Consider the following dataset: [math: X=[−π,−0.5π,0,0.5π,π]]X=[-\pi,-0.5\pi,0,0.5\pi,\pi] with corresponding labels [math: y=[1,−1,−1,−1,1]]y=[1,-1,-1,-1,1]. Which of the following transformations would make the data linearly separable? A. [math: ϕ(x)=(x,cos(x))]\phi(x)=(x,cos( x)) B. [math: ϕ(x)=(x,sin(x))]\phi(x)=(x,sin(x)) C. [math: ϕ(x)=(x,cos(0.5x))]\phi(x)=(x,cos(0.5 x)) D. [math: ϕ(x)=(x,sin(0.5x))]\phi(x)=(x,sin(0.5 x))

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We’re given a small 1D dataset with X = [−π, −0.5π, 0, 0.5π, π] and labels y = [1, −1, −1, −1, 1]. The question asks which feature mappings φ(x) would make the data linearly separable in the new feature space.

Option A: φ(x) = (x, cos(x))
- The original x coordinate preserves order, and adding cos(x) introduces a smooth, bounded nonlinear dimension that correlates with the sign pattern of y across the symmetric range. Since cos(x) is even and x is odd, the combination can create a decision boundary that splits the endpoints (where y flips sign) more easily than using x alone. This pairing often helps separate the two end points (−π and π) with a linear threshold in the (x, cos(x)) plane, making the data linearly separable in that feature space.
- Intuition: cos(x) captures the periodic variation around 0, providing a second axis that can align a single linear separator to separate the positive labels from the negative ones given the symmetry of the labels around 0.

Option B: φ(x) = (x, sin(x))
- Here we add sin(x), which is an odd function, sharing the same odd symmetry as x itself. The combination (x, sin(x)) may not provide enough independent variation to separate the two outer positive points (−π and π) from the inner negative cluster (−0.5π to 0.5π) with a single linear decision boundary. Since both coordinates vary similarly with x and both are near antisymmetric, the resulting 2D plot could still require a nonlinear boundary to separate the labeled groups, leaving linear separability unlikely.

Option C: φ(x) = (x, cos(0.5x))
- This mapping uses cos(0.5x), i.e., a cosine wave with longer period than cos(x). It adds a nonlinear feature that can rotate and bend the decision boundary in a way that aligns with the label pattern across the five points. The combination of x and cos(0.5x) can yield a line in the (x, cos(0.5x)) space that separates the two end points from the middle cluster, making the data linearly separable in this transformed space.
- The key is that cos(0.5x) introduces a different curvature profile than cos(x), which can produce a more favorable linear separator given the specific spacing of the provided x-values and their labels.

Option D: φ(x) = (x, sin(0.5x))
- Similar reasoning to option C, but with sin(0.5x) instead of cos. While sin(0.5x) is a valid nonlinear feature, its phase relative to x may produce a decision boundary that still fails to separate the endpoints cleanly for this particular dataset. Depending on the exact arrangement of the five points, the (x, sin(0.5x)) pair might not yield a line that cleanly separates y = 1 at both ends from the middle y = −1 values, so linear separability is not guaranteed.

Putting it together, each option’s viability hinges on whether the 2D feature space spanned by the chosen φ(x) allows a straight line to split the positive and negative labels. A and C provide feature pairs that can yield such a separating line given the sample positions and label pattern, whereas B and D do not consistently produce a linearly separable arrangement for this exact dataset.

COMP9417-Machine Learning & Data Mining - T3 2025

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