Vector database operations (CRUD) in Prompt Engineering / GenAI - Practice Problems & Coding Challenges

Challenge - 5 Problems

🧠 Conceptual

intermediate

Understanding Vector Database Insert Operation

Which statement best describes what happens during the insert operation in a vector database?

A The vector database adds new vectors along with their metadata to the storage, making them available for future queries.

B The vector database removes vectors that match a given query from the storage permanently.

C The vector database compresses all stored vectors to save space without changing their content.

D The vector database updates the similarity scores of existing vectors without changing the stored vectors themselves.

💡 Hint

Think about what it means to add new data to a database.

Inserting means adding new vectors and their metadata into the database so they can be searched later.

❓ Predict Output

intermediate

Output of Vector Deletion Code

What will be the output of the following code snippet that deletes a vector by ID and then queries the database?

Prompt Engineering / GenAI

vectors = {'v1': [0.1, 0.2], 'v2': [0.4, 0.5]}

# Delete vector with ID 'v1'
del vectors['v1']

# Query remaining keys
print(list(vectors.keys()))

A ['v1', 'v2']

B []

C ['v2']

D KeyError

💡 Hint

Deleting a key removes it from the dictionary.

Deleting 'v1' removes it from the dictionary, so only 'v2' remains.

❓ Model Choice

advanced

Choosing the Best Model for Vector Similarity Search

You want to build a vector database that supports fast similarity search on high-dimensional image embeddings. Which model type is best suited for this task?

A A pre-trained transformer model fine-tuned for text generation

B A vector embedding model trained to produce compact, meaningful image feature vectors

C A dense neural network trained for classification tasks

D A clustering algorithm like K-means used directly as a model

💡 Hint

Think about which model creates vectors that capture image features well.

Embedding models trained on images produce vectors that represent image features well for similarity search.

❓ Hyperparameter

advanced

Effect of Changing Vector Search 'k' Parameter

In a vector database query, the parameter k controls how many nearest neighbors are returned. What happens if you increase k from 5 to 20?

A The query will fail because <code>k</code> must be less than 10.

B The query returns fewer vectors, making the search faster but less comprehensive.

C The query ignores the <code>k</code> value and returns all vectors in the database.

D The query returns more vectors, increasing the chance of finding relevant results but taking more time.

💡 Hint

Think about what increasing the number of neighbors means for results and speed.

Increasing k returns more neighbors, which can improve recall but slows down the query.

🔧 Debug

expert

Debugging Vector Update Operation

Given the following code to update a vector in a vector database, what error will it raise?

Prompt Engineering / GenAI

vectors = {'v1': [0.1, 0.2], 'v2': [0.4, 0.5]}

# Attempt to update vector 'v3'
vectors['v3'][0] = 0.9

print(vectors)

A KeyError because 'v3' does not exist in the dictionary

B IndexError because the vector has no elements

C TypeError because list elements cannot be assigned

D No error; the vector 'v3' is created and updated

💡 Hint

Consider what happens when you try to access a key that is not in a dictionary.

Accessing vectors['v3'] raises a KeyError because 'v3' is not a key in the dictionary.

Practice

1. What does the CRUD acronym stand for in vector database operations?

easy

A. Connect, Run, Undo, Deploy

B. Compute, Retrieve, Upload, Download

C. Create, Read, Update, Delete

D. Cache, Refresh, Use, Drop

Solution

1. Step 1: Understand CRUD basics

   CRUD is a common term in databases meaning the four basic operations you can do with data.

2. Step 2: Match each letter to its meaning

   C stands for Create (add new data), R for Read (get data), U for Update (change data), and D for Delete (remove data).

3. Final Answer:

   Create, Read, Update, Delete -> Option C

4. Quick Check:

   CRUD = Create, Read, Update, Delete [OK]

Hint: Remember CRUD as basic data actions: add, get, change, remove [OK]

Common Mistakes:

• Confusing CRUD with unrelated terms
• Mixing up the order of operations
• Thinking CRUD only applies to files, not vectors

2. Which of the following is the correct syntax to add a vector with ID 'vec1' and values [0.1, 0.2, 0.3] to a vector database named db?

easy

A. db.push_vector(['vec1', 0.1, 0.2, 0.3])

B. db.insert('vec1', [0.1, 0.2, 0.3])

C. db.create_vector('vec1', 0.1, 0.2, 0.3)

D. db.add_vector('vec1', [0.1, 0.2, 0.3])

Solution

1. Step 1: Identify the common method for adding vectors

   Most vector databases use a method like add_vector with an ID and a list of numbers.

2. Step 2: Check method parameters

   The method should take the vector ID as a string and the vector values as a list or array.

3. Final Answer:

   db.add_vector('vec1', [0.1, 0.2, 0.3]) -> Option D

4. Quick Check:

   Add vector syntax = db.add_vector(id, vector) [OK]

Hint: Add vectors with add_vector(id, vector_list) method [OK]

Common Mistakes:

• Using wrong method names like insert or push_vector
• Passing vector values as separate arguments instead of a list
• Mixing ID and vector in one list

3. Given the following code snippet, what will be the output?

db = VectorDB()
db.add_vector('v1', [1, 0, 0])
db.add_vector('v2', [0, 1, 0])
results = db.search([0.9, 0.1, 0], top_k=1)
print(results)

medium

A. [('v1', 0.9)]

B. [('v2', 0.9)]

C. [('v1', 0.1)]

D. [('v2', 0.1)]

Solution

1. Step 1: Understand the vectors and query

   Vectors 'v1' = [1,0,0], 'v2' = [0,1,0], query = [0.9,0.1,0].

2. Step 2: Calculate similarity or distance

   Assuming cosine similarity, 'v1' is closer to query (dot product ~0.9), 'v2' is less similar (~0.1).

3. Final Answer:

   [('v1', 0.9)] -> Option A

4. Quick Check:

   Closest vector = v1 with similarity 0.9 [OK]

Hint: Closest vector has highest dot product with query [OK]

Common Mistakes:

• Confusing similarity with distance
• Mixing up vector IDs in output
• Assuming lower score means closer

4. The following code tries to update a vector but throws an error. What is the likely cause?

db = VectorDB()
db.add_vector('v1', [0.5, 0.5, 0.5])
db.update_vector('v2', [0.1, 0.1, 0.1])

medium

A. Vector 'v2' does not exist, so update fails

B. The update_vector method requires 4 arguments

C. Vector values must be integers, not floats

D. The add_vector method was not called before update_vector

Solution

1. Step 1: Check vector existence before update

   Updating a vector requires it to exist in the database first.

2. Step 2: Identify the error cause

   Since 'v2' was never added, trying to update it causes an error.

3. Final Answer:

   Vector 'v2' does not exist, so update fails -> Option A

4. Quick Check:

   Update needs existing vector [OK]

Hint: Update only existing vectors, else error occurs [OK]

Common Mistakes:

• Assuming update creates new vectors
• Thinking data type mismatch causes error
• Ignoring vector existence check

5. You want to delete vectors with similarity less than 0.5 to a query vector [0, 1, 0] from your vector database. Which sequence of operations correctly achieves this?

hard

A. Delete all vectors, then add only those with similarity >= 0.5

B. Search vectors with similarity < 0.5, then delete each by ID

C. Update vectors with similarity < 0.5 to zero vectors

D. Add new vectors with similarity >= 0.5, ignoring deletion

Solution

1. Step 1: Find vectors below similarity threshold

   Use a search or filter operation to get IDs of vectors with similarity less than 0.5.

2. Step 2: Delete vectors by their IDs

   Use the delete operation on each vector ID found to remove them from the database.

3. Final Answer:

   Search vectors with similarity < 0.5, then delete each by ID -> Option B

4. Quick Check:

   Filter then delete unwanted vectors [OK]

Hint: Filter vectors first, then delete by ID [OK]

Common Mistakes:

• Deleting all vectors instead of selective ones
• Trying to update instead of delete
• Ignoring the similarity filter step