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Let $S_n$ be the interior of the unitary $n$-simplex, i.e $ S_n =\{{\bf x} \in \mathbb{R}^n \mid x_i\ge0 \wedge \sum_{i=1}^n x_i\le1\}$

Let $T_n({\bf y})$ be the reversed simplex with origin at ${\bf y}$, ie

$T_n({\bf y}) = \{{\bf x} \in \mathbb{R}^n \mid y_i-x_i\ge 0 \wedge \sum_{i=1}^n y_i-x_i\le1\}$

I want to compute the intersection volume $V({\bf y})=S_n \cap T_n({{\bf y}})$ for given ${\bf y}$ (with $y_i\ge 0$).

Clearly, $V({\bf y})=0$ if $\sum_{i=1}^n y_i>2$. Also, $V({\bf y})=\prod_{i=1}^n y_i$ if $\sum_{i=1}^n y_i\le1$ (the intersection is a rectangular parallelepiped). But in the range $1<\sum_{i=1}^n y_i\le2$ it gets more difficult.

Given the symmetry of the problem, we could asssume WLOG $y_1 \le y_2\le \cdots \le y_n$

leonbloy
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  • Won't $V(y)$ be another simplex? (My powers of imagination are limited to the plane :-).) – copper.hat Feb 26 '16 at 17:27
  • I don't understand your notation. It seems that $S_n(\mathbf x)$ doesn't depend on $\mathbf x$, but rather $\mathbf x$ is a variable point in $S_n$? But then how is $S_n({{\bf x}})\cap S_n({{\bf y-x}})$ defined? – joriki Feb 26 '16 at 17:34
  • I assumed he meant $\Sigma + {y}$ or similar, but may be way off... – copper.hat Feb 26 '16 at 17:35
  • @joriki, Yes, that was bad notation. Fixed now - feel free to improve it – leonbloy Feb 26 '16 at 17:51
  • Very interesting question. The elementary cases (1D triangle, and 2D, pyramid) invite to think to a piecewise linear solution in nD, but, in order to give a more effective answer, I would turn to barycentrical coordinates. – Jean Marie Feb 26 '16 at 18:40

1 Answers1

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I am restricted now to $1\leq\sum_{i=1}^n y_i<2$

First case: triangles in $\mathbb{R}^2$. Suppose that $y_i\leq 1$ for $i=1,2$. Then: \begin{equation} \begin{split} A(S_n\cap T_n(y))&=\frac{1}{2}-\frac{1}{2}(1-y_1)^2-\frac{1}{2}(1-y_2)^2-\frac{1}{2}(y_1+y_2-1)^2=\\ &=-1+y_1+y_2-y_1^2-y_2^2-y_1y_2. \end{split} \end{equation}

Now WLOG $y_1\leq 1$ and $y_2>1$. Then the intersection is a parallelogram and the answer is: \begin{equation} A(S_n\cap T_n(y))=y_1\cdot(2-y_1-y_2). \end{equation}

The $2$-dimensional case is completely solved.

General case: symplexes in $\mathbb{R}^n$. If for any $(n-1)$-uple $\{y_{i_1},\ldots, y_{i_{n-1}}\}$ we have that $\sum_{j=1}^{n-1}y_{i_j}\leq 1$ then:

\begin{equation} \begin{split} Vol(S_n\cap T_n(y))&=\frac{1}{n!}-\frac{1}{n!}\sum_{i=1}^n(1-y_i)^n-\frac{1}{n!}(\sum_{i=1}^n y_i-1)^n=\\ &=\frac{1}{n!}\left(1-\sum_{i=1}^n \sum_{k=0}^n\binom{n}{k}y_i^k(-1)^{n-k}-\sum_{k=0}^n\binom{n}{k}\left(\sum_{i=1}^n y_i\right)^k(-1)^{n-k}\right)=\\ &=\frac{1}{n!}\left(1- \sum_{k=0}^n\binom{n}{k}\left(\sum_{i=1}^n y_i^k\right)(-1)^{n-k}-\sum_{k=0}^n\binom{n}{k}\left(\sum_{i=1}^n y_i\right)^k(-1)^{n-k}\right)=\\ &=\frac{1}{n!}\left(1- \sum_{k=0}^n\binom{n}{k}\left(\sum_{i=1}^n y_i^k+(\sum_{i=1}^n y_i)^k\right)(-1)^{n-k}\right). \end{split} \end{equation}

Suppose there exists a $(n-1)$-uple $\{y_{i_1},\ldots, y_{i_{n-1}}\}$ such that $\sum_{j=1}^{n-1}y_{i_j}>1$ Like you noticed we can WLOG assume now that $y_1\leq y_2\leq\ldots\leq y_n$.

Therefore we have two cases: either $y_1\in\{y_{i_1},\ldots, y_{i_{n-1}}\}$ or $y_1\not\in\{y_{i_1},\ldots, y_{i_{n-1}}\}$.

Thanks to the order and to the fact that we are dealing with positive numbers we can assume that $every$ $(n-1)$-uple is such that $\sum_{j=1}^n y_{i_j}>1$.

In the second case you can assume $\sum_{i=2}^n y_i>1$ and $y_1+\sum_{j=1}^{n-2} y_{i_j}>1$ for any $(n-2)$-uple $\{y_{i_1},\ldots,y_{i_{n-2}}\}$.

Note that for the limit case $\sum y_i=1$ your result and this general case coincides.

NOTE: As far as I understand at the moment there is not a smart way of computing such a volume. I am afraid one is forced to integrate. Still not sure. I neem to think more.

Diesirae92
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  • For formula for $n=2$ does not look right. For example, for $y_i=0$ it should give 0 and it gives $-1$. Perhaps you are considering only the case where the excluded volumen $T_n \setminus S_n$ is a tetrahedron? – leonbloy Mar 05 '16 at 03:45
  • I restricted myself to the case you didn't solved. However you are right, the answer is not completely correct. Now I will be more precise. – Diesirae92 Mar 05 '16 at 10:09
  • Even after the editions, the answer looks unclear (where does the first general formula comes from?) and probably wrong (I still suspect that this is computing the volume only for the case where the excluded volume is a tetrahedron - this is ok for $n=2$ but not necessarily for $n >2$) – leonbloy Mar 05 '16 at 14:55
  • That's what I am exactly saying in the answer above. And I point out that if for any projection on any $(n-1)$ coordinate plane, you still get that the excluded volume is a union of tetraedrons. In what follows I moreover explain that you have only 2 cases left. Probably, in the second case, you get again a parallelepiped which volume is easy to compute. However I think that you cannot get a closed formula for every case. – Diesirae92 Mar 05 '16 at 15:06
  • In the first case for $n=2$, it should be $-1 + 2(y_1+y_2)-(y_1^2+y_1 y_2 +y_2^2)$, shouldn't it? – Hans Lundmark Mar 05 '16 at 15:33