Timeline for A specific collection of subgraphs in $K_{70, 70}$
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Sep 17, 2019 at 7:07 | comment | added | Aaron Meyerowitz | I should have said “consisting of those I” I tried a longer description. | |
| Sep 17, 2019 at 7:05 | history | edited | Aaron Meyerowitz | CC BY-SA 4.0 |
added 88 characters in body
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| Sep 17, 2019 at 5:39 | comment | added | Brendan McKay | I understand the situation, but I still think the sentence I quoted is nearly incomprehensible and I suggest you reword it. By the way, there are 47 partitions of 46 into distinct parts from [8,...,24]. | |
| Sep 17, 2019 at 4:37 | comment | added | Aaron Meyerowitz | And when this assignment is done, each vertex on that side is labelled by some partition. Likewise each vertex on the other side is labelled by some partition. Then we know which $2j$ vertices to use for the one copy of $K_{jj}.$ The result is (isomorphic to) the original graph. Perhaps it was originally presented in a nice manner that makes it clear it arises from a squared square (if it does). After extracting the partitions and arbitrarily reassigning them it is as if we acted on the graph by some $\sigma \in S_{70} \times S_{70} .$ It might be less nice, but still the same decomposition. | |
| Sep 17, 2019 at 2:21 | comment | added | Brendan McKay | "Assign each vertex the partition corresponding to the $i$ such that some $𝐾_{𝑖𝑖}$ uses that vertex." I don't understand that or why it is needed. The 70 partitions on each side can be assigned to vertices arbitrarily. | |
| Sep 16, 2019 at 20:25 | history | answered | Aaron Meyerowitz | CC BY-SA 4.0 |