following number of buckets and hash function: m = 100; hash(id) = first two digits of id Starting with an empty hash table, show the effect of successively adding the following student identifiers: 000014, 990021, 990019, 970036, 000015, 970012, 970023. Use following function as second hash function hash(id) = last digit of id 2. Show the effect of deleting 000014 from the hash table
Given an empty hash table of size 7 that uses open addressing, the following sequence of keys is to be inserted: 15 17 8 23 3 5 Insert these keys using each of the following approaches. h(x) = x % 7; linear probing 0 1 2 3 4 5 6 2. h(x) = x % 7; quadratic probing 0 1 2 3 4 5 6 3.h(x) = x % 7; double hashing with stepSize = 5 - (x % 5); 0 1 2 3 4 5 6
1. Give the contents of the resulting hash table when you insert items with the keys L A G U N I V E R S in that order into an initially empty table of size M = 16 using linear probing. Use the hash function 11k mod M to transform the kth letter of the alphabet into a table index. Example: hash(J) hash(10) = 110 % 16 = 14 Show the detailed solution for each key and plot the items in the hash table.
Consider a open bucket hash table and closed bucket hash table in which the keys are student identifiers (strings of 6 digits). Assume the following number of buckets and hash function: m = 100; hash(id) = first two digits of id #. Starting with an empty hash table, show the effect of successively adding the following student identifiers: 000014, 990021, 990019, 970036, 000015, 970012, 970023. Use following function as second hash function hash(id) = last digit of id # Show the effect of deleting 000014 from the hash table. (Java Programming DSA)
20 a) Given a hashing function hash(x) = ((x * x) + x) % 11, Insert the value 17 into each hash table using the rules specified below. Note that some of the boxes in each hash table are already full. If there’s a collision, start the probing at i = 0 and increment from there if you collide. b) After we insert our new data (that 17) into these hash tables, should we resize and rehash the values? Why or Why not? c) If this was a hash table using separate chaining instead of linear probing, would it be time to rehash after we insert 17?
Given is the hash function h(k) = k mod 10. How many different insertion sequences of keys are there to generate the following hash table if closed hashing with linear probing is used? B0 = ∅ B1 = ∅ B2 = {32} B3 = {43} B4 = {54} B5 = {12} B6 = {76} B7 = {23} B8 = ∅ B9 = ∅
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Show the result when we insert the keys 5; 28; 19; 15; 20; 33; 12; 17; 10 into a hash table with collisions resolved by linked list at each slot. Let the hash table have 9 slots, and let the hash function be h(x) = (2x+1) mod 9. (You are expected to draw the final hash table)
Given a sequence of 5 keys < 53, 66, 32, 41, 88 > for searching task: a) Given the hash function H(k) = (3.k) mod 7, insert the keys above according to its original order (from left to right) into a hash table of 7 slots. Indicate the cases of collision. Clearly show the steps and calculation with a table. b) In case of any collisions found above in a) part, determine the new slot for each collided case using Linear Probing to solve the collision problem. Clearly show your answer for each identified case. No step of calculation required. o (Answer with “NO collision found”, in case there is no collisions found above) c) In case of any collisions found above in a) part, determine the new slot ONLY for the first collided case using Double-Hashing (with functions below) to solve the collision problem. Clearly show the steps of your calculation. d1 = H(k) = (3.k) mod 7 di = (di−1 + ((5·k) mod 6) + 1) mod 7 , i ≥ 2 o (Answer with “NO collision found”, in case there is no collisions…