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Aug 28, 2022 at 3:53 vote accept John
Aug 28, 2022 at 3:52 answer added Harry timeline score: 5
Nov 23, 2018 at 22:12 comment added Mahdi - Free Palestine Yes. Indeed, A convex function (over a polytope) attains its maximum at an extreme point.
Nov 23, 2018 at 21:59 comment added John @Mahdi, did you mean to get a continuous $P$ first via optimization, then make it discrete with {0,1} elements?
Nov 23, 2018 at 21:54 comment added Mahdi - Free Palestine Yes, that was a typo. Since $\|Q_1^T P Q_2 \|_F$ is a convex function, the optimum value didn't change if we get optimum over the convex hull of all permutation matrices, which is equal to doubly stochastic matrices. I am not sure, is there any exact algorithm, for maximizing that quadratic function over some linear constraints.
Nov 23, 2018 at 21:31 comment added John @Mahdi, I think you meant $Q_{ij}=b_ja_i^T$. So, we need to maximize $\sum_{1\le i,j,\le k}tr(PQ_{ij})^2$. Could you explain a little bit more to maximize it? Here, $P$ is a permutation matrix.
Nov 22, 2018 at 18:17 comment added Mahdi - Free Palestine Also, we have $\|Q_1^T P Q_2 \|_F^2 = \sum_{1\leq i,j\leq k}(a_i^TPb_j)^2 = \sum_{1\leq i,j\leq k} \tr(PQ_{ij})^2 $, where $a_i$'s and $b_i$'s are respectively columns of $Q_1$ and $Q_2$ and $Q_{ij}=a_ib_j^T$.
Nov 22, 2018 at 18:17 comment added Mahdi - Free Palestine As a cheap observation, we can equivalently maximize $\|Q_1^T P Q_2 \|_F$ over doubly stochastic matrices.
Nov 22, 2018 at 5:57 review Suggested edits
Nov 22, 2018 at 11:22
Nov 21, 2018 at 21:28 history edited John CC BY-SA 4.0
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Nov 21, 2018 at 20:47 history asked John CC BY-SA 4.0