Question

Question 13 Insert the values 15, 8, 10, 4, 2, 6,3,7,19,24, and 32 into an open addressing hash table of size 13 with hash function h(x) = x mod 13 and quadratic probing collision resolution. Show the steps taken to insert each number, including collision resolution, as well as the final table. It is not necessary to include all the intermediate tables, only the arithmetic used in insertions. B I VA-A-I EX5 11 xX, EE - ? Vx 12pt Paragraph 7

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Answer 1

13) From the question Size of the hash table - 13 Numbers need to insert 15, 8, 10,4, 2, 6, 3, 7, 19, 24, and 32, Hash function hex) = x mod 13 2015} = 15 mod 13=2 his) 8 mod 13 = 8 huol 10 mod 13 =10 h (4) - 4 mod 13 = 4 h (2) 2 mod 13 = 2 Since 2 is already occupied collision occur. Apply quadratic Probing (hck) +i) mod size . 3 il as (2+ 12) mod 13 Insert at "3" position, h (6) = 6 mod 13 = 6

h(3) = 3 mod 13 3 Collision occur. (hck) tin mod 13 22 121 =3 (3+17) mod 13 =4 4 is already occupied 1:2=) ( 3+22) mod 13 = 7 7 is not occupied insert 7th position Since 3 at h17) = 7 mod 13 OCCU, collision (h (7) +1%) mod 13 =(7+1) Mod 13 28 It is already occupied. (hc 7) + 2) mod 13 = (7+4) mod 13 = 11 II is not occupied insert at lyth position. since he19 = 19 mod 13 - 6 Occur Collision mod 13 - (4219)+") (6+1) mod 13=7 already occupied (1119)+2") mod 13 = (6+4) med 13516 already occupied

(6 (19) + 3%) mod 13 = (6+9) mod 13 - 2 already occupied. (h (19)+4%) mod 13 = (6+16) mod 13 : 9 Insert at 9th position, hc24) = 24 mod 13 = 11 already occupied. Chez4)+19) mod 13 = (1+1) mod 13 =12 Insert at 12th position. h(32) = 32 mod 13 = 6 already occupied. (2132) + 10) mod 13 = (6+1) mod 13 Ch 0322 +2”) mod 13 = (6+4) mod 13.210 (h(32) +3") mod 13 (649) mod 13 =2 (h (32) +47) mod 13 = (6+16) mod 13=9 (6+25) mod 13 = 5 Insert at 5th position, (2132)+5) mod 13

Hash Table : 15 2 3 4 5 7 2. 4 32 6 3 8 19 10 7 24 9 10 12 The collision resowed by using quadratic probing and all the elements inserted successfully.