Prove algorithm to find a tour in an even degree graph by induction.
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Suppose I have an even degree undirected graph G. Given an algorithm to find a tour in G FINDTOUR(G,s), where s is the starting vertex, prove by induction on the length of the walk that the only place the algorithm will get stuck is at s.
FINDTOUR(G,s): start by walking from a vertex s, at each step choose an untraversed edge incident to the current vertex until you get stuck.
Proof by induction on n, the length of the walk in an even degree graph G=(V,E)
Base Case: for n=0 the claim is trivially true as there is no tour.
Inductive Hypothesis: Suppose that for a walk of length $n = k geq 0$ the algorithm gets stuck at the starting vertex.
*Inductive Step: This is where I'm unsure of how to proceed.
Intuitively the claim makes sense because if the graph has an even degree then every vertex $v neq s$ will have an odd number of untraversed incident edges to it.
graph-theory
asked Jan 5 at 4:52
Matteo CiccozziMatteo Ciccozzi
469
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add a comment |
$begingroup$
Suppose I have an even degree undirected graph G. Given an algorithm to find a tour in G FINDTOUR(G,s), where s is the starting vertex, prove by induction on the length of the walk that the only place the algorithm will get stuck is at s.
FINDTOUR(G,s): start by walking from a vertex s, at each step choose an untraversed edge incident to the current vertex until you get stuck.
Proof by induction on n, the length of the walk in an even degree graph G=(V,E)
Base Case: for n=0 the claim is trivially true as there is no tour.
Inductive Hypothesis: Suppose that for a walk of length $n = k geq 0$ the algorithm gets stuck at the starting vertex.
*Inductive Step: This is where I'm unsure of how to proceed.
Intuitively the claim makes sense because if the graph has an even degree then every vertex $v neq s$ will have an odd number of untraversed incident edges to it.
graph-theory
asked Jan 5 at 4:52
Matteo CiccozziMatteo Ciccozzi
469
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2
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It is a strange task, because it doesn't feel like induction will be useful at all here. It's reasonably easy to prove directly, but it doesn't seem to become appreciably easier if one imagines one already has the result for smaller ... something? If the algorithm gets stuck after $n$ steps it doesn't help us to know that if it had gotten stuck after $n-1$ (or fewer) steps it would have been at $s$ -- because we're assuming it only gets stuck at step $n$ it did not in fact get stuck earlier, so the induction hypothesis is useless.
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– Henning Makholm
Jan 5 at 5:11
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Is this for a particular type of graph? There does not exist a tour of every even degree connected graph, for an example, see the Meredith Graph. Given a family of graphs, a general solution can possibly be constructed.
$endgroup$
– Zachary Hunter
Jan 5 at 7:14
add a comment |
$begingroup$
Suppose I have an even degree undirected graph G. Given an algorithm to find a tour in G FINDTOUR(G,s), where s is the starting vertex, prove by induction on the length of the walk that the only place the algorithm will get stuck is at s.
FINDTOUR(G,s): start by walking from a vertex s, at each step choose an untraversed edge incident to the current vertex until you get stuck.
Proof by induction on n, the length of the walk in an even degree graph G=(V,E)
Base Case: for n=0 the claim is trivially true as there is no tour.
Inductive Hypothesis: Suppose that for a walk of length $n = k geq 0$ the algorithm gets stuck at the starting vertex.
*Inductive Step: This is where I'm unsure of how to proceed.
Intuitively the claim makes sense because if the graph has an even degree then every vertex $v neq s$ will have an odd number of untraversed incident edges to it.
graph-theory
asked Jan 5 at 4:52
Matteo CiccozziMatteo Ciccozzi
469
$endgroup$
Suppose I have an even degree undirected graph G. Given an algorithm to find a tour in G FINDTOUR(G,s), where s is the starting vertex, prove by induction on the length of the walk that the only place the algorithm will get stuck is at s.
FINDTOUR(G,s): start by walking from a vertex s, at each step choose an untraversed edge incident to the current vertex until you get stuck.
Proof by induction on n, the length of the walk in an even degree graph G=(V,E)
Base Case: for n=0 the claim is trivially true as there is no tour.
Inductive Hypothesis: Suppose that for a walk of length $n = k geq 0$ the algorithm gets stuck at the starting vertex.
*Inductive Step: This is where I'm unsure of how to proceed.
Intuitively the claim makes sense because if the graph has an even degree then every vertex $v neq s$ will have an odd number of untraversed incident edges to it.
graph-theory
graph-theory
asked Jan 5 at 4:52
Matteo CiccozziMatteo Ciccozzi
469
asked Jan 5 at 4:52
Matteo CiccozziMatteo Ciccozzi
469
asked Jan 5 at 4:52
Matteo CiccozziMatteo Ciccozzi
469
asked Jan 5 at 4:52
Matteo CiccozziMatteo Ciccozzi
469
asked Jan 5 at 4:52
Matteo CiccozziMatteo Ciccozzi
469
469
2
$begingroup$
It is a strange task, because it doesn't feel like induction will be useful at all here. It's reasonably easy to prove directly, but it doesn't seem to become appreciably easier if one imagines one already has the result for smaller ... something? If the algorithm gets stuck after $n$ steps it doesn't help us to know that if it had gotten stuck after $n-1$ (or fewer) steps it would have been at $s$ -- because we're assuming it only gets stuck at step $n$ it did not in fact get stuck earlier, so the induction hypothesis is useless.
$endgroup$
– Henning Makholm
Jan 5 at 5:11
$begingroup$
Is this for a particular type of graph? There does not exist a tour of every even degree connected graph, for an example, see the Meredith Graph. Given a family of graphs, a general solution can possibly be constructed.
$endgroup$
– Zachary Hunter
Jan 5 at 7:14
add a comment |
2
$begingroup$
It is a strange task, because it doesn't feel like induction will be useful at all here. It's reasonably easy to prove directly, but it doesn't seem to become appreciably easier if one imagines one already has the result for smaller ... something? If the algorithm gets stuck after $n$ steps it doesn't help us to know that if it had gotten stuck after $n-1$ (or fewer) steps it would have been at $s$ -- because we're assuming it only gets stuck at step $n$ it did not in fact get stuck earlier, so the induction hypothesis is useless.
$endgroup$
– Henning Makholm
Jan 5 at 5:11
$begingroup$
Is this for a particular type of graph? There does not exist a tour of every even degree connected graph, for an example, see the Meredith Graph. Given a family of graphs, a general solution can possibly be constructed.
$endgroup$
– Zachary Hunter
Jan 5 at 7:14
2
2
$begingroup$
It is a strange task, because it doesn't feel like induction will be useful at all here. It's reasonably easy to prove directly, but it doesn't seem to become appreciably easier if one imagines one already has the result for smaller ... something? If the algorithm gets stuck after $n$ steps it doesn't help us to know that if it had gotten stuck after $n-1$ (or fewer) steps it would have been at $s$ -- because we're assuming it only gets stuck at step $n$ it did not in fact get stuck earlier, so the induction hypothesis is useless.
$endgroup$
– Henning Makholm
Jan 5 at 5:11
$begingroup$
It is a strange task, because it doesn't feel like induction will be useful at all here. It's reasonably easy to prove directly, but it doesn't seem to become appreciably easier if one imagines one already has the result for smaller ... something? If the algorithm gets stuck after $n$ steps it doesn't help us to know that if it had gotten stuck after $n-1$ (or fewer) steps it would have been at $s$ -- because we're assuming it only gets stuck at step $n$ it did not in fact get stuck earlier, so the induction hypothesis is useless.
$endgroup$
– Henning Makholm
Jan 5 at 5:11
$begingroup$
Is this for a particular type of graph? There does not exist a tour of every even degree connected graph, for an example, see the Meredith Graph. Given a family of graphs, a general solution can possibly be constructed.
$endgroup$
– Zachary Hunter
Jan 5 at 7:14
$begingroup$
Is this for a particular type of graph? There does not exist a tour of every even degree connected graph, for an example, see the Meredith Graph. Given a family of graphs, a general solution can possibly be constructed.
$endgroup$
– Zachary Hunter
Jan 5 at 7:14
add a comment |
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$begingroup$
It is a strange task, because it doesn't feel like induction will be useful at all here. It's reasonably easy to prove directly, but it doesn't seem to become appreciably easier if one imagines one already has the result for smaller ... something? If the algorithm gets stuck after $n$ steps it doesn't help us to know that if it had gotten stuck after $n-1$ (or fewer) steps it would have been at $s$ -- because we're assuming it only gets stuck at step $n$ it did not in fact get stuck earlier, so the induction hypothesis is useless.
$endgroup$
– Henning Makholm
Jan 5 at 5:11
$begingroup$
Is this for a particular type of graph? There does not exist a tour of every even degree connected graph, for an example, see the Meredith Graph. Given a family of graphs, a general solution can possibly be constructed.
$endgroup$
– Zachary Hunter
Jan 5 at 7:14