$x$ is a binary vector which means the elements in $x$ are 0 or 1, and $p$ is another vector with the same length. How to express the product of elements in $p$ whose corresponding elements are 1 in $x$ and transform the product into quadratic form? An additional condition is $x^Tx=3$. For example, $x = (1,0,0,1,0,1)$ and $p = (p_1,p_2,p_3,p_4,p_5,p_6)$. How to express $p_1p_4p_6$ with $x$ and $p$, and how to transform the equation into quadratic form like $x^TQx$?
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1$\begingroup$ It doesn't seem likely that this is possible. For fixed $p$, if you want to find a $Q$ such that $x^T Q x = p_ip_jp_k$ whenever $x$ is the vector with $1$'s at $i,j,k$, you're looking at $\binom{n}{3}$ linear equations constraining the $n^2$ many entries of $Q$. And they look completely independent. I bet once $n\geq 9$ (where $\binom{n}{3}>n^2$) there is no such $Q$ for generic $p$ anymore. $\endgroup$– Achim KrauseApr 14 at 13:23
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1$\begingroup$ Would $e^{x\cdot\ln(p)}$ do? $\endgroup$– LSpiceApr 14 at 13:39
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