Index 1 2 3 4 Value using Quadratic Overflow is processed Index 0 1 2 3 4 Value 0 5 6 probing. 5 6 7 8 9 10 10 7 8 9
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- 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 742Draw 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.]Insert the following data into a hash table implemented using linear openaddressing. Assume that the buckets have three slots each. Make use of the hashfunction h(X) = X mod 9.{17, 09, 34, 56, 11, 71, 86, 55, 22, 10, 4, 39, 49, 52, 82, 13, 40, 31, 35, 28, 44}
- 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)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.1. There are n numbers of students in your class. Your class teacher wants to search a particular student information based on student ID. Consider the records are already arranged in ascending order. Explain the steps for the following operations using binary search. • Search for any record which is available in the list. • Search for a record which is not available in the list. 2. Use any Hashing Technique to explain the following for the above scenario. • Place all the records into the hash table. • Use any probing /chaining technique to eliminate collision. Rubrics: No. Criteria Marks Binary Search 1 Search for any available record 2 Search for any unavailable record Hashing Technique 3 Placing records into the list with proper collision avoiding 1.0 technique 3.0 1.0 Total Marks 5.0
- Given values below: 66 47 87 900 126 140 145 500 177 285 393 395 467 566 620 735 Store the values into a hash table with ten buckets, each containing three slots. If a bucket is full, use the next (sequential) bucket that contains a free slot. And Store the values into a hash table that uses the hash function key % 10 to determine into which of ten chains (separate chaining) to put the value?Using the values provided below, show the hash key and what the hash table will look like when using the two different techniques indicated. The hash function used is based upon the first letter of the person's last name. The array size is 27 Index 0 is not used The letter 'A' will go to index 1, 'B' to index 2, etc Separate chaining Open addressing with linear probing i_table = ( h(k) + j ) mod S where i is the index of the underlying array,h is the hash function,j is the iteration of the probe, andS is the size of the table Values needing to be hashed (in order from top to bottom) Big Bird The Penguin Jackie Chan The Joker The Riddler Lewis Carroll Arthur Doyle Clark Kent Clive Lewis Lois Lane Yogi Bear Charles Dickens Geoffrey Chaucer Mr Ranger William Shakespeare John Tolkien Fred RogersQuestion 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
- Five keys 8, 25, 10, 15, 18 have been added to a hash table of size 4 that uses Separate Chaining using hash function h(x) resultant hash-table. = x % 4. Show the You have to insert two more keys 43 and 92 into the hash table. The designer of the hash table decided to resize the table by doubling its size when the load factor reaches or exceeds 1.5. Draw the new hash table after the insertions. The keys 2222, 1118, 1323, 2, 343, 2133, 155 and 95 are inserted into an initially empty hash table of length 10 using open addressing with hash function h(k) = k mod 10 with linear probing. What is the resultant hash table?Consider the following cuckoo hashing with 2 tables. Both tables have asize of 11.The hashing function of the first table is:h1 (k) = kmod 11The hashing function of the second table is:ha(k) = 141 mod 11(Note: mod is the modulo operation. It returns the remainder of a division.For example, "3 mod 11 = 3" and "20 mod 11 = 9")The following sequence of insert operations are performed:• Insert 50• Insert 53• Insert 75Which of the following is TRUE after the above operations? 75 is in the first table 50 is in the second tableI NEED HELP IN THIS QUESTION OF DATA STRUCUTRE AND ALGO ASAP 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