in Graph Theory recategorized by
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62 votes
62 votes

The number of distinct simple graphs with up to three nodes is

  1. $15$
  2. $10$
  3. $7$
  4. $9$
in Graph Theory recategorized by
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4 Comments

@shaktisingh

Generalized solution for this question .....since graph with n nodes can have n(n-1)/2 edges hence  total simple  graph possible for unlabeled vertices are  n(n-1)/2 +1…..(here +1 is because graph with no edges) 

example-- for 3 vertices   max  edges  can be 3*2/2=3 hence 4 graphs possible ..(graph with no edge+ graph with 1 edge +graph with 2 edge +graph with 3 edge)

Question is for upto n nodes  hence 

for upto 1  graph possible only 1

for upto 2 nodes    (1+2)

for 3 nodes (1+2+4)

for n nodes  (1+2+4+ 7+11+15+….+upto n terms)

nth term of this series =n(n-1)/2 +1

hence for n nodes

sum of this series =$n(n^{^{2}}+5)$/6… are total  simple graphs possible for unlabelled nodes

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4

They have played with words quite well up to should be in bold

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@rupesh17 Your Generalized solution is not correct. For 4 unlabeled vertices we have 11 simple graphs (not 7). See below references.

https://mathworld.wolfram.com/SimpleGraph.html

https://garsia.math.yorku.ca/~zabrocki/math3260w03/nall.html

 

As per my current understanding no general solution for unlabeled vertices simple graph.If anyone knows please comment.

 

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12 Answers

108 votes
108 votes
Best answer

Answer is (C)

edited by

4 Comments

@abir_banerjee 

Also from Wikipedia:

"Some authors exclude K0 from consideration as a graph (either by definition, or more simply as a matter of convenience). Whether including K0 as a valid graph is useful depends on context."

Same thing I told in previous message. 

Half-reading is misleading & dangerous. 

IIT, IISc, NPTEL professors consider graphs to have at least one vertex. You can follow that. 

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This is a screenshot from the very famous graph theory book by Reinhard Diestel, often called as "Diestel's graph theory book" which is considered as a standard textbook and followed by many people all over the world.
Now,  we can see vertex set can be empty and according to this 8 should be the correct answer provided graph is unlabeled. So,”theoretically” both 7 and 8 are the "possible" answers for an unlabeled graph. It should not be like that: 7 is correct and 8 is not correct though it is another case that $8$ is not given as the option.

To avoid ambiguity in the questions, it would be very helpful if more details would be added in the questions. It might create long questions but chances of ambiguity will be less.

Graph Theory is considered as a young subject and so, terms should be properly defined in the question. For example, a complete graph is defined as a graph whose vertices are pairwise adjacent, while a clique is a set of pairwise adjacent vertices in a graph. Many authors use the terms interchangeably. And, there are other examples too. So, to avoid ambiguity, it would be much helpful if things are defined properly and as much as details should be added in the question.  

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edited by

@ankitgupta.1729

That’s well written.

It depends on the author whether they allow empty vertex set or not in the graph definition. Particularly, in IISc CSA Graph Theory course also, Diestel is followed, But Prof doesn’t allow empty vertex set in the definition of Graph.

Conclusion: 7, 8 both are correct answers, But since 8 is Not given in the options, so, 7 is unambiguously correct answer.

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48 votes
48 votes
Answer: C

The number of max edges a simple graph can have is $\frac{n×(n−1)}{2}$.

So, for a graph with $3$ nodes the max number of edges is $3.$
Now there can be $0$ edges, $1$ edge, $2$ edges or $3$ edges in a $3$ node simple graph.
So the total number of unlabeled simple graphs on $3$ nodes will be  $4.$
Similarly for two node graph we have option of $0$ or $1$ edge and for one node graph we have option of $0$ edge.
So the total number of simple graphs upto three nodes $=4+2+1=7.$

4 Comments

can i say it is $2^n - 1$?
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Sourabh Kumar according to this http://oeis.org/A005470/list 8 is the answer

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No you cannot Generalize 2^n−1.

As Up to 4 nodes Answer will be ,

Up to 3 nodes= 8 + for 4 nodes= 11,

So 8+11=19.
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33 votes
33 votes
With n number of nodes, max edges are possible $\frac{n(n-1)}{2}$ in a simple graph. each edge has two choices, it'll be taken in a graph or it won't be taken. so total no of graphs are possible
$2^{\frac{n(n-1)}{2}}$.
in the question, it is asked upto three nodes:
with 1 node total graph possible = 1
with two nodes graph possible = 2
with three nodes, possible graphs= $2^{\frac{3(3-1)}{2}}  = 2^3 = 8$   but there are 3 non distinct graphs which are isomorphic to each other, so with n=3 nodes total graphs possible = 8 but total unique graph possible = 4

upto 3 nodes $1 + 2 +4 = 7 $hence answer is (C)
edited by

3 Comments

in 7,you have not counted null graph? why?

total unique graph means?
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" but there are 3 non distinct graphs which are isomorphic to each other, so with n=3 nodes total graphs possible = 8 but total unique graph possible = 4"

did not understand this part

 

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Good explanation, here if we take labeled nodes then answer is 11 which is not in the option so go with the 7 because if we take unlabeled nodes then 7 simple graphs will be possible.
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13 votes
13 votes
A graph is an ordered pair (V,E). So, given a set V, graph will be different if E is different. Lets see how many different E we can get when |V| = 3.

For |V| = 3, we can have |E| = 0, 1, 2 or 3. So, 4 possible graphs.

Now, the question asks for |V| upto 3. So, we have to consider |V| = 2 and |V| = 1 also. When |V| = 2, we can have |E| = 1 or 0, so 2 possibilities. For |V| = 1, |E| can be only 0 and hence only one possibility.  So, total number of possibilities is $$4 + 2 + 1  = 7$$.
edited by

2 Comments

this won't work, having an edge on one node means there is some other node with one edge also.
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Yes. That was a bad mistake. I have corrected it now.
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Answer: