Is “induced subgraph” necessary to define an $H$-free graph?
$begingroup$
I am looking at what it means for a graph to be $H$-free, if we let $H$ be some arbitrary graph. So far it seems to me that a graph $G$ is $H$-free if it does not contain $H$ as an induced subgraph.
Is the induced condition necessary? It doesn't seem to me like this is the case, since an induced subgraph is also a subgraph. Why do we not just say that $G$ is $H$-free if it does not contain $H$ as a subgraph?
I see that having an induced subgraph is a stronger condition and is more in-line with what we might expect out of subgraphs (at least to me, the notion of a subgraph not necessarily inheriting the edge connections of the original graph is a bit strange), but is this condition so much better that we need to specify "induced"?
graph-theory forbidden-subgraphs
asked Jan 23 at 5:15
RichardRichard
497
$endgroup$
add a comment |
$begingroup$
I am looking at what it means for a graph to be $H$-free, if we let $H$ be some arbitrary graph. So far it seems to me that a graph $G$ is $H$-free if it does not contain $H$ as an induced subgraph.
Is the induced condition necessary? It doesn't seem to me like this is the case, since an induced subgraph is also a subgraph. Why do we not just say that $G$ is $H$-free if it does not contain $H$ as a subgraph?
I see that having an induced subgraph is a stronger condition and is more in-line with what we might expect out of subgraphs (at least to me, the notion of a subgraph not necessarily inheriting the edge connections of the original graph is a bit strange), but is this condition so much better that we need to specify "induced"?
graph-theory forbidden-subgraphs
asked Jan 23 at 5:15
RichardRichard
497
$endgroup$
add a comment |
$begingroup$
I am looking at what it means for a graph to be $H$-free, if we let $H$ be some arbitrary graph. So far it seems to me that a graph $G$ is $H$-free if it does not contain $H$ as an induced subgraph.
Is the induced condition necessary? It doesn't seem to me like this is the case, since an induced subgraph is also a subgraph. Why do we not just say that $G$ is $H$-free if it does not contain $H$ as a subgraph?
I see that having an induced subgraph is a stronger condition and is more in-line with what we might expect out of subgraphs (at least to me, the notion of a subgraph not necessarily inheriting the edge connections of the original graph is a bit strange), but is this condition so much better that we need to specify "induced"?
graph-theory forbidden-subgraphs
asked Jan 23 at 5:15
RichardRichard
497
$endgroup$
I am looking at what it means for a graph to be $H$-free, if we let $H$ be some arbitrary graph. So far it seems to me that a graph $G$ is $H$-free if it does not contain $H$ as an induced subgraph.
Is the induced condition necessary? It doesn't seem to me like this is the case, since an induced subgraph is also a subgraph. Why do we not just say that $G$ is $H$-free if it does not contain $H$ as a subgraph?
I see that having an induced subgraph is a stronger condition and is more in-line with what we might expect out of subgraphs (at least to me, the notion of a subgraph not necessarily inheriting the edge connections of the original graph is a bit strange), but is this condition so much better that we need to specify "induced"?
graph-theory forbidden-subgraphs
graph-theory forbidden-subgraphs
asked Jan 23 at 5:15
RichardRichard
497
asked Jan 23 at 5:15
RichardRichard
497
asked Jan 23 at 5:15
RichardRichard
497
asked Jan 23 at 5:15
RichardRichard
497
asked Jan 23 at 5:15
RichardRichard
497
497
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Why do we not just say that $G$ is $H$-free if it does not contain $H$ as a subgraph?
In my experience, this is generally what being $H$-free means. If we want to say that $G$ does not contain $H$ as an induced subgraph, we say that $G$ is induced $H$-free.
Certainly, discussion of $H$-free graphs is very common when discussing the forbidden subgraph problem. This problem is generally only considered with the ordinary notion of subgraphs, not the induced kind. (If you're asking for the largest number of edges in an $n$-vertex induced $H$-free graph, then unless $H$ is a clique, you will get a boring answer: the best graph is $K_n$ with $binom n2$ edges. This is why we don't talk about induced subgraphs for this problem.)
Both notions are useful, and there is terminology for both. Sometimes people are lazy about this terminology, and don't specify that $G$ is induced $H$-free when they should. Sometimes, it's hard to tell which convention is being used, and sometimes it doesn't matter. (For example, if we're counting $C_5$'s in a triangle-free graph, the answer doesn't change if we add the word "induced" in both places.)
answered Jan 23 at 5:37
Misha LavrovMisha Lavrov
47k657107
$endgroup$
$begingroup$
Thanks for the reply. I am a little confused on your example with induced subgraphs in the forbidden subgraph problem. I understand that induced subgraphs contain edges between vertices iff the original graph contains those same edges. Is the point of your example just to drive home this iff condition? It seems to me that if $H$ is a clique we still have the same upper bound on edges.
$endgroup$
– Richard
Jan 24 at 0:07
$begingroup$
If $H$ is a clique, then $K_n$ does contain an induced copy of $H$, and so we have to do actual work to determine the largest graph that doesn't. My point is that if $H$ is missing any edges, $K_n$ does not have any induced copies of $H$, so it is trivially the largest graph. This is not an interesting result.
$endgroup$
– Misha Lavrov
Jan 24 at 0:13
$begingroup$
On the other hand, the problem of the largest (non-induced) $H$-free graph is interesting for every $H$.
$endgroup$
– Misha Lavrov
Jan 24 at 0:13
add a comment |
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$begingroup$
Why do we not just say that $G$ is $H$-free if it does not contain $H$ as a subgraph?
In my experience, this is generally what being $H$-free means. If we want to say that $G$ does not contain $H$ as an induced subgraph, we say that $G$ is induced $H$-free.
Certainly, discussion of $H$-free graphs is very common when discussing the forbidden subgraph problem. This problem is generally only considered with the ordinary notion of subgraphs, not the induced kind. (If you're asking for the largest number of edges in an $n$-vertex induced $H$-free graph, then unless $H$ is a clique, you will get a boring answer: the best graph is $K_n$ with $binom n2$ edges. This is why we don't talk about induced subgraphs for this problem.)
Both notions are useful, and there is terminology for both. Sometimes people are lazy about this terminology, and don't specify that $G$ is induced $H$-free when they should. Sometimes, it's hard to tell which convention is being used, and sometimes it doesn't matter. (For example, if we're counting $C_5$'s in a triangle-free graph, the answer doesn't change if we add the word "induced" in both places.)
answered Jan 23 at 5:37
Misha LavrovMisha Lavrov
47k657107
$endgroup$
$begingroup$
Thanks for the reply. I am a little confused on your example with induced subgraphs in the forbidden subgraph problem. I understand that induced subgraphs contain edges between vertices iff the original graph contains those same edges. Is the point of your example just to drive home this iff condition? It seems to me that if $H$ is a clique we still have the same upper bound on edges.
$endgroup$
– Richard
Jan 24 at 0:07
$begingroup$
If $H$ is a clique, then $K_n$ does contain an induced copy of $H$, and so we have to do actual work to determine the largest graph that doesn't. My point is that if $H$ is missing any edges, $K_n$ does not have any induced copies of $H$, so it is trivially the largest graph. This is not an interesting result.
$endgroup$
– Misha Lavrov
Jan 24 at 0:13
$begingroup$
On the other hand, the problem of the largest (non-induced) $H$-free graph is interesting for every $H$.
$endgroup$
– Misha Lavrov
Jan 24 at 0:13
add a comment |
$begingroup$
Why do we not just say that $G$ is $H$-free if it does not contain $H$ as a subgraph?
In my experience, this is generally what being $H$-free means. If we want to say that $G$ does not contain $H$ as an induced subgraph, we say that $G$ is induced $H$-free.
Certainly, discussion of $H$-free graphs is very common when discussing the forbidden subgraph problem. This problem is generally only considered with the ordinary notion of subgraphs, not the induced kind. (If you're asking for the largest number of edges in an $n$-vertex induced $H$-free graph, then unless $H$ is a clique, you will get a boring answer: the best graph is $K_n$ with $binom n2$ edges. This is why we don't talk about induced subgraphs for this problem.)
Both notions are useful, and there is terminology for both. Sometimes people are lazy about this terminology, and don't specify that $G$ is induced $H$-free when they should. Sometimes, it's hard to tell which convention is being used, and sometimes it doesn't matter. (For example, if we're counting $C_5$'s in a triangle-free graph, the answer doesn't change if we add the word "induced" in both places.)
answered Jan 23 at 5:37
Misha LavrovMisha Lavrov
47k657107
$endgroup$
$begingroup$
Thanks for the reply. I am a little confused on your example with induced subgraphs in the forbidden subgraph problem. I understand that induced subgraphs contain edges between vertices iff the original graph contains those same edges. Is the point of your example just to drive home this iff condition? It seems to me that if $H$ is a clique we still have the same upper bound on edges.
$endgroup$
– Richard
Jan 24 at 0:07
$begingroup$
If $H$ is a clique, then $K_n$ does contain an induced copy of $H$, and so we have to do actual work to determine the largest graph that doesn't. My point is that if $H$ is missing any edges, $K_n$ does not have any induced copies of $H$, so it is trivially the largest graph. This is not an interesting result.
$endgroup$
– Misha Lavrov
Jan 24 at 0:13
$begingroup$
On the other hand, the problem of the largest (non-induced) $H$-free graph is interesting for every $H$.
$endgroup$
– Misha Lavrov
Jan 24 at 0:13
add a comment |
$begingroup$
Why do we not just say that $G$ is $H$-free if it does not contain $H$ as a subgraph?
In my experience, this is generally what being $H$-free means. If we want to say that $G$ does not contain $H$ as an induced subgraph, we say that $G$ is induced $H$-free.
Certainly, discussion of $H$-free graphs is very common when discussing the forbidden subgraph problem. This problem is generally only considered with the ordinary notion of subgraphs, not the induced kind. (If you're asking for the largest number of edges in an $n$-vertex induced $H$-free graph, then unless $H$ is a clique, you will get a boring answer: the best graph is $K_n$ with $binom n2$ edges. This is why we don't talk about induced subgraphs for this problem.)
Both notions are useful, and there is terminology for both. Sometimes people are lazy about this terminology, and don't specify that $G$ is induced $H$-free when they should. Sometimes, it's hard to tell which convention is being used, and sometimes it doesn't matter. (For example, if we're counting $C_5$'s in a triangle-free graph, the answer doesn't change if we add the word "induced" in both places.)
answered Jan 23 at 5:37
Misha LavrovMisha Lavrov
47k657107
$endgroup$
Why do we not just say that $G$ is $H$-free if it does not contain $H$ as a subgraph?
In my experience, this is generally what being $H$-free means. If we want to say that $G$ does not contain $H$ as an induced subgraph, we say that $G$ is induced $H$-free.
Certainly, discussion of $H$-free graphs is very common when discussing the forbidden subgraph problem. This problem is generally only considered with the ordinary notion of subgraphs, not the induced kind. (If you're asking for the largest number of edges in an $n$-vertex induced $H$-free graph, then unless $H$ is a clique, you will get a boring answer: the best graph is $K_n$ with $binom n2$ edges. This is why we don't talk about induced subgraphs for this problem.)
Both notions are useful, and there is terminology for both. Sometimes people are lazy about this terminology, and don't specify that $G$ is induced $H$-free when they should. Sometimes, it's hard to tell which convention is being used, and sometimes it doesn't matter. (For example, if we're counting $C_5$'s in a triangle-free graph, the answer doesn't change if we add the word "induced" in both places.)
answered Jan 23 at 5:37
Misha LavrovMisha Lavrov
47k657107
edited Jan 24 at 1:13
answered Jan 23 at 5:37
Misha LavrovMisha Lavrov
47k657107
answered Jan 23 at 5:37
Misha LavrovMisha Lavrov
47k657107
answered Jan 23 at 5:37
Misha LavrovMisha Lavrov
47k657107
47k657107
$begingroup$
Thanks for the reply. I am a little confused on your example with induced subgraphs in the forbidden subgraph problem. I understand that induced subgraphs contain edges between vertices iff the original graph contains those same edges. Is the point of your example just to drive home this iff condition? It seems to me that if $H$ is a clique we still have the same upper bound on edges.
$endgroup$
– Richard
Jan 24 at 0:07
$begingroup$
If $H$ is a clique, then $K_n$ does contain an induced copy of $H$, and so we have to do actual work to determine the largest graph that doesn't. My point is that if $H$ is missing any edges, $K_n$ does not have any induced copies of $H$, so it is trivially the largest graph. This is not an interesting result.
$endgroup$
– Misha Lavrov
Jan 24 at 0:13
$begingroup$
On the other hand, the problem of the largest (non-induced) $H$-free graph is interesting for every $H$.
$endgroup$
– Misha Lavrov
Jan 24 at 0:13
add a comment |
$begingroup$
Thanks for the reply. I am a little confused on your example with induced subgraphs in the forbidden subgraph problem. I understand that induced subgraphs contain edges between vertices iff the original graph contains those same edges. Is the point of your example just to drive home this iff condition? It seems to me that if $H$ is a clique we still have the same upper bound on edges.
$endgroup$
– Richard
Jan 24 at 0:07
$begingroup$
If $H$ is a clique, then $K_n$ does contain an induced copy of $H$, and so we have to do actual work to determine the largest graph that doesn't. My point is that if $H$ is missing any edges, $K_n$ does not have any induced copies of $H$, so it is trivially the largest graph. This is not an interesting result.
$endgroup$
– Misha Lavrov
Jan 24 at 0:13
$begingroup$
On the other hand, the problem of the largest (non-induced) $H$-free graph is interesting for every $H$.
$endgroup$
– Misha Lavrov
Jan 24 at 0:13
$begingroup$
Thanks for the reply. I am a little confused on your example with induced subgraphs in the forbidden subgraph problem. I understand that induced subgraphs contain edges between vertices iff the original graph contains those same edges. Is the point of your example just to drive home this iff condition? It seems to me that if $H$ is a clique we still have the same upper bound on edges.
$endgroup$
– Richard
Jan 24 at 0:07
$begingroup$
Thanks for the reply. I am a little confused on your example with induced subgraphs in the forbidden subgraph problem. I understand that induced subgraphs contain edges between vertices iff the original graph contains those same edges. Is the point of your example just to drive home this iff condition? It seems to me that if $H$ is a clique we still have the same upper bound on edges.
$endgroup$
– Richard
Jan 24 at 0:07
$begingroup$
If $H$ is a clique, then $K_n$ does contain an induced copy of $H$, and so we have to do actual work to determine the largest graph that doesn't. My point is that if $H$ is missing any edges, $K_n$ does not have any induced copies of $H$, so it is trivially the largest graph. This is not an interesting result.
$endgroup$
– Misha Lavrov
Jan 24 at 0:13
$begingroup$
If $H$ is a clique, then $K_n$ does contain an induced copy of $H$, and so we have to do actual work to determine the largest graph that doesn't. My point is that if $H$ is missing any edges, $K_n$ does not have any induced copies of $H$, so it is trivially the largest graph. This is not an interesting result.
$endgroup$
– Misha Lavrov
Jan 24 at 0:13
$begingroup$
On the other hand, the problem of the largest (non-induced) $H$-free graph is interesting for every $H$.
$endgroup$
– Misha Lavrov
Jan 24 at 0:13
$begingroup$
On the other hand, the problem of the largest (non-induced) $H$-free graph is interesting for every $H$.
$endgroup$
– Misha Lavrov
Jan 24 at 0:13
add a comment |
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