Graph theory in disguise?












2












$begingroup$


There are $2n-1$ two-element subsets of set ${1,2,...,n}$. Prove that one can choose $n$ out of these such that their union contains no more than $frac{2}{3}n+1$ elements.




I was trying this one too and with no success. Then I read the official solution and it is a kind of not natural to me. Can someone try to walk another way?






combinatorics graph-theory optimization extremal-combinatorics probabilistic-method







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    There are $2n-1$ two-element subsets of set ${1,2,...,n}$. Prove that one can choose $n$ out of these such that their union contains no more than $frac{2}{3}n+1$ elements.




    I was trying this one too and with no success. Then I read the official solution and it is a kind of not natural to me. Can someone try to walk another way?






    combinatorics graph-theory optimization extremal-combinatorics probabilistic-method







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      0



      $begingroup$


      There are $2n-1$ two-element subsets of set ${1,2,...,n}$. Prove that one can choose $n$ out of these such that their union contains no more than $frac{2}{3}n+1$ elements.




      I was trying this one too and with no success. Then I read the official solution and it is a kind of not natural to me. Can someone try to walk another way?






      combinatorics graph-theory optimization extremal-combinatorics probabilistic-method







      share|cite|improve this question











      $endgroup$




      There are $2n-1$ two-element subsets of set ${1,2,...,n}$. Prove that one can choose $n$ out of these such that their union contains no more than $frac{2}{3}n+1$ elements.




      I was trying this one too and with no success. Then I read the official solution and it is a kind of not natural to me. Can someone try to walk another way?






      combinatorics graph-theory optimization extremal-combinatorics probabilistic-method




      combinatorics graph-theory optimization extremal-combinatorics probabilistic-method






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 11 '18 at 15:30







      greedoid

















      asked Jul 15 '17 at 16:10









      greedoidgreedoid

      44k1155109




      44k1155109






















          1 Answer
          1






          active

          oldest

          votes


















          3












          $begingroup$

          We are given a graph $G = (V, E)$ where $V = {v_1,dots,v_n}$ and $|E| = 2n - 1$. What we are trying to do is to remove as few edges as possible to leave as many isolated vertices as possible. To do this, we try to find a vertex of least degree and remove its edges.



          The degree formula says that



          $$ sum_{i = 1}^n deg(v_i) = 2|E| = 4n - 2. $$



          It follows that there is a vertex in $G$ of degree $le 3$. If every vertex had degree $ge 4$ then $sum_{i = 1}^n deg(v_i) ge 4n$, which is impossible.



          Without loss of generality, let this vertex be $v_n$. Then form the graph $G_1 = Gsetminus v = (V_1, E_1)$. Now $|E_1| ge 2n - 4$. Again by the degree formula we have



          $$ sum_{i = 1}^{n - 1} deg(v_i) = 2|E_1| ge 4(n - 1) - 4 tag{1}$$



          We want to conclude that there is a vertex of degree $le 3$. To conclude this, we would like to have an equality in $(1)$. To achieve this we allow ourselves to remove exactly three edges at each step (which might be more than the degree of the vertex we removed). Doing this, we have $2|E_1| = 4(n - 1) - 4$ and we get a vertex of degree $le 3$.



          Inductively, suppose we remove the vertices $v_{n - k + 1}, dots, v_n$ (after possibly relabeling) and $3k$ edges to get the graph $G_k = (V_k, E_k)$ where $|E_k| = 2n - 1 - 3k$. Then we have



          $$ sum_{i = 1}^{n - k} deg(v_i) = 2|E_k| = 4(n - k) - 2 - 2k $$



          so there is a vertex of degree $le 3$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            How can you deduce this: $$ sum_{i = 1}^{n - 1} deg(v_i) = 2|E_1| ge 4(n - 1) - 4 $$ so again there is a vertex of degree $le 3$.
            $endgroup$
            – greedoid
            Jul 15 '17 at 17:51












          • $begingroup$
            And, what then? Where is a contradiction? With what?
            $endgroup$
            – greedoid
            Jul 15 '17 at 17:56












          • $begingroup$
            @guest It should make sense now.
            $endgroup$
            – Trevor Gunn
            Jul 15 '17 at 18:03






          • 1




            $begingroup$
            You didn't finished explicitly but I can see where are you going (everybody does). So a process will terminate when $|E_k|le 0$ and this is when $nle {3k+1over 2}$. So when $k = {2n+1over 3}$ we get the desired union. Great solution. Thanks! Do you have any idea how to approach with the probabilistic method?
            $endgroup$
            – greedoid
            Jul 15 '17 at 18:12












          • $begingroup$
            @guest It's the same solution as given in the booklet but with graph terminology. Unfortunately I don't know a lot about probabilistic methods in combinatorics.
            $endgroup$
            – Trevor Gunn
            Jul 15 '17 at 18:19











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          $begingroup$

          We are given a graph $G = (V, E)$ where $V = {v_1,dots,v_n}$ and $|E| = 2n - 1$. What we are trying to do is to remove as few edges as possible to leave as many isolated vertices as possible. To do this, we try to find a vertex of least degree and remove its edges.



          The degree formula says that



          $$ sum_{i = 1}^n deg(v_i) = 2|E| = 4n - 2. $$



          It follows that there is a vertex in $G$ of degree $le 3$. If every vertex had degree $ge 4$ then $sum_{i = 1}^n deg(v_i) ge 4n$, which is impossible.



          Without loss of generality, let this vertex be $v_n$. Then form the graph $G_1 = Gsetminus v = (V_1, E_1)$. Now $|E_1| ge 2n - 4$. Again by the degree formula we have



          $$ sum_{i = 1}^{n - 1} deg(v_i) = 2|E_1| ge 4(n - 1) - 4 tag{1}$$



          We want to conclude that there is a vertex of degree $le 3$. To conclude this, we would like to have an equality in $(1)$. To achieve this we allow ourselves to remove exactly three edges at each step (which might be more than the degree of the vertex we removed). Doing this, we have $2|E_1| = 4(n - 1) - 4$ and we get a vertex of degree $le 3$.



          Inductively, suppose we remove the vertices $v_{n - k + 1}, dots, v_n$ (after possibly relabeling) and $3k$ edges to get the graph $G_k = (V_k, E_k)$ where $|E_k| = 2n - 1 - 3k$. Then we have



          $$ sum_{i = 1}^{n - k} deg(v_i) = 2|E_k| = 4(n - k) - 2 - 2k $$



          so there is a vertex of degree $le 3$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            How can you deduce this: $$ sum_{i = 1}^{n - 1} deg(v_i) = 2|E_1| ge 4(n - 1) - 4 $$ so again there is a vertex of degree $le 3$.
            $endgroup$
            – greedoid
            Jul 15 '17 at 17:51












          • $begingroup$
            And, what then? Where is a contradiction? With what?
            $endgroup$
            – greedoid
            Jul 15 '17 at 17:56












          • $begingroup$
            @guest It should make sense now.
            $endgroup$
            – Trevor Gunn
            Jul 15 '17 at 18:03






          • 1




            $begingroup$
            You didn't finished explicitly but I can see where are you going (everybody does). So a process will terminate when $|E_k|le 0$ and this is when $nle {3k+1over 2}$. So when $k = {2n+1over 3}$ we get the desired union. Great solution. Thanks! Do you have any idea how to approach with the probabilistic method?
            $endgroup$
            – greedoid
            Jul 15 '17 at 18:12












          • $begingroup$
            @guest It's the same solution as given in the booklet but with graph terminology. Unfortunately I don't know a lot about probabilistic methods in combinatorics.
            $endgroup$
            – Trevor Gunn
            Jul 15 '17 at 18:19
















          3












          $begingroup$

          We are given a graph $G = (V, E)$ where $V = {v_1,dots,v_n}$ and $|E| = 2n - 1$. What we are trying to do is to remove as few edges as possible to leave as many isolated vertices as possible. To do this, we try to find a vertex of least degree and remove its edges.



          The degree formula says that



          $$ sum_{i = 1}^n deg(v_i) = 2|E| = 4n - 2. $$



          It follows that there is a vertex in $G$ of degree $le 3$. If every vertex had degree $ge 4$ then $sum_{i = 1}^n deg(v_i) ge 4n$, which is impossible.



          Without loss of generality, let this vertex be $v_n$. Then form the graph $G_1 = Gsetminus v = (V_1, E_1)$. Now $|E_1| ge 2n - 4$. Again by the degree formula we have



          $$ sum_{i = 1}^{n - 1} deg(v_i) = 2|E_1| ge 4(n - 1) - 4 tag{1}$$



          We want to conclude that there is a vertex of degree $le 3$. To conclude this, we would like to have an equality in $(1)$. To achieve this we allow ourselves to remove exactly three edges at each step (which might be more than the degree of the vertex we removed). Doing this, we have $2|E_1| = 4(n - 1) - 4$ and we get a vertex of degree $le 3$.



          Inductively, suppose we remove the vertices $v_{n - k + 1}, dots, v_n$ (after possibly relabeling) and $3k$ edges to get the graph $G_k = (V_k, E_k)$ where $|E_k| = 2n - 1 - 3k$. Then we have



          $$ sum_{i = 1}^{n - k} deg(v_i) = 2|E_k| = 4(n - k) - 2 - 2k $$



          so there is a vertex of degree $le 3$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            How can you deduce this: $$ sum_{i = 1}^{n - 1} deg(v_i) = 2|E_1| ge 4(n - 1) - 4 $$ so again there is a vertex of degree $le 3$.
            $endgroup$
            – greedoid
            Jul 15 '17 at 17:51












          • $begingroup$
            And, what then? Where is a contradiction? With what?
            $endgroup$
            – greedoid
            Jul 15 '17 at 17:56












          • $begingroup$
            @guest It should make sense now.
            $endgroup$
            – Trevor Gunn
            Jul 15 '17 at 18:03






          • 1




            $begingroup$
            You didn't finished explicitly but I can see where are you going (everybody does). So a process will terminate when $|E_k|le 0$ and this is when $nle {3k+1over 2}$. So when $k = {2n+1over 3}$ we get the desired union. Great solution. Thanks! Do you have any idea how to approach with the probabilistic method?
            $endgroup$
            – greedoid
            Jul 15 '17 at 18:12












          • $begingroup$
            @guest It's the same solution as given in the booklet but with graph terminology. Unfortunately I don't know a lot about probabilistic methods in combinatorics.
            $endgroup$
            – Trevor Gunn
            Jul 15 '17 at 18:19














          3












          3








          3





          $begingroup$

          We are given a graph $G = (V, E)$ where $V = {v_1,dots,v_n}$ and $|E| = 2n - 1$. What we are trying to do is to remove as few edges as possible to leave as many isolated vertices as possible. To do this, we try to find a vertex of least degree and remove its edges.



          The degree formula says that



          $$ sum_{i = 1}^n deg(v_i) = 2|E| = 4n - 2. $$



          It follows that there is a vertex in $G$ of degree $le 3$. If every vertex had degree $ge 4$ then $sum_{i = 1}^n deg(v_i) ge 4n$, which is impossible.



          Without loss of generality, let this vertex be $v_n$. Then form the graph $G_1 = Gsetminus v = (V_1, E_1)$. Now $|E_1| ge 2n - 4$. Again by the degree formula we have



          $$ sum_{i = 1}^{n - 1} deg(v_i) = 2|E_1| ge 4(n - 1) - 4 tag{1}$$



          We want to conclude that there is a vertex of degree $le 3$. To conclude this, we would like to have an equality in $(1)$. To achieve this we allow ourselves to remove exactly three edges at each step (which might be more than the degree of the vertex we removed). Doing this, we have $2|E_1| = 4(n - 1) - 4$ and we get a vertex of degree $le 3$.



          Inductively, suppose we remove the vertices $v_{n - k + 1}, dots, v_n$ (after possibly relabeling) and $3k$ edges to get the graph $G_k = (V_k, E_k)$ where $|E_k| = 2n - 1 - 3k$. Then we have



          $$ sum_{i = 1}^{n - k} deg(v_i) = 2|E_k| = 4(n - k) - 2 - 2k $$



          so there is a vertex of degree $le 3$.






          share|cite|improve this answer











          $endgroup$



          We are given a graph $G = (V, E)$ where $V = {v_1,dots,v_n}$ and $|E| = 2n - 1$. What we are trying to do is to remove as few edges as possible to leave as many isolated vertices as possible. To do this, we try to find a vertex of least degree and remove its edges.



          The degree formula says that



          $$ sum_{i = 1}^n deg(v_i) = 2|E| = 4n - 2. $$



          It follows that there is a vertex in $G$ of degree $le 3$. If every vertex had degree $ge 4$ then $sum_{i = 1}^n deg(v_i) ge 4n$, which is impossible.



          Without loss of generality, let this vertex be $v_n$. Then form the graph $G_1 = Gsetminus v = (V_1, E_1)$. Now $|E_1| ge 2n - 4$. Again by the degree formula we have



          $$ sum_{i = 1}^{n - 1} deg(v_i) = 2|E_1| ge 4(n - 1) - 4 tag{1}$$



          We want to conclude that there is a vertex of degree $le 3$. To conclude this, we would like to have an equality in $(1)$. To achieve this we allow ourselves to remove exactly three edges at each step (which might be more than the degree of the vertex we removed). Doing this, we have $2|E_1| = 4(n - 1) - 4$ and we get a vertex of degree $le 3$.



          Inductively, suppose we remove the vertices $v_{n - k + 1}, dots, v_n$ (after possibly relabeling) and $3k$ edges to get the graph $G_k = (V_k, E_k)$ where $|E_k| = 2n - 1 - 3k$. Then we have



          $$ sum_{i = 1}^{n - k} deg(v_i) = 2|E_k| = 4(n - k) - 2 - 2k $$



          so there is a vertex of degree $le 3$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jul 15 '17 at 18:03

























          answered Jul 15 '17 at 17:19









          Trevor GunnTrevor Gunn

          14.7k32047




          14.7k32047












          • $begingroup$
            How can you deduce this: $$ sum_{i = 1}^{n - 1} deg(v_i) = 2|E_1| ge 4(n - 1) - 4 $$ so again there is a vertex of degree $le 3$.
            $endgroup$
            – greedoid
            Jul 15 '17 at 17:51












          • $begingroup$
            And, what then? Where is a contradiction? With what?
            $endgroup$
            – greedoid
            Jul 15 '17 at 17:56












          • $begingroup$
            @guest It should make sense now.
            $endgroup$
            – Trevor Gunn
            Jul 15 '17 at 18:03






          • 1




            $begingroup$
            You didn't finished explicitly but I can see where are you going (everybody does). So a process will terminate when $|E_k|le 0$ and this is when $nle {3k+1over 2}$. So when $k = {2n+1over 3}$ we get the desired union. Great solution. Thanks! Do you have any idea how to approach with the probabilistic method?
            $endgroup$
            – greedoid
            Jul 15 '17 at 18:12












          • $begingroup$
            @guest It's the same solution as given in the booklet but with graph terminology. Unfortunately I don't know a lot about probabilistic methods in combinatorics.
            $endgroup$
            – Trevor Gunn
            Jul 15 '17 at 18:19


















          • $begingroup$
            How can you deduce this: $$ sum_{i = 1}^{n - 1} deg(v_i) = 2|E_1| ge 4(n - 1) - 4 $$ so again there is a vertex of degree $le 3$.
            $endgroup$
            – greedoid
            Jul 15 '17 at 17:51












          • $begingroup$
            And, what then? Where is a contradiction? With what?
            $endgroup$
            – greedoid
            Jul 15 '17 at 17:56












          • $begingroup$
            @guest It should make sense now.
            $endgroup$
            – Trevor Gunn
            Jul 15 '17 at 18:03






          • 1




            $begingroup$
            You didn't finished explicitly but I can see where are you going (everybody does). So a process will terminate when $|E_k|le 0$ and this is when $nle {3k+1over 2}$. So when $k = {2n+1over 3}$ we get the desired union. Great solution. Thanks! Do you have any idea how to approach with the probabilistic method?
            $endgroup$
            – greedoid
            Jul 15 '17 at 18:12












          • $begingroup$
            @guest It's the same solution as given in the booklet but with graph terminology. Unfortunately I don't know a lot about probabilistic methods in combinatorics.
            $endgroup$
            – Trevor Gunn
            Jul 15 '17 at 18:19
















          $begingroup$
          How can you deduce this: $$ sum_{i = 1}^{n - 1} deg(v_i) = 2|E_1| ge 4(n - 1) - 4 $$ so again there is a vertex of degree $le 3$.
          $endgroup$
          – greedoid
          Jul 15 '17 at 17:51






          $begingroup$
          How can you deduce this: $$ sum_{i = 1}^{n - 1} deg(v_i) = 2|E_1| ge 4(n - 1) - 4 $$ so again there is a vertex of degree $le 3$.
          $endgroup$
          – greedoid
          Jul 15 '17 at 17:51














          $begingroup$
          And, what then? Where is a contradiction? With what?
          $endgroup$
          – greedoid
          Jul 15 '17 at 17:56






          $begingroup$
          And, what then? Where is a contradiction? With what?
          $endgroup$
          – greedoid
          Jul 15 '17 at 17:56














          $begingroup$
          @guest It should make sense now.
          $endgroup$
          – Trevor Gunn
          Jul 15 '17 at 18:03




          $begingroup$
          @guest It should make sense now.
          $endgroup$
          – Trevor Gunn
          Jul 15 '17 at 18:03




          1




          1




          $begingroup$
          You didn't finished explicitly but I can see where are you going (everybody does). So a process will terminate when $|E_k|le 0$ and this is when $nle {3k+1over 2}$. So when $k = {2n+1over 3}$ we get the desired union. Great solution. Thanks! Do you have any idea how to approach with the probabilistic method?
          $endgroup$
          – greedoid
          Jul 15 '17 at 18:12






          $begingroup$
          You didn't finished explicitly but I can see where are you going (everybody does). So a process will terminate when $|E_k|le 0$ and this is when $nle {3k+1over 2}$. So when $k = {2n+1over 3}$ we get the desired union. Great solution. Thanks! Do you have any idea how to approach with the probabilistic method?
          $endgroup$
          – greedoid
          Jul 15 '17 at 18:12














          $begingroup$
          @guest It's the same solution as given in the booklet but with graph terminology. Unfortunately I don't know a lot about probabilistic methods in combinatorics.
          $endgroup$
          – Trevor Gunn
          Jul 15 '17 at 18:19




          $begingroup$
          @guest It's the same solution as given in the booklet but with graph terminology. Unfortunately I don't know a lot about probabilistic methods in combinatorics.
          $endgroup$
          – Trevor Gunn
          Jul 15 '17 at 18:19


















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          Musa

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          Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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