Please do fast don't use ai to answer 7 1 pointStarting with an empty hash table with a fixed size of 11, insert the following keys in order into it using the specified collision mechanism:[12, 9, 3, 0, 42, 98, 70, 1}.Suppose linear probing is used and the hash function is defined as follows:H_i(key)=(h(key) + probe(i)) %tablesize, where h(key)= key* key + 3 and probe(i) = (i + 1)Show the final result of this hash table. If a collision resolution mechanism is unable to successfully resolve, simply record the state of the last successful insert.01423122299870094270123456789109831012221678910012981942700123450H3231123456789104298970012345678910

Hash Table DetailsTable Size: 11Hash Function:…

Answered: 7 1 point Starting with an empty hash…

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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hash table
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.]
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)
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…
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
b) Consider an empty hash table of size 7. Draw the table that results after inserting, in the given order, the following key values: {19, 26, 13, 48, 17} when collisions are managed and handled by double hashing method with function: h(k, i) = [h,(k) + i * hz(k)] mod 7, where h, (k) = k mod 7 and h, (k) = 5 – (k mod 5). I. 812 3 4 5 6
b) Consider an empty hash table of size 7. Draw the table that results after inserting, in the given order, the following key values: {19, 26, 13, 48, 17} when collisions are managed and handled by double hashing method with function: h(k,i) = [h,(k) + i * h2(k)] mod 7, where h,(k) = k mod 7 and h, (k) = 5 – (k mod 5). [4 Marks] 01 2 3 45 6
b) Consider an empty hash table of size 7. Draw the table that results after inserting, in the given order, the following key values: {19, 26, 13, 48, 17} when collisions are managed and handled by double hashing method with function: h(k, i) = [h,(k) + i * h2(k)] mod 7, where h, (k) =k mod 7 and h, (k) = 5 – (k mod 5). 0 1 2 3 4 5 6
6. Hashing Given a hash table with m = 11 entries and the following hash function h1 and step function h2: Show the status of the hash table after inserting the numbers using each method in Table 2. Table 2: The status of the hash table 0 #Linear Probing 1 2 3 4 LO 5 6 7 8 hl(key)= key mod m h2(key)= {key mod (m-1)} +1 Insert the keys {22, 1, 13, 11, 24, 33, 18, 42, 31} in the given order (from left to right) into the hash table using each of the following hash methods: 9 Linear-Probing with h(k, i) = (h1(k)+ i) mod m • Double-Hashing with h1 as the hash function and h2 as the step function h(k, i) = (h1(k) + i + h2(k)) mod m Chaining with h(k) = h1(k) 10 Double Hashing Chaining
1.Question: Consider a hash table of size m=7. Draw the table that results after inserting the following keys: 19, 26, 13, 48 and 17 for each case: a. collision resolved with chaining, with hash function h(k)=k%7 b. collision resolved by linear probing, with hash function h1(k,i)=[h(k)+i] % 7 (i=0,1,..,m-1) c. collision resolved by quadratic probing, with hash function h1(k,i)=[h(k)+i²] % 7 (i=0,1,..,m-1) d. collisions resolved by double hashing, with h2(k,i)=[h(k)+i*{5-(k%5)}]%7 (i=0,1,..,m-1)
1. A hash table with 10 buckets uses the hash function f(x) = x % 10, and linear probing to resolve collisions. If the keys [232, 227, 11, 106, 107] are inserted in that order into the hash table, what is the greatest number of steps a number can be from its natural position in the table?
8. Insert the following keys one-by-one into an initially empty hash table of size 7. Use the hash function h(x) = x mod 7 Show the result for {10, 1, 18, 15, 26, 11, 19} (a) a separate chaining hash table (do not rehash if the load factor becomes too large). (b) a table with linear probing, i.e. using the probing function f(i) = i. For parts (c) and (d), it is possible that you will not be able to insert some of the keys into the table. If you find such a key, rehash the table into a new table of size 11. You need to modify the hash function h(x) accordingly to adjust for the new table size. You do not have to change g(x) for part (d). Do not rehash earlier for any of the tables (even if the load factor becomes too large). (c) a table with quadratic probing, i.e. using the probing function f(i) = i² (d) a table that uses double hashing with a secondary hash function, i.e. f(i) = i* g(x), where the secondary hash function is g(x) = 5 - (x mod 5).

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