Proof that if all vertices have degree at least two then $G$ contains a cycle












1












$begingroup$



Prove that if all vertices have degree at least two then $G$ contains a cycle.




Here is the proof, but please correct me if wrong :



We assume $G$ is simple and let $P$ be the longest path $=v_0v_1v_2ldots v_{a-1}v_a$.



As it is given that the degree of $v_a$ is even ,then $v_a$ must have $mathbf{2}$ neigbors; one of them is $v_{a-1}$ and the other must be any $v$ in the Graph $G$, so $v=v_i$, where $i$ is any integer between $0$ and $a-2$ : $0leq i leq(a-2)$ #



Please correct me if I am wrong, and sorry for any mistakes.






combinatorics discrete-mathematics graph-theory trees







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$endgroup$












  • $begingroup$
    You can omit the [incorrect] claim that the degree of $v_a$ is even... you already know it has at least two neighbors
    $endgroup$
    – angryavian
    Jun 22 '15 at 17:32










  • $begingroup$
    Put the whole subscript in curly braces: to get $v_{a-1}$ use v_{a-1}.
    $endgroup$
    – Brian M. Scott
    Jun 22 '15 at 17:37
















1












$begingroup$



Prove that if all vertices have degree at least two then $G$ contains a cycle.




Here is the proof, but please correct me if wrong :



We assume $G$ is simple and let $P$ be the longest path $=v_0v_1v_2ldots v_{a-1}v_a$.



As it is given that the degree of $v_a$ is even ,then $v_a$ must have $mathbf{2}$ neigbors; one of them is $v_{a-1}$ and the other must be any $v$ in the Graph $G$, so $v=v_i$, where $i$ is any integer between $0$ and $a-2$ : $0leq i leq(a-2)$ #



Please correct me if I am wrong, and sorry for any mistakes.






combinatorics discrete-mathematics graph-theory trees







share|cite|improve this question











$endgroup$












  • $begingroup$
    You can omit the [incorrect] claim that the degree of $v_a$ is even... you already know it has at least two neighbors
    $endgroup$
    – angryavian
    Jun 22 '15 at 17:32










  • $begingroup$
    Put the whole subscript in curly braces: to get $v_{a-1}$ use v_{a-1}.
    $endgroup$
    – Brian M. Scott
    Jun 22 '15 at 17:37














1












1








1





$begingroup$



Prove that if all vertices have degree at least two then $G$ contains a cycle.




Here is the proof, but please correct me if wrong :



We assume $G$ is simple and let $P$ be the longest path $=v_0v_1v_2ldots v_{a-1}v_a$.



As it is given that the degree of $v_a$ is even ,then $v_a$ must have $mathbf{2}$ neigbors; one of them is $v_{a-1}$ and the other must be any $v$ in the Graph $G$, so $v=v_i$, where $i$ is any integer between $0$ and $a-2$ : $0leq i leq(a-2)$ #



Please correct me if I am wrong, and sorry for any mistakes.






combinatorics discrete-mathematics graph-theory trees







share|cite|improve this question











$endgroup$





Prove that if all vertices have degree at least two then $G$ contains a cycle.




Here is the proof, but please correct me if wrong :



We assume $G$ is simple and let $P$ be the longest path $=v_0v_1v_2ldots v_{a-1}v_a$.



As it is given that the degree of $v_a$ is even ,then $v_a$ must have $mathbf{2}$ neigbors; one of them is $v_{a-1}$ and the other must be any $v$ in the Graph $G$, so $v=v_i$, where $i$ is any integer between $0$ and $a-2$ : $0leq i leq(a-2)$ #



Please correct me if I am wrong, and sorry for any mistakes.






combinatorics discrete-mathematics graph-theory trees




combinatorics discrete-mathematics graph-theory trees






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 2:05









Batominovski

33.2k33293




33.2k33293










asked Jun 22 '15 at 17:28









John KantiJohn Kanti

217




217












  • $begingroup$
    You can omit the [incorrect] claim that the degree of $v_a$ is even... you already know it has at least two neighbors
    $endgroup$
    – angryavian
    Jun 22 '15 at 17:32










  • $begingroup$
    Put the whole subscript in curly braces: to get $v_{a-1}$ use v_{a-1}.
    $endgroup$
    – Brian M. Scott
    Jun 22 '15 at 17:37


















  • $begingroup$
    You can omit the [incorrect] claim that the degree of $v_a$ is even... you already know it has at least two neighbors
    $endgroup$
    – angryavian
    Jun 22 '15 at 17:32










  • $begingroup$
    Put the whole subscript in curly braces: to get $v_{a-1}$ use v_{a-1}.
    $endgroup$
    – Brian M. Scott
    Jun 22 '15 at 17:37
















$begingroup$
You can omit the [incorrect] claim that the degree of $v_a$ is even... you already know it has at least two neighbors
$endgroup$
– angryavian
Jun 22 '15 at 17:32




$begingroup$
You can omit the [incorrect] claim that the degree of $v_a$ is even... you already know it has at least two neighbors
$endgroup$
– angryavian
Jun 22 '15 at 17:32












$begingroup$
Put the whole subscript in curly braces: to get $v_{a-1}$ use v_{a-1}.
$endgroup$
– Brian M. Scott
Jun 22 '15 at 17:37




$begingroup$
Put the whole subscript in curly braces: to get $v_{a-1}$ use v_{a-1}.
$endgroup$
– Brian M. Scott
Jun 22 '15 at 17:37










2 Answers
2






active

oldest

votes


















4












$begingroup$

The idea is right, but there’s a small mistake, and there are a couple of places where you could be a little clearer. You don’t know that the degree of $v_a$ is even: it’s quite possible that $deg v_a$ is odd. You do know, however, that $deg v_age 2$, which is all that you need. Then you can say that $v_a$ must be adjacent to at least one vertex $v$ that is not $v_{a-1}$. At that point you really should say a little more than you did: you should point out that if $vnotin{v_0,ldots,v_{a-2}}$, then we could extent the path $P$ to $v$. However, $P$ was chosen to be of maximal length, so this is impossible, and therefore $vin{v_0ldots,v_{a-2}}$. Thus, if $v=v_i$, then $v_iv_{i+1}ldots v_av_i$ is a cycle.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I just I assume it is a string of nodes <contradiction> at the end ? is not it ?
    $endgroup$
    – John Kanti
    Jun 23 '15 at 10:46










  • $begingroup$
    @John: I’m not sure what you mean. The only place where a contradiction is involved is when we observe that $v$ is one of the vertices $v_0,ldots,v_{n-2}$: if it were not, the path $v_0ldots v_nv$ would contradict the choice of $P$ as a maximal path. There’s no contradiction at the end, though: there we simply have the desired cycle.
    $endgroup$
    – Brian M. Scott
    Jun 23 '15 at 19:36



















0












$begingroup$

I guess you assume that $G$ is a finite graph. What is the maximum possible number of edges that a spanning forest of $G$ can have? What is the minimum possible number of edges can $G$ have?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    if you know the definition of a Simple Graph then you do not need to ask about the Graph limitation ? why you need the maximum possible number ? suppose maximum is X or any +number
    $endgroup$
    – John Kanti
    Jun 22 '15 at 17:35






  • 2




    $begingroup$
    Let $n$ be the number of vertices of $G$. If $G$ has no cycle, then $G$ is a forest, then $G$ has at most $n-1$ edges. If every vertex of $G$ is of degree at least $2$, then $G$ has at least $n$ edges. Do you see my point now?
    $endgroup$
    – Batominovski
    Jun 23 '15 at 3:56










  • $begingroup$
    not clear, why you need # of nodes, the theory does not tell about how many nodes we need, it says it is true whatever the number of nodes is. but at least each node must have at least 2 degrees. Did you get it ?!
    $endgroup$
    – John Kanti
    Jun 23 '15 at 10:49






  • 1




    $begingroup$
    No, I don't get what you said. Actually, I think it is you who doesn't "get it!" Unless you can make a coherent counter-argument, I will stop my conversation with you.
    $endgroup$
    – Batominovski
    Jun 23 '15 at 10:58










  • $begingroup$
    please read the Theory carefully, It does not talk about the number of nodes? so why you assume that we need to know the number of nodes? it is not the key point, Stopping conversation is your decision , I do not care about it
    $endgroup$
    – John Kanti
    Jun 23 '15 at 11:01














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2 Answers
2






active

oldest

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2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

The idea is right, but there’s a small mistake, and there are a couple of places where you could be a little clearer. You don’t know that the degree of $v_a$ is even: it’s quite possible that $deg v_a$ is odd. You do know, however, that $deg v_age 2$, which is all that you need. Then you can say that $v_a$ must be adjacent to at least one vertex $v$ that is not $v_{a-1}$. At that point you really should say a little more than you did: you should point out that if $vnotin{v_0,ldots,v_{a-2}}$, then we could extent the path $P$ to $v$. However, $P$ was chosen to be of maximal length, so this is impossible, and therefore $vin{v_0ldots,v_{a-2}}$. Thus, if $v=v_i$, then $v_iv_{i+1}ldots v_av_i$ is a cycle.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I just I assume it is a string of nodes <contradiction> at the end ? is not it ?
    $endgroup$
    – John Kanti
    Jun 23 '15 at 10:46










  • $begingroup$
    @John: I’m not sure what you mean. The only place where a contradiction is involved is when we observe that $v$ is one of the vertices $v_0,ldots,v_{n-2}$: if it were not, the path $v_0ldots v_nv$ would contradict the choice of $P$ as a maximal path. There’s no contradiction at the end, though: there we simply have the desired cycle.
    $endgroup$
    – Brian M. Scott
    Jun 23 '15 at 19:36
















4












$begingroup$

The idea is right, but there’s a small mistake, and there are a couple of places where you could be a little clearer. You don’t know that the degree of $v_a$ is even: it’s quite possible that $deg v_a$ is odd. You do know, however, that $deg v_age 2$, which is all that you need. Then you can say that $v_a$ must be adjacent to at least one vertex $v$ that is not $v_{a-1}$. At that point you really should say a little more than you did: you should point out that if $vnotin{v_0,ldots,v_{a-2}}$, then we could extent the path $P$ to $v$. However, $P$ was chosen to be of maximal length, so this is impossible, and therefore $vin{v_0ldots,v_{a-2}}$. Thus, if $v=v_i$, then $v_iv_{i+1}ldots v_av_i$ is a cycle.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I just I assume it is a string of nodes <contradiction> at the end ? is not it ?
    $endgroup$
    – John Kanti
    Jun 23 '15 at 10:46










  • $begingroup$
    @John: I’m not sure what you mean. The only place where a contradiction is involved is when we observe that $v$ is one of the vertices $v_0,ldots,v_{n-2}$: if it were not, the path $v_0ldots v_nv$ would contradict the choice of $P$ as a maximal path. There’s no contradiction at the end, though: there we simply have the desired cycle.
    $endgroup$
    – Brian M. Scott
    Jun 23 '15 at 19:36














4












4








4





$begingroup$

The idea is right, but there’s a small mistake, and there are a couple of places where you could be a little clearer. You don’t know that the degree of $v_a$ is even: it’s quite possible that $deg v_a$ is odd. You do know, however, that $deg v_age 2$, which is all that you need. Then you can say that $v_a$ must be adjacent to at least one vertex $v$ that is not $v_{a-1}$. At that point you really should say a little more than you did: you should point out that if $vnotin{v_0,ldots,v_{a-2}}$, then we could extent the path $P$ to $v$. However, $P$ was chosen to be of maximal length, so this is impossible, and therefore $vin{v_0ldots,v_{a-2}}$. Thus, if $v=v_i$, then $v_iv_{i+1}ldots v_av_i$ is a cycle.






share|cite|improve this answer









$endgroup$



The idea is right, but there’s a small mistake, and there are a couple of places where you could be a little clearer. You don’t know that the degree of $v_a$ is even: it’s quite possible that $deg v_a$ is odd. You do know, however, that $deg v_age 2$, which is all that you need. Then you can say that $v_a$ must be adjacent to at least one vertex $v$ that is not $v_{a-1}$. At that point you really should say a little more than you did: you should point out that if $vnotin{v_0,ldots,v_{a-2}}$, then we could extent the path $P$ to $v$. However, $P$ was chosen to be of maximal length, so this is impossible, and therefore $vin{v_0ldots,v_{a-2}}$. Thus, if $v=v_i$, then $v_iv_{i+1}ldots v_av_i$ is a cycle.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jun 22 '15 at 17:34









Brian M. ScottBrian M. Scott

460k40517918




460k40517918












  • $begingroup$
    I just I assume it is a string of nodes <contradiction> at the end ? is not it ?
    $endgroup$
    – John Kanti
    Jun 23 '15 at 10:46










  • $begingroup$
    @John: I’m not sure what you mean. The only place where a contradiction is involved is when we observe that $v$ is one of the vertices $v_0,ldots,v_{n-2}$: if it were not, the path $v_0ldots v_nv$ would contradict the choice of $P$ as a maximal path. There’s no contradiction at the end, though: there we simply have the desired cycle.
    $endgroup$
    – Brian M. Scott
    Jun 23 '15 at 19:36


















  • $begingroup$
    I just I assume it is a string of nodes <contradiction> at the end ? is not it ?
    $endgroup$
    – John Kanti
    Jun 23 '15 at 10:46










  • $begingroup$
    @John: I’m not sure what you mean. The only place where a contradiction is involved is when we observe that $v$ is one of the vertices $v_0,ldots,v_{n-2}$: if it were not, the path $v_0ldots v_nv$ would contradict the choice of $P$ as a maximal path. There’s no contradiction at the end, though: there we simply have the desired cycle.
    $endgroup$
    – Brian M. Scott
    Jun 23 '15 at 19:36
















$begingroup$
I just I assume it is a string of nodes <contradiction> at the end ? is not it ?
$endgroup$
– John Kanti
Jun 23 '15 at 10:46




$begingroup$
I just I assume it is a string of nodes <contradiction> at the end ? is not it ?
$endgroup$
– John Kanti
Jun 23 '15 at 10:46












$begingroup$
@John: I’m not sure what you mean. The only place where a contradiction is involved is when we observe that $v$ is one of the vertices $v_0,ldots,v_{n-2}$: if it were not, the path $v_0ldots v_nv$ would contradict the choice of $P$ as a maximal path. There’s no contradiction at the end, though: there we simply have the desired cycle.
$endgroup$
– Brian M. Scott
Jun 23 '15 at 19:36




$begingroup$
@John: I’m not sure what you mean. The only place where a contradiction is involved is when we observe that $v$ is one of the vertices $v_0,ldots,v_{n-2}$: if it were not, the path $v_0ldots v_nv$ would contradict the choice of $P$ as a maximal path. There’s no contradiction at the end, though: there we simply have the desired cycle.
$endgroup$
– Brian M. Scott
Jun 23 '15 at 19:36











0












$begingroup$

I guess you assume that $G$ is a finite graph. What is the maximum possible number of edges that a spanning forest of $G$ can have? What is the minimum possible number of edges can $G$ have?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    if you know the definition of a Simple Graph then you do not need to ask about the Graph limitation ? why you need the maximum possible number ? suppose maximum is X or any +number
    $endgroup$
    – John Kanti
    Jun 22 '15 at 17:35






  • 2




    $begingroup$
    Let $n$ be the number of vertices of $G$. If $G$ has no cycle, then $G$ is a forest, then $G$ has at most $n-1$ edges. If every vertex of $G$ is of degree at least $2$, then $G$ has at least $n$ edges. Do you see my point now?
    $endgroup$
    – Batominovski
    Jun 23 '15 at 3:56










  • $begingroup$
    not clear, why you need # of nodes, the theory does not tell about how many nodes we need, it says it is true whatever the number of nodes is. but at least each node must have at least 2 degrees. Did you get it ?!
    $endgroup$
    – John Kanti
    Jun 23 '15 at 10:49






  • 1




    $begingroup$
    No, I don't get what you said. Actually, I think it is you who doesn't "get it!" Unless you can make a coherent counter-argument, I will stop my conversation with you.
    $endgroup$
    – Batominovski
    Jun 23 '15 at 10:58










  • $begingroup$
    please read the Theory carefully, It does not talk about the number of nodes? so why you assume that we need to know the number of nodes? it is not the key point, Stopping conversation is your decision , I do not care about it
    $endgroup$
    – John Kanti
    Jun 23 '15 at 11:01


















0












$begingroup$

I guess you assume that $G$ is a finite graph. What is the maximum possible number of edges that a spanning forest of $G$ can have? What is the minimum possible number of edges can $G$ have?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    if you know the definition of a Simple Graph then you do not need to ask about the Graph limitation ? why you need the maximum possible number ? suppose maximum is X or any +number
    $endgroup$
    – John Kanti
    Jun 22 '15 at 17:35






  • 2




    $begingroup$
    Let $n$ be the number of vertices of $G$. If $G$ has no cycle, then $G$ is a forest, then $G$ has at most $n-1$ edges. If every vertex of $G$ is of degree at least $2$, then $G$ has at least $n$ edges. Do you see my point now?
    $endgroup$
    – Batominovski
    Jun 23 '15 at 3:56










  • $begingroup$
    not clear, why you need # of nodes, the theory does not tell about how many nodes we need, it says it is true whatever the number of nodes is. but at least each node must have at least 2 degrees. Did you get it ?!
    $endgroup$
    – John Kanti
    Jun 23 '15 at 10:49






  • 1




    $begingroup$
    No, I don't get what you said. Actually, I think it is you who doesn't "get it!" Unless you can make a coherent counter-argument, I will stop my conversation with you.
    $endgroup$
    – Batominovski
    Jun 23 '15 at 10:58










  • $begingroup$
    please read the Theory carefully, It does not talk about the number of nodes? so why you assume that we need to know the number of nodes? it is not the key point, Stopping conversation is your decision , I do not care about it
    $endgroup$
    – John Kanti
    Jun 23 '15 at 11:01
















0












0








0





$begingroup$

I guess you assume that $G$ is a finite graph. What is the maximum possible number of edges that a spanning forest of $G$ can have? What is the minimum possible number of edges can $G$ have?






share|cite|improve this answer









$endgroup$



I guess you assume that $G$ is a finite graph. What is the maximum possible number of edges that a spanning forest of $G$ can have? What is the minimum possible number of edges can $G$ have?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jun 22 '15 at 17:32









BatominovskiBatominovski

33.2k33293




33.2k33293












  • $begingroup$
    if you know the definition of a Simple Graph then you do not need to ask about the Graph limitation ? why you need the maximum possible number ? suppose maximum is X or any +number
    $endgroup$
    – John Kanti
    Jun 22 '15 at 17:35






  • 2




    $begingroup$
    Let $n$ be the number of vertices of $G$. If $G$ has no cycle, then $G$ is a forest, then $G$ has at most $n-1$ edges. If every vertex of $G$ is of degree at least $2$, then $G$ has at least $n$ edges. Do you see my point now?
    $endgroup$
    – Batominovski
    Jun 23 '15 at 3:56










  • $begingroup$
    not clear, why you need # of nodes, the theory does not tell about how many nodes we need, it says it is true whatever the number of nodes is. but at least each node must have at least 2 degrees. Did you get it ?!
    $endgroup$
    – John Kanti
    Jun 23 '15 at 10:49






  • 1




    $begingroup$
    No, I don't get what you said. Actually, I think it is you who doesn't "get it!" Unless you can make a coherent counter-argument, I will stop my conversation with you.
    $endgroup$
    – Batominovski
    Jun 23 '15 at 10:58










  • $begingroup$
    please read the Theory carefully, It does not talk about the number of nodes? so why you assume that we need to know the number of nodes? it is not the key point, Stopping conversation is your decision , I do not care about it
    $endgroup$
    – John Kanti
    Jun 23 '15 at 11:01




















  • $begingroup$
    if you know the definition of a Simple Graph then you do not need to ask about the Graph limitation ? why you need the maximum possible number ? suppose maximum is X or any +number
    $endgroup$
    – John Kanti
    Jun 22 '15 at 17:35






  • 2




    $begingroup$
    Let $n$ be the number of vertices of $G$. If $G$ has no cycle, then $G$ is a forest, then $G$ has at most $n-1$ edges. If every vertex of $G$ is of degree at least $2$, then $G$ has at least $n$ edges. Do you see my point now?
    $endgroup$
    – Batominovski
    Jun 23 '15 at 3:56










  • $begingroup$
    not clear, why you need # of nodes, the theory does not tell about how many nodes we need, it says it is true whatever the number of nodes is. but at least each node must have at least 2 degrees. Did you get it ?!
    $endgroup$
    – John Kanti
    Jun 23 '15 at 10:49






  • 1




    $begingroup$
    No, I don't get what you said. Actually, I think it is you who doesn't "get it!" Unless you can make a coherent counter-argument, I will stop my conversation with you.
    $endgroup$
    – Batominovski
    Jun 23 '15 at 10:58










  • $begingroup$
    please read the Theory carefully, It does not talk about the number of nodes? so why you assume that we need to know the number of nodes? it is not the key point, Stopping conversation is your decision , I do not care about it
    $endgroup$
    – John Kanti
    Jun 23 '15 at 11:01


















$begingroup$
if you know the definition of a Simple Graph then you do not need to ask about the Graph limitation ? why you need the maximum possible number ? suppose maximum is X or any +number
$endgroup$
– John Kanti
Jun 22 '15 at 17:35




$begingroup$
if you know the definition of a Simple Graph then you do not need to ask about the Graph limitation ? why you need the maximum possible number ? suppose maximum is X or any +number
$endgroup$
– John Kanti
Jun 22 '15 at 17:35




2




2




$begingroup$
Let $n$ be the number of vertices of $G$. If $G$ has no cycle, then $G$ is a forest, then $G$ has at most $n-1$ edges. If every vertex of $G$ is of degree at least $2$, then $G$ has at least $n$ edges. Do you see my point now?
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– Batominovski
Jun 23 '15 at 3:56




$begingroup$
Let $n$ be the number of vertices of $G$. If $G$ has no cycle, then $G$ is a forest, then $G$ has at most $n-1$ edges. If every vertex of $G$ is of degree at least $2$, then $G$ has at least $n$ edges. Do you see my point now?
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– Batominovski
Jun 23 '15 at 3:56












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not clear, why you need # of nodes, the theory does not tell about how many nodes we need, it says it is true whatever the number of nodes is. but at least each node must have at least 2 degrees. Did you get it ?!
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– John Kanti
Jun 23 '15 at 10:49




$begingroup$
not clear, why you need # of nodes, the theory does not tell about how many nodes we need, it says it is true whatever the number of nodes is. but at least each node must have at least 2 degrees. Did you get it ?!
$endgroup$
– John Kanti
Jun 23 '15 at 10:49




1




1




$begingroup$
No, I don't get what you said. Actually, I think it is you who doesn't "get it!" Unless you can make a coherent counter-argument, I will stop my conversation with you.
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– Batominovski
Jun 23 '15 at 10:58




$begingroup$
No, I don't get what you said. Actually, I think it is you who doesn't "get it!" Unless you can make a coherent counter-argument, I will stop my conversation with you.
$endgroup$
– Batominovski
Jun 23 '15 at 10:58












$begingroup$
please read the Theory carefully, It does not talk about the number of nodes? so why you assume that we need to know the number of nodes? it is not the key point, Stopping conversation is your decision , I do not care about it
$endgroup$
– John Kanti
Jun 23 '15 at 11:01






$begingroup$
please read the Theory carefully, It does not talk about the number of nodes? so why you assume that we need to know the number of nodes? it is not the key point, Stopping conversation is your decision , I do not care about it
$endgroup$
– John Kanti
Jun 23 '15 at 11:01




















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