<p>I believe the answer to your question is "no" because an equivalent condition would imply a polynomial time solution to GI.</p>

<p>For $A$, the adjacency matrix of the graph $G$, note that the number of paths from $i$ to $j$ of length $k$ is $(A^k)_{i, j}$ (with repetition of vertices and edges allowed).  For two graphs $G_1$ and $G_2$ (with adjacency matrices $A_1$ and $A_2$) and $k \ge 1$, if you sorted the elements of $A_1^k$ and $A_2^k$ then in order for $G_1$ to be isomorphic to $G_2$, it is a necessary condition that the lists be identical for all $k$.</p>

<p>I believe your conjecture is equivalent to:</p>

<p>If the sorted lists of elements of $A_1^k$ and $A_2^d$ are identical for $k = 1$ to $n - 1$ (upperbound on the longest path with non-repeating vertices) then $G_1$ and $G_2$ are isomorphic.</p>

<p>So to solve GI, one only has to perform $n - 1$ multiplications of $n \times n$ matrices (and a little extra time to sort and compare $n^2$ elements).  This would take less than $n^4$ time.</p>

<p>I admit two possible flaws in my argument.  First, it is totally possible that GI has a polynomial time algorithm and that we just discovered it together, just now (hooray, we're famous!).  I find this highly unlikely.  Second (and much more probable), what I've proposed is not actually equivalent to your conjecture.</p>

<p>Final thought.  Have you tried this out for all, say, 3-regular graphs for size 8 or so?  I would think that if your conjecture is false, that there should be a counter example in 3-regular graphs of fairly small size.</p>

CC BY-SA 3.0
2011-04-28T14:47:59.400Z
I did not know that the number of distinct paths from i to j of length k is $(A^{k})_{i,j}$. If so, and if I am understanding well what you are doing, then my initial hypothesis is answered.
N27
CC BY-SA 3.0
2011-04-28T15:16:36.847Z
@N27: It can be proved using the definition of matrix multiplication and induction.
Tomek Tarczynski
CC BY-SA 3.0
2011-04-28T15:38:40.073Z
Yes, easily, in fact...
N27
CC BY-SA 3.0
2011-05-04T19:37:03.147Z
Ah, it appears that once again my intuition led me astray. Counting the number of distinct simple paths in a graph (or even just between 2 nodes) is #P-complete. So my argument is wrong because it says that a polynomial time algorithm is equivalent to counting simple paths. I'm also now completely unsure if your conjecture is correct or not. However, its a bit moot point because you're not likely to choose to solve a #P-complete problem over GI.
bbejot
CC BY-SA 3.0
2011-04-28T14:15:37.470Z
2011-04-28T14:15:37.470Z
bbejot
<p>Since you're only comparing the lengths of the paths (and in the meantime forgetting what pair of nodes they correspond to if I understood you well), I think that very similar graphs should provide a counterexample : in the end you're just counting the number of paths of a fixed length and independently of the vertices they link. For example I think these graphs are a counterexample : <a href="http://www.mathe2.uni-bayreuth.de/markus/REGGRAPHS/GIF/06_3_3-2.gif" rel="nofollow">http://www.mathe2.uni-bayreuth.de/markus/REGGRAPHS/GIF/06_3_3-2.gif</a> and <a href="http://www.mathe2.uni-bayreuth.de/markus/REGGRAPHS/GIF/06_3_3-1.gif" rel="nofollow">http://www.mathe2.uni-bayreuth.de/markus/REGGRAPHS/GIF/06_3_3-1.gif</a></p>

<p>If I'm not mistaken (counting paths is tedious), they both have 9 paths of length 1, 18 paths of length 2, 48 paths of length 3, 30 paths of length 4 and 36 paths of length 5</p>

CC BY-SA 3.0
2011-05-02T02:31:40.970Z
I count 36 paths of length 3 in the first graph and 30 graphs of length 3 in the second graph. The problem is that the second graph has cycles of length 3 where the first graph does not. I still agree, however, that there should be a relatively small graph as a counterexample. I havent found one yet, though.
bbejot
CC BY-SA 3.0
2011-05-02T11:19:39.653Z
I agree with you, writing a program testing all the small graphs would probably give a quick answer.
Arnaud
CC BY-SA 3.0
2011-04-28T15:04:43.590Z
2011-04-28T15:04:43.590Z
Arnaud
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CC BY-SA 3.0
2011-04-28T10:29:55.100Z
in all of these graphs lambda = mu
trg787
CC BY-SA 3.0
2011-04-28T10:35:46.470Z
it's the 3 simplest pairs (non-isomorphic)
trg787
CC BY-SA 3.0
2011-04-28T10:38:28.697Z
what is that?!! and how do you know that there is at least one different path?
N27
CC BY-SA 3.0
2011-04-28T10:47:24.337Z
I mean how do you know that the lists of all possible paths between every pair of nodes are identical?
N27
CC BY-SA 3.0
2011-04-28T11:13:56.110Z
Hmm... I meant only lengths. Of course, paths are different http://funkybee.narod.ru/griso_praust.htm
trg787
CC BY-SA 3.0
2011-04-28T11:20:48.803Z
Anyway, sorry, I dont understand what you have tested or trying to say... My question was whether the 2 lists of all lengths of distinct paths between all pairs of nodes are NOT the same for 2 non-isomorphic graphs.
N27
CC BY-SA 3.0
2011-04-28T10:23:48.460Z
2011-04-28T10:23:48.460Z
trg787
<p>I would like to ask whether there is an already published result on that:</p>

<p>We take all possible different paths between <em>each pair of nodes</em> of two connected regular (with degree $d$ let's say, and number of nodes $n$) graphs and write down their lengths. Of course this number of distinct paths is exponential. My  question is, if we sort the lengths and compare them (the lists obtained by the two graphs) and they are exactly the same, can we say that the two graphs are isomorphic?</p>

<p>Of course, even if this is a result we cannot use it to reply for Graph Isomorphism, since the number of distinct paths is exponential, as said</p>

<p>By <em>distinct paths</em>, I refer to paths having at least one different node, obviously.</p>

<p>Thanks in priori for your help.</p>

CC BY-SA 3.0
2011-04-28T05:47:47.543Z
in 2-regular graphs there is a very small number of different paths, as a 2-regular graph is a disjoint union of cycles. Hence you either have 2 or 0 paths between each pair of vertices.
Nathann Cohen
CC BY-SA 3.0
2011-04-28T06:09:17.367Z
This question, while interesting, seems better suited for [MathOverflow](http://mathoverflow.net/) to me.
Niel de Beaudrap
<graph-theory><co.combinatorics><graph-algorithms><graph-isomorphism>
Regular Graphs and Isomorphism
N27
# Regular Graphs and Isomorphism

(Asked by: username_0 on 2011-04-28)

I would like to ask whether there is an already published result on that:

We take all possible different paths between *each pair of nodes* of two connected regular (with degree $d$ let's say, and number of nodes $n$) graphs and write down their lengths. Of course this number of distinct paths is exponential. My question is, if we sort the lengths and compare them (the lists obtained by the two graphs) and they are exactly the same, can we say that the two graphs are isomorphic?

Of course, even if this is a result we cannot use it to reply for Graph Isomorphism, since the number of distinct paths is exponential, as said

By *distinct paths*, I refer to paths having at least one different node, obviously.

Thanks in priori for your help.

**Tags:** graph-theory, co.combinatorics, graph-algorithms, graph-isomorphism

> **Comments:**
> **username_1**: in 2-regular graphs there is a very small number of different paths, as a 2-regular graph is a disjoint union of cycles. Hence you either have 2 or 0 paths between each pair of vertices.  
> **username_2**: This question, while interesting, seems better suited for [MathOverflow](http://mathoverflow.net/) to me.  

---

## 3 answers

### Answer by username_3 on 2011-04-28. Score: 6 | Accepted answer

I believe the answer to your question is "no" because an equivalent condition would imply a polynomial time solution to GI.

For $A$, the adjacency matrix of the graph $G$, note that the number of paths from $i$ to $j$ of length $k$ is $(A^k)\_{i, j}$ (with repetition of vertices and edges allowed). For two graphs $G\_1$ and $G\_2$ (with adjacency matrices $A\_1$ and $A\_2$) and $k \ge 1$, if you sorted the elements of $A\_1^k$ and $A\_2^k$ then in order for $G\_1$ to be isomorphic to $G\_2$, it is a necessary condition that the lists be identical for all $k$.

I believe your conjecture is equivalent to:

If the sorted lists of elements of $A\_1^k$ and $A\_2^d$ are identical for $k = 1$ to $n - 1$ (upperbound on the longest path with non-repeating vertices) then $G\_1$ and $G\_2$ are isomorphic.

So to solve GI, one only has to perform $n - 1$ multiplications of $n \times n$ matrices (and a little extra time to sort and compare $n^2$ elements). This would take less than $n^4$ time.

I admit two possible flaws in my argument. First, it is totally possible that GI has a polynomial time algorithm and that we just discovered it together, just now (hooray, we're famous!). I find this highly unlikely. Second (and much more probable), what I've proposed is not actually equivalent to your conjecture.

Final thought. Have you tried this out for all, say, 3-regular graphs for size 8 or so? I would think that if your conjecture is false, that there should be a counter example in 3-regular graphs of fairly small size.

> **Comments:**
> **username_0**: I did not know that the number of distinct paths from i to j of length k is $(A^{k})_{i,j}$. If so, and if I am understanding well what you are doing, then my initial hypothesis is answered.  
> **username_4**: @username_0: It can be proved using the definition of matrix multiplication and induction.  
> **username_0**: Yes, easily, in fact...  
> **username_3**: Ah, it appears that once again my intuition led me astray. Counting the number of distinct simple paths in a graph (or even just between 2 nodes) is #P-complete. So my argument is wrong because it says that a polynomial time algorithm is equivalent to counting simple paths. I'm also now completely unsure if your conjecture is correct or not. However, its a bit moot point because you're not likely to choose to solve a #P-complete problem over GI.  

---

### Answer by username_5 on 2011-04-28. Score: 6

Since you're only comparing the lengths of the paths (and in the meantime forgetting what pair of nodes they correspond to if I understood you well), I think that very similar graphs should provide a counterexample : in the end you're just counting the number of paths of a fixed length and independently of the vertices they link. For example I think these graphs are a counterexample : <http://www.mathe2.uni-bayreuth.de/markus/REGGRAPHS/GIF/06_3_3-2.gif> and <http://www.mathe2.uni-bayreuth.de/markus/REGGRAPHS/GIF/06_3_3-1.gif>

If I'm not mistaken (counting paths is tedious), they both have 9 paths of length 1, 18 paths of length 2, 48 paths of length 3, 30 paths of length 4 and 36 paths of length 5

> **Comments:**
> **username_3**: I count 36 paths of length 3 in the first graph and 30 graphs of length 3 in the second graph. The problem is that the second graph has cycles of length 3 where the first graph does not. I still agree, however, that there should be a relatively small graph as a counterexample. I havent found one yet, though.  
> **username_5**: I agree with you, writing a program testing all the small graphs would probably give a quick answer.  

---

### Answer by username_6 on 2011-04-28. Score: -2

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011111111111111100000000000000000000
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> **Comments:**
> **username_6**: in all of these graphs lambda = mu  
> **username_6**: it's the 3 simplest pairs (non-isomorphic)  
> **username_0**: what is that?!! and how do you know that there is at least one different path?  
> **username_0**: I mean how do you know that the lists of all possible paths between every pair of nodes are identical?  
> **username_6**: Hmm... I meant only lengths. Of course, paths are different http://funkybee.narod.ru/griso_praust.htm  
> **username_0**: Anyway, sorry, I dont understand what you have tested or trying to say... My question was whether the 2 lists of all lengths of distinct paths between all pairs of nodes are NOT the same for 2 non-isomorphic graphs.  

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