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Nov 16, 2017 at 22:51 history edited Seva CC BY-SA 3.0
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Jul 15, 2017 at 7:49 history edited Seva CC BY-SA 3.0
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Apr 13, 2017 at 12:58 history edited CommunityBot
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Apr 6, 2017 at 19:21 comment added Wolfgang @FedericoPoloni Yes don't worry, I agree. I just wanted to point out that it's not THE adjacency matrix. :) And to note that without the -1's, the problem might be much easier.
Apr 6, 2017 at 18:32 comment added Seva "Triangle-free" is just the term that came to my mind. Let's say, the support of $A$ is free of some triangles. Anyway, the linguistics is not that much of an issue here, the math is...
Apr 6, 2017 at 18:26 comment added Federico Poloni @Wolfgang What you write is true, but still "triangle-free" seems like the perfect name to describe this concept to me.
Apr 6, 2017 at 16:22 comment added Tony Huynh @Wolfgang Thanks you're right! Deleted my silly comment.
Apr 6, 2017 at 15:30 comment added Wolfgang @TonyHuynh The indices in $a_{ij}a_{jk}a_{ki}$ are cyclic. It is rather triangles which are not right ones - at least not with sides parallel to the axis - and only some of those. Try $\{i,j,k\}=\{1,2,3\}$.
Apr 6, 2017 at 15:21 comment added Wolfgang @FedericoPoloni If we have two matrices, one with (for a certain pair i,j) $a_{ij}=1,a_{ji}=0$ and one with $a_{ij}=0,a_{ji}=-1$, both describe the same directed graph, don't they? But in terms of ranks, they may have different ranks. I suppose this possibility is intended by the OP, otherwise it would be stated "all elements in $\{0,1\}$".
Apr 6, 2017 at 12:47 comment added Federico Poloni @AmirSagiv Because it is the adjacency matrix of a triangle-free directed graph.
Apr 6, 2017 at 11:15 comment added Amir Sagiv Just out of curiousity, why do you call it "triangle free"?
Apr 6, 2017 at 11:05 history asked Seva CC BY-SA 3.0