How many labeled graphs there are of order $n$ without any isolated vertices?
By the inclusion-exclusion principle, we have:
$$\sum_{i=0}^n (-1)^i{n\choose i}2^{n-i \choose 2} $$
But is there any closed form for it? Or is there any better solution?
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4What is the meaning of acronym IEP ? – Jean Marie Dec 27 '16 at 14:33
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1To count how many graphs there are, are the vertices considered different or indistinguishable? That is, is $1-2-3$ the same graph as $1-3-2$? – ajotatxe Dec 27 '16 at 14:38
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2@JeanMarie Inclusion-Exclusion Principle – MR_BD Dec 27 '16 at 14:39
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@ajotatxe No, they are not the same... probably it would be much harder if they are indistinguishable... – MR_BD Dec 27 '16 at 14:40
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1Shouldn't there be a $(-1)^i$ factor in there? – Zestylemonzi Dec 27 '16 at 14:51
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@Zestylemonzi Of Course! Done! – MR_BD Dec 27 '16 at 14:52
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By the way, is your question about all labeled graphs or only simple graphs? – rgm Dec 29 '16 at 19:41
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@shn just simple. Is it easier for general graphs? – MR_BD Dec 29 '16 at 19:50
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The numbers are given in OEIS A006219. They start $$1, 0, 1, 4, 41, 768, 27449, 1887284, 252522481, 66376424160, 34509011894545, 35645504882731588, 73356937912127722841, 301275024444053951967648, 2471655539737552842139838345, 40527712706903544101000417059892, 1328579255614092968399503598175745633$$ No clean formula is given. I found it by computing the value for $n=8$ and searching for $252522481$. They grow approximately as $2^{n \choose 2}$
Ross Millikan
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can you prove this formula please? I can't understand one moment, let n = 4, then total number of possible graphs is 64, now we should exclude all graphs with property that at least 1 vertex isolated. But, using this formula we can select 4 vertices to be isolated, for each one we can build 8 graphs on 3 others vertices. So some cases we count multiple times, for example when all vertices isolated it is 4 cases of 32. So, what is the correct rule in inclusion-exclusion formula we should apply? – Evgeny Mar 08 '21 at 05:48
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No, I can't prove it. I don't remember how I did the $n=8$ case. For $n=4$ you can isolate a vertex by cutting all $3$ edges incident, so each vertex can be isolated in $2^3$ ways because there are $3$ other edges which might or might not be there. If you isolate two vertices this way you have eliminate $5$ edges so there are only $2$ choices. You can't isolate three vertices without isolating all four. This would give $64-4\cdot 8 +{4 \choose 2}2=38$ But you can also have graphs with two pairs of connected vertices, there are $3$, which gives $41$. – Ross Millikan Mar 08 '21 at 06:09
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Then how they give this formula?) If there is no proof for that. Do you mean that I should do any case separately??? – Evgeny Mar 08 '21 at 06:15
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It depends whether you want an approximate or exact answer. Often we can approximate the growth of a function more easily than we can find the exact values. I was just reporting what I found in OEIS. You can follow the links to see where it comes from. – Ross Millikan Mar 08 '21 at 14:38