Components of graph with edges given by congruence condition
Let $G_n = (V_n, E_n)$ denote the graph with vertex set $V_n = {0, 1, 2, dotsc, n – 1}$ and edge set $E_n = left{{i, j}: i + j equiv 1 pmod{n}right}$ where $n ge 3$. How many components does $G_n$ have for $n ge 3$?
The answer is $lceil n/2 rceil$ if we allow loops. I've gotten the answer by making graphs for different values of $n$. Can anyone help with a constructive proof?
combinatorics graph-theory
edited Nov 29 '16 at 6:21
Gerry Myerson
146k8147298
asked Nov 29 '16 at 3:04
Mr.Sigma.Mr.Sigma.
1719
|
show 8 more comments
Let $G_n = (V_n, E_n)$ denote the graph with vertex set $V_n = {0, 1, 2, dotsc, n – 1}$ and edge set $E_n = left{{i, j}: i + j equiv 1 pmod{n}right}$ where $n ge 3$. How many components does $G_n$ have for $n ge 3$?
The answer is $lceil n/2 rceil$ if we allow loops. I've gotten the answer by making graphs for different values of $n$. Can anyone help with a constructive proof?
combinatorics graph-theory
edited Nov 29 '16 at 6:21
Gerry Myerson
146k8147298
asked Nov 29 '16 at 3:04
Mr.Sigma.Mr.Sigma.
1719
If you've been constructing the graphs by hand, you should see a pattern. What's the pattern you see?
– Mike Pierce
Nov 29 '16 at 3:16
Sir +Mike, Thanks for you comment. What I see is - this question is reduced to the combinatorial problem of finding number of combination (x,y) = n+1.
– Mr.Sigma.
Nov 29 '16 at 4:04
Basically, yeah, ... and what's the pattern you see when drawing the graphs? What led you to conclude that the answer should be $lceil n/2 rceil$?
– Mike Pierce
Nov 29 '16 at 4:07
Sir +Mike, I've just tried with n=3,4,5 & got the answer. But the pattern I see somewhat is like (n/2,n/2+1), (n/2-1,n/2+2),(n/2-2,n/2+3) ... Am I correct? Apart from trivial (0,1) edge.
– Mr.Sigma.
Nov 29 '16 at 4:10
Thanks Sir +Mike for helping me, I got it counting terms of this series alone. :)
– Mr.Sigma.
Nov 29 '16 at 4:19
|
show 8 more comments
Let $G_n = (V_n, E_n)$ denote the graph with vertex set $V_n = {0, 1, 2, dotsc, n – 1}$ and edge set $E_n = left{{i, j}: i + j equiv 1 pmod{n}right}$ where $n ge 3$. How many components does $G_n$ have for $n ge 3$?
The answer is $lceil n/2 rceil$ if we allow loops. I've gotten the answer by making graphs for different values of $n$. Can anyone help with a constructive proof?
combinatorics graph-theory
edited Nov 29 '16 at 6:21
Gerry Myerson
146k8147298
asked Nov 29 '16 at 3:04
Mr.Sigma.Mr.Sigma.
1719
Let $G_n = (V_n, E_n)$ denote the graph with vertex set $V_n = {0, 1, 2, dotsc, n – 1}$ and edge set $E_n = left{{i, j}: i + j equiv 1 pmod{n}right}$ where $n ge 3$. How many components does $G_n$ have for $n ge 3$?
The answer is $lceil n/2 rceil$ if we allow loops. I've gotten the answer by making graphs for different values of $n$. Can anyone help with a constructive proof?
combinatorics graph-theory
combinatorics graph-theory
edited Nov 29 '16 at 6:21
Gerry Myerson
146k8147298
asked Nov 29 '16 at 3:04
Mr.Sigma.Mr.Sigma.
1719
edited Nov 29 '16 at 6:21
Gerry Myerson
146k8147298
asked Nov 29 '16 at 3:04
Mr.Sigma.Mr.Sigma.
1719
edited Nov 29 '16 at 6:21
Gerry Myerson
146k8147298
edited Nov 29 '16 at 6:21
Gerry Myerson
146k8147298
edited Nov 29 '16 at 6:21
Gerry Myerson
146k8147298
146k8147298
asked Nov 29 '16 at 3:04
Mr.Sigma.Mr.Sigma.
1719
asked Nov 29 '16 at 3:04
Mr.Sigma.Mr.Sigma.
1719
asked Nov 29 '16 at 3:04
Mr.Sigma.Mr.Sigma.
1719
1719
If you've been constructing the graphs by hand, you should see a pattern. What's the pattern you see?
– Mike Pierce
Nov 29 '16 at 3:16
Sir +Mike, Thanks for you comment. What I see is - this question is reduced to the combinatorial problem of finding number of combination (x,y) = n+1.
– Mr.Sigma.
Nov 29 '16 at 4:04
Basically, yeah, ... and what's the pattern you see when drawing the graphs? What led you to conclude that the answer should be $lceil n/2 rceil$?
– Mike Pierce
Nov 29 '16 at 4:07
Sir +Mike, I've just tried with n=3,4,5 & got the answer. But the pattern I see somewhat is like (n/2,n/2+1), (n/2-1,n/2+2),(n/2-2,n/2+3) ... Am I correct? Apart from trivial (0,1) edge.
– Mr.Sigma.
Nov 29 '16 at 4:10
Thanks Sir +Mike for helping me, I got it counting terms of this series alone. :)
– Mr.Sigma.
Nov 29 '16 at 4:19
|
show 8 more comments
If you've been constructing the graphs by hand, you should see a pattern. What's the pattern you see?
– Mike Pierce
Nov 29 '16 at 3:16
Sir +Mike, Thanks for you comment. What I see is - this question is reduced to the combinatorial problem of finding number of combination (x,y) = n+1.
– Mr.Sigma.
Nov 29 '16 at 4:04
Basically, yeah, ... and what's the pattern you see when drawing the graphs? What led you to conclude that the answer should be $lceil n/2 rceil$?
– Mike Pierce
Nov 29 '16 at 4:07
Sir +Mike, I've just tried with n=3,4,5 & got the answer. But the pattern I see somewhat is like (n/2,n/2+1), (n/2-1,n/2+2),(n/2-2,n/2+3) ... Am I correct? Apart from trivial (0,1) edge.
– Mr.Sigma.
Nov 29 '16 at 4:10
Thanks Sir +Mike for helping me, I got it counting terms of this series alone. :)
– Mr.Sigma.
Nov 29 '16 at 4:19
If you've been constructing the graphs by hand, you should see a pattern. What's the pattern you see?
– Mike Pierce
Nov 29 '16 at 3:16
If you've been constructing the graphs by hand, you should see a pattern. What's the pattern you see?
– Mike Pierce
Nov 29 '16 at 3:16
Sir +Mike, Thanks for you comment. What I see is - this question is reduced to the combinatorial problem of finding number of combination (x,y) = n+1.
– Mr.Sigma.
Nov 29 '16 at 4:04
Sir +Mike, Thanks for you comment. What I see is - this question is reduced to the combinatorial problem of finding number of combination (x,y) = n+1.
– Mr.Sigma.
Nov 29 '16 at 4:04
Basically, yeah, ... and what's the pattern you see when drawing the graphs? What led you to conclude that the answer should be $lceil n/2 rceil$?
– Mike Pierce
Nov 29 '16 at 4:07
Basically, yeah, ... and what's the pattern you see when drawing the graphs? What led you to conclude that the answer should be $lceil n/2 rceil$?
– Mike Pierce
Nov 29 '16 at 4:07
Sir +Mike, I've just tried with n=3,4,5 & got the answer. But the pattern I see somewhat is like (n/2,n/2+1), (n/2-1,n/2+2),(n/2-2,n/2+3) ... Am I correct? Apart from trivial (0,1) edge.
– Mr.Sigma.
Nov 29 '16 at 4:10
Sir +Mike, I've just tried with n=3,4,5 & got the answer. But the pattern I see somewhat is like (n/2,n/2+1), (n/2-1,n/2+2),(n/2-2,n/2+3) ... Am I correct? Apart from trivial (0,1) edge.
– Mr.Sigma.
Nov 29 '16 at 4:10
Thanks Sir +Mike for helping me, I got it counting terms of this series alone. :)
– Mr.Sigma.
Nov 29 '16 at 4:19
Thanks Sir +Mike for helping me, I got it counting terms of this series alone. :)
– Mr.Sigma.
Nov 29 '16 at 4:19
|
show 8 more comments
2 Answers
2
active
oldest
votes
Since we are working $bmod n$, I think it would be easier to name the vertices ${1, dotsc, n}$ instead; the problem is the same. We see pretty quickly by playing around that the edges
$$(1,n)quad(2,n!-!1)quaddotsbquadleft(lfloor n/2rfloor, 1+lfloor n/2rfloorright)$$
must be in our graph. So there are at most $lceil n/2 rceil$ connected components, each component having one edge except for a single component with an isolated vertex if $n$ is odd. This is also all the components. On the contrary if we suppose that our graph has some edges $(a,b)$ and $(b,c)$ for distinct vertices $a$ and $c$, we would have
$$a+b equiv 1 quadtext{and}quad b+c equiv 1;.$$
Subtracting these two congruence we get that $aequiv c pmod n$. But since our vertices are just ${1,dotsc,n}$ this is only possible if $a=c$.
answered Nov 30 '16 at 1:15
Mike PierceMike Pierce
11.4k103583
add a comment |
Here is how I solved the problem.
Since sum would be $n+1$. Patterns of edges would be like this:
$(lfloor{(n/2)}rfloor, lfloor{(n/2)}rfloor +1), (lfloor{(n/2)}rfloor-1, lfloor{(n/2)}rfloor +2),...,(2,lfloor{(n/2)}rfloor+lfloor{(n/2)}rfloor
-1)$
Therefore total pairs $= lfloor{(n/2)}rfloor -1 -1+1 = lfloor{(n/2)}rfloor -1$
Considering trivial $(0,1)$ Total pairs or edges are $= lfloor{(n/2)}rfloor$.
Now, with $n$ vertices & $k$ components a forest has $n-k$ edges.
$implies n - k = lfloor{(n/2)}rfloor$
$implies k = n - lfloor{(n/2)}rfloor =
lceil{(n/2)}rceil$
answered Nov 29 '16 at 6:09
Mr.Sigma.Mr.Sigma.
1719
1
You should check out this guide on how to nicely typeset mathematics on this site.
– Mike Pierce
Nov 29 '16 at 6:15
😷😷😷😂😂😂 Will try sir.
– Mr.Sigma.
Nov 29 '16 at 6:19
add a comment |
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2 Answers
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2 Answers
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Since we are working $bmod n$, I think it would be easier to name the vertices ${1, dotsc, n}$ instead; the problem is the same. We see pretty quickly by playing around that the edges
$$(1,n)quad(2,n!-!1)quaddotsbquadleft(lfloor n/2rfloor, 1+lfloor n/2rfloorright)$$
must be in our graph. So there are at most $lceil n/2 rceil$ connected components, each component having one edge except for a single component with an isolated vertex if $n$ is odd. This is also all the components. On the contrary if we suppose that our graph has some edges $(a,b)$ and $(b,c)$ for distinct vertices $a$ and $c$, we would have
$$a+b equiv 1 quadtext{and}quad b+c equiv 1;.$$
Subtracting these two congruence we get that $aequiv c pmod n$. But since our vertices are just ${1,dotsc,n}$ this is only possible if $a=c$.
answered Nov 30 '16 at 1:15
Mike PierceMike Pierce
11.4k103583
add a comment |
Since we are working $bmod n$, I think it would be easier to name the vertices ${1, dotsc, n}$ instead; the problem is the same. We see pretty quickly by playing around that the edges
$$(1,n)quad(2,n!-!1)quaddotsbquadleft(lfloor n/2rfloor, 1+lfloor n/2rfloorright)$$
must be in our graph. So there are at most $lceil n/2 rceil$ connected components, each component having one edge except for a single component with an isolated vertex if $n$ is odd. This is also all the components. On the contrary if we suppose that our graph has some edges $(a,b)$ and $(b,c)$ for distinct vertices $a$ and $c$, we would have
$$a+b equiv 1 quadtext{and}quad b+c equiv 1;.$$
Subtracting these two congruence we get that $aequiv c pmod n$. But since our vertices are just ${1,dotsc,n}$ this is only possible if $a=c$.
answered Nov 30 '16 at 1:15
Mike PierceMike Pierce
11.4k103583
add a comment |
Since we are working $bmod n$, I think it would be easier to name the vertices ${1, dotsc, n}$ instead; the problem is the same. We see pretty quickly by playing around that the edges
$$(1,n)quad(2,n!-!1)quaddotsbquadleft(lfloor n/2rfloor, 1+lfloor n/2rfloorright)$$
must be in our graph. So there are at most $lceil n/2 rceil$ connected components, each component having one edge except for a single component with an isolated vertex if $n$ is odd. This is also all the components. On the contrary if we suppose that our graph has some edges $(a,b)$ and $(b,c)$ for distinct vertices $a$ and $c$, we would have
$$a+b equiv 1 quadtext{and}quad b+c equiv 1;.$$
Subtracting these two congruence we get that $aequiv c pmod n$. But since our vertices are just ${1,dotsc,n}$ this is only possible if $a=c$.
answered Nov 30 '16 at 1:15
Mike PierceMike Pierce
11.4k103583
Since we are working $bmod n$, I think it would be easier to name the vertices ${1, dotsc, n}$ instead; the problem is the same. We see pretty quickly by playing around that the edges
$$(1,n)quad(2,n!-!1)quaddotsbquadleft(lfloor n/2rfloor, 1+lfloor n/2rfloorright)$$
must be in our graph. So there are at most $lceil n/2 rceil$ connected components, each component having one edge except for a single component with an isolated vertex if $n$ is odd. This is also all the components. On the contrary if we suppose that our graph has some edges $(a,b)$ and $(b,c)$ for distinct vertices $a$ and $c$, we would have
$$a+b equiv 1 quadtext{and}quad b+c equiv 1;.$$
Subtracting these two congruence we get that $aequiv c pmod n$. But since our vertices are just ${1,dotsc,n}$ this is only possible if $a=c$.
answered Nov 30 '16 at 1:15
Mike PierceMike Pierce
11.4k103583
answered Nov 30 '16 at 1:15
Mike PierceMike Pierce
11.4k103583
answered Nov 30 '16 at 1:15
Mike PierceMike Pierce
11.4k103583
answered Nov 30 '16 at 1:15
Mike PierceMike Pierce
11.4k103583
11.4k103583
add a comment |
add a comment |
Here is how I solved the problem.
Since sum would be $n+1$. Patterns of edges would be like this:
$(lfloor{(n/2)}rfloor, lfloor{(n/2)}rfloor +1), (lfloor{(n/2)}rfloor-1, lfloor{(n/2)}rfloor +2),...,(2,lfloor{(n/2)}rfloor+lfloor{(n/2)}rfloor
-1)$
Therefore total pairs $= lfloor{(n/2)}rfloor -1 -1+1 = lfloor{(n/2)}rfloor -1$
Considering trivial $(0,1)$ Total pairs or edges are $= lfloor{(n/2)}rfloor$.
Now, with $n$ vertices & $k$ components a forest has $n-k$ edges.
$implies n - k = lfloor{(n/2)}rfloor$
$implies k = n - lfloor{(n/2)}rfloor =
lceil{(n/2)}rceil$
answered Nov 29 '16 at 6:09
Mr.Sigma.Mr.Sigma.
1719
1
You should check out this guide on how to nicely typeset mathematics on this site.
– Mike Pierce
Nov 29 '16 at 6:15
😷😷😷😂😂😂 Will try sir.
– Mr.Sigma.
Nov 29 '16 at 6:19
add a comment |
Here is how I solved the problem.
Since sum would be $n+1$. Patterns of edges would be like this:
$(lfloor{(n/2)}rfloor, lfloor{(n/2)}rfloor +1), (lfloor{(n/2)}rfloor-1, lfloor{(n/2)}rfloor +2),...,(2,lfloor{(n/2)}rfloor+lfloor{(n/2)}rfloor
-1)$
Therefore total pairs $= lfloor{(n/2)}rfloor -1 -1+1 = lfloor{(n/2)}rfloor -1$
Considering trivial $(0,1)$ Total pairs or edges are $= lfloor{(n/2)}rfloor$.
Now, with $n$ vertices & $k$ components a forest has $n-k$ edges.
$implies n - k = lfloor{(n/2)}rfloor$
$implies k = n - lfloor{(n/2)}rfloor =
lceil{(n/2)}rceil$
answered Nov 29 '16 at 6:09
Mr.Sigma.Mr.Sigma.
1719
1
You should check out this guide on how to nicely typeset mathematics on this site.
– Mike Pierce
Nov 29 '16 at 6:15
😷😷😷😂😂😂 Will try sir.
– Mr.Sigma.
Nov 29 '16 at 6:19
add a comment |
Here is how I solved the problem.
Since sum would be $n+1$. Patterns of edges would be like this:
$(lfloor{(n/2)}rfloor, lfloor{(n/2)}rfloor +1), (lfloor{(n/2)}rfloor-1, lfloor{(n/2)}rfloor +2),...,(2,lfloor{(n/2)}rfloor+lfloor{(n/2)}rfloor
-1)$
Therefore total pairs $= lfloor{(n/2)}rfloor -1 -1+1 = lfloor{(n/2)}rfloor -1$
Considering trivial $(0,1)$ Total pairs or edges are $= lfloor{(n/2)}rfloor$.
Now, with $n$ vertices & $k$ components a forest has $n-k$ edges.
$implies n - k = lfloor{(n/2)}rfloor$
$implies k = n - lfloor{(n/2)}rfloor =
lceil{(n/2)}rceil$
answered Nov 29 '16 at 6:09
Mr.Sigma.Mr.Sigma.
1719
Here is how I solved the problem.
Since sum would be $n+1$. Patterns of edges would be like this:
$(lfloor{(n/2)}rfloor, lfloor{(n/2)}rfloor +1), (lfloor{(n/2)}rfloor-1, lfloor{(n/2)}rfloor +2),...,(2,lfloor{(n/2)}rfloor+lfloor{(n/2)}rfloor
-1)$
Therefore total pairs $= lfloor{(n/2)}rfloor -1 -1+1 = lfloor{(n/2)}rfloor -1$
Considering trivial $(0,1)$ Total pairs or edges are $= lfloor{(n/2)}rfloor$.
Now, with $n$ vertices & $k$ components a forest has $n-k$ edges.
$implies n - k = lfloor{(n/2)}rfloor$
$implies k = n - lfloor{(n/2)}rfloor =
lceil{(n/2)}rceil$
answered Nov 29 '16 at 6:09
Mr.Sigma.Mr.Sigma.
1719
edited Dec 1 '18 at 5:34
answered Nov 29 '16 at 6:09
Mr.Sigma.Mr.Sigma.
1719
answered Nov 29 '16 at 6:09
Mr.Sigma.Mr.Sigma.
1719
answered Nov 29 '16 at 6:09
Mr.Sigma.Mr.Sigma.
1719
1719
1
You should check out this guide on how to nicely typeset mathematics on this site.
– Mike Pierce
Nov 29 '16 at 6:15
😷😷😷😂😂😂 Will try sir.
– Mr.Sigma.
Nov 29 '16 at 6:19
add a comment |
1
You should check out this guide on how to nicely typeset mathematics on this site.
– Mike Pierce
Nov 29 '16 at 6:15
😷😷😷😂😂😂 Will try sir.
– Mr.Sigma.
Nov 29 '16 at 6:19
1
1
You should check out this guide on how to nicely typeset mathematics on this site.
– Mike Pierce
Nov 29 '16 at 6:15
You should check out this guide on how to nicely typeset mathematics on this site.
– Mike Pierce
Nov 29 '16 at 6:15
😷😷😷😂😂😂 Will try sir.
– Mr.Sigma.
Nov 29 '16 at 6:19
😷😷😷😂😂😂 Will try sir.
– Mr.Sigma.
Nov 29 '16 at 6:19
add a comment |
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If you've been constructing the graphs by hand, you should see a pattern. What's the pattern you see?
– Mike Pierce
Nov 29 '16 at 3:16
Sir +Mike, Thanks for you comment. What I see is - this question is reduced to the combinatorial problem of finding number of combination (x,y) = n+1.
– Mr.Sigma.
Nov 29 '16 at 4:04
Basically, yeah, ... and what's the pattern you see when drawing the graphs? What led you to conclude that the answer should be $lceil n/2 rceil$?
– Mike Pierce
Nov 29 '16 at 4:07
Sir +Mike, I've just tried with n=3,4,5 & got the answer. But the pattern I see somewhat is like (n/2,n/2+1), (n/2-1,n/2+2),(n/2-2,n/2+3) ... Am I correct? Apart from trivial (0,1) edge.
– Mr.Sigma.
Nov 29 '16 at 4:10
Thanks Sir +Mike for helping me, I got it counting terms of this series alone. :)
– Mr.Sigma.
Nov 29 '16 at 4:19