Proof - A tree $T$ has a vertex of degree n and the others have degree $<n$. T has at least n leaves












0












$begingroup$


I tried to think about some characteristics of the trees, for example, they have a number of edges equal to $|text{Vertices}|-1$ and of course the handshaking lemma but I couldn't find properly logical reasoning. Any help?






discrete-mathematics graph-theory trees







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$endgroup$








  • 1




    $begingroup$
    Hint: if you delete the vertex of degree $n$, then $n$ disjoint trees remain. How many leaves do each of those trees have?
    $endgroup$
    – Mike Earnest
    Dec 24 '18 at 17:53










  • $begingroup$
    @MikeEarnest sorry... i saw your comment after I posted my answer
    $endgroup$
    – Ankit Kumar
    Dec 24 '18 at 17:58
















0












$begingroup$


I tried to think about some characteristics of the trees, for example, they have a number of edges equal to $|text{Vertices}|-1$ and of course the handshaking lemma but I couldn't find properly logical reasoning. Any help?






discrete-mathematics graph-theory trees







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Hint: if you delete the vertex of degree $n$, then $n$ disjoint trees remain. How many leaves do each of those trees have?
    $endgroup$
    – Mike Earnest
    Dec 24 '18 at 17:53










  • $begingroup$
    @MikeEarnest sorry... i saw your comment after I posted my answer
    $endgroup$
    – Ankit Kumar
    Dec 24 '18 at 17:58














0












0








0





$begingroup$


I tried to think about some characteristics of the trees, for example, they have a number of edges equal to $|text{Vertices}|-1$ and of course the handshaking lemma but I couldn't find properly logical reasoning. Any help?






discrete-mathematics graph-theory trees







share|cite|improve this question











$endgroup$




I tried to think about some characteristics of the trees, for example, they have a number of edges equal to $|text{Vertices}|-1$ and of course the handshaking lemma but I couldn't find properly logical reasoning. Any help?






discrete-mathematics graph-theory trees




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 24 '18 at 18:01









EdOverflow

24719




24719










asked Dec 24 '18 at 17:44









PCNFPCNF

1338




1338








  • 1




    $begingroup$
    Hint: if you delete the vertex of degree $n$, then $n$ disjoint trees remain. How many leaves do each of those trees have?
    $endgroup$
    – Mike Earnest
    Dec 24 '18 at 17:53










  • $begingroup$
    @MikeEarnest sorry... i saw your comment after I posted my answer
    $endgroup$
    – Ankit Kumar
    Dec 24 '18 at 17:58














  • 1




    $begingroup$
    Hint: if you delete the vertex of degree $n$, then $n$ disjoint trees remain. How many leaves do each of those trees have?
    $endgroup$
    – Mike Earnest
    Dec 24 '18 at 17:53










  • $begingroup$
    @MikeEarnest sorry... i saw your comment after I posted my answer
    $endgroup$
    – Ankit Kumar
    Dec 24 '18 at 17:58








1




1




$begingroup$
Hint: if you delete the vertex of degree $n$, then $n$ disjoint trees remain. How many leaves do each of those trees have?
$endgroup$
– Mike Earnest
Dec 24 '18 at 17:53




$begingroup$
Hint: if you delete the vertex of degree $n$, then $n$ disjoint trees remain. How many leaves do each of those trees have?
$endgroup$
– Mike Earnest
Dec 24 '18 at 17:53












$begingroup$
@MikeEarnest sorry... i saw your comment after I posted my answer
$endgroup$
– Ankit Kumar
Dec 24 '18 at 17:58




$begingroup$
@MikeEarnest sorry... i saw your comment after I posted my answer
$endgroup$
– Ankit Kumar
Dec 24 '18 at 17:58










2 Answers
2






active

oldest

votes


















3












$begingroup$

Let $v_i$ be the degrees of the vertices and suppose the graph has $v$ nodes.



Notice $sumlimits_{i=1}^v (v_i-1) = v-2$



Without loss of generality $v_1=n$



This means $sumlimits_{i=2}^v (v_i-1) = v-n-1$



So at least $n$ of the summands are equal to $0$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Let the tree be $T$, vertex with degree $n$ be $u$. Let the vertices adjacent to $u$ be $v_1, v_2,cdots v_n$. Delete the vertex $u$. You'll get $n$ new components(trees), $T_1, T_2,cdots T_n$. This is because no vertex in $T_i$ is adjacent to any vertex in $T_j$, $ineq j$. WHY? Suppose they were adjacent. Let "that" vertex in $T_i$ be a, and in $T_j$ be $b$ $implies$ you can go from $u$ to $a$, then $a$ to $b$ and then from $b$ to $u$$implies$ cycle, which isn't possible. Hence, $T_i$s are components. Further, since $T$ was a tree, $T_i$s will also be trees.



    Note that if $T_i$ has only a single vertex, it was already a leaf in $T$. Suppose there are $k$ such $i$$implies $we get $k$ leaves from them $implies $we need $n-k$ more leaves. We need not consider "those" $T_i$s anymore. Finally, every tree has at least leaf, and $T$ has $n-k$ disjoint sub-trees, ( with $n(T_i)>1 )$ $implies T$ has at least $(n-k)+k=n$ leaves!






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thanks, but I didn't understand the last part. It's clear that you can divide $T$ into $T_n$ components which are trees. It is not clear to me how you go from these considerations to saying that T has n leaves.
      $endgroup$
      – PCNF
      Dec 24 '18 at 18:20










    • $begingroup$
      (1) Do you agree deleting u won't affect the number of leaves? (2) Do you agree that every tree has at least one leaf? (3) Do you agree that deleting u will lead to n (sub-)trees? If it's all yes, you're done ;)
      $endgroup$
      – Ankit Kumar
      Dec 24 '18 at 18:21








    • 1




      $begingroup$
      Apparently, everything is clear but I can not apply what I said to a simple example. For example, I have the tree 1-2-3 and the vertex 2 is connected to a vertex 4. So, if I delete the vertex 2 which has the maximum degree, I obtain the vertex 1,3,4 without the vertex 2 that they shared. How do I continue now?
      $endgroup$
      – PCNF
      Dec 24 '18 at 18:42






    • 2




      $begingroup$
      I think if you want your proof to be complete, you should replace the phrase "every tree has at least a leaf" with "every tree has at least two leaves." A leaf in $T_i$ might not be a leaf in $T$, if that leaf was originally joined to $u$.
      $endgroup$
      – Mike Earnest
      Dec 24 '18 at 18:52










    • $begingroup$
      Fixed everything now. Thank you for comments @MikeEarnest and PCNF
      $endgroup$
      – Ankit Kumar
      Dec 24 '18 at 19:04











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    2 Answers
    2






    active

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    2 Answers
    2






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    Let $v_i$ be the degrees of the vertices and suppose the graph has $v$ nodes.



    Notice $sumlimits_{i=1}^v (v_i-1) = v-2$



    Without loss of generality $v_1=n$



    This means $sumlimits_{i=2}^v (v_i-1) = v-n-1$



    So at least $n$ of the summands are equal to $0$.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      Let $v_i$ be the degrees of the vertices and suppose the graph has $v$ nodes.



      Notice $sumlimits_{i=1}^v (v_i-1) = v-2$



      Without loss of generality $v_1=n$



      This means $sumlimits_{i=2}^v (v_i-1) = v-n-1$



      So at least $n$ of the summands are equal to $0$.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        Let $v_i$ be the degrees of the vertices and suppose the graph has $v$ nodes.



        Notice $sumlimits_{i=1}^v (v_i-1) = v-2$



        Without loss of generality $v_1=n$



        This means $sumlimits_{i=2}^v (v_i-1) = v-n-1$



        So at least $n$ of the summands are equal to $0$.






        share|cite|improve this answer









        $endgroup$



        Let $v_i$ be the degrees of the vertices and suppose the graph has $v$ nodes.



        Notice $sumlimits_{i=1}^v (v_i-1) = v-2$



        Without loss of generality $v_1=n$



        This means $sumlimits_{i=2}^v (v_i-1) = v-n-1$



        So at least $n$ of the summands are equal to $0$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 24 '18 at 18:24









        Jorge FernándezJorge Fernández

        75.4k1191192




        75.4k1191192























            1












            $begingroup$

            Let the tree be $T$, vertex with degree $n$ be $u$. Let the vertices adjacent to $u$ be $v_1, v_2,cdots v_n$. Delete the vertex $u$. You'll get $n$ new components(trees), $T_1, T_2,cdots T_n$. This is because no vertex in $T_i$ is adjacent to any vertex in $T_j$, $ineq j$. WHY? Suppose they were adjacent. Let "that" vertex in $T_i$ be a, and in $T_j$ be $b$ $implies$ you can go from $u$ to $a$, then $a$ to $b$ and then from $b$ to $u$$implies$ cycle, which isn't possible. Hence, $T_i$s are components. Further, since $T$ was a tree, $T_i$s will also be trees.



            Note that if $T_i$ has only a single vertex, it was already a leaf in $T$. Suppose there are $k$ such $i$$implies $we get $k$ leaves from them $implies $we need $n-k$ more leaves. We need not consider "those" $T_i$s anymore. Finally, every tree has at least leaf, and $T$ has $n-k$ disjoint sub-trees, ( with $n(T_i)>1 )$ $implies T$ has at least $(n-k)+k=n$ leaves!






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thanks, but I didn't understand the last part. It's clear that you can divide $T$ into $T_n$ components which are trees. It is not clear to me how you go from these considerations to saying that T has n leaves.
              $endgroup$
              – PCNF
              Dec 24 '18 at 18:20










            • $begingroup$
              (1) Do you agree deleting u won't affect the number of leaves? (2) Do you agree that every tree has at least one leaf? (3) Do you agree that deleting u will lead to n (sub-)trees? If it's all yes, you're done ;)
              $endgroup$
              – Ankit Kumar
              Dec 24 '18 at 18:21








            • 1




              $begingroup$
              Apparently, everything is clear but I can not apply what I said to a simple example. For example, I have the tree 1-2-3 and the vertex 2 is connected to a vertex 4. So, if I delete the vertex 2 which has the maximum degree, I obtain the vertex 1,3,4 without the vertex 2 that they shared. How do I continue now?
              $endgroup$
              – PCNF
              Dec 24 '18 at 18:42






            • 2




              $begingroup$
              I think if you want your proof to be complete, you should replace the phrase "every tree has at least a leaf" with "every tree has at least two leaves." A leaf in $T_i$ might not be a leaf in $T$, if that leaf was originally joined to $u$.
              $endgroup$
              – Mike Earnest
              Dec 24 '18 at 18:52










            • $begingroup$
              Fixed everything now. Thank you for comments @MikeEarnest and PCNF
              $endgroup$
              – Ankit Kumar
              Dec 24 '18 at 19:04
















            1












            $begingroup$

            Let the tree be $T$, vertex with degree $n$ be $u$. Let the vertices adjacent to $u$ be $v_1, v_2,cdots v_n$. Delete the vertex $u$. You'll get $n$ new components(trees), $T_1, T_2,cdots T_n$. This is because no vertex in $T_i$ is adjacent to any vertex in $T_j$, $ineq j$. WHY? Suppose they were adjacent. Let "that" vertex in $T_i$ be a, and in $T_j$ be $b$ $implies$ you can go from $u$ to $a$, then $a$ to $b$ and then from $b$ to $u$$implies$ cycle, which isn't possible. Hence, $T_i$s are components. Further, since $T$ was a tree, $T_i$s will also be trees.



            Note that if $T_i$ has only a single vertex, it was already a leaf in $T$. Suppose there are $k$ such $i$$implies $we get $k$ leaves from them $implies $we need $n-k$ more leaves. We need not consider "those" $T_i$s anymore. Finally, every tree has at least leaf, and $T$ has $n-k$ disjoint sub-trees, ( with $n(T_i)>1 )$ $implies T$ has at least $(n-k)+k=n$ leaves!






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thanks, but I didn't understand the last part. It's clear that you can divide $T$ into $T_n$ components which are trees. It is not clear to me how you go from these considerations to saying that T has n leaves.
              $endgroup$
              – PCNF
              Dec 24 '18 at 18:20










            • $begingroup$
              (1) Do you agree deleting u won't affect the number of leaves? (2) Do you agree that every tree has at least one leaf? (3) Do you agree that deleting u will lead to n (sub-)trees? If it's all yes, you're done ;)
              $endgroup$
              – Ankit Kumar
              Dec 24 '18 at 18:21








            • 1




              $begingroup$
              Apparently, everything is clear but I can not apply what I said to a simple example. For example, I have the tree 1-2-3 and the vertex 2 is connected to a vertex 4. So, if I delete the vertex 2 which has the maximum degree, I obtain the vertex 1,3,4 without the vertex 2 that they shared. How do I continue now?
              $endgroup$
              – PCNF
              Dec 24 '18 at 18:42






            • 2




              $begingroup$
              I think if you want your proof to be complete, you should replace the phrase "every tree has at least a leaf" with "every tree has at least two leaves." A leaf in $T_i$ might not be a leaf in $T$, if that leaf was originally joined to $u$.
              $endgroup$
              – Mike Earnest
              Dec 24 '18 at 18:52










            • $begingroup$
              Fixed everything now. Thank you for comments @MikeEarnest and PCNF
              $endgroup$
              – Ankit Kumar
              Dec 24 '18 at 19:04














            1












            1








            1





            $begingroup$

            Let the tree be $T$, vertex with degree $n$ be $u$. Let the vertices adjacent to $u$ be $v_1, v_2,cdots v_n$. Delete the vertex $u$. You'll get $n$ new components(trees), $T_1, T_2,cdots T_n$. This is because no vertex in $T_i$ is adjacent to any vertex in $T_j$, $ineq j$. WHY? Suppose they were adjacent. Let "that" vertex in $T_i$ be a, and in $T_j$ be $b$ $implies$ you can go from $u$ to $a$, then $a$ to $b$ and then from $b$ to $u$$implies$ cycle, which isn't possible. Hence, $T_i$s are components. Further, since $T$ was a tree, $T_i$s will also be trees.



            Note that if $T_i$ has only a single vertex, it was already a leaf in $T$. Suppose there are $k$ such $i$$implies $we get $k$ leaves from them $implies $we need $n-k$ more leaves. We need not consider "those" $T_i$s anymore. Finally, every tree has at least leaf, and $T$ has $n-k$ disjoint sub-trees, ( with $n(T_i)>1 )$ $implies T$ has at least $(n-k)+k=n$ leaves!






            share|cite|improve this answer











            $endgroup$



            Let the tree be $T$, vertex with degree $n$ be $u$. Let the vertices adjacent to $u$ be $v_1, v_2,cdots v_n$. Delete the vertex $u$. You'll get $n$ new components(trees), $T_1, T_2,cdots T_n$. This is because no vertex in $T_i$ is adjacent to any vertex in $T_j$, $ineq j$. WHY? Suppose they were adjacent. Let "that" vertex in $T_i$ be a, and in $T_j$ be $b$ $implies$ you can go from $u$ to $a$, then $a$ to $b$ and then from $b$ to $u$$implies$ cycle, which isn't possible. Hence, $T_i$s are components. Further, since $T$ was a tree, $T_i$s will also be trees.



            Note that if $T_i$ has only a single vertex, it was already a leaf in $T$. Suppose there are $k$ such $i$$implies $we get $k$ leaves from them $implies $we need $n-k$ more leaves. We need not consider "those" $T_i$s anymore. Finally, every tree has at least leaf, and $T$ has $n-k$ disjoint sub-trees, ( with $n(T_i)>1 )$ $implies T$ has at least $(n-k)+k=n$ leaves!







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 24 '18 at 19:03

























            answered Dec 24 '18 at 17:57









            Ankit KumarAnkit Kumar

            1,479221




            1,479221












            • $begingroup$
              Thanks, but I didn't understand the last part. It's clear that you can divide $T$ into $T_n$ components which are trees. It is not clear to me how you go from these considerations to saying that T has n leaves.
              $endgroup$
              – PCNF
              Dec 24 '18 at 18:20










            • $begingroup$
              (1) Do you agree deleting u won't affect the number of leaves? (2) Do you agree that every tree has at least one leaf? (3) Do you agree that deleting u will lead to n (sub-)trees? If it's all yes, you're done ;)
              $endgroup$
              – Ankit Kumar
              Dec 24 '18 at 18:21








            • 1




              $begingroup$
              Apparently, everything is clear but I can not apply what I said to a simple example. For example, I have the tree 1-2-3 and the vertex 2 is connected to a vertex 4. So, if I delete the vertex 2 which has the maximum degree, I obtain the vertex 1,3,4 without the vertex 2 that they shared. How do I continue now?
              $endgroup$
              – PCNF
              Dec 24 '18 at 18:42






            • 2




              $begingroup$
              I think if you want your proof to be complete, you should replace the phrase "every tree has at least a leaf" with "every tree has at least two leaves." A leaf in $T_i$ might not be a leaf in $T$, if that leaf was originally joined to $u$.
              $endgroup$
              – Mike Earnest
              Dec 24 '18 at 18:52










            • $begingroup$
              Fixed everything now. Thank you for comments @MikeEarnest and PCNF
              $endgroup$
              – Ankit Kumar
              Dec 24 '18 at 19:04


















            • $begingroup$
              Thanks, but I didn't understand the last part. It's clear that you can divide $T$ into $T_n$ components which are trees. It is not clear to me how you go from these considerations to saying that T has n leaves.
              $endgroup$
              – PCNF
              Dec 24 '18 at 18:20










            • $begingroup$
              (1) Do you agree deleting u won't affect the number of leaves? (2) Do you agree that every tree has at least one leaf? (3) Do you agree that deleting u will lead to n (sub-)trees? If it's all yes, you're done ;)
              $endgroup$
              – Ankit Kumar
              Dec 24 '18 at 18:21








            • 1




              $begingroup$
              Apparently, everything is clear but I can not apply what I said to a simple example. For example, I have the tree 1-2-3 and the vertex 2 is connected to a vertex 4. So, if I delete the vertex 2 which has the maximum degree, I obtain the vertex 1,3,4 without the vertex 2 that they shared. How do I continue now?
              $endgroup$
              – PCNF
              Dec 24 '18 at 18:42






            • 2




              $begingroup$
              I think if you want your proof to be complete, you should replace the phrase "every tree has at least a leaf" with "every tree has at least two leaves." A leaf in $T_i$ might not be a leaf in $T$, if that leaf was originally joined to $u$.
              $endgroup$
              – Mike Earnest
              Dec 24 '18 at 18:52










            • $begingroup$
              Fixed everything now. Thank you for comments @MikeEarnest and PCNF
              $endgroup$
              – Ankit Kumar
              Dec 24 '18 at 19:04
















            $begingroup$
            Thanks, but I didn't understand the last part. It's clear that you can divide $T$ into $T_n$ components which are trees. It is not clear to me how you go from these considerations to saying that T has n leaves.
            $endgroup$
            – PCNF
            Dec 24 '18 at 18:20




            $begingroup$
            Thanks, but I didn't understand the last part. It's clear that you can divide $T$ into $T_n$ components which are trees. It is not clear to me how you go from these considerations to saying that T has n leaves.
            $endgroup$
            – PCNF
            Dec 24 '18 at 18:20












            $begingroup$
            (1) Do you agree deleting u won't affect the number of leaves? (2) Do you agree that every tree has at least one leaf? (3) Do you agree that deleting u will lead to n (sub-)trees? If it's all yes, you're done ;)
            $endgroup$
            – Ankit Kumar
            Dec 24 '18 at 18:21






            $begingroup$
            (1) Do you agree deleting u won't affect the number of leaves? (2) Do you agree that every tree has at least one leaf? (3) Do you agree that deleting u will lead to n (sub-)trees? If it's all yes, you're done ;)
            $endgroup$
            – Ankit Kumar
            Dec 24 '18 at 18:21






            1




            1




            $begingroup$
            Apparently, everything is clear but I can not apply what I said to a simple example. For example, I have the tree 1-2-3 and the vertex 2 is connected to a vertex 4. So, if I delete the vertex 2 which has the maximum degree, I obtain the vertex 1,3,4 without the vertex 2 that they shared. How do I continue now?
            $endgroup$
            – PCNF
            Dec 24 '18 at 18:42




            $begingroup$
            Apparently, everything is clear but I can not apply what I said to a simple example. For example, I have the tree 1-2-3 and the vertex 2 is connected to a vertex 4. So, if I delete the vertex 2 which has the maximum degree, I obtain the vertex 1,3,4 without the vertex 2 that they shared. How do I continue now?
            $endgroup$
            – PCNF
            Dec 24 '18 at 18:42




            2




            2




            $begingroup$
            I think if you want your proof to be complete, you should replace the phrase "every tree has at least a leaf" with "every tree has at least two leaves." A leaf in $T_i$ might not be a leaf in $T$, if that leaf was originally joined to $u$.
            $endgroup$
            – Mike Earnest
            Dec 24 '18 at 18:52




            $begingroup$
            I think if you want your proof to be complete, you should replace the phrase "every tree has at least a leaf" with "every tree has at least two leaves." A leaf in $T_i$ might not be a leaf in $T$, if that leaf was originally joined to $u$.
            $endgroup$
            – Mike Earnest
            Dec 24 '18 at 18:52












            $begingroup$
            Fixed everything now. Thank you for comments @MikeEarnest and PCNF
            $endgroup$
            – Ankit Kumar
            Dec 24 '18 at 19:04




            $begingroup$
            Fixed everything now. Thank you for comments @MikeEarnest and PCNF
            $endgroup$
            – Ankit Kumar
            Dec 24 '18 at 19:04


















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