How many Hamiltonian cycles in a complete graph cover edges that don't share vertices?
$begingroup$
Consider a complete graph, K, that has n vertices.
There is a set of edges within K that have a common property, which is that they do not share vertices anywhere on the graph. Let's call these set of edges "M".
The amount of edges in set M is limited: |M| <= n/2
Question: How many Hamiltonian cycles in graph K contain all the edges in set M? Give your answer in terms of n and |M|.
For the sake of this exercise, let's pretend graph K has 6 vertices.
My answer is that only one Hamiltonian cycle gets to cover all the edges in set M.
The reason for this is that it takes n/2 edges (in this case, 3) to surround the outer circuit of any complete graph.
Once these edges take hold you can't really make any edges inside the graph part of set M because they'd be sharing vertice with those on the outer layer.
Likewise, if you make the edges inside the graph members of set M, you won't be able to make edges on the outer circuit part of set M.
Either way, it seems to me only one Hamiltonian cycle can contain all members of set M.
Am I interpreting this correctly?
graph-theory hamiltonian-path
asked Nov 30 '18 at 7:40
potatoguypotatoguy
525
$endgroup$
add a comment |
$begingroup$
Consider a complete graph, K, that has n vertices.
There is a set of edges within K that have a common property, which is that they do not share vertices anywhere on the graph. Let's call these set of edges "M".
The amount of edges in set M is limited: |M| <= n/2
Question: How many Hamiltonian cycles in graph K contain all the edges in set M? Give your answer in terms of n and |M|.
For the sake of this exercise, let's pretend graph K has 6 vertices.
My answer is that only one Hamiltonian cycle gets to cover all the edges in set M.
The reason for this is that it takes n/2 edges (in this case, 3) to surround the outer circuit of any complete graph.
Once these edges take hold you can't really make any edges inside the graph part of set M because they'd be sharing vertice with those on the outer layer.
Likewise, if you make the edges inside the graph members of set M, you won't be able to make edges on the outer circuit part of set M.
Either way, it seems to me only one Hamiltonian cycle can contain all members of set M.
Am I interpreting this correctly?
graph-theory hamiltonian-path
asked Nov 30 '18 at 7:40
potatoguypotatoguy
525
$endgroup$
add a comment |
$begingroup$
Consider a complete graph, K, that has n vertices.
There is a set of edges within K that have a common property, which is that they do not share vertices anywhere on the graph. Let's call these set of edges "M".
The amount of edges in set M is limited: |M| <= n/2
Question: How many Hamiltonian cycles in graph K contain all the edges in set M? Give your answer in terms of n and |M|.
For the sake of this exercise, let's pretend graph K has 6 vertices.
My answer is that only one Hamiltonian cycle gets to cover all the edges in set M.
The reason for this is that it takes n/2 edges (in this case, 3) to surround the outer circuit of any complete graph.
Once these edges take hold you can't really make any edges inside the graph part of set M because they'd be sharing vertice with those on the outer layer.
Likewise, if you make the edges inside the graph members of set M, you won't be able to make edges on the outer circuit part of set M.
Either way, it seems to me only one Hamiltonian cycle can contain all members of set M.
Am I interpreting this correctly?
graph-theory hamiltonian-path
asked Nov 30 '18 at 7:40
potatoguypotatoguy
525
$endgroup$
Consider a complete graph, K, that has n vertices.
There is a set of edges within K that have a common property, which is that they do not share vertices anywhere on the graph. Let's call these set of edges "M".
The amount of edges in set M is limited: |M| <= n/2
Question: How many Hamiltonian cycles in graph K contain all the edges in set M? Give your answer in terms of n and |M|.
For the sake of this exercise, let's pretend graph K has 6 vertices.
My answer is that only one Hamiltonian cycle gets to cover all the edges in set M.
The reason for this is that it takes n/2 edges (in this case, 3) to surround the outer circuit of any complete graph.
Once these edges take hold you can't really make any edges inside the graph part of set M because they'd be sharing vertice with those on the outer layer.
Likewise, if you make the edges inside the graph members of set M, you won't be able to make edges on the outer circuit part of set M.
Either way, it seems to me only one Hamiltonian cycle can contain all members of set M.
Am I interpreting this correctly?
graph-theory hamiltonian-path
graph-theory hamiltonian-path
asked Nov 30 '18 at 7:40
potatoguypotatoguy
525
asked Nov 30 '18 at 7:40
potatoguypotatoguy
525
asked Nov 30 '18 at 7:40
potatoguypotatoguy
525
asked Nov 30 '18 at 7:40
potatoguypotatoguy
525
asked Nov 30 '18 at 7:40
potatoguypotatoguy
525
525
add a comment |
add a comment |
1 Answer
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Not quite. The reasoning here is that you can change both the order these special edges appear in the cycle, as well as the order in which they are traversed (which vertex is seen first). In fact, if we let $|M| = m$, the answer should be $boxed{2^{m-1}(n-m-1)!}$; a proof can be found here.
The counting argument there can be summarized as follows: Fix a particular edge $e in M$ and fix a direction to traverse this edge. Construct the rest of the Hamiltonian cycle by permuting the set of edges in $M setminus {e}$ and the vertices which are not endpoints of any edge in $M$. There are $n - m - 1$ such entities ($n-2m$ vertices which are not endpoints of edges in $M$ and $m-1$ edges in $M setminus {e}$), so $(n-m-1)!$ ways to permute them. But we also have to choose the direction each edge in $M$ is traversed (other than the one which we have already fixed); this adds the additional factor of $2^{m-1}$.
answered Nov 30 '18 at 8:36
plattyplatty
3,370320
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1 Answer
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active
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1 Answer
1
active
oldest
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$begingroup$
Not quite. The reasoning here is that you can change both the order these special edges appear in the cycle, as well as the order in which they are traversed (which vertex is seen first). In fact, if we let $|M| = m$, the answer should be $boxed{2^{m-1}(n-m-1)!}$; a proof can be found here.
The counting argument there can be summarized as follows: Fix a particular edge $e in M$ and fix a direction to traverse this edge. Construct the rest of the Hamiltonian cycle by permuting the set of edges in $M setminus {e}$ and the vertices which are not endpoints of any edge in $M$. There are $n - m - 1$ such entities ($n-2m$ vertices which are not endpoints of edges in $M$ and $m-1$ edges in $M setminus {e}$), so $(n-m-1)!$ ways to permute them. But we also have to choose the direction each edge in $M$ is traversed (other than the one which we have already fixed); this adds the additional factor of $2^{m-1}$.
answered Nov 30 '18 at 8:36
plattyplatty
3,370320
$endgroup$
add a comment |
$begingroup$
Not quite. The reasoning here is that you can change both the order these special edges appear in the cycle, as well as the order in which they are traversed (which vertex is seen first). In fact, if we let $|M| = m$, the answer should be $boxed{2^{m-1}(n-m-1)!}$; a proof can be found here.
The counting argument there can be summarized as follows: Fix a particular edge $e in M$ and fix a direction to traverse this edge. Construct the rest of the Hamiltonian cycle by permuting the set of edges in $M setminus {e}$ and the vertices which are not endpoints of any edge in $M$. There are $n - m - 1$ such entities ($n-2m$ vertices which are not endpoints of edges in $M$ and $m-1$ edges in $M setminus {e}$), so $(n-m-1)!$ ways to permute them. But we also have to choose the direction each edge in $M$ is traversed (other than the one which we have already fixed); this adds the additional factor of $2^{m-1}$.
answered Nov 30 '18 at 8:36
plattyplatty
3,370320
$endgroup$
add a comment |
$begingroup$
Not quite. The reasoning here is that you can change both the order these special edges appear in the cycle, as well as the order in which they are traversed (which vertex is seen first). In fact, if we let $|M| = m$, the answer should be $boxed{2^{m-1}(n-m-1)!}$; a proof can be found here.
The counting argument there can be summarized as follows: Fix a particular edge $e in M$ and fix a direction to traverse this edge. Construct the rest of the Hamiltonian cycle by permuting the set of edges in $M setminus {e}$ and the vertices which are not endpoints of any edge in $M$. There are $n - m - 1$ such entities ($n-2m$ vertices which are not endpoints of edges in $M$ and $m-1$ edges in $M setminus {e}$), so $(n-m-1)!$ ways to permute them. But we also have to choose the direction each edge in $M$ is traversed (other than the one which we have already fixed); this adds the additional factor of $2^{m-1}$.
answered Nov 30 '18 at 8:36
plattyplatty
3,370320
$endgroup$
Not quite. The reasoning here is that you can change both the order these special edges appear in the cycle, as well as the order in which they are traversed (which vertex is seen first). In fact, if we let $|M| = m$, the answer should be $boxed{2^{m-1}(n-m-1)!}$; a proof can be found here.
The counting argument there can be summarized as follows: Fix a particular edge $e in M$ and fix a direction to traverse this edge. Construct the rest of the Hamiltonian cycle by permuting the set of edges in $M setminus {e}$ and the vertices which are not endpoints of any edge in $M$. There are $n - m - 1$ such entities ($n-2m$ vertices which are not endpoints of edges in $M$ and $m-1$ edges in $M setminus {e}$), so $(n-m-1)!$ ways to permute them. But we also have to choose the direction each edge in $M$ is traversed (other than the one which we have already fixed); this adds the additional factor of $2^{m-1}$.
answered Nov 30 '18 at 8:36
plattyplatty
3,370320
answered Nov 30 '18 at 8:36
plattyplatty
3,370320
answered Nov 30 '18 at 8:36
plattyplatty
3,370320
answered Nov 30 '18 at 8:36
plattyplatty
3,370320
3,370320
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