HT_Rev_6

A key 48 is to be inserted into a hash table using the hash function
h(k)=k mod 10,
open addressing, and linear probing for collision handling.
Given the current state of the hash table below, where will key 48 be stored?

Question

HT_Rev_6

A key 48 is to be inserted into a hash table using the hash function 
h(k)=k mod  10, 
open addressing, and linear probing for collision handling. 
Given the current state of the hash table below, where will key 48 be stored?​

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BlackTom AI · Accepted Answer

Question restatement: A key 48 is inserted into a hash table using h(k) = k mod 10 with open addressing and linear probing. We need to determine where 48 will be stored given the current state of the hash table. Answer options: Cell 3, Cell 1, Cell 6, Cell 2.

Step-by-step reasoning for the insertion process:
- First compute the initial hash index for 48: h(48) = 48 mod 10 = 8. This tells us the first cell to check is Cell 8.
- Since the table uses linear probing for collision resolution, if Cell 8 is occupied, we examine the next cell in sequence: 9, then 0, then 1, then 2, and so on, wrapping around after Cell 9 back to Cell 0.
- The exact final resting cell for 48 depends on which cells along this probe sequence are already occupied in the current table state. Without seeing the diagram of which cells are filled, we cannot definitively state which cell among the options will be the first empty encountered.

Analysis of each option as a potential placement:
- Option: Cell 3
  - If during probing the sequence 8, 9, 0, 1, 2, 3 finds Cell 3 to be the first empty slot, then 48 would be placed there. This would require Cells 8, 9, 0, 1, and 2 to be occupied prior to insertion.
  - If any of those earlier cells (8, 9, 0, 1, or 2) are empty, the algorithm would have stopped earlier, so Cell 3 would not be chosen in that case.
- Option: Cell 1
  - For 48 to land in Cell 1, Cells 8, 9, 0 would need to be occupied (in that probing order) and Cell 1 must be the first empty encountered. If Cell 2 were empty, the algorithm would have stopped at Cell 2 instead, so this outcome requires Cells 8, 9, and 0 to be occupied while Cell 1 is the first empty after them.
- Option: Cell 6
  - Reaching Cell 6 would require that Cells 8, 9, 0, 1, 2, 3, 4, and 5 are all occupied in sequence up to Cell 5, with Cell 6 being the first empty. This is a long chain of occupied cells prior to 6.
- Option: Cell 2
  - For 48 to be placed at Cell 2, Cells 8 and 9 (and possibly 0 and 1, depending on their occupancy) must lead to Cell 2 being the first empty slot in the probing sequence. Specifically, if 8 and 9 are occupied and 0 and 1 are also occupied, then Cell 2 would be the first free cell encountered.

Important caveat: The actual determination hinges on the current contents of Cells 0 through 9 in the hash table, which are not provided in the prompt’s diagram. Therefore, while the process above shows how you would decide the placement, an exact answer requires knowing which cells along the probe path are already occupied.

Summary of the reasoning pattern:
- Compute h(48) = 8 to identify the starting probe index.
- Use linear probing to examine subsequent cells in the order 8, 9, 0, 1, 2, 3, 4, 5, 6, 7 until an empty slot is found.
- Identify which of the listed cells (3, 1, 6, 2) would be the first empty along that sequence, given the actual occupancy pattern.
- Since the occupancy info is missing, we cannot conclusively pick one option; the correct choice depends on which cells before the target are occupied.

HT_Rev_6 A key 48 is to be inserted into a hash table using the hash function h(k)=k mod 10, open addressing, and linear probing for collision handling. Given the current state of the hash table below, where will key 48 be stored?

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HT_Rev_6 A key 48 is to be inserted into a hash table using the hash function h(k)=k mod 10, open addressing, and linear probing for collision handling. Given the current state of the hash table below, where will key 48 be stored?​ ​