Do any two spanning trees of a simple graph always have some common edges?











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I tried few cases and found any two spanning tree of a simple graph has some common edges. I mean I couldn't find any counter example so far. But I couldn't prove or disprove this either. How to prove or disprove this conjecture?






graphs graph-theory spanning-trees







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  • Assuming the weights of the graph are distinct, the edge with the minimum weight will be present in all the minimum spanning trees.
    – Gokul
    4 hours ago










  • @Gokul minimum spanning tree? What?...
    – Mr. Sigma.
    3 hours ago










  • Oh, apologies. I read the title as whether minimum spanning tree have common edges.
    – Gokul
    3 hours ago















up vote
2
down vote

favorite
1












I tried few cases and found any two spanning tree of a simple graph has some common edges. I mean I couldn't find any counter example so far. But I couldn't prove or disprove this either. How to prove or disprove this conjecture?






graphs graph-theory spanning-trees







share|cite|improve this question






















  • Assuming the weights of the graph are distinct, the edge with the minimum weight will be present in all the minimum spanning trees.
    – Gokul
    4 hours ago










  • @Gokul minimum spanning tree? What?...
    – Mr. Sigma.
    3 hours ago










  • Oh, apologies. I read the title as whether minimum spanning tree have common edges.
    – Gokul
    3 hours ago













up vote
2
down vote

favorite
1









up vote
2
down vote

favorite
1






1





I tried few cases and found any two spanning tree of a simple graph has some common edges. I mean I couldn't find any counter example so far. But I couldn't prove or disprove this either. How to prove or disprove this conjecture?






graphs graph-theory spanning-trees







share|cite|improve this question













I tried few cases and found any two spanning tree of a simple graph has some common edges. I mean I couldn't find any counter example so far. But I couldn't prove or disprove this either. How to prove or disprove this conjecture?






graphs graph-theory spanning-trees




graphs graph-theory spanning-trees






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 4 hours ago









Mr. Sigma.

355116




355116












  • Assuming the weights of the graph are distinct, the edge with the minimum weight will be present in all the minimum spanning trees.
    – Gokul
    4 hours ago










  • @Gokul minimum spanning tree? What?...
    – Mr. Sigma.
    3 hours ago










  • Oh, apologies. I read the title as whether minimum spanning tree have common edges.
    – Gokul
    3 hours ago


















  • Assuming the weights of the graph are distinct, the edge with the minimum weight will be present in all the minimum spanning trees.
    – Gokul
    4 hours ago










  • @Gokul minimum spanning tree? What?...
    – Mr. Sigma.
    3 hours ago










  • Oh, apologies. I read the title as whether minimum spanning tree have common edges.
    – Gokul
    3 hours ago
















Assuming the weights of the graph are distinct, the edge with the minimum weight will be present in all the minimum spanning trees.
– Gokul
4 hours ago




Assuming the weights of the graph are distinct, the edge with the minimum weight will be present in all the minimum spanning trees.
– Gokul
4 hours ago












@Gokul minimum spanning tree? What?...
– Mr. Sigma.
3 hours ago




@Gokul minimum spanning tree? What?...
– Mr. Sigma.
3 hours ago












Oh, apologies. I read the title as whether minimum spanning tree have common edges.
– Gokul
3 hours ago




Oh, apologies. I read the title as whether minimum spanning tree have common edges.
– Gokul
3 hours ago










2 Answers
2






active

oldest

votes

















up vote
6
down vote



accepted










EDIT: This is incorrect as pointed out in the comments. As the other answer says, a spanning tree for $K_4$ can be done without sharing edges.



No, it's not true that any two spanning trees of a graph have common edges.



Consider the wheel graph:



enter image description here



You can make a spanning tree with edges "inside" the loop and another one from the outer loop.






share|cite|improve this answer























  • but the outer loop doesn't reach the center node
    – amI
    1 hour ago










  • You're right, I'll delete this answer as the other one suffices.
    – Gokul
    1 hour ago






  • 1




    You can modify this by taking the out loop minus some "chord" plus some "radius" and its complement.
    – boboquack
    38 mins ago


















up vote
5
down vote













No, consider the complete graph $K_4$:



It has the following edge-disjoint spanning trees:
enter image description here






share|cite|improve this answer



















  • 2




    Oh! Elegant. Why I couldn't come upon this solution. ':O.
    – Mr. Sigma.
    2 hours ago











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
6
down vote



accepted










EDIT: This is incorrect as pointed out in the comments. As the other answer says, a spanning tree for $K_4$ can be done without sharing edges.



No, it's not true that any two spanning trees of a graph have common edges.



Consider the wheel graph:



enter image description here



You can make a spanning tree with edges "inside" the loop and another one from the outer loop.






share|cite|improve this answer























  • but the outer loop doesn't reach the center node
    – amI
    1 hour ago










  • You're right, I'll delete this answer as the other one suffices.
    – Gokul
    1 hour ago






  • 1




    You can modify this by taking the out loop minus some "chord" plus some "radius" and its complement.
    – boboquack
    38 mins ago















up vote
6
down vote



accepted










EDIT: This is incorrect as pointed out in the comments. As the other answer says, a spanning tree for $K_4$ can be done without sharing edges.



No, it's not true that any two spanning trees of a graph have common edges.



Consider the wheel graph:



enter image description here



You can make a spanning tree with edges "inside" the loop and another one from the outer loop.






share|cite|improve this answer























  • but the outer loop doesn't reach the center node
    – amI
    1 hour ago










  • You're right, I'll delete this answer as the other one suffices.
    – Gokul
    1 hour ago






  • 1




    You can modify this by taking the out loop minus some "chord" plus some "radius" and its complement.
    – boboquack
    38 mins ago













up vote
6
down vote



accepted







up vote
6
down vote



accepted






EDIT: This is incorrect as pointed out in the comments. As the other answer says, a spanning tree for $K_4$ can be done without sharing edges.



No, it's not true that any two spanning trees of a graph have common edges.



Consider the wheel graph:



enter image description here



You can make a spanning tree with edges "inside" the loop and another one from the outer loop.






share|cite|improve this answer














EDIT: This is incorrect as pointed out in the comments. As the other answer says, a spanning tree for $K_4$ can be done without sharing edges.



No, it's not true that any two spanning trees of a graph have common edges.



Consider the wheel graph:



enter image description here



You can make a spanning tree with edges "inside" the loop and another one from the outer loop.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 1 hour ago

























answered 3 hours ago









Gokul

296110




296110












  • but the outer loop doesn't reach the center node
    – amI
    1 hour ago










  • You're right, I'll delete this answer as the other one suffices.
    – Gokul
    1 hour ago






  • 1




    You can modify this by taking the out loop minus some "chord" plus some "radius" and its complement.
    – boboquack
    38 mins ago


















  • but the outer loop doesn't reach the center node
    – amI
    1 hour ago










  • You're right, I'll delete this answer as the other one suffices.
    – Gokul
    1 hour ago






  • 1




    You can modify this by taking the out loop minus some "chord" plus some "radius" and its complement.
    – boboquack
    38 mins ago
















but the outer loop doesn't reach the center node
– amI
1 hour ago




but the outer loop doesn't reach the center node
– amI
1 hour ago












You're right, I'll delete this answer as the other one suffices.
– Gokul
1 hour ago




You're right, I'll delete this answer as the other one suffices.
– Gokul
1 hour ago




1




1




You can modify this by taking the out loop minus some "chord" plus some "radius" and its complement.
– boboquack
38 mins ago




You can modify this by taking the out loop minus some "chord" plus some "radius" and its complement.
– boboquack
38 mins ago










up vote
5
down vote













No, consider the complete graph $K_4$:



It has the following edge-disjoint spanning trees:
enter image description here






share|cite|improve this answer



















  • 2




    Oh! Elegant. Why I couldn't come upon this solution. ':O.
    – Mr. Sigma.
    2 hours ago















up vote
5
down vote













No, consider the complete graph $K_4$:



It has the following edge-disjoint spanning trees:
enter image description here






share|cite|improve this answer



















  • 2




    Oh! Elegant. Why I couldn't come upon this solution. ':O.
    – Mr. Sigma.
    2 hours ago













up vote
5
down vote










up vote
5
down vote









No, consider the complete graph $K_4$:



It has the following edge-disjoint spanning trees:
enter image description here






share|cite|improve this answer














No, consider the complete graph $K_4$:



It has the following edge-disjoint spanning trees:
enter image description here







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 hours ago

























answered 3 hours ago









Bjørn Kjos-Hanssen

27417




27417








  • 2




    Oh! Elegant. Why I couldn't come upon this solution. ':O.
    – Mr. Sigma.
    2 hours ago














  • 2




    Oh! Elegant. Why I couldn't come upon this solution. ':O.
    – Mr. Sigma.
    2 hours ago








2




2




Oh! Elegant. Why I couldn't come upon this solution. ':O.
– Mr. Sigma.
2 hours ago




Oh! Elegant. Why I couldn't come upon this solution. ':O.
– Mr. Sigma.
2 hours ago


















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