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### #ActualÁlvaro

Posted 22 February 2013 - 03:33 PM

I don't have inequalities in my solution. What I suggested disregards direction because it already assumes that the feasible region of the half spaces defines a convex polyhedron. If we'd flip one of the inequalities, the feasible region would be empty.

No, that's not how it works. Imagine you have a bunch of half-spaces that already define a convex polyhedron. Now take a plane that cuts that convex polyhedron through the middle, and either add one half-space or the other to the list. In both situations the feasible region is not empty, and your proposed solution will return the same answer.

### #2Álvaro

Posted 22 February 2013 - 03:33 PM

I don't have inequalities in my solution. What I suggested disregards direction because it already assumes that the feasible region of the half spaces defines a convex polyhedron. If we'd flip one of the inequalities, the feasible region would be empty.

No, that's not how it works. Imagine you have a bunch of half-spaces that already define a convex polyhedron. Now take a plane that cuts that convex polyhedron through the middle, and either pick one half-space or the other. In both situations the feasible region is not empty, and your proposed solution will return the same answer.

### #1Álvaro

Posted 22 February 2013 - 03:32 PM

I don't have inequalities in my solution. What I suggested disregards direction because it already assumes that the feasible region of the half spaces defines a convex polyhedron. If we'd flip one of the inequalities, the feasible region would be empty.

No, that's not how it works. Imagine you have a bunch of half-spaces that already define a convex region. Now take a plane that cuts that convex region through the middle, and either pick one half-space or the other. In both situations the feasible region is not empty, and your proposed solution will return the same answer.

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