VC联盟维度?

发布于 2024-10-10 17:14:02 字数 81 浏览 4 评论 0原文

假设我有两个概念类:C1 和 C2。 假设C1的VC维为d,C2的VC维为d。

C1 和 C2 并集的 VC 维数的最大值是多少?

Suppose I have two concept classes: C1 and C2.
Suppose that the VC Dimension of C1 is d and the VC Dimension of C2 is d.

What is the biggest value of the VC dimension of the union of C1 and C2?

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西瓜 2024-10-17 17:14:02

请参阅 Eisenstat 和 Angluin 的论文“The VC Dimension of k-fold union”,他们在其中表明 VC 维度随着 Theta(klogk) 渐近增加。

StompChicken 的答案不可能是正确的,因为它意味着 k 重并集的 VC 维数是 O(k)。我相信他对 d_1+d_2 的下限的论证是正确的。

See Eisenstat and Angluin's paper, "The VC Dimension of k-fold union", where they show that the VC dimension increases asymptotically as Theta(klogk).

The StompChicken answer can not be correct, because it implies the VC dimension of k-fold union is O(k). I believe he argues correctly for a lower bound of d_1+d_2.

嗫嚅 2024-10-17 17:14:02

下面我假设您并不是要指定 C1 和 C2 具有相同的 VC 尺寸 d,而是指定不同的 VC 尺寸 d1 和 d2。我还将假设(不失一般性)d1 >= d2。

这取决于“C1 和 C2 的并集”的含义。由C1和C2并集形成的概念类的VC维具有VC维d1。这非常简单,因为要粉碎 d1 或更少的点,只需使用 C1 中的东西即可。然而,根据定义,C1 或 C2 都不会破坏 d1 + 1 点。

编辑 - 下段中的论点是错误的,请参阅HRJ对所谓“k-fold union”真实故事的回答。

因为这很无聊,也许你的意思是概念类
你可以根据一个元素的并集形成一个假设
C1 和 C2 的一个元素。这里的 VC 维数是 d1 + d2。查看
这样,将任何 d1+d2 点划分为两个子集并将它们粉碎
分别包含 C1 和 C2 中的元素。这样做的一个后果是
对于 2D 线性分离器,VC 维度将是
3+3=6 你可以从以下事实看出这一点:
明显的六边形标记不能被两条线打碎。

不同意 HRJ 的观点,我认为这甚至不是工会的正确下限。例如,让 X = {x1,x2,x3,x4}C = {{x1,x3},{x2,x4}} 那么 C 可以粉碎任何大小为 1 的子集,但不是,例如 {x1,x2},因此 C 的 VC 维度为 1。但是,C 的 2 重并集是 C^ 2={{x1,x3},{x2,x4},{x1,x2,x3,x4}} 仍然是 VC 维度 1。进一步的并集最终会得到同样的结果。所以,我认为 k 重并集的下界是 d。话又说回来,我可能是错的。

In the following I am going to assume you didn't mean to specify that the C1 and C2 have the same VC dimension d, but rather different VC dimensions d1 and d2. I'm also going to assume (without loss of generality) that d1 >= d2.

It depends what you mean by 'the union of C1 and C2'. The VC dimension of the concept class formed by the union of C1 and C2 has VC dimension d1. This is pretty straightforward, since to shatter d1 or fewer points, just use something from C1. However, neither C1 or C2 will shatter d1 + 1 points, pretty much by definition.

EDIT - The argument in the following paragraph is wrong, see HRJ's answer for the real story of what is apparently called the 'k-fold union'.

Since that's quite boring, maybe you meant the the concept class where
you are allowed to form a hypothesis from the union of one element of
C1 and one element of C2. The VC dimension here is d1 + d2. To see
this, partition any d1+d2 points into two subsets and shatter them
with elements in C1 and C2 respectively. A consequence of this is
that, for linear separators in 2D, the VC dimension is going to be
3+3=6 and you can see this from the fact that there is a fairly
obvious labelling of a hexagon that cannot be shattered by two lines.

To disagree with HRJ I don't think it's even a correct lower bound of the union. For example, let X = {x1,x2,x3,x4} and C = {{x1,x3},{x2,x4}} then C can shatter any subset of size 1 but not, for example {x1,x2} so C has VC dimension 1. But, the 2-fold union of C is C^2={{x1,x3},{x2,x4},{x1,x2,x3,x4}} which is still of VC dimension 1. Further unions are just going to end up with the same thing. So, I think the lower bound of the k-fold union is d. Then again, I could be wrong.

野稚 2024-10-17 17:14:02

如果 VC(H_1)=d_1 且 VC(H_2)=d_2 且 d=max(d_1,d_2),则并集 VC 上的一般界限为 2d+1。请参阅附图,因为我找不到插入乳胶的方法。
VC(H_1 \CUP H_2)\le 2d+1

If VC(H_1)=d_1 and VC(H_2)=d_2 and d=max(d_1,d_2) the general bound on the VC of the union is 2d+1. See attached img since i could not find a way to insert latex.
VC(H_1 \CUP H_2)\le 2d+1

浪荡不羁 2024-10-17 17:14:02

如果 VCdim(A) = d_A 且 VCdim(B) = d_B,则 A 和 B 的并集(我们称之为并集 C)的 VCdim 至多为 d_A + d_B + 1。 这是我的证明:

在此处输入图像描述

If VCdim(A) = d_A and VCdim(B) = d_B, VCdim of union of A and B (We call the union C) is at most d_A + d_B + 1. Here is my proof:

enter image description here

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