Consider the following scenario/łassumptions: . 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 stored in their respective home addresses?

Consider the following scenario/łassumptions: . 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 stored in their respective home addresses?

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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Similar questions

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.
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?
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)
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
Let 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%7
B) Suppose a hash table has 11 locations, keys are placed in the table using the hash function f (x) = x mod 11, and linear chaining is used to resolve collisions Draw a picture of the result of storing the following keys in the table: 0, 12, 42, 18, 6, 22, 8, 105, 97
Given a hash table of size M=10 and a linear probing method, following are the contents of the hash table: [-1, 41, 71, -1, 64, 75, 84, -1, 88, 9]. -1 indicates slot is empty. Use the standard hash function: h(x)=x%M Match the index on the left with the value on the right for the hash table after the following operations are carried out in the given order: delete(88) delete(41) delete(44) insert(108) insert(71)
3. (A) Show the result when putting the keys 19,12,14,13,9 into a hash table of size 7, usingHashing with Linear Probing. The hash function is hash(k) = k mod 7. Don’t resize.(B) Continuing from part A, show the result after deleting 19.
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.
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
Suppose we have a hash map that uses the standard “mod” hash function shown in the chapter and uses linear probing for collision resolution. The starting hash table length is 5, and the table chooses to rehash to twice its former size once the load factor reaches or exceeds 0.5. If we begin with an empty map, what will be the final state of the hash table after the following key/value pairs are added and removed? Draw the entire array and the contents of each index, including any resizing and rehashing necessary. Write “X” in any index in which an element is removed and not replaced by another element. Also write the size, capacity, and load factor of the final hash table. HashMap<Integer, String> map = new HashMap<Integer, String>(); map.put(7, "Jessica"); map.put(34, "Tyler"); map.put(17, "Ryan"); map.put(15, "Tina"); map.put(84, "Saptarshi"); map.remove("Tyler"); map.put(7, "Meghan"); map.put(33, "Kona"); map.remove(17); map.put(6, "Tina"); map.remove(84); map.put(15,…
The 9-entry hash table that results from using the hash function, h(i) = (2i+7) mod 9, to hash the keys 12, 5, 13, 4, 10, 19, 8, 3, and 28, assuming collisions are handled by linear probing as follows: 0 1 2 3 4 5 6 7 8 10 19 3 28 12 8 13 4 5 group answer choice a) True b) False