What is in the minimum number of simple paths in a forest of k trees with n vertices?












1












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I'm stumped on this one.



Let G be a forest containing 6 trees with 27 total vertices. What is the minimum number of simple paths for G?



I know how to compute this for an even number of vertices. For example, a forest with 4 trees and 18 total vertices would have at least 82 paths. You partition the total number of vertices:



$4^2 + 4^2 + 5^2 + 5^2 = 82 $



I'm not sure how to adapt this formula to an odd number of total vertices.






graph-theory trees







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    1












    $begingroup$


    I'm stumped on this one.



    Let G be a forest containing 6 trees with 27 total vertices. What is the minimum number of simple paths for G?



    I know how to compute this for an even number of vertices. For example, a forest with 4 trees and 18 total vertices would have at least 82 paths. You partition the total number of vertices:



    $4^2 + 4^2 + 5^2 + 5^2 = 82 $



    I'm not sure how to adapt this formula to an odd number of total vertices.






    graph-theory trees







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I'm stumped on this one.



      Let G be a forest containing 6 trees with 27 total vertices. What is the minimum number of simple paths for G?



      I know how to compute this for an even number of vertices. For example, a forest with 4 trees and 18 total vertices would have at least 82 paths. You partition the total number of vertices:



      $4^2 + 4^2 + 5^2 + 5^2 = 82 $



      I'm not sure how to adapt this formula to an odd number of total vertices.






      graph-theory trees







      share|cite|improve this question









      $endgroup$




      I'm stumped on this one.



      Let G be a forest containing 6 trees with 27 total vertices. What is the minimum number of simple paths for G?



      I know how to compute this for an even number of vertices. For example, a forest with 4 trees and 18 total vertices would have at least 82 paths. You partition the total number of vertices:



      $4^2 + 4^2 + 5^2 + 5^2 = 82 $



      I'm not sure how to adapt this formula to an odd number of total vertices.






      graph-theory trees




      graph-theory trees






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 25 '18 at 17:47









      shattershatter

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          2 Answers
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          $begingroup$

          I don't see why it makes a difference if you have an odd number of vertices. You do exactly the same thing: divide the vertices up as evenly as possible.



          You seem to already know this, but the total number of simple paths (counting a path and its reversal separately) in any tree with $k$ vertices is $k^2$. Every vertex is a path of length $0$, and for every pair of distinct vertices $(a,b)$ there is exactly one path from $a$ to $b$.



          So if a forest has six trees with $k_1,...,k_6$ vertices respectively there are $k_1^2+cdots+k_6^2$ paths, and you have to choose $k_1,...,k_6$ all positive with $k_1+cdots+k_6=27$, to make $k_1^2+cdots+k_6^2$ as small as possible. $27/6=4.5$, so this is achieved by taking three components of $4$ vertices and three of $5$, for $4^2+4^2+4^2+5^2+5^2+5^2=123$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I guess my confusion had to do with an odd number of vertices. Thank you for the clarification.
            $endgroup$
            – shatter
            Nov 25 '18 at 19:56





















          0












          $begingroup$

          From the example you give one is lead to the conjecture that the total number of paths in a tree with $j$ vertices is minimal when the tree is a path graph of length $j-1$. The number of (directed) subpaths then is $j^2$. Assuming this it seems that the total number of paths in a forest is minimal if the all trees in this forest are path graphs, and have about the same size.



          Im not going into the proof of the first of these conjectures (if true). For the second it suffices to consider two path graphs $Gamma$ and $Gamma'$ with $j$ and $j'$ vertices, and to show that, if $j'geq j+2$ you can make the total number of subpaths smaller by making $Gamma$ one edge longer and $Gamma'$ one edge shorter.






          share|cite|improve this answer









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            2 Answers
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            2 Answers
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            0












            $begingroup$

            I don't see why it makes a difference if you have an odd number of vertices. You do exactly the same thing: divide the vertices up as evenly as possible.



            You seem to already know this, but the total number of simple paths (counting a path and its reversal separately) in any tree with $k$ vertices is $k^2$. Every vertex is a path of length $0$, and for every pair of distinct vertices $(a,b)$ there is exactly one path from $a$ to $b$.



            So if a forest has six trees with $k_1,...,k_6$ vertices respectively there are $k_1^2+cdots+k_6^2$ paths, and you have to choose $k_1,...,k_6$ all positive with $k_1+cdots+k_6=27$, to make $k_1^2+cdots+k_6^2$ as small as possible. $27/6=4.5$, so this is achieved by taking three components of $4$ vertices and three of $5$, for $4^2+4^2+4^2+5^2+5^2+5^2=123$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I guess my confusion had to do with an odd number of vertices. Thank you for the clarification.
              $endgroup$
              – shatter
              Nov 25 '18 at 19:56


















            0












            $begingroup$

            I don't see why it makes a difference if you have an odd number of vertices. You do exactly the same thing: divide the vertices up as evenly as possible.



            You seem to already know this, but the total number of simple paths (counting a path and its reversal separately) in any tree with $k$ vertices is $k^2$. Every vertex is a path of length $0$, and for every pair of distinct vertices $(a,b)$ there is exactly one path from $a$ to $b$.



            So if a forest has six trees with $k_1,...,k_6$ vertices respectively there are $k_1^2+cdots+k_6^2$ paths, and you have to choose $k_1,...,k_6$ all positive with $k_1+cdots+k_6=27$, to make $k_1^2+cdots+k_6^2$ as small as possible. $27/6=4.5$, so this is achieved by taking three components of $4$ vertices and three of $5$, for $4^2+4^2+4^2+5^2+5^2+5^2=123$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I guess my confusion had to do with an odd number of vertices. Thank you for the clarification.
              $endgroup$
              – shatter
              Nov 25 '18 at 19:56
















            0












            0








            0





            $begingroup$

            I don't see why it makes a difference if you have an odd number of vertices. You do exactly the same thing: divide the vertices up as evenly as possible.



            You seem to already know this, but the total number of simple paths (counting a path and its reversal separately) in any tree with $k$ vertices is $k^2$. Every vertex is a path of length $0$, and for every pair of distinct vertices $(a,b)$ there is exactly one path from $a$ to $b$.



            So if a forest has six trees with $k_1,...,k_6$ vertices respectively there are $k_1^2+cdots+k_6^2$ paths, and you have to choose $k_1,...,k_6$ all positive with $k_1+cdots+k_6=27$, to make $k_1^2+cdots+k_6^2$ as small as possible. $27/6=4.5$, so this is achieved by taking three components of $4$ vertices and three of $5$, for $4^2+4^2+4^2+5^2+5^2+5^2=123$.






            share|cite|improve this answer









            $endgroup$



            I don't see why it makes a difference if you have an odd number of vertices. You do exactly the same thing: divide the vertices up as evenly as possible.



            You seem to already know this, but the total number of simple paths (counting a path and its reversal separately) in any tree with $k$ vertices is $k^2$. Every vertex is a path of length $0$, and for every pair of distinct vertices $(a,b)$ there is exactly one path from $a$ to $b$.



            So if a forest has six trees with $k_1,...,k_6$ vertices respectively there are $k_1^2+cdots+k_6^2$ paths, and you have to choose $k_1,...,k_6$ all positive with $k_1+cdots+k_6=27$, to make $k_1^2+cdots+k_6^2$ as small as possible. $27/6=4.5$, so this is achieved by taking three components of $4$ vertices and three of $5$, for $4^2+4^2+4^2+5^2+5^2+5^2=123$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 25 '18 at 19:41









            Especially LimeEspecially Lime

            21.9k22858




            21.9k22858












            • $begingroup$
              I guess my confusion had to do with an odd number of vertices. Thank you for the clarification.
              $endgroup$
              – shatter
              Nov 25 '18 at 19:56




















            • $begingroup$
              I guess my confusion had to do with an odd number of vertices. Thank you for the clarification.
              $endgroup$
              – shatter
              Nov 25 '18 at 19:56


















            $begingroup$
            I guess my confusion had to do with an odd number of vertices. Thank you for the clarification.
            $endgroup$
            – shatter
            Nov 25 '18 at 19:56






            $begingroup$
            I guess my confusion had to do with an odd number of vertices. Thank you for the clarification.
            $endgroup$
            – shatter
            Nov 25 '18 at 19:56













            0












            $begingroup$

            From the example you give one is lead to the conjecture that the total number of paths in a tree with $j$ vertices is minimal when the tree is a path graph of length $j-1$. The number of (directed) subpaths then is $j^2$. Assuming this it seems that the total number of paths in a forest is minimal if the all trees in this forest are path graphs, and have about the same size.



            Im not going into the proof of the first of these conjectures (if true). For the second it suffices to consider two path graphs $Gamma$ and $Gamma'$ with $j$ and $j'$ vertices, and to show that, if $j'geq j+2$ you can make the total number of subpaths smaller by making $Gamma$ one edge longer and $Gamma'$ one edge shorter.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              From the example you give one is lead to the conjecture that the total number of paths in a tree with $j$ vertices is minimal when the tree is a path graph of length $j-1$. The number of (directed) subpaths then is $j^2$. Assuming this it seems that the total number of paths in a forest is minimal if the all trees in this forest are path graphs, and have about the same size.



              Im not going into the proof of the first of these conjectures (if true). For the second it suffices to consider two path graphs $Gamma$ and $Gamma'$ with $j$ and $j'$ vertices, and to show that, if $j'geq j+2$ you can make the total number of subpaths smaller by making $Gamma$ one edge longer and $Gamma'$ one edge shorter.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                From the example you give one is lead to the conjecture that the total number of paths in a tree with $j$ vertices is minimal when the tree is a path graph of length $j-1$. The number of (directed) subpaths then is $j^2$. Assuming this it seems that the total number of paths in a forest is minimal if the all trees in this forest are path graphs, and have about the same size.



                Im not going into the proof of the first of these conjectures (if true). For the second it suffices to consider two path graphs $Gamma$ and $Gamma'$ with $j$ and $j'$ vertices, and to show that, if $j'geq j+2$ you can make the total number of subpaths smaller by making $Gamma$ one edge longer and $Gamma'$ one edge shorter.






                share|cite|improve this answer









                $endgroup$



                From the example you give one is lead to the conjecture that the total number of paths in a tree with $j$ vertices is minimal when the tree is a path graph of length $j-1$. The number of (directed) subpaths then is $j^2$. Assuming this it seems that the total number of paths in a forest is minimal if the all trees in this forest are path graphs, and have about the same size.



                Im not going into the proof of the first of these conjectures (if true). For the second it suffices to consider two path graphs $Gamma$ and $Gamma'$ with $j$ and $j'$ vertices, and to show that, if $j'geq j+2$ you can make the total number of subpaths smaller by making $Gamma$ one edge longer and $Gamma'$ one edge shorter.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 25 '18 at 19:27









                Christian BlatterChristian Blatter

                172k7113326




                172k7113326






























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