In a double hashing scheme, \(β_1 (π) = π \ mod \ 11\) and \(β_2 (π) = 1 +β¦
2025
In a double hashing scheme,Β \(β_1 (π) = π \ mod \ 11\) andΒ \(β_2 (π) = 1 + (π \ mod \ 7)\) are the auxiliary hash functions. The sizeΒ \(m\) of the hash table is 11. The hash function for the \(i\)-th probe in the open address table is \([β_1 (π) + π \ β_2(π)] \ mod \ π\). The following keys are inserted in the given order: 63, 50, 25, 79, 67, 24.
The slot at which key 24 gets stored is ___________. (Answer in integer)
Attempted by 33 students.
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Correct answer: 10
Insert the keys in the given order and compute h1 and h2 for each to find the probe sequence.
Key 63: h1 = 63 mod 11 = 8; h2 = 1 + (63 mod 7) = 1. Probe i=0 -> slot (8 + 0*1) mod 11 = 8. Place 63 at slot 8.
Key 50: h1 = 50 mod 11 = 6; h2 = 1 + (50 mod 7) = 2. Probe i=0 -> slot 6. Place 50 at slot 6.
Key 25: h1 = 25 mod 11 = 3; h2 = 1 + (25 mod 7) = 5. Probe i=0 -> slot 3. Place 25 at slot 3.
Key 79: h1 = 79 mod 11 = 2; h2 = 1 + (79 mod 7) = 3. Probe i=0 -> slot 2. Place 79 at slot 2.
Key 67: h1 = 67 mod 11 = 1; h2 = 1 + (67 mod 7) = 5. Probe i=0 -> slot 1. Place 67 at slot 1.
Key 24: h1 = 24 mod 11 = 2; h2 = 1 + (24 mod 7) = 4.
Probe i=0 -> slot (2 + 0*4) mod 11 = 2 (occupied by 79).
Probe i=1 -> slot (2 + 1*4) mod 11 = 6 (occupied by 50).
Probe i=2 -> slot (2 + 2*4) mod 11 = (2 + 8) mod 11 = 10 (free). Place 24 at slot 10.
Final answer: 10 (slot 10)
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