4. a) Insert the following keys sequentially into an extensible hash structure using the Hash Function h(key)=(sum of the digits of the key) mod 8. Assume each bucket can hold at most 2 records. Search Keys: 44, 98, 79, 57, 69, 85, 39 b) Demonstrate two scenarios when we should use Ordered (B-tree) and Unordered (Extensible Hashing) indexing

4. a) Insert the following keys sequentially into an extensible hash structure using the Hash Function h(key)=(sum of the digits of the key) mod 8. Assume each bucket can hold at most 2 records. Search Keys: 44, 98, 79, 57, 69, 85, 39 b) Demonstrate two scenarios when we should use Ordered (B-tree) and Unordered (Extensible Hashing) indexing

Computer Networking: A Top-Down Approach (7th Edition)

7th Edition

ISBN:9780133594140

Author:James Kurose, Keith Ross

Publisher:James Kurose, Keith Ross

Chapter1: Computer Networks And The Internet

Section: Chapter Questions

Problem R1RQ: What is the difference between a host and an end system? List several different types of end...

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2. Give the use of a hash function, Draw the hash table that results using the hash function: hik)-k mod 13 to hash the keys 2. 7.4, 41, 15, 32. 25, 11, 30. Assuming collisions are handled by Quadratic probing. 3. Draw the hash table that results using the hash function: h(k)-k mod 13 to hash the keys 2,7,4, 41, 15, 32, 25. 11, 30. Assuming collisions are handled by Linear probing 4. Draw the hash table that results using the hash function: h(k)-kmod7 to hash the keys 76. 93. 40, 47, 10. 55. Assuming collisions are handied by Double hashing.
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…
CSC 472 Introduction to Database Systems Project Consider the hashing approach for computing aggregations. If the size of the hash table is too large to fit in memory, then the DBMS has to spill it to disk. During the Partition phase, a hash function h1 is used to split tuples into partitions on disk based on targethash key. During the ReHash phase, the DBMS can store pairs of the form (GroupByKey -> RunningValue) to computethe aggregation. Which of the following is FALSE ? Group of answer choices The Partition phase will put all tuples that match (using h1) into the same partition. To insert a new tuple into the hash table, a new (GroupByKey -> RunningValue) pair is inserted if it finds a matching GroupByKey. A second hash function (e.g., h2) is used in the ReHash phase. The RunningValue could be updated during the ReHash phase.
Insert the following sequence of keys in the hash table.Keys = {4, 2, 1, 3, 5, 6, 8}Use linear probing technique for collision resolution: h(k, i) = [h(k) + i] mod mHash function, h(k) = 3k + 1Table size, M = 10 Draw the hash table, compute the index where the hashvalue will be inserted for each key, and show how you would use the technique when a collisionoccurs.
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valsTable: o Empty-since-start 1 Empty-after-removal Occupied Hash table valsTable uses double probing with the hash functions hash1(key): key % 11 hash2(key): 5 - key % 5 and a table size of 11. 15 5 27 6. 7 HashInsert(valsTable, item 70) inserts item 70 into bucket Ex: 10 8 9 Hashlnsert(valsTable, item 62) inserts item 62 into bucket 10 Hashlnsert(valsTable, item 59) inserts item 59 into bucket 2. 3.
2. Consider the Linear Hashing index. Linear Hashing | h1 000 001 010 011 ho ठ 00 01 10 11 Mod 8 (%68) Mod 4 (%4) 16* 21* 40* 37* 14* 18* 6* Primary Page 13* A. Show the resulting index after inserting the entry 9* C. Show the resulting index after inserting entry 7* 31* 19* 11* 23* B. Show the resulting index after inserting the entry 3* 30* D. Show the resulting index after inserting the entry 22* Overflow Page
vals Table: o 1 2 3 4 5 6 7 8 9 10 30 21 Empty-since-start Empty-after-removal Occupied Hash table valsTable uses double probing with the hash functions hash1(key): key % 11 hash2(key): 5-key % 5 and a table size of 11. HashInsert(vals Table, item 63) inserts item 63 into bucket Ex: 10 HashInsert(vals Table, item 62) inserts item 62 into bucket HashInsert(vals Table, item 32) inserts item 32 into bucket
1) Suppose hashing is being done by using open addressing with linear probing. The hash function tobe used for an element with key k is h(k) = k mod 11. Show the contents of the hash table afterinserting the following keys in order into the hash table:27 40 22 15 31 18 36 19 20
(e) Assume a hash table with 6 buckets is initially empty. What are the contents of the hash table after insertion of the numbers 1, 18, 11, 4, 14 using the hash function h(k) = k%6?
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