Collision handling (open addressing) in Data Structures Theory - Mini Project: Build & Apply

Collision Handling with Open Addressing

📖 Scenario: Imagine you are designing a simple hash table to store student IDs and their names. Sometimes, two student IDs might hash to the same position. To handle this, you will use open addressing, which means if a spot is taken, you look for the next empty spot.

🎯 Goal: You will build a step-by-step example of how open addressing works to resolve collisions in a hash table.

📋 What You'll Learn

Create an initial hash table with empty slots

Set a hash function and a starting index

Implement linear probing to find the next empty slot

Insert a new student ID and name into the hash table using open addressing

💡 Why This Matters

🌍 Real World

Hash tables are used in many applications like databases, caches, and dictionaries to store and quickly find data.

💼 Career

Understanding collision handling is important for software developers and data engineers who work with efficient data storage and retrieval.

1

Create the initial hash table

Create a list called hash_table with 5 empty slots represented by None.

Data Structures Theory

# Create a list called hash_table with 5 None values
# Your code here

Hint

Use a list with 5 None elements to represent empty slots.

2

Set the hash function and starting index

Create a variable called student_id with the value 12. Then create a variable called start_index that stores the hash value of student_id modulo the length of hash_table.

Data Structures Theory

hash_table = [None, None, None, None, None]
# Create student_id = 12 and start_index using modulo
# Your code here

Hint

Use the modulo operator % to find the starting index.

3

Implement linear probing to find an empty slot

Create a variable called index and set it equal to start_index. Then create a while loop that continues as long as hash_table[index] is not None. Inside the loop, update index to (index + 1) % len(hash_table) to move to the next slot.

Data Structures Theory

hash_table = [None, None, None, None, None]
student_id = 12
start_index = student_id % len(hash_table)
# Use linear probing to find an empty slot
# Your code here

Hint

Use a while loop and update index with modulo to wrap around.

4

Insert the student ID and name into the hash table

Create a variable called student_name with the value "Alice". Then assign a tuple (student_id, student_name) to hash_table[index] to insert the student data at the found empty slot.

Data Structures Theory

hash_table = [None, None, None, None, None]
student_id = 12
start_index = student_id % len(hash_table)
index = start_index
while hash_table[index] is not None:
    index = (index + 1) % len(hash_table)
# Insert the student data at hash_table[index]
# Your code here

Hint

Store the student ID and name as a tuple in the empty slot.

Practice

1. What is the main idea behind open addressing in collision handling?

easy

A. Find another empty spot in the hash table when a collision occurs

B. Store collided elements in a separate linked list

C. Use multiple hash functions to avoid collisions

D. Resize the hash table immediately after a collision

Solution

1. Step 1: Understand collision handling methods

   Open addressing handles collisions by searching for another empty slot inside the hash table itself.

2. Step 2: Compare with other methods

   Unlike chaining (which uses linked lists), open addressing keeps all data inside the array by probing for free spots.

3. Final Answer:

   Find another empty spot in the hash table when a collision occurs -> Option A

4. Quick Check:

   Open addressing = find empty spot inside table [OK]

Hint: Open addressing means searching inside the table for free spots [OK]

Common Mistakes:

• Confusing open addressing with chaining
• Thinking collisions are ignored
• Assuming resizing happens immediately on collision

2. Which of the following is the correct syntax to start linear probing from index i in a hash table of size m?

easy

A. next_index = (i - 1) % m

B. next_index = (i * 2) % m

C. next_index = (i + 1) % m

D. next_index = (i + m) % 2

Solution

1. Step 1: Recall linear probing formula

   Linear probing moves to the next slot by adding 1 to the current index and wrapping around using modulo.

2. Step 2: Apply modulo with table size

   To stay within the table size, use (i + 1) % m, where m is the table size.

3. Final Answer:

   next_index = (i + 1) % m -> Option C

4. Quick Check:

   Linear probing = increment index by 1 modulo table size [OK]

Hint: Linear probing adds 1 and wraps with modulo table size [OK]

Common Mistakes:

• Using multiplication instead of addition
• Incorrect modulo divisor
• Subtracting instead of adding

3. Given a hash table of size 5 and a hash function h(k) = k % 5, if keys 7, 12, and 17 are inserted using linear probing, what will be the final index of key 17?

medium

A. 3

B. 2

C. 0

D. 4

Solution

1. Step 1: Calculate initial hash indices

   h(7) = 7 % 5 = 2, h(12) = 12 % 5 = 2, h(17) = 17 % 5 = 2.

2. Step 2: Insert keys using linear probing

   Insert 7 at index 2. For 12, index 2 is taken, so try 3 (free). Insert 12 at 3. For 17, indices 2 and 3 are taken, try 4 (free). Insert 17 at 4.

3. Final Answer:

   4 -> Option D

4. Quick Check:

   Linear probing moves to next free slot: 17 at index 4 [OK]

Hint: Check initial hash, then probe next slots until free [OK]

Common Mistakes:

• Assuming all keys go to same index
• Not probing after collision
• Miscomputing modulo

4. What is the main problem with this linear probing insertion code snippet?

table = [None]*5
key = 12
index = key % 5
while table[index] is not None:
    index = (index + 1) % 5
# Missing insertion step here

medium

A. It never inserts the key into the table

B. It causes an infinite loop even if the table has empty slots

C. It uses wrong modulo operation

D. It overwrites existing keys without checking

Solution

1. Step 1: Analyze the loop behavior

   The loop finds the next free slot by incrementing index until it finds None.

2. Step 2: Check for insertion after loop

   After the loop, the code does not insert the key into the found free slot, so the key is never stored.

3. Final Answer:

   It never inserts the key into the table -> Option A

4. Quick Check:

   Missing insertion step means key not stored [OK]

Hint: Always insert key after finding free slot [OK]

Common Mistakes:

• Forgetting to insert after probing
• Ignoring full table infinite loop
• Wrong modulo usage

5. In a hash table using open addressing with linear probing, what happens if the table becomes completely full and you try to insert a new key?

hard

A. The new key overwrites the oldest key in the table

B. The insertion fails because no empty slot is available

C. The table automatically resizes and rehashes all keys

D. The insertion loops infinitely without stopping

Solution

1. Step 1: Understand open addressing limits

   Open addressing requires an empty slot to insert a new key by probing.

2. Step 2: Consider full table scenario

   If the table is full, no empty slot exists, so insertion cannot proceed and must fail or trigger resizing externally.

3. Final Answer:

   The insertion fails because no empty slot is available -> Option B

4. Quick Check:

   Full table means no free slot, insertion fails [OK]

Hint: Full table means no free slot, insertion fails [OK]

Common Mistakes:

• Assuming automatic resizing always happens
• Thinking keys get overwritten silently
• Ignoring infinite loop possibility