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

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We are given a dataset with X = [-π, -0.5π, 0, 0.5π, π] and labels y = [1, -1, -1, -1, 1]. We must evaluate which feature mappings φ(x) make the data linearly separable in the transformed feature space. Below, I analyze each option and explain why it would or would not yield linear separability.

Option A: φ(x) = (x, cos(x))
- Inspect the transformed points: the first coordinate is x, which takes values [-π, -0.5π, 0, 0.5π, π], and the second coordinate is cos(x), which yields [cos(-π)=-1, cos(-0.5π)=0, cos(0)=1, cos(0.5π)=0, cos(π)=-1]. So the transformed set is {(-π, -1), (-0.5π, 0), (0, 1), (0.5π, 0), (π, -1)} with labels {+1, -1, -1, -1, +1} respectively.
- A linear separator in 2D is a line ax + by + c = 0 that assigns +1 to some points and -1 to others. Visually, the two positive points lie at the far left and far right, both with second coordinates -1, while three negative points lie near the center with second coordinates 0 or 1. It is plausible that a line can separate the outer positives from the inner negatives by choosing a slope that weights x more heavily and uses the -1 at the ends to cross the line above the center points. The key is that cos(x) provides a non-monotonic, symmetric feature that creates a distinctive pattern across the x-values, enabling a separating line to exist.
- Conclusion: It is feasible for a linear classifier in (x, cos(x)) space to separate the end-point positives from the center negatives, making A a valid choice.

Option B: φ(x) = (x, sin(x))
- Here the second feature is sin(x), which takes values sin(-π)=0, sin(-0.5π)=-1, sin(0)=0, sin(0.5π)=1, sin(π)=0. So the transformed points are {(-π, 0), (-0.5π, -1), (0, 0), (0.5π, 1), (π, 0)} with labels {+1, -1, -1, -1, +1}.
- The center negatives are located at (-0.5π, -1), (0,0), and (0.5π, 1). The positives are at the far left and far right with y-values 0. It is unlikely that a single straight line can cleanly separate the two end-point positives from the central cluster of negatives, because the line would need to discriminate based on a combination of x and sin(x) that places the middle negatives on one side and both ends on the other, while keeping the two end points with 0 as positives. The scatter in the second feature (especially the negative at x = -0.5π with y = -1 and the positive at x = 0.5π with y = +1) creates a non-separable configuration for a single linear boundary.
- Conclusion: B is unlikely to yield linear separability in this feature space.

Option C: φ(x) = (x, cos(0.5x))
- Now the second feature is cos(0.5x). Compute 0.5x for each x: x ∈ { -π, -0.5π, 0, 0.5π, π } gives 0.5x ∈ { -0.5π, -0.25π, 0, 0.25π, 0.5π }. Then cos(0.5x) values are {cos(-0.5π)=0, cos(-0.25π)≈0.707, cos(0)=1, cos(0.25π)≈0.707, cos(0.5π)=0}. So the transformed points are {(-π, 0), (-0.5π, 0.707), (0, 1), (0.5π, 0.707), (π, 0)} with labels {+1, -1, -1, -1, +1}.
- This arrangement places the positives at the far left and far right with a second feature near 0, while the negatives cluster toward the middle with higher second-feature values around 0.707 to 1. A straight line can be drawn roughly across the bottom of the central cluster, separating the outer two positives from the interior negatives. The symmetry and the monotonic decline away from the center in the second feature create a linearly separable pattern in this 2D space.
- Conclusion: C is a viable choice for linear separability in the transformed space.

Option D: φ(x) = (x, sin(0.5x))
- For sin(0.5x), using the same 0.5x values as in C, we get sin(-0.5π)= -1, sin(-0.25π)≈ -0.707, sin(0)=0, sin(0.25π)≈ 0.707, sin(0.5π)=1. The transformed points are {(-π, -1), (-0.5π, -0.707), (0, 0), (0.5π, 0.707), (π, 1)} with labels {+1, -1, -1, -1, +1}.
- In this configuration, the positive points lie at the far left with y = -1 and far right with y = +1, while the negative points span a rising sequence from -0.707 to 0.707 to 1 as x moves from left to right. A single linear decision boundary would have to separate these two outer positives from the central negatives, but the central negatives straddle a broad range of the sin(0.5x) values, making a clean linear separation unlikely. This arrangement tends to produce a non-separable pattern for a single line in (x, sin(0.5x)) space.
- Conclusion: D is unlikely to yield linear separability.

Summary of options: A and C provide transformed representations in which a single linear boundary can plausibly separate the classes, whereas B and D create configurations that are less amenable to a single hyperplane separation in the respective 2D spaces. Therefore, the correct transformations are A and C.

COMP9417-Machine Learning & Data Mining - T3 2025

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