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CS 559 – Quiz 3: Linear Classification exam with questions and verified answers graded A+ new!!

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CS 559 – Quiz 3: Linear Classification exam with questions and verified answers graded A+ new!!

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CS 559 – Quiz 3: Linear Classification
exam with questions and verified
answers graded A+ new!!

SECTION A: FOUNDATIONS & DECISION BOUNDARIES (Q1–Q20)



Q1. A linear classifier assigns a class label based on:



A) A non-linear combination of input features

B) A linear combination of input features compared to a threshold

C) The Euclidean distance to every training point

D) A polynomial kernel of degree ≥ 2



Answer B

Rationale: A linear classifier computes w·x + b and assigns a class based on the sign (or comparison
to a threshold) of this linear combination.



Q2. The decision boundary of a linear classifier in ℝ² is:



A) A parabola

B) A circle

C) A straight line

D) An ellipse



Answer: C

Rationale: In 2D, the equation w₁x₁ + w₂x₂ + b = 0 defines a straight line that separates the two
classes.



Q3. In d-dimensional feature space, the decision boundary of a linear classifier is a:

,A) (d+1)-dimensional surface

B) (d−1)-dimensional hyperplane

C) d-dimensional hyperplane

D) (d−2)-dimensional manifold



Answer: B

Rationale: A hyperplane in ℝᵈ has dimensionality d−1. For example, in ℝ³ the boundary is a 2D
plane.



Q4. The weight vector w in a linear classifier w·x + b = 0 is:



A) Parallel to the decision boundary

B) Perpendicular (normal) to the decision boundary

C) Tangent to the decision boundary

D) Unrelated to the decision boundary orientation



Answer: B

Rationale: The weight vector w is the normal to the hyperplane w·x + b = 0, determining its
orientation.



Q5. The bias term b in a linear classifier affects:



A) The orientation of the decision boundary

B) The position (offset) of the decision boundary from the origin

C) The dimensionality of the feature space

D) The number of support vectors



Answer: B

Rationale: The bias b shifts the hyperplane toward or away from the origin without changing its
orientation.



Q6. If a dataset in ℝ² is linearly separable, this means:

,A) No classifier can separate the classes

B) A straight line can perfectly separate the two classes

C) Only a neural network can classify it

D) The classes overlap completely



Answer: B

Rationale: Linear separability means there exists a hyperplane (line in 2D) that separates all positive
examples from all negative examples without error.



Q7. Which of the following datasets is NOT linearly separable?



A) AND gate truth table

B) OR gate truth table

C) XOR gate truth table

D) NAND gate truth table



Answer: C

Rationale: XOR is the classic example of a non-linearly-separable problem. No single line can
separate the outputs.



Q8. The signed distance from a point x₀ to the decision boundary w·x + b = 0 is:



A) (w·x₀ + b) / ||w||

B) (w·x₀ + b) × ||w||

C) ||x₀|| / ||w||

D) w·x₀



Answer: A

Rationale: The signed distance from a point to a hyperplane is given by (w·x₀ + b) / ||w||, a standard
result from analytic geometry.

, Q9. A linear classifier can represent which type of decision region?



A) Convex half-spaces

B) Arbitrary non-convex regions

C) Circular regions

D) Disconnected regions



Answer: A

Rationale: A single linear classifier divides the space into two half-spaces, both of which are convex.



Q10. Increasing the magnitude of w while keeping the direction constant:



A) Rotates the decision boundary

B) Does not change the decision boundary (if b is scaled proportionally)

C) Moves the boundary farther from the origin

D) Makes the model non-linear



Answer: B

Rationale: The hyperplane w·x + b = 0 and cw·x + cb = 0 (c > 0) define the same set of points, hence
the same decision boundary.



Q11. To use a linear classifier for a problem with non-linear boundaries, one common approach is:



A) Remove features

B) Use feature mapping (basis function expansion) to a higher-dimensional space

C) Reduce the learning rate to zero

D) Increase the bias term



Answer: B

Rationale: Mapping inputs to a higher-dimensional space via basis functions (e.g., polynomial
features) can make a non-linearly-separable problem become linearly separable.

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