Open avagreyyy4 opened 2 weeks ago
I am not sure I am right The first, I think, yes. Assume d = 2. If we want the first quadrant to be negative, the rest positive, we can use (-1,-1) as weights for input (x1,x2). But it's also strange to me that if a point is in the second quadrant, and really close to the y axis, say (-1,6), the sign would be negative (-1 -1 + -1 6 = -5) whereas if the point in the second axis is close to x axis (-10, 1), the sign would be positive (-1 -10 + -1 1) . So my guess is that multiaxis2 could ensure that one specific axis is classified to a certain sign, but it can't guarantee the rest would be classified to an opposite sign.
The second, d = 2, the weight could be (0,0), which is the extra hypothesis besides axis2 and multiaxis2.
These are all good questions, but I think they're complicated enough I'd have to go over them on a whiteboard in person and not just in the comments section here, sorry.
Hi! I was wondering if anybody had gone through the dvc in 2D of the finite hypothesis classes we went over. Specifically, for multiaxis2 and multiaxis3. I understand for multiaxis2 it is quadrants, so one positive quadrant and then the rest are negative. I was wondering if we are able to flip, like have one negative quadrant and 3 positive ones? Similarly, I may be confused on multiaxis3's size. We have M = 3^d, so = 9 in 2D space, but since it's the union of axis2 and mutliaxis2 that adds up to 8. What is that additional hypothesis?
Thanks!