Question 17076 – NTA UGC NET JUNE-2023 Paper-2

To insert the sequence 1, 3, 8, 10, and 5 into the hash table using closed hashing, we’ll use the given hash function (6x + 3) mod 4. Here’s how the table would look after each insertion:

Insert 1:
Hash value = (6 * 1 + 3) mod 4 = (6 + 3) mod 4 = 9 mod 4 = 1
Table: _, 1, _, _, _, _, _

Insert 3:
Hash value = (6 * 3 + 3) mod 4 = (18 + 3) mod 4 = 21 mod 4 = 1 (collision)
Since the slot at index 1 is already occupied, we’ll perform linear probing (moving to the next slot until we find an empty one).
Table: _, 1, 3, _, _, _, _

Insert 8:
Hash value = (6 * 8 + 3) mod 4 = (48 + 3) mod 4 = 51 mod 4 = 3
Table: _, 1, 3, 8, _, _, _

Insert 10:
Hash value = (6 * 10 + 3) mod 4 = (60 + 3) mod 4 = 63 mod 4 = 3 (collision)
Linear probing again.
Table: _, 1, 3, 8, 10, _, _

Insert 5:
Hash value = (6 * 5 + 3) mod 4 = (30 + 3) mod 4 = 33 mod 4 = 1 (collision)
Linear probing.
Table: _, 1, 3, 8, 10, 5, _

So, the content of the table after inserting the sequence 1, 3, 8, 10, and 5 is:

_, 1, 3, 8, 10, 5, _