Timeline for Topology of directed graph $G$ with non-singular adjacency matrix
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 8 at 17:16 | history | edited | KhashF | CC BY-SA 4.0 |
Finalized the solution
|
| Apr 8 at 16:04 | vote | accept | ABB | ||
| Apr 8 at 8:13 | comment | added | ABB | After finding the set of disjoint cycles containing all nodes of $G$, the desired subgraph $H$ is formed by the union of these disjoint cycles. | |
| Apr 8 at 8:05 | comment | added | ABB | Firstly, thank you very much for your insightful arguments and amendments. However, I believe the procedure is not yet complete. Both algorithms indicate that for a given node, one can definitely find a cycle containing it. I'm struggling to understand how one can arrange the set of disjoint cycles so that all nodes can feasibly be included in one of them. | |
| Apr 8 at 0:57 | history | edited | KhashF | CC BY-SA 4.0 |
added 73 characters in body
|
| Apr 7 at 22:55 | history | edited | KhashF | CC BY-SA 4.0 |
Updated by adding a stronger result
|
| Apr 7 at 22:30 | comment | added | KhashF | @ABB The way your question reads, it says subgraph. Do you want the subgraph to include all vertices of $G$? | |
| Apr 7 at 21:00 | comment | added | ABB | If I understand correctly, you are trying to find a disjoint union of cycles containing all nodes of $g$. If so, how can we form other cycles? | |
| Apr 7 at 20:32 | history | answered | KhashF | CC BY-SA 4.0 |