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brownrice
brownrice
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I agree with D.W.'s comment; please include more information! I believe the algorithm you're referring to outputs $A(S)$ as a membership indicator of the smallest rectangle containing all positive samples. This would not be an arbitrary rectangle;rectangle in the case of only negative rectangles, rather $\emptyset$. So, it would always output 0 as $A(S)(x) = 1 (x \in \emptyset) = 0$ where $\emptyset \subset R$. So it wouldn't merit a special case.

I agree with D.W.'s comment; please include more information! I believe the algorithm you're referring to outputs $A(S)$ as a membership indicator of the smallest rectangle containing all positive samples. This would not be an arbitrary rectangle; it would always output 0 as $A(S)(x) = 1 (x \in \emptyset) = 0$ where $\emptyset \subset R$. So it wouldn't merit a special case.

I agree with D.W.'s comment; please include more information! I believe the algorithm you're referring to outputs $A(S)$ as a membership indicator of the smallest rectangle containing all positive samples. This would not be an arbitrary rectangle in the case of only negative rectangles, rather $\emptyset$. So, it would always output 0 as $A(S)(x) = 1 (x \in \emptyset) = 0$ where $\emptyset \subset R$. So it wouldn't merit a special case.

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brownrice
brownrice
  • 46
  • 5

I agree with D.W.'s comment; please include more information! I believe the algorithm you're referring to outputs $A(S)$ as a membership indicator of the smallest rectangle containing all positive samples. This would not be an arbitrary rectangle; it would always output 0 as $A(S)(x) = 1 (x \in \emptyset) = 0$ where $\emptyset \subset R$. So it wouldn't merit a special case.