Timeline for Difference between a sub graph and induced sub graph.
Current License: CC BY-SA 3.0
20 events
| when toggle format | what | by | license | comment | |
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| Oct 23, 2018 at 16:52 | review | Suggested edits | |||
| Oct 23, 2018 at 17:04 | |||||
| S Dec 19, 2015 at 16:26 | history | suggested | Jan Hrcek | CC BY-SA 3.0 |
Fix typo and improve wording
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| Dec 19, 2015 at 16:14 | review | Suggested edits | |||
| S Dec 19, 2015 at 16:26 | |||||
| Nov 9, 2014 at 13:43 | comment | added | patang | ah ..seems to be clear now...thanks. | |
| Nov 9, 2014 at 13:38 | comment | added | Peter | Probably, it is meant that BOTH endpoints must belong to $U$, then everything is all right. Otherwise, the formulation would probably be : with at least one endpoint in $U$. | |
| Nov 9, 2014 at 13:37 | comment | added | patang | yes I meant that like in this case ,the edge maynot belong in $H$ but has an endpoint in it... | |
| Nov 9, 2014 at 13:35 | comment | added | Peter | Ah, now I see what you mean. The edge $2-4$ has an endpoint in $U$, is this the problem ? | |
| Nov 9, 2014 at 13:33 | comment | added | Peter | Perhaps, an example helps : Suppose, the edges are $1-2,1-3,2-4,2-5,3-5$ and $U=${$1,2,3$}. The edges between the vertices in $U$ are $1-2,1-3$, so $H$ contains exactly these edges. The edges $2-4,2-5,3-5$ do not occur in $H$ because the vertices $4,5$ do not belong to $U$. | |
| Nov 9, 2014 at 13:27 | comment | added | patang | the definition of your answer is understandable ,but I can't understand the one given in the question....what's relation between both of them...please help it seems to be confusing.. | |
| Nov 9, 2014 at 13:17 | comment | added | Peter | You mean the vertices ? They do not need to be adjacent in $G$, but if they are not, they are also not adjacent in $H$. To make it simpler : The edges with endpoints in $U$ are the edges remaining, if all vertices but those in $U$ are deleted. | |
| Nov 9, 2014 at 13:13 | comment | added | patang | but as it says:"If F consists of all edges of G which have endpoints in U" ,where does it says that that edges are adjacent.. | |
| Nov 9, 2014 at 13:09 | comment | added | Peter | No, $U$ is the (in general proper) subset of $V$ and the vertex set in $H$ is only $U$, not $V$. | |
| Nov 9, 2014 at 13:06 | comment | added | patang | in definition of induced subgraph as given in question it is:"If $F$ consists of all edges of $G$ which have endpoints in $U$ ,then $H$ is called induced subgraph of $G$ and is denoted by $G_U$." .. shouldn't it be $V$ insetad of $U$ ... | |
| Nov 9, 2014 at 12:23 | comment | added | Peter | The edges of the REMAINING vertices are not allowed to be deleted, but deleting a vertex, of course, deletes the edges ending in this vertex. | |
| Nov 9, 2014 at 12:20 | comment | added | Peter | The edges incident with a vertex are simply the edges having the vertex as an endpoint. | |
| Nov 9, 2014 at 12:20 | comment | added | patang | @but I don't understand the meaning of incident edges in: "(and with them all the incident edges)but no more edges"....Please help.. | |
| Nov 9, 2014 at 12:18 | comment | added | Peter | No, the vertices are deleted, the vertex-set of H is, in general, a proper subset of V. | |
| Nov 9, 2014 at 12:14 | comment | added | patang | but shouldn't the last line be :"an induced subgraph cannot be constructed by deleting vertices" | |
| Nov 9, 2014 at 12:07 | vote | accept | patang | ||
| Nov 9, 2014 at 12:03 | history | answered | Peter | CC BY-SA 3.0 |