For the input 40, 60, 37, 83, 42, 18 and hash function h(K) = K mod 11, a. Construct both the open and closed hash tables. b. Find the largest number of key comparisons in a succes... For the input 40, 60, 37, 83, 42, 18 and hash function h(K) = K mod 11, a. Construct both the open and closed hash tables. b. Find the largest number of key comparisons in a successful search in both tables. c. Find the average number of key comparisons in a successful search in both tables. Read more

Steps to Solve

1. Identify the keys and hash function

The keys provided for the hash table construction are: 15, 28, 33, 10, 17, 3, 5. The hash function is defined as $h(K) = K \mod 11$.

2. Hash the keys into the closed hash table

Using the hash function, calculate the index for each key:

• For key 15: $h(15) = 15 \mod 11 = 4$
• For key 28: $h(28) = 28 \mod 11 = 6$
• For key 33: $h(33) = 33 \mod 11 = 0$
• For key 10: $h(10) = 10 \mod 11 = 10$
• For key 17: $h(17) = 17 \mod 11 = 5$
• For key 3: $h(3) = 3 \mod 11 = 3$
• For key 5: $h(5) = 5 \mod 11 = 5$

The closed hash table will look like this, where each bucket can only store one key:

• Index 0: 33
• Index 1:
• Index 2:
• Index 3: 3
• Index 4: 15
• Index 5: 5 (collision, will replace with 17)
• Index 6: 28
• Index 7:
• Index 8:
• Index 9:
• Index 10: 10

3. Hash the keys into the open hash table

To manage collisions, the open hash table uses a probing method:

• Index 0: 33
• Index 1:
• Index 2:
• Index 3: 3
• Index 4: 15
• Index 5: 5 (insert 17 to the next free space)
• Index 6: 28
• Index 7: (insert 17 to the next free space after 5)
• Index 8:
• Index 9:
• Index 10: 10

4. Determine searches in closed hash table

Now we need to find the largest and average number of key comparisons:

• For index 4 (key 15): 1 comparison
• For index 6 (key 28): 1 comparison
• For index 5 (key 17): 1 comparison
• For index 10 (key 10): 1 comparison
• Index 0 (key 33): 1 comparison
• Index 3 (key 3): 1 comparison
• The largest number of comparisons: 1
• Average number is: $\frac{1 + 1 + 1 + 1 + 1 + 1 + 1}{7} = 1$

5. Determine searches in open hash table

In the open hash table, we may have more comparisons due to the probing:

• Keys 15, 28, and 33: 1 comparison each
• Key 3: 1 comparison
• Key 5: 1 comparison, must check for 17 next
• For key 10: 1 comparison
• Key 17: 2 comparisons (due to probing)
• Average number of comparisons: $\frac{6 + 2}{7} = \frac{8}{7} \approx 1.14$, Largest: 2