Well, something that comes to mind is this... Suppose your facet-defining hyperplanes have equations $v^i \cdot x = a_i$ and you're working in $n$ dimensions. Suppose you have the additional information that $v^i \cdot x < a_i$ inside $P$ P$. (Hopefully this is reasonable for you). Which way the inequality goes can be determined by testing the vertices of $P$). Then a vertex of the polytope defined by your hyperplanes is a point where $v^i \cdot x = a_i$ for $n$ independent $v^i$ and where $v^i \cdot x \leq a_i$ for the other $v^i$. You can find each such vertex by solving some linear equations and then checking some inequalities hold at the solution. If the resulting set of vertices obtained is exactly the vertex set of $P$, you're in business. Otherwise, you're not.
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Well, something that comes to mind is this... Suppose your facet-defining hyperplanes have equations $v^i \cdot x = a_i$ and you're working in $n$ dimensions. Suppose you have the additional information that $v^i \cdot x < a_i$ inside $P$ (Hopefully this is reasonable for you). Then a vertex of the polytope defined by your hyperplanes is a point where $v^i \cdot x = a_i$ for $n$ independent $v^i$ and where $v^i \cdot x \leq a_i$ for the other $v^i$. You can find each such vertex by solving some linear equations and then checking some inequalities hold at the solution. If the resulting set of vertices obtained is exactly the vertex set of $P$, you're in business. Otherwise, you're not. |
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