Hamilton Paths in Complete graph $K_n$
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In complete graph $K_n$, is it true that we can have at least $2*n$ Hamilton paths?
graph-theory algorithms hamiltonian-path
asked Dec 18 '18 at 22:01
JuneJune
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0
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In complete graph $K_n$, is it true that we can have at least $2*n$ Hamilton paths?
graph-theory algorithms hamiltonian-path
asked Dec 18 '18 at 22:01
JuneJune
122
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add a comment |
0
0
0
$begingroup$
In complete graph $K_n$, is it true that we can have at least $2*n$ Hamilton paths?
graph-theory algorithms hamiltonian-path
asked Dec 18 '18 at 22:01
JuneJune
122
$endgroup$
In complete graph $K_n$, is it true that we can have at least $2*n$ Hamilton paths?
graph-theory algorithms hamiltonian-path
graph-theory algorithms hamiltonian-path
asked Dec 18 '18 at 22:01
JuneJune
122
asked Dec 18 '18 at 22:01
JuneJune
122
asked Dec 18 '18 at 22:01
JuneJune
122
asked Dec 18 '18 at 22:01
JuneJune
122
asked Dec 18 '18 at 22:01
JuneJune
122
122
add a comment |
add a comment |
1 Answer
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Since in a complete graph $K_n$ each two vertices are adjacent, the set of Hamiltonian paths of $K_n$ is exactly a set $S_n$ of permutations of its vertices. But $|S_n|=n!$, which is at least $2n$ when $nge 3$, but it is less than $2n$ for $nle 2$.
answered Dec 19 '18 at 6:45
Alex RavskyAlex Ravsky
41.4k32282
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I know it’s stupid, but you have to divide by two since the reverse of a permutation gives the same Hamiltonian path.
$endgroup$
– Bob Krueger
Dec 19 '18 at 8:50
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@BobKrueger I was aware of this point, so I have checked it (in “Chromatic Graph Theory” by Chartrand and Zhang). A path is defined as a sequence of vertices such that... . So, for instance, since the sequences $(1,2,3)$ and $(3,2,1)$ are distinct, these paths are distinct too.
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– Alex Ravsky
Dec 21 '18 at 7:02
1
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Fair enough. Then with that, the OP can prove their claim for any graph with a Hamiltonian cycle.
$endgroup$
– Bob Krueger
Dec 21 '18 at 9:05
add a comment |
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1 Answer
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1 Answer
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1
$begingroup$
Since in a complete graph $K_n$ each two vertices are adjacent, the set of Hamiltonian paths of $K_n$ is exactly a set $S_n$ of permutations of its vertices. But $|S_n|=n!$, which is at least $2n$ when $nge 3$, but it is less than $2n$ for $nle 2$.
answered Dec 19 '18 at 6:45
Alex RavskyAlex Ravsky
41.4k32282
$endgroup$
$begingroup$
I know it’s stupid, but you have to divide by two since the reverse of a permutation gives the same Hamiltonian path.
$endgroup$
– Bob Krueger
Dec 19 '18 at 8:50
$begingroup$
@BobKrueger I was aware of this point, so I have checked it (in “Chromatic Graph Theory” by Chartrand and Zhang). A path is defined as a sequence of vertices such that... . So, for instance, since the sequences $(1,2,3)$ and $(3,2,1)$ are distinct, these paths are distinct too.
$endgroup$
– Alex Ravsky
Dec 21 '18 at 7:02
1
$begingroup$
Fair enough. Then with that, the OP can prove their claim for any graph with a Hamiltonian cycle.
$endgroup$
– Bob Krueger
Dec 21 '18 at 9:05
add a comment |
1
$begingroup$
Since in a complete graph $K_n$ each two vertices are adjacent, the set of Hamiltonian paths of $K_n$ is exactly a set $S_n$ of permutations of its vertices. But $|S_n|=n!$, which is at least $2n$ when $nge 3$, but it is less than $2n$ for $nle 2$.
answered Dec 19 '18 at 6:45
Alex RavskyAlex Ravsky
41.4k32282
$endgroup$
$begingroup$
I know it’s stupid, but you have to divide by two since the reverse of a permutation gives the same Hamiltonian path.
$endgroup$
– Bob Krueger
Dec 19 '18 at 8:50
$begingroup$
@BobKrueger I was aware of this point, so I have checked it (in “Chromatic Graph Theory” by Chartrand and Zhang). A path is defined as a sequence of vertices such that... . So, for instance, since the sequences $(1,2,3)$ and $(3,2,1)$ are distinct, these paths are distinct too.
$endgroup$
– Alex Ravsky
Dec 21 '18 at 7:02
1
$begingroup$
Fair enough. Then with that, the OP can prove their claim for any graph with a Hamiltonian cycle.
$endgroup$
– Bob Krueger
Dec 21 '18 at 9:05
add a comment |
1
1
1
$begingroup$
Since in a complete graph $K_n$ each two vertices are adjacent, the set of Hamiltonian paths of $K_n$ is exactly a set $S_n$ of permutations of its vertices. But $|S_n|=n!$, which is at least $2n$ when $nge 3$, but it is less than $2n$ for $nle 2$.
answered Dec 19 '18 at 6:45
Alex RavskyAlex Ravsky
41.4k32282
$endgroup$
Since in a complete graph $K_n$ each two vertices are adjacent, the set of Hamiltonian paths of $K_n$ is exactly a set $S_n$ of permutations of its vertices. But $|S_n|=n!$, which is at least $2n$ when $nge 3$, but it is less than $2n$ for $nle 2$.
answered Dec 19 '18 at 6:45
Alex RavskyAlex Ravsky
41.4k32282
answered Dec 19 '18 at 6:45
Alex RavskyAlex Ravsky
41.4k32282
answered Dec 19 '18 at 6:45
Alex RavskyAlex Ravsky
41.4k32282
answered Dec 19 '18 at 6:45
Alex RavskyAlex Ravsky
41.4k32282
41.4k32282
$begingroup$
I know it’s stupid, but you have to divide by two since the reverse of a permutation gives the same Hamiltonian path.
$endgroup$
– Bob Krueger
Dec 19 '18 at 8:50
$begingroup$
@BobKrueger I was aware of this point, so I have checked it (in “Chromatic Graph Theory” by Chartrand and Zhang). A path is defined as a sequence of vertices such that... . So, for instance, since the sequences $(1,2,3)$ and $(3,2,1)$ are distinct, these paths are distinct too.
$endgroup$
– Alex Ravsky
Dec 21 '18 at 7:02
1
$begingroup$
Fair enough. Then with that, the OP can prove their claim for any graph with a Hamiltonian cycle.
$endgroup$
– Bob Krueger
Dec 21 '18 at 9:05
add a comment |
$begingroup$
I know it’s stupid, but you have to divide by two since the reverse of a permutation gives the same Hamiltonian path.
$endgroup$
– Bob Krueger
Dec 19 '18 at 8:50
$begingroup$
@BobKrueger I was aware of this point, so I have checked it (in “Chromatic Graph Theory” by Chartrand and Zhang). A path is defined as a sequence of vertices such that... . So, for instance, since the sequences $(1,2,3)$ and $(3,2,1)$ are distinct, these paths are distinct too.
$endgroup$
– Alex Ravsky
Dec 21 '18 at 7:02
1
$begingroup$
Fair enough. Then with that, the OP can prove their claim for any graph with a Hamiltonian cycle.
$endgroup$
– Bob Krueger
Dec 21 '18 at 9:05
$begingroup$
I know it’s stupid, but you have to divide by two since the reverse of a permutation gives the same Hamiltonian path.
$endgroup$
– Bob Krueger
Dec 19 '18 at 8:50
$begingroup$
I know it’s stupid, but you have to divide by two since the reverse of a permutation gives the same Hamiltonian path.
$endgroup$
– Bob Krueger
Dec 19 '18 at 8:50
$begingroup$
@BobKrueger I was aware of this point, so I have checked it (in “Chromatic Graph Theory” by Chartrand and Zhang). A path is defined as a sequence of vertices such that... . So, for instance, since the sequences $(1,2,3)$ and $(3,2,1)$ are distinct, these paths are distinct too.
$endgroup$
– Alex Ravsky
Dec 21 '18 at 7:02
$begingroup$
@BobKrueger I was aware of this point, so I have checked it (in “Chromatic Graph Theory” by Chartrand and Zhang). A path is defined as a sequence of vertices such that... . So, for instance, since the sequences $(1,2,3)$ and $(3,2,1)$ are distinct, these paths are distinct too.
$endgroup$
– Alex Ravsky
Dec 21 '18 at 7:02
1
1
$begingroup$
Fair enough. Then with that, the OP can prove their claim for any graph with a Hamiltonian cycle.
$endgroup$
– Bob Krueger
Dec 21 '18 at 9:05
$begingroup$
Fair enough. Then with that, the OP can prove their claim for any graph with a Hamiltonian cycle.
$endgroup$
– Bob Krueger
Dec 21 '18 at 9:05
add a comment |
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