Finding paths in a graph with n vertices
Let n ≥ 2 be a natural number. Consider the graph G = (V, E) where
V ={0,1,2,...,n} and E=({0,1},{0,2},...,{0,n}) ∪ ({1,2},...,{n−1,n}) ∪ ({n,1})
For paths, it's a sequence of (non-repeating) vertices.
For cycles, we only distinguish them if they form different subgraphs.
How many paths of length 2 are there in G?
How many paths of length 3 are there in G?
How many cycles are there in G?
I can obviously draw out the first couple cases and count this, but there has to be a summation formula or something I'm missing...
discrete-mathematics graph-theory
asked Apr 3 '14 at 2:41
ConfusedGraphTheorist
11
bumped to the homepage by Community♦ 11 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
Let n ≥ 2 be a natural number. Consider the graph G = (V, E) where
V ={0,1,2,...,n} and E=({0,1},{0,2},...,{0,n}) ∪ ({1,2},...,{n−1,n}) ∪ ({n,1})
For paths, it's a sequence of (non-repeating) vertices.
For cycles, we only distinguish them if they form different subgraphs.
How many paths of length 2 are there in G?
How many paths of length 3 are there in G?
How many cycles are there in G?
I can obviously draw out the first couple cases and count this, but there has to be a summation formula or something I'm missing...
discrete-mathematics graph-theory
asked Apr 3 '14 at 2:41
ConfusedGraphTheorist
11
bumped to the homepage by Community♦ 11 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
If $d_0,d_1,dots,d_n$ is the degree sequence, then the number of paths of length $2$ (paths with $2$ edges, $3$ vertices) is $$sum_{i=0}^nbinom{d_i}2 = binom n2+nbinom32=frac{n(n+5)}2$$
– bof
Feb 8 '17 at 12:57
The number of cycles is $$1+2binom n2=n^2-n+1$$
– bof
Feb 8 '17 at 13:05
add a comment |
Let n ≥ 2 be a natural number. Consider the graph G = (V, E) where
V ={0,1,2,...,n} and E=({0,1},{0,2},...,{0,n}) ∪ ({1,2},...,{n−1,n}) ∪ ({n,1})
For paths, it's a sequence of (non-repeating) vertices.
For cycles, we only distinguish them if they form different subgraphs.
How many paths of length 2 are there in G?
How many paths of length 3 are there in G?
How many cycles are there in G?
I can obviously draw out the first couple cases and count this, but there has to be a summation formula or something I'm missing...
discrete-mathematics graph-theory
asked Apr 3 '14 at 2:41
ConfusedGraphTheorist
11
Let n ≥ 2 be a natural number. Consider the graph G = (V, E) where
V ={0,1,2,...,n} and E=({0,1},{0,2},...,{0,n}) ∪ ({1,2},...,{n−1,n}) ∪ ({n,1})
For paths, it's a sequence of (non-repeating) vertices.
For cycles, we only distinguish them if they form different subgraphs.
How many paths of length 2 are there in G?
How many paths of length 3 are there in G?
How many cycles are there in G?
I can obviously draw out the first couple cases and count this, but there has to be a summation formula or something I'm missing...
discrete-mathematics graph-theory
discrete-mathematics graph-theory
asked Apr 3 '14 at 2:41
ConfusedGraphTheorist
11
asked Apr 3 '14 at 2:41
ConfusedGraphTheorist
11
asked Apr 3 '14 at 2:41
ConfusedGraphTheorist
11
asked Apr 3 '14 at 2:41
ConfusedGraphTheorist
11
asked Apr 3 '14 at 2:41
ConfusedGraphTheorist
11
11
bumped to the homepage by Community♦ 11 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ 11 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
If $d_0,d_1,dots,d_n$ is the degree sequence, then the number of paths of length $2$ (paths with $2$ edges, $3$ vertices) is $$sum_{i=0}^nbinom{d_i}2 = binom n2+nbinom32=frac{n(n+5)}2$$
– bof
Feb 8 '17 at 12:57
The number of cycles is $$1+2binom n2=n^2-n+1$$
– bof
Feb 8 '17 at 13:05
add a comment |
If $d_0,d_1,dots,d_n$ is the degree sequence, then the number of paths of length $2$ (paths with $2$ edges, $3$ vertices) is $$sum_{i=0}^nbinom{d_i}2 = binom n2+nbinom32=frac{n(n+5)}2$$
– bof
Feb 8 '17 at 12:57
The number of cycles is $$1+2binom n2=n^2-n+1$$
– bof
Feb 8 '17 at 13:05
If $d_0,d_1,dots,d_n$ is the degree sequence, then the number of paths of length $2$ (paths with $2$ edges, $3$ vertices) is $$sum_{i=0}^nbinom{d_i}2 = binom n2+nbinom32=frac{n(n+5)}2$$
– bof
Feb 8 '17 at 12:57
If $d_0,d_1,dots,d_n$ is the degree sequence, then the number of paths of length $2$ (paths with $2$ edges, $3$ vertices) is $$sum_{i=0}^nbinom{d_i}2 = binom n2+nbinom32=frac{n(n+5)}2$$
– bof
Feb 8 '17 at 12:57
The number of cycles is $$1+2binom n2=n^2-n+1$$
– bof
Feb 8 '17 at 13:05
The number of cycles is $$1+2binom n2=n^2-n+1$$
– bof
Feb 8 '17 at 13:05
add a comment |
2 Answers
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Assume that $nge4$: smaller cases are sometimes a bit different and can be investigated individually.
Paths of length $2$ are just (directed) edges: there are $2n$ edges and allowing for the direction gives $4n$ paths.
For paths of length $3$
- without the centre vertex, choose the first vertex and the "direction of travel": $2n$ possibilities;
- starting with the centre vertex, choose the second vertex and one of its two neighbours: $2n$ possibilities;
- ending with the centre vertex: same as the previous case;
- with the centre vertex in the middle, choose the first and last vertex: they must not be the same: $n(n-1)$ possibilities.
So the total number is $n^2+5n$.
Since $nge4$, a cycle of length $3$ can only consist of two adjacent vertices on the "circumference", together with the centre: to look at it another way, one of the edges on the "circumference" together with the two edges joining it to the centre. There are $n$ possibilities.
answered Apr 3 '14 at 4:08
David
67.7k663126
add a comment |
The graph you are describing is a Wheel graph: http://en.wikipedia.org/wiki/Wheel_graph
To get the number of $P_{3}$ (a path of length $2$, which has $3$ vertices) in the graph, you consider the paths along the exterior of the graph. There are $n$ such paths, where $n = |V|$. Then you look at the interior paths (only interior edges are used) through the center vertex, which forms an arithmetic progression $sum_{i=1}^{n-1} i$. Finally, you count paths using an exterior and an interior edge. You again get $n$ such paths.
To count cycles, you have $C_{i}$, for $i in {3, ..., n}$, for $n geq 3$.
answered Apr 3 '14 at 3:15
ml0105
11.4k21538
add a comment |
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Assume that $nge4$: smaller cases are sometimes a bit different and can be investigated individually.
Paths of length $2$ are just (directed) edges: there are $2n$ edges and allowing for the direction gives $4n$ paths.
For paths of length $3$
- without the centre vertex, choose the first vertex and the "direction of travel": $2n$ possibilities;
- starting with the centre vertex, choose the second vertex and one of its two neighbours: $2n$ possibilities;
- ending with the centre vertex: same as the previous case;
- with the centre vertex in the middle, choose the first and last vertex: they must not be the same: $n(n-1)$ possibilities.
So the total number is $n^2+5n$.
Since $nge4$, a cycle of length $3$ can only consist of two adjacent vertices on the "circumference", together with the centre: to look at it another way, one of the edges on the "circumference" together with the two edges joining it to the centre. There are $n$ possibilities.
answered Apr 3 '14 at 4:08
David
67.7k663126
add a comment |
Assume that $nge4$: smaller cases are sometimes a bit different and can be investigated individually.
Paths of length $2$ are just (directed) edges: there are $2n$ edges and allowing for the direction gives $4n$ paths.
For paths of length $3$
- without the centre vertex, choose the first vertex and the "direction of travel": $2n$ possibilities;
- starting with the centre vertex, choose the second vertex and one of its two neighbours: $2n$ possibilities;
- ending with the centre vertex: same as the previous case;
- with the centre vertex in the middle, choose the first and last vertex: they must not be the same: $n(n-1)$ possibilities.
So the total number is $n^2+5n$.
Since $nge4$, a cycle of length $3$ can only consist of two adjacent vertices on the "circumference", together with the centre: to look at it another way, one of the edges on the "circumference" together with the two edges joining it to the centre. There are $n$ possibilities.
answered Apr 3 '14 at 4:08
David
67.7k663126
add a comment |
Assume that $nge4$: smaller cases are sometimes a bit different and can be investigated individually.
Paths of length $2$ are just (directed) edges: there are $2n$ edges and allowing for the direction gives $4n$ paths.
For paths of length $3$
- without the centre vertex, choose the first vertex and the "direction of travel": $2n$ possibilities;
- starting with the centre vertex, choose the second vertex and one of its two neighbours: $2n$ possibilities;
- ending with the centre vertex: same as the previous case;
- with the centre vertex in the middle, choose the first and last vertex: they must not be the same: $n(n-1)$ possibilities.
So the total number is $n^2+5n$.
Since $nge4$, a cycle of length $3$ can only consist of two adjacent vertices on the "circumference", together with the centre: to look at it another way, one of the edges on the "circumference" together with the two edges joining it to the centre. There are $n$ possibilities.
answered Apr 3 '14 at 4:08
David
67.7k663126
Assume that $nge4$: smaller cases are sometimes a bit different and can be investigated individually.
Paths of length $2$ are just (directed) edges: there are $2n$ edges and allowing for the direction gives $4n$ paths.
For paths of length $3$
- without the centre vertex, choose the first vertex and the "direction of travel": $2n$ possibilities;
- starting with the centre vertex, choose the second vertex and one of its two neighbours: $2n$ possibilities;
- ending with the centre vertex: same as the previous case;
- with the centre vertex in the middle, choose the first and last vertex: they must not be the same: $n(n-1)$ possibilities.
So the total number is $n^2+5n$.
Since $nge4$, a cycle of length $3$ can only consist of two adjacent vertices on the "circumference", together with the centre: to look at it another way, one of the edges on the "circumference" together with the two edges joining it to the centre. There are $n$ possibilities.
answered Apr 3 '14 at 4:08
David
67.7k663126
answered Apr 3 '14 at 4:08
David
67.7k663126
answered Apr 3 '14 at 4:08
David
67.7k663126
answered Apr 3 '14 at 4:08
David
67.7k663126
67.7k663126
add a comment |
add a comment |
The graph you are describing is a Wheel graph: http://en.wikipedia.org/wiki/Wheel_graph
To get the number of $P_{3}$ (a path of length $2$, which has $3$ vertices) in the graph, you consider the paths along the exterior of the graph. There are $n$ such paths, where $n = |V|$. Then you look at the interior paths (only interior edges are used) through the center vertex, which forms an arithmetic progression $sum_{i=1}^{n-1} i$. Finally, you count paths using an exterior and an interior edge. You again get $n$ such paths.
To count cycles, you have $C_{i}$, for $i in {3, ..., n}$, for $n geq 3$.
answered Apr 3 '14 at 3:15
ml0105
11.4k21538
add a comment |
The graph you are describing is a Wheel graph: http://en.wikipedia.org/wiki/Wheel_graph
To get the number of $P_{3}$ (a path of length $2$, which has $3$ vertices) in the graph, you consider the paths along the exterior of the graph. There are $n$ such paths, where $n = |V|$. Then you look at the interior paths (only interior edges are used) through the center vertex, which forms an arithmetic progression $sum_{i=1}^{n-1} i$. Finally, you count paths using an exterior and an interior edge. You again get $n$ such paths.
To count cycles, you have $C_{i}$, for $i in {3, ..., n}$, for $n geq 3$.
answered Apr 3 '14 at 3:15
ml0105
11.4k21538
add a comment |
The graph you are describing is a Wheel graph: http://en.wikipedia.org/wiki/Wheel_graph
To get the number of $P_{3}$ (a path of length $2$, which has $3$ vertices) in the graph, you consider the paths along the exterior of the graph. There are $n$ such paths, where $n = |V|$. Then you look at the interior paths (only interior edges are used) through the center vertex, which forms an arithmetic progression $sum_{i=1}^{n-1} i$. Finally, you count paths using an exterior and an interior edge. You again get $n$ such paths.
To count cycles, you have $C_{i}$, for $i in {3, ..., n}$, for $n geq 3$.
answered Apr 3 '14 at 3:15
ml0105
11.4k21538
The graph you are describing is a Wheel graph: http://en.wikipedia.org/wiki/Wheel_graph
To get the number of $P_{3}$ (a path of length $2$, which has $3$ vertices) in the graph, you consider the paths along the exterior of the graph. There are $n$ such paths, where $n = |V|$. Then you look at the interior paths (only interior edges are used) through the center vertex, which forms an arithmetic progression $sum_{i=1}^{n-1} i$. Finally, you count paths using an exterior and an interior edge. You again get $n$ such paths.
To count cycles, you have $C_{i}$, for $i in {3, ..., n}$, for $n geq 3$.
answered Apr 3 '14 at 3:15
ml0105
11.4k21538
edited Apr 3 '14 at 4:14
answered Apr 3 '14 at 3:15
ml0105
11.4k21538
answered Apr 3 '14 at 3:15
ml0105
11.4k21538
answered Apr 3 '14 at 3:15
ml0105
11.4k21538
11.4k21538
add a comment |
add a comment |
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If $d_0,d_1,dots,d_n$ is the degree sequence, then the number of paths of length $2$ (paths with $2$ edges, $3$ vertices) is $$sum_{i=0}^nbinom{d_i}2 = binom n2+nbinom32=frac{n(n+5)}2$$
– bof
Feb 8 '17 at 12:57
The number of cycles is $$1+2binom n2=n^2-n+1$$
– bof
Feb 8 '17 at 13:05