Skip to main content
14 events
when toggle format what by license comment
Mar 20, 2014 at 22:57 vote accept Felix Goldberg
Oct 25, 2013 at 0:28 history edited Brendan McKay CC BY-SA 3.0
fix typo (missing brace)
Oct 24, 2013 at 23:03 history edited leonid gurvits CC BY-SA 3.0
new inequalities added
Oct 24, 2013 at 19:31 comment added leonid gurvits I forgot to mention new, very cool and accurate lower bound: let $RI(k,n$ be the set of integer $n \times n$ matrices with row/column sums equal $k$ $n(l)$ be the number of entries in $A \in RI(k,n)$ equal $l$, $1 \leq l \leq k$. Then $Per(A) \geq
Oct 20, 2013 at 1:08 comment added leonid gurvits The rate of convergence: $1 - \frac{a_n}{n} \leq O(n^{-1} \log(n))$ with some universal easily computable constant.
Oct 19, 2013 at 22:52 comment added Suvrit You're welcome! Also, if you click on the 'edit' link towards the bottom left of your original answer, you can see the markup corresponding to the edited answer.
Oct 19, 2013 at 22:50 history edited Suvrit CC BY-SA 3.0
Incorporated Leonid's comments into his answer.
Oct 19, 2013 at 21:47 comment added leonid gurvits Here $Prod_A$ is the product polynomial $\prod_{1 \leq i \leq n}\sum_{1 \leq j \leq n} A(i,j) x_j$; $G(k) = (\frac{k-1}{k})^{k-1}, ECCC paper: eccc.hpi-web.de/report/2013/141; and the original paper: arxiv.org/abs/0711.3496 . Thanks for the editing!
Oct 19, 2013 at 21:31 comment added Suvrit Dear Leonid, we (j.c. and i) made edits to this post; hope, you are ok with them.
Oct 19, 2013 at 21:30 history edited Suvrit CC BY-SA 3.0
fixed some formatting...
Oct 19, 2013 at 21:28 history edited j.c. CC BY-SA 3.0
latexify
S Oct 19, 2013 at 21:13 review Late answers
Oct 19, 2013 at 21:28
S Oct 19, 2013 at 21:13 review First posts
Oct 19, 2013 at 21:35
Oct 19, 2013 at 20:55 history answered leonid gurvits CC BY-SA 3.0