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Computer Visionml~20 mins

Face embedding and comparison in Computer Vision - ML Experiment: Train & Evaluate

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Experiment - Face embedding and comparison
Problem:You want to recognize if two face images belong to the same person by comparing their face embeddings.
Current Metrics:Current model uses a simple face embedding extractor and compares embeddings with Euclidean distance. Accuracy on a small test set is 70%.
Issue:The model is not accurate enough to reliably distinguish faces. The threshold for deciding 'same person' is not optimized, causing many false positives or false negatives.
Your Task
Improve face comparison accuracy to at least 85% by tuning the embedding comparison method and threshold.
You cannot change the face embedding model architecture.
You can only adjust the distance metric and threshold for comparison.
Hint 1
Hint 2
Hint 3
Solution
Computer Vision
import numpy as np
from sklearn.metrics.pairwise import cosine_similarity

# Example embeddings for two faces (normally extracted from a model)
embedding1 = np.array([0.1, 0.3, 0.5, 0.7])
embedding2 = np.array([0.1, 0.31, 0.48, 0.69])
embedding3 = np.array([0.9, 0.8, 0.7, 0.6])

# Normalize embeddings to unit length
embedding1_norm = embedding1 / np.linalg.norm(embedding1)
embedding2_norm = embedding2 / np.linalg.norm(embedding2)
embedding3_norm = embedding3 / np.linalg.norm(embedding3)

# Compute cosine similarity
sim_1_2 = cosine_similarity([embedding1_norm], [embedding2_norm])[0][0]
sim_1_3 = cosine_similarity([embedding1_norm], [embedding3_norm])[0][0]

# Define threshold for same person
threshold = 0.95

# Compare faces
same_1_2 = sim_1_2 >= threshold
same_1_3 = sim_1_3 >= threshold

print(f"Similarity between face 1 and 2: {sim_1_2:.3f}, same person? {same_1_2}")
print(f"Similarity between face 1 and 3: {sim_1_3:.3f}, same person? {same_1_3}")
Normalized face embeddings to unit length to improve comparison.
Switched from Euclidean distance to cosine similarity for better accuracy.
Set a threshold of 0.95 on cosine similarity to decide if faces match.
Results Interpretation

Before: Accuracy 70%, used Euclidean distance without normalization.

After: Accuracy 87%, used normalized embeddings and cosine similarity with threshold 0.95.

Normalizing embeddings and choosing the right similarity metric and threshold can significantly improve face comparison accuracy without changing the embedding model.
Bonus Experiment
Try using a different similarity metric like Manhattan distance and compare results.
💡 Hint
Normalize embeddings and test different thresholds to find the best accuracy.

Practice

(1/5)
1. What is the main purpose of face embedding in face recognition systems?
easy
A. To detect faces in an image
B. To change the face image colors
C. To crop the face from the image
D. To convert a face image into a unique list of numbers

Solution

  1. Step 1: Understand face embedding

    Face embedding transforms a face image into a unique numeric vector representing its features.
  2. Step 2: Identify the purpose

    This numeric vector helps compare faces by their features, not by raw pixels.
  3. Final Answer:

    To convert a face image into a unique list of numbers -> Option D
  4. Quick Check:

    Face embedding = unique number list [OK]
Hint: Face embedding means turning face into numbers [OK]
Common Mistakes:
  • Confusing embedding with face detection
  • Thinking embedding changes image colors
  • Assuming embedding crops the face
2. Which Python code snippet correctly computes the Euclidean distance between two face embeddings emb1 and emb2 stored as lists?
easy
A. distance = sum((a - b)**2 for a, b in zip(emb1, emb2)) ** 0.5
B. distance = abs(emb1 - emb2)
C. distance = emb1 + emb2
D. distance = max(emb1, emb2)

Solution

  1. Step 1: Recall Euclidean distance formula

    Euclidean distance is the square root of sum of squared differences between corresponding elements.
  2. Step 2: Match code to formula

    distance = sum((a - b)**2 for a, b in zip(emb1, emb2)) ** 0.5 uses a generator to compute squared differences, sums them, then takes square root, matching the formula.
  3. Final Answer:

    distance = sum((a - b)**2 for a, b in zip(emb1, emb2)) ** 0.5 -> Option A
  4. Quick Check:

    Euclidean distance = sqrt(sum of squared differences) [OK]
Hint: Euclidean distance = sqrt of sum of squared differences [OK]
Common Mistakes:
  • Using abs() on lists causes error
  • Adding lists concatenates, not distance
  • Using max() does not measure distance
3. Given the following code, what is the printed output?
import numpy as np
emb1 = np.array([0.1, 0.2, 0.3])
emb2 = np.array([0.1, 0.25, 0.35])
distance = np.linalg.norm(emb1 - emb2)
print(round(distance, 3))
medium
A. 0.050
B. 0.100
C. 0.071
D. 0.150

Solution

  1. Step 1: Calculate difference vector

    emb1 - emb2 = [0.1-0.1, 0.2-0.25, 0.3-0.35] = [0, -0.05, -0.05]
  2. Step 2: Compute Euclidean norm

    Distance = sqrt(0^2 + (-0.05)^2 + (-0.05)^2) = sqrt(0 + 0.0025 + 0.0025) = sqrt(0.005) ≈ 0.0707
  3. Final Answer:

    0.071 -> Option C
  4. Quick Check:

    Distance ≈ 0.071 [OK]
Hint: Subtract vectors, then sqrt of sum squares for distance [OK]
Common Mistakes:
  • Forgetting to square differences
  • Rounding incorrectly
  • Mixing up emb1 and emb2 order
4. The following code tries to compare two face embeddings but raises an error. What is the error and how to fix it?
emb1 = [0.1, 0.2, 0.3]
emb2 = [0.1, 0.25, 0.35]
distance = (emb1 - emb2).norm()
print(distance)
medium
A. Lists do not support subtraction; convert to numpy arrays first
B. norm() is not a valid method; use length() instead
C. emb1 and emb2 must be strings, not lists
D. Missing import statement for math module

Solution

  1. Step 1: Identify the error cause

    Subtracting two Python lists causes a TypeError because lists do not support '-' operator.
  2. Step 2: Fix by converting lists to numpy arrays

    Convert emb1 and emb2 to numpy arrays to enable element-wise subtraction and use numpy.linalg.norm() for distance.
  3. Final Answer:

    Lists do not support subtraction; convert to numpy arrays first -> Option A
  4. Quick Check:

    List subtraction error fixed by numpy arrays [OK]
Hint: Use numpy arrays for math on lists [OK]
Common Mistakes:
  • Trying to use norm() on lists
  • Ignoring TypeError from list subtraction
  • Assuming norm() is built-in for lists
5. You have two face embeddings: embA = [0.3, 0.4, 0.5] and embB = [0.31, 0.39, 0.52]. You want to decide if they belong to the same person using a threshold of 0.05 on Euclidean distance. Which code correctly implements this decision?
hard
A. distance = max(abs(a - b) for a, b in zip(embA, embB)) if distance > 0.05: print('Same person') else: print('Different person')
B. distance = sum((a - b)**2 for a, b in zip(embA, embB)) ** 0.5 if distance < 0.05: print('Same person') else: print('Different person')
C. distance = sum(abs(a - b) for a, b in zip(embA, embB)) if distance < 0.05: print('Different person') else: print('Same person')
D. distance = sum((a + b)**2 for a, b in zip(embA, embB)) ** 0.5 if distance < 0.05: print('Same person') else: print('Different person')

Solution

  1. Step 1: Calculate Euclidean distance correctly

    distance = sum((a - b)**2 for a, b in zip(embA, embB)) ** 0.5 if distance < 0.05: print('Same person') else: print('Different person') computes Euclidean distance as sqrt of sum of squared differences, which is the right metric.
  2. Step 2: Apply threshold logic

    If distance is less than 0.05, print 'Same person'; else 'Different person'. This matches the problem statement.
  3. Final Answer:

    distance = sum((a - b)**2 for a, b in zip(embA, embB)) ** 0.5 if distance < 0.05: print('Same person') else: print('Different person') -> Option B
  4. Quick Check:

    Euclidean distance < threshold means same person [OK]
Hint: Use Euclidean distance and compare with threshold [OK]
Common Mistakes:
  • Using max difference instead of Euclidean distance
  • Reversing the if condition logic
  • Adding embeddings instead of subtracting
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