给定下列 1-D 数据集，包含 7 个点 {-3, -2, -1, 0, 1, 2, 3} ，其中 + 和 - 是对应样本的分类，我们不能使用一个阈值对所有点进行分类（因为 + 类样本两边都被 - 类样本包围）。我们可以定义一个从 1 维到 2 维的映射，从而使这两个类变为线性可分离的。下列哪个特征映射可以实现上述目的（即，确保映射样本变为线性可分离的）？

Given the following 1-D dataset of 7 points {-3, -2, -1, 0, 1, 2, 3}, where + and - are the labels for the corresponding samples, we cannot classify all the points by using a single threshold (since the + class samples are surrounded by the - class samples from both sides). We may define a mapping from 1-D to 2-D, so as to make the two classes linearly separable. Which of the following feature mappings may achieve that goal (i.e., ensuring the mapped samples to be linearly separable)?

Question

给定下列  1-D 数据集，包含  7  个点  {-3, -2, -1, 0, 1, 2, 3} ，其中  + 和  -  是对应样本的分类，我们不能使用一个阈值对所有点进行分类（因为  +  类样本两边都被  -   类样本包围）。我们可以定义一个从  1 维 到  2 维 的映射，从而使这两个类变为线性可分离的。  下列哪个特征映射可以实现上述目的（即，确保映射样本变为线性可分离的）？

Given the following 1-D dataset of 7 points {-3, -2, -1, 0, 1, 2, 3}, where + and - are the labels for the corresponding samples, we cannot classify all the points by using a single threshold (since the + class samples are surrounded by the - class samples from both sides). We may define a mapping from 1-D to 2-D, so as to make the two classes linearly separable. Which of the following feature mappings may achieve that goal (i.e., ensuring the mapped samples to be linearly separable)?

BlackTom AI · Accepted Answer

Mapping to the second coordinate x^2 − 1 creates a parabola that pushes the outer points up and the center points down, so a single straight line (horizontal) in the 2-D plane can separate the two classes.

1) Relevant concepts
- A 1-D dataset that is not separable by a single threshold can become linearly separable after a nonlinear feature mapping into higher dimension. 
- Choosing φ(x) = (x, x^2 − 1) maps points onto a parabola; a linear separator in the 2-D feature space corresponds to a line (ax + by + c = 0).

2) Step-by-step reasoning
Assume the positive class consists of the central points (-1, 0, 1) (the usual setup for this example) and negatives are the outer points (-3, -2, 2, 3). Apply φ(x) = (x, x^2 − 1) to every input and inspect the second coordinate (the decisive one here).

3) Calculations (intermediate steps)
Compute x^2 − 1 for x ∈ {−3, −2, −1, 0, 1, 2, 3}:
- x = −3: x^2 − 1 = 9 − 1 = 8
- x = −2: 4 − 1 = 3
- x = −1: 1 − 1 = 0
- x =  0: 0 − 1 = −1
- x =  1: 1 − 1 = 0
- x =  2: 4 − 1 = 3
- x =  3: 9 − 1 = 8

So mapped points (x, x^2 − 1) are:
(-3,8), (-2,3), (-1,0), (0,-1), (1,0), (2,3), (3,8).

If positives are the central three x = −1,0,1, their y-values are {0, −1, 0}; negatives’ y-values are {3,8,3,8}. A horizontal line y = 1 (or any value between 0 and 3, e.g. y = 1.5) gives a linear separator: all positives have y ≤ 0 < 1.5 and all negatives have y ≥ 3 > 1.5. Thus a single line separates the classes in 2-D.

4) Why the correct answer is correct
φ(x) = (x, x^2 − 1) groups the central points low on the parabola and pushes outer points high. Because the separation can be done by a horizontal line (a linear classifier in 2-D), the mapping makes the classes linearly separable.

5) Why the other options are incorrect
- x -> {x, (x/2)^3}: cubic (odd) preserves sign symmetry and does not lift the outer points away from the center in a way that isolates central positives; it won’t create the clear gap needed.
- x -> {x, (x/2)^3 − (x/2)^2}: complicated polynomial but does not generate the simple “center low, edges high” pattern guaranteed by x^2 − 1; also may mix signs and fail to produce a single linear gap.
- x -> {x, sqrt(x)}: sqrt(x) is undefined for negative x (our dataset contains negatives), so this mapping is invalid for the whole dataset.

6) Conclusion
Therefore the mapping x -> {x, x^2 − 1} is the mapping that achieves linear separability; this matches the provided correct answer.

类似问题

Let [math: x=(x1,x2)]x=(x_1,x_2) and [math: y=(y1,y2)]y=(y_1,y_2) and let kernel k be defined as follows: [math: k(x,y)=ex1x2+y1y2+2x1y1x2y2+0.25x13y13]k(x,y) = e^{x_1x_2+y_1y_2} +2 \frac{x_1 y_1}{x_2y_2} + 0.25 x_1^3 y_1^3 which transformation [math: ϕ]\phi does this kernel correspond to?

Let x=(x_1,x_2) and y=(y_1,y_2) and let kernel k be defined as follows: k(x,y) = e^{x_1x_2+y_1y_2} +2 \frac{x_1 y_1}{x_2y_2} + 0.25 x_1^3 y_1^3 which transformation \phi does this kernel correspond to?

Let [math: x=(x1,x2)]x=(x_1,x_2) and [math: y=(y1,y2)]y=(y_1,y_2) and let kernel k be defined as follows: [math: k(x,y)=ex1x2+y1y2+2x1y1x2y2+0.25x13y13]k(x,y) = e^{x_1x_2+y_1y_2} +2 \frac{x_1 y_1}{x_2y_2} + 0.25 x_1^3 y_1^3 which transformation [math: ϕ]\phi does this kernel correspond to?

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

Let [math: x=(x1,x2)]x=(x_1,x_2) and [math: y=(y1,y2)]y=(y_1,y_2) and let kernel k be defined as follows: [math: k(x,y)=ex1x2+y1y2+2x1y1x2y2+0.25x13y13]k(x,y) = e^{x_1x_2+y_1y_2} +2 \frac{x_1 y_1}{x_2y_2} + 0.25 x_1^3 y_1^3 which transformation [math: ϕ]\phi does this kernel correspond to?

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