1 / 9

Hash Table (Separate Chaining)

0. 1. 2. 3. NULL. 2. 5. Hash Table (Separate Chaining). Assume we have 3 buckets, and the hash function is f(x) = (x % 3), sorting key is the integer itself After inserting 2,3,5, the hash table will become. 0. 1. 2. 0. 3. 4. 2. 5. Hash Table (Separate Chaining).

rogan-humphrey
Télécharger la présentation

Hash Table (Separate Chaining)

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 0 1 2 3 NULL 2 5 Hash Table (Separate Chaining) • Assume we have 3 buckets, and the hash function is f(x) = (x % 3), sorting key is the integer itself • After inserting 2,3,5, the hash table will become

  2. 0 1 2 0 3 4 2 5 Hash Table (Separate Chaining) • If we further insert 0 and 4, then the hash table become

  3. Student Table • Assume we have 3 buckets, and the hash function is the sum of the digits of StudentID and then mod 3 • Also, the sorting key is the alphabetic order of the StudentID

  4. 0 1 2 [“03123123”, “John”, Year 2, Male] [“03123456”, “Tom”, Year 1, Male] NULL [“03000002”, “Mary”, Year 3, Female] Student Table • Inserting the following records • [“03123456”, “Tom”, Year 1, Male] • [“03000002”, “Mary”, Year 3, Female] • [“03123123”, “John”, Year 2, Male]

  5. Root [“03000000”, “COMP104”, 30] [“03123455”, “COMP151”, 50] [“03123455”, “COMP171”, 10] Registration Table • Sorting key is firstly the StudentID, and then secondly the CourseCode • Inserting the following records: • [“03123455”, “COMP171”, 10] • [“03123455”, “COMP151”, 50] • [“03000000”, “COMP104”, 30]

  6. 0 1 2 [“03000000”, pointer = 0x12345678] NULL [“03123455”, pointer = 0x111222AA] [“03123455”, pointer = 0x00001234] 0x12345678 0x00001234 0x111222AA Root [“03000000”, “COMP104”, 30] [“03123455”, “COMP151”, 50] [“03123455”, “COMP171”, 10] Index (key: StudentID) • The address are shown above the nodes (Same key, either one first)

  7. 0 1 2 [“COMP171”, pointer = 0x111222AA] [“COMP151”, pointer = 0x00001234] [“COMP104”, pointer = 0x12345678] 0x12345678 0x00001234 0x111222AA Root [“03000000”, “COMP104”, 30] [“03123455”, “COMP151”, 50] [“03123455”, “COMP171”, 10] Index (key: CourseCode) • Assume the hash function is the sum of the three digits in the course code mod 3

  8. Example • Assume there are 3 student [ID] (A,B,C) and 3 courses [code] (X,Y,Z) • Now there are the following registrations: • Student A takes course X • Student B takes course Y • Student C takes course X • Student A takes course Z • Assume the hash value of students (2 buckets) are • Student A = 0 • Student B = 1 • Student C = 0 • Assume the hash value of courses (2 buckets) are • Course X = 0 • Course Y = 1 • Course Z = 1 • Further assume the linked list is a singly-linked list (for illustration only)

  9. Course Selection Linked List 0x12345678 0x12341234 0x111222AA 0x00001234 [A, X, 10] [A, Z, 40] [B, Y, 20] [C, X, 30] Root Student Index Hash Table [C, pointer= 0x00001234] [A, pointer= 0x12345678] 0 [A, pointer= 0x12341234] 1 [B, pointer= 0x111222AA] Course Index Hash Table [X, pointer= 0x12345678] 0 [X, pointer= 0x00001234] 1 [Z, pointer= 0x12341234] [Y, pointer= 0x111222AA] • Insertion (add registration): Example (Cont’) • Student A takes course X • Student B takes course Y • Student C takes course X • Student A takes course Z [Same sorting key, any ordering] [Same sorting key, any ordering]

More Related