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Title: Monotone Paths in Graphs Post by william wu on Nov 8th, 2011, 1:45pm Given an undirected connected graph with E edges, arbitrarily each edge with a unique number between 1 and E. Show that there exists a path with monotone labels whose length is at least the average degree of the graph. |
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Title: Re: Monotone Paths in Graphs Post by Michael Dagg on Nov 9th, 2011, 8:40am Your problem "hints" at a Ramsey theory solution using the Erdos result on increasing or decreasing sequences of a certain length. |
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Title: Re: Monotone Paths in Graphs Post by Grimbal on Nov 10th, 2011, 6:58am Hm... I must be missing something. Is the "average degree" the average number of edges that meet at a vertex? By saying that the statement is true for a random placement of labels, do you mean it is true for all placements? |
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Title: Re: Monotone Paths in Graphs Post by william wu on Nov 10th, 2011, 11:15am Yes, the average degree is the average # of edges that meet at a vertex. Also it is more clear to say arbitrary labeling rather than random labeling, so I will revise that in the problem statement. The labels are also being assigned to edges, not to vertices. A path is considered monotone if the *edges* in that path are monotone. Attached is a visualization of the bipartite graph you mentioned, where the vertices {1,2,3,4} are complete to the vertices {5,6,7,8}. We see there are many monotone paths of length >= 4, such as the sequence of edges {1,3,7,9,11}. The sixteen distinct edge labels were assigned randomly without replacement. |
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Title: Re: Monotone Paths in Graphs Post by william wu on Nov 10th, 2011, 11:19am This problem may be too difficult without a hint, as I was simply told of the result: Hint: [hide] Imagine having a person standing on each vertex. A loudspeaker shouts the edges in sequence, and with each shout, the people move along the graph in a certain way. [/hide] |
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Title: Re: Monotone Paths in Graphs Post by Grimbal on Nov 10th, 2011, 1:24pm on 11/10/11 at 11:15:41, william wu wrote:
I see. Sorry. I don't know why, I always mix up the words "edge" and "vertex" in english. Really annoying. I must have learnt it wrong the first time. OK, Now I can start thinking about the problem. |
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Title: Re: Monotone Paths in Graphs Post by rmsgrey on Nov 11th, 2011, 4:43am on 11/10/11 at 11:19:11, william wu wrote:
Well, that hint makes it obvious: [hideb] Start with an empty pile of counters on each node. For each arc in order, swap the two piles at either end of the arc and add a counter to each. At the end of the process, each pile has followed a monotone path, of length equal to the number of counters in that pile. Since the total number of counters is twice the number of arcs, which is also the sum of the degrees of the individual nodes, and the number of piles is the number of nodes, the average number of counters per pile is exactly equal to the average degree. By the pigeonhole principle, at least one pile must have at least the average number of counters, so at least one pile must have traversed a path of length at least equal to the average degree.[/hideb] |
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Title: Re: Monotone Paths in Graphs Post by Michael Dagg on Nov 11th, 2011, 7:41am Excellent rmsgrey! Ramsey theory generalizes the pigeonhole principle, guaranteeing the existence of the required monotonic path. There are lots of fruitful results in graph/group theory similarly derived. |
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Title: Re: Monotone Paths in Graphs Post by Grimbal on Nov 11th, 2011, 9:49am Very elegant proof, indeed! |
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Title: Re: Monotone Paths in Graphs Post by mgccl on Feb 4th, 2013, 5:37pm on 11/11/11 at 04:43:24, rmsgrey wrote:
It followed a monotone walk instead of a path. Usually path means a simple path. So it is still unclear if the original statement is true. |
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Title: Re: Monotone Paths in Graphs Post by rmsgrey on Feb 5th, 2013, 4:34am on 02/04/13 at 17:37:32, mgccl wrote:
True, there's no guarantee against loops in that proof. |
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