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Isomorphisms between regular graphs of same degree

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2

$begingroup$

Are all $n$-vertex regular graphs of degree $d$ isomorphic?

Can someone provide an example of two non-isomorphic graphs $G_1$ and $G_2$ which are both regular with degree $d$ and have the same number of vertices (i.e., $|G_1| = |G_2|$)?

graph-theory graph-isomorphism

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edited 2 days ago

dkaeae

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asked 2 days ago

Jim NewtonJim Newton

1107

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes indeed, but clearly regular graphs of degree 2 are not isomorphic to regular graphs of degree 3. So I'm asking about regular graphs of the same degree, if they have the same number of vertices, are they necessarily isomorphic?
  $endgroup$
  – Jim Newton
  2 days ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Are all $n$-vertex regular graphs of degree $d$ isomorphic?

Can someone provide an example of two non-isomorphic graphs $G_1$ and $G_2$ which are both regular with degree $d$ and have the same number of vertices (i.e., $|G_1| = |G_2|$)?

graph-theory graph-isomorphism

share|cite|improve this question

edited 2 days ago

dkaeae

1,981721

asked 2 days ago

Jim NewtonJim Newton

1107

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes indeed, but clearly regular graphs of degree 2 are not isomorphic to regular graphs of degree 3. So I'm asking about regular graphs of the same degree, if they have the same number of vertices, are they necessarily isomorphic?
  $endgroup$
  – Jim Newton
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

Are all $n$-vertex regular graphs of degree $d$ isomorphic?

Can someone provide an example of two non-isomorphic graphs $G_1$ and $G_2$ which are both regular with degree $d$ and have the same number of vertices (i.e., $|G_1| = |G_2|$)?

graph-theory graph-isomorphism

share|cite|improve this question

edited 2 days ago

dkaeae

1,981721

asked 2 days ago

Jim NewtonJim Newton

1107

$endgroup$

share|cite|improve this question

edited 2 days ago

dkaeae

1,981721

asked 2 days ago

Jim NewtonJim Newton

1107

share|cite|improve this question

edited 2 days ago

dkaeae

1,981721

asked 2 days ago

Jim NewtonJim Newton

1107

edited 2 days ago

dkaeae

1,981721

edited 2 days ago

dkaeae

1,981721

asked 2 days ago

Jim NewtonJim Newton

1107

asked 2 days ago

Jim NewtonJim Newton

1107

2 Answers
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4

$begingroup$

Playing around with a pencil and paper for a few minutes, it should be easy to come up with non-isomorphic $d$-regular graphs with the same number of vertices, for small $d$. For example, take two cycles of length $2n$ and connect chords across them in different ways.

However, there is a polynomial-time isomorphism algorithm for any class of graphs of bounded degree, which includes the $d$-regular graphs for any fixed $d$. It's due to Luks (Isomorphism of graphs of bounded valence can be tested in polynomial time, Journal of Computer and System Sciences 25(1):42–65, 1982) and uses a bunch of group theory.

share|cite|improve this answer

answered 2 days ago

David RicherbyDavid Richerby

68k15102193

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4

$begingroup$

Of course not.

Consider, for example, the cycle $C_6$ with six vertices and the graph obtained by the union of two copies of $C_3$. Then both are 2-regular, but they are obviously not isomorphic.

This is also the case if we restrict the question to connected graphs. Consider, for instance, the following two 3-regular graphs:

                                                

You can see they are not isomorphic because the second one contains cycles with six vertices that have chords; this is impossible in the first graph since it has precisely four six-cycles and you can see none of them have chords. (Or even easier: The second one has a five cycle, whereas the first one has only cycles with three, four, six, or more vertices.)

What you might be able to prove is that all 2-regular and connected graphs are isomorphic (see, e.g., this), but this is a big restriction compared to the original question.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

dkaeaedkaeae

1,981721

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4

$begingroup$

Playing around with a pencil and paper for a few minutes, it should be easy to come up with non-isomorphic $d$-regular graphs with the same number of vertices, for small $d$. For example, take two cycles of length $2n$ and connect chords across them in different ways.

However, there is a polynomial-time isomorphism algorithm for any class of graphs of bounded degree, which includes the $d$-regular graphs for any fixed $d$. It's due to Luks (Isomorphism of graphs of bounded valence can be tested in polynomial time, Journal of Computer and System Sciences 25(1):42–65, 1982) and uses a bunch of group theory.

share|cite|improve this answer

answered 2 days ago

David RicherbyDavid Richerby

68k15102193

$endgroup$

add a comment | 

4

$begingroup$

Playing around with a pencil and paper for a few minutes, it should be easy to come up with non-isomorphic $d$-regular graphs with the same number of vertices, for small $d$. For example, take two cycles of length $2n$ and connect chords across them in different ways.

However, there is a polynomial-time isomorphism algorithm for any class of graphs of bounded degree, which includes the $d$-regular graphs for any fixed $d$. It's due to Luks (Isomorphism of graphs of bounded valence can be tested in polynomial time, Journal of Computer and System Sciences 25(1):42–65, 1982) and uses a bunch of group theory.

share|cite|improve this answer

answered 2 days ago

David RicherbyDavid Richerby

68k15102193

$endgroup$

add a comment | 

$begingroup$

Playing around with a pencil and paper for a few minutes, it should be easy to come up with non-isomorphic $d$-regular graphs with the same number of vertices, for small $d$. For example, take two cycles of length $2n$ and connect chords across them in different ways.

However, there is a polynomial-time isomorphism algorithm for any class of graphs of bounded degree, which includes the $d$-regular graphs for any fixed $d$. It's due to Luks (Isomorphism of graphs of bounded valence can be tested in polynomial time, Journal of Computer and System Sciences 25(1):42–65, 1982) and uses a bunch of group theory.

share|cite|improve this answer

answered 2 days ago

David RicherbyDavid Richerby

68k15102193

$endgroup$

share|cite|improve this answer

answered 2 days ago

David RicherbyDavid Richerby

68k15102193

answered 2 days ago

David RicherbyDavid Richerby

68k15102193

answered 2 days ago

David RicherbyDavid Richerby

68k15102193

4

$begingroup$

Of course not.

Consider, for example, the cycle $C_6$ with six vertices and the graph obtained by the union of two copies of $C_3$. Then both are 2-regular, but they are obviously not isomorphic.

This is also the case if we restrict the question to connected graphs. Consider, for instance, the following two 3-regular graphs:

                                                

You can see they are not isomorphic because the second one contains cycles with six vertices that have chords; this is impossible in the first graph since it has precisely four six-cycles and you can see none of them have chords. (Or even easier: The second one has a five cycle, whereas the first one has only cycles with three, four, six, or more vertices.)

What you might be able to prove is that all 2-regular and connected graphs are isomorphic (see, e.g., this), but this is a big restriction compared to the original question.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

dkaeaedkaeae

1,981721

$endgroup$

add a comment | 

4

$begingroup$

Of course not.

Consider, for example, the cycle $C_6$ with six vertices and the graph obtained by the union of two copies of $C_3$. Then both are 2-regular, but they are obviously not isomorphic.

This is also the case if we restrict the question to connected graphs. Consider, for instance, the following two 3-regular graphs:

                                                

You can see they are not isomorphic because the second one contains cycles with six vertices that have chords; this is impossible in the first graph since it has precisely four six-cycles and you can see none of them have chords. (Or even easier: The second one has a five cycle, whereas the first one has only cycles with three, four, six, or more vertices.)

What you might be able to prove is that all 2-regular and connected graphs are isomorphic (see, e.g., this), but this is a big restriction compared to the original question.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

dkaeaedkaeae

1,981721

$endgroup$

add a comment | 

$begingroup$

Of course not.

Consider, for example, the cycle $C_6$ with six vertices and the graph obtained by the union of two copies of $C_3$. Then both are 2-regular, but they are obviously not isomorphic.

This is also the case if we restrict the question to connected graphs. Consider, for instance, the following two 3-regular graphs:

                                                

You can see they are not isomorphic because the second one contains cycles with six vertices that have chords; this is impossible in the first graph since it has precisely four six-cycles and you can see none of them have chords. (Or even easier: The second one has a five cycle, whereas the first one has only cycles with three, four, six, or more vertices.)

What you might be able to prove is that all 2-regular and connected graphs are isomorphic (see, e.g., this), but this is a big restriction compared to the original question.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

dkaeaedkaeae

1,981721

$endgroup$

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

dkaeaedkaeae

1,981721

answered 2 days ago

dkaeaedkaeae

1,981721

answered 2 days ago

dkaeaedkaeae

1,981721