Questions

CMPSC 132 Spring 2025 Module 6.2 Checkpoint

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Assume you have a hash table of size 11. What is the final status of the hash table after inserting 10, 22, 31, 4, 15, 28, 17, 88, 59 (in that order) using the open addressing techniques linear probing with probe +1, quadratic probing, and double hashing to resolve collision? For double hashing, H2(x) = 7- (x % 7). If the bucket remains empty after all insertions, select None Linear Probing +1 0 [ Select ] 59 88 28 31 15 4 None 22 10 17 1 [ Select ] 10 28 17 22 15 None 4 88 59 31 2 [ Select ] 59 4 17 15 22 None 10 88 28 31 3 [ Select ] 88 10 4 15 22 28 31 17 None 59 4 [ Select ] 17 31 22 59 15 4 88 28 10 None 5 [ Select ] 17 59 88 4 22 10 15 31 None 28 6 [ Select ] 28 None 88 22 17 31 59 4 15 10 7 [ Select ] 31 10 None 88 17 22 28 59 15 4 8 [ Select ] 88 22 None 31 4 10 59 17 15 28 9 [ Select ] 4 10 59 17 31 28 15 22 88 None 10 [ Select ] 88 None 17 10 28 31 4 59 22 15 Quadratic Probing 0 [ Select ] 4 17 88 None 10 15 31 22 28 1 [ Select ] 59 31 4 22 15 28 10 88 None 17 2 [ Select ] 15 22 17 59 28 88 31 10 4 None 3 [ Select ] 17 None 15 31 10 88 59 4 22 28 4 [ Select ] None 17 28 59 88 4 10 31 22 15 5 [ Select ] 28 4 22 59 88 31 17 15 None 10 6 [ Select ] 15 22 4 None 17 10 28 59 31 88 7 [ Select ] 10 28 17 4 88 22 None 15 31 59 8 [ Select ] 15 59 17 4 22 88 10 None 31 9 [ Select ] None 59 22 28 10 4 31 15 17 88 10 [ Select ] 31 4 15 59 28 88 22 17 10 None Double Hashing 0 [ Select ] 22 31 15 28 4 10 17 None 59 88 1 [ Select ] 10 22 17 None 28 88 59 15 31 4 2 [ Select ] 31 None 28 22 59 88 10 17 15 4 3 [ Select ] 4 28 10 22 88 17 31 None 59 15 4 [ Select ] 4 None 15 28 17 88 22 59 31 10 5 [ Select ] 4 10 15 17 88 31 59 None 22 28 6 [ Select ] 10 15 31 88 59 None 22 17 28 4 7 [ Select ] 59 28 17 88 31 15 22 None 4 10 8 [ Select ] 4 10 31 88 None 15 17 59 28 22 9 [ Select ] 15 10 17 31 None 88 22 28 59 4 10 [ Select ] 88 17 59 22 4 31 15 28 None 10

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Step-by-Step Analysis

We will analyze how each collision-resolution method handles the same sequence of insertions and then describe the final table state for each method. First, compute the primary hash h1(x) = x mod 11 for all values to know the initial target slots: - 10 -> 10 - 22 -> 0 - 31 -> 9 - 4 -> 4 - 15 -> 4 - 28 -> 6 - 17 -> 6 - 88 -> 0 - 59 -> 4 Thus, the initial slots attempted before probing are: 10, 0, 9, 4, 4, 6, 6, 0, 4. Linear Probing (probe +1): - Insert 10 at index 10 (empty) -> slot10 = 10. - Insert 22 at index 0 (empty) -> slot0 = 22. - Insert 31 at index 9 (empty) -> slot9 = 31. - Insert 4 at index 4 (empty) -> slot4 = 4. - Insert 15 at index 4 (occupied) -> probe to 5 (empty) -> slot5 = 15. - Insert 28 at index 6 (empty) -> slot6 = 28. - Insert 17 at index 6 (occupie......Login to view full explanation

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