Graph Problem: Reconstructing Spanning Trees

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Homework Help Overview

The discussion revolves around a graph theory problem concerning spanning trees and their properties. The original poster presents a matrix representation of spanning trees and poses a question about reconstructing these trees into columns that also represent spanning trees.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the definition of spanning trees and discuss the implications of a theorem related to edge replacement in trees. Questions arise about the validity of certain assumptions regarding distinct edges and their relationship to spanning trees.

Discussion Status

The discussion has progressed with participants sharing definitions and theorems that may assist in proving the original poster's claim. There is an indication that some participants are actively working on proofs, and at least one participant claims to have completed their proof.

Contextual Notes

Participants are navigating definitions and theorems related to spanning trees, with some uncertainty about specific conditions such as the presence of loops and the uniqueness of edges in their proofs.

user1616
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Graph problem!

Proove that:

If we have a matrix
|e11 e12 ... e1n|
|e21 e22 ... e2n|
| ..... |
|en1 en2 ... enn|

(eij edges of a graph G)
where every row is a spanning tree of G

then
there is a recomposition of every row so that the columns are also spanning trees of G.
 
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It might be good to start with the definition of a spanning tree...
 
well
A spanning tree is a cohesive connected graph (T) with no loops that contains all the vertices of G...
...any idea?
 
Refresh my memory. Is there somewhere a theorem stating that, in a graph of n+1 vertices, any set of n (distinct) edges is a spanning tree?

Edit: Oh, sorry, I should've said "any tree of n distinct edges is spanning". If the set contains a loop, it will not comprise n+1 distinct vertices.
 
Last edited:
Yes there should be no loops.
Anyway
I read a theorem that may help. It says:

We have 2 trees T,T'.
For every edge (x) that belongs to T there is an edge (Y) that belongs to T' so that:
(T-{x})U{y} and (T'-{y})U{x} are also trees.

I'm trying to use it...
 
Well using it I just prooved it for the first column.
For the next columns, in the previews theorem I should know that for every x' != x
there is a y' != y. (!= is different)

Can anybody tell me whether this is true or not?
Also if it is, does anybody have any proof?
 
Thanx anyway.
I prooved it.
 

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