The absolute value of the product of x1 and x2 (the two axis) is upper-bounded by c = 1. If you pick any two numbers and the absolute value of their product happens to be <= 1 it will be a red dot, everything else will be blue.
I would add that you can remove the ones using only one variable.
It also can't be x1² + x2² because it would be an ellipse.
Then you need to choose between |x1x2|, 1/|x1 + x2| and log2(x1+x2).
Let's take 2 points on the decision boundary. They should have the same value.
(1,1) = 1 for |x1x2| and 1/2 for 1/|x1 + x2| and log2(2)
(2,0.5) = 1 for |x1x2| and 1/2.5 for 1/|x1 + x2| and log2(2.5)
Since only |x1x2| keeps the same value, it must be it
Exam 2016 Question 9
How to see that the answer makes sense? I have absolutely no idea why it makes sense I just don’t see it.
The absolute value of the product of x1 and x2 (the two axis) is upper-bounded by c = 1. If you pick any two numbers and the absolute value of their product happens to be <= 1 it will be a red dot, everything else will be blue.
3
I would add that you can remove the ones using only one variable.
It also can't be x1² + x2² because it would be an ellipse.
Then you need to choose between |x1x2|, 1/|x1 + x2| and log2(x1+x2).
Let's take 2 points on the decision boundary. They should have the same value.
(1,1) = 1 for |x1x2| and 1/2 for 1/|x1 + x2| and log2(2)
(2,0.5) = 1 for |x1x2| and 1/2.5 for 1/|x1 + x2| and log2(2.5)
Since only |x1x2| keeps the same value, it must be it
3
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