Catalan numbers and triangulations

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The number of ways to parenthesize an $n$ fold product is a Catalan number in the list $1,1,2,5,14,cdots$ where these are in order of the number of terms in the product. The $n$th such number is also the numbr of ways to triangulate an $n+1$-gon.



I'm wondering whether there is a simple translation between a specific parenthesized $n$ fold product to a specific triangulation of an $n+1$-gon.



For example one parenthesized 5-fold product is $(12)(3(45))$ This then would (hopefully) be translatable to one of the $14$ triangulations of a $6$-gon.



I have tried various ways to label the vertices of the $n+1$ gon to see which parenthesized $n$ fold product a given triangulation goes with, but no luck.




combinatorics triangulation catalan-numbers




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  • See also Theorem 1.19 on page 9 of Chapter 1 of Discrete and Computational Geometry by Devadoss and O'Rourke. This chapter is freely available.
    – lhf
    Jul 30 at 16:23







  • 1




    I believe your definition of $C_n$ is off by one from the usual one. That is, $C_n$ counts ways to multiply $n+1$ numbers and triangulate an $(n+2)$-gon.
    – Mike Earnest
    Jul 30 at 16:31










  • @MikeEarnest Yes it's off by 1, however I kept it that way so the n-th number would go with n things to be multiplied. This way is also cleaner for the recursion $c(n)=sum c(I)c(j)$ where $1 le i,j le n-1, i+j=n.$ There my be a better way of looking at things...
    – coffeemath
    Jul 30 at 16:42














up vote
9
down vote

favorite
1












The number of ways to parenthesize an $n$ fold product is a Catalan number in the list $1,1,2,5,14,cdots$ where these are in order of the number of terms in the product. The $n$th such number is also the numbr of ways to triangulate an $n+1$-gon.



I'm wondering whether there is a simple translation between a specific parenthesized $n$ fold product to a specific triangulation of an $n+1$-gon.



For example one parenthesized 5-fold product is $(12)(3(45))$ This then would (hopefully) be translatable to one of the $14$ triangulations of a $6$-gon.



I have tried various ways to label the vertices of the $n+1$ gon to see which parenthesized $n$ fold product a given triangulation goes with, but no luck.




combinatorics triangulation catalan-numbers




share|cite|improve this question





















  • See also Theorem 1.19 on page 9 of Chapter 1 of Discrete and Computational Geometry by Devadoss and O'Rourke. This chapter is freely available.
    – lhf
    Jul 30 at 16:23







  • 1




    I believe your definition of $C_n$ is off by one from the usual one. That is, $C_n$ counts ways to multiply $n+1$ numbers and triangulate an $(n+2)$-gon.
    – Mike Earnest
    Jul 30 at 16:31










  • @MikeEarnest Yes it's off by 1, however I kept it that way so the n-th number would go with n things to be multiplied. This way is also cleaner for the recursion $c(n)=sum c(I)c(j)$ where $1 le i,j le n-1, i+j=n.$ There my be a better way of looking at things...
    – coffeemath
    Jul 30 at 16:42












up vote
9
down vote

favorite
1









up vote
9
down vote

favorite
1






1





The number of ways to parenthesize an $n$ fold product is a Catalan number in the list $1,1,2,5,14,cdots$ where these are in order of the number of terms in the product. The $n$th such number is also the numbr of ways to triangulate an $n+1$-gon.



I'm wondering whether there is a simple translation between a specific parenthesized $n$ fold product to a specific triangulation of an $n+1$-gon.



For example one parenthesized 5-fold product is $(12)(3(45))$ This then would (hopefully) be translatable to one of the $14$ triangulations of a $6$-gon.



I have tried various ways to label the vertices of the $n+1$ gon to see which parenthesized $n$ fold product a given triangulation goes with, but no luck.




combinatorics triangulation catalan-numbers




share|cite|improve this question













The number of ways to parenthesize an $n$ fold product is a Catalan number in the list $1,1,2,5,14,cdots$ where these are in order of the number of terms in the product. The $n$th such number is also the numbr of ways to triangulate an $n+1$-gon.



I'm wondering whether there is a simple translation between a specific parenthesized $n$ fold product to a specific triangulation of an $n+1$-gon.



For example one parenthesized 5-fold product is $(12)(3(45))$ This then would (hopefully) be translatable to one of the $14$ triangulations of a $6$-gon.



I have tried various ways to label the vertices of the $n+1$ gon to see which parenthesized $n$ fold product a given triangulation goes with, but no luck.





combinatorics triangulation catalan-numbers





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 30 at 16:36









Arnaud Mortier

18.1k21958




18.1k21958









asked Jul 30 at 15:10









coffeemath

1,2291212




1,2291212











  • See also Theorem 1.19 on page 9 of Chapter 1 of Discrete and Computational Geometry by Devadoss and O'Rourke. This chapter is freely available.
    – lhf
    Jul 30 at 16:23







  • 1




    I believe your definition of $C_n$ is off by one from the usual one. That is, $C_n$ counts ways to multiply $n+1$ numbers and triangulate an $(n+2)$-gon.
    – Mike Earnest
    Jul 30 at 16:31










  • @MikeEarnest Yes it's off by 1, however I kept it that way so the n-th number would go with n things to be multiplied. This way is also cleaner for the recursion $c(n)=sum c(I)c(j)$ where $1 le i,j le n-1, i+j=n.$ There my be a better way of looking at things...
    – coffeemath
    Jul 30 at 16:42
















  • See also Theorem 1.19 on page 9 of Chapter 1 of Discrete and Computational Geometry by Devadoss and O'Rourke. This chapter is freely available.
    – lhf
    Jul 30 at 16:23







  • 1




    I believe your definition of $C_n$ is off by one from the usual one. That is, $C_n$ counts ways to multiply $n+1$ numbers and triangulate an $(n+2)$-gon.
    – Mike Earnest
    Jul 30 at 16:31










  • @MikeEarnest Yes it's off by 1, however I kept it that way so the n-th number would go with n things to be multiplied. This way is also cleaner for the recursion $c(n)=sum c(I)c(j)$ where $1 le i,j le n-1, i+j=n.$ There my be a better way of looking at things...
    – coffeemath
    Jul 30 at 16:42















See also Theorem 1.19 on page 9 of Chapter 1 of Discrete and Computational Geometry by Devadoss and O'Rourke. This chapter is freely available.
– lhf
Jul 30 at 16:23





See also Theorem 1.19 on page 9 of Chapter 1 of Discrete and Computational Geometry by Devadoss and O'Rourke. This chapter is freely available.
– lhf
Jul 30 at 16:23





1




1




I believe your definition of $C_n$ is off by one from the usual one. That is, $C_n$ counts ways to multiply $n+1$ numbers and triangulate an $(n+2)$-gon.
– Mike Earnest
Jul 30 at 16:31




I believe your definition of $C_n$ is off by one from the usual one. That is, $C_n$ counts ways to multiply $n+1$ numbers and triangulate an $(n+2)$-gon.
– Mike Earnest
Jul 30 at 16:31












@MikeEarnest Yes it's off by 1, however I kept it that way so the n-th number would go with n things to be multiplied. This way is also cleaner for the recursion $c(n)=sum c(I)c(j)$ where $1 le i,j le n-1, i+j=n.$ There my be a better way of looking at things...
– coffeemath
Jul 30 at 16:42




@MikeEarnest Yes it's off by 1, however I kept it that way so the n-th number would go with n things to be multiplied. This way is also cleaner for the recursion $c(n)=sum c(I)c(j)$ where $1 le i,j le n-1, i+j=n.$ There my be a better way of looking at things...
– coffeemath
Jul 30 at 16:42










2 Answers
2






active

oldest

votes

















up vote
6
down vote



accepted










There is a nice bijection between both objects and binary trees:



  • Given a triangulation on the polygon with vertices numbered $0$ to $n$, build a tree whose vertices are the edges of the figure, including the $n+1$ edges of the polygon and the $n-2$ internal edges. The root is the external edge connected vertex $0$ to vertex $1$. For each internal edge $e$, which borders two triangles $T_1$ and $T_2$, then supposing $T_1$ is further from the root edge $(0,1)$, then the right (left) child of $e$ is the edge of $T_1$ which is clockwise (anti-clockwise) adjacent to $e$.


  • Given a parenthesization of $x_1cdots x_n$, build a tree with one vertex for each intermediate result when performing all of the multiplications. The result $a$ has left and right children $b$ and $c$ if $a$ was computed as the product of $b$ and $c$ in that order.


For example, with $(1,2)(3,(4,5))$, the tree is



 120
 / 
 / 60
 2 / 
 / 3 20
1 2 / 
 4 5


This tree corresponds to the triangultion



 1–––––0
 /| /| 
 / | / | 
2 | / | 5
 | / | /
 |/ |/
 3–––––4


whose tree is



 (01)
 / 
 / (03)
 (13) / 
 / (34) (04)
(12) (23) / 
 (45) (05)





share|cite|improve this answer























  • I'll likely accept, after I absorb it better. Thanks, +1 for now.
    – coffeemath
    Jul 30 at 16:43






  • 2




    Here’s an excellent series of blog posts about this correspondence with tons of cool applications to binary search trees.
    – templatetypedef
    Jul 30 at 17:53

















up vote
3
down vote













Here is a possible way, which makes use of the intrinsically similar recursive property of both numbers.



I'm assuming the vertices of the $(n+1)$-gon are numbered counterclockwise from $1$ to $n+1$.



Starting from a triangulation.



In every triangulation, there is one triangle containing the edge $1to 2$. Call the third vertex $k$. The corresponding parenthesizing is going to start with $$(1 ldots k-2)(k-1 ldots n)$$
Now "on the right" of the edge $2to k$, you have a triangulated $(k-1)$-gon. Rename its vertices from $1$ to $k-1$ starting from the lowest and iterate.



Similarly "on the left" of the edge $1to k$ there is a triangulated $(n-k+3)$-gon. Rename its vertices from $k-1$ to $n+1$ starting from the lowest and iterate.



Starting from a parenthesized product.



  • Consider the most external product, say $$(1 ldots k-2)(k-1 ldots n)$$

  • Draw a triangle $1 2 k$

  • Iterate after renumbering the vertices accordingly as before.





share|cite|improve this answer























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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    6
    down vote



    accepted










    There is a nice bijection between both objects and binary trees:



    • Given a triangulation on the polygon with vertices numbered $0$ to $n$, build a tree whose vertices are the edges of the figure, including the $n+1$ edges of the polygon and the $n-2$ internal edges. The root is the external edge connected vertex $0$ to vertex $1$. For each internal edge $e$, which borders two triangles $T_1$ and $T_2$, then supposing $T_1$ is further from the root edge $(0,1)$, then the right (left) child of $e$ is the edge of $T_1$ which is clockwise (anti-clockwise) adjacent to $e$.


    • Given a parenthesization of $x_1cdots x_n$, build a tree with one vertex for each intermediate result when performing all of the multiplications. The result $a$ has left and right children $b$ and $c$ if $a$ was computed as the product of $b$ and $c$ in that order.


    For example, with $(1,2)(3,(4,5))$, the tree is



     120
     / 
     / 60
     2 / 
     / 3 20
    1 2 / 
     4 5


    This tree corresponds to the triangultion



     1–––––0
     /| /| 
     / | / | 
    2 | / | 5
     | / | /
     |/ |/
     3–––––4


    whose tree is



     (01)
     / 
     / (03)
     (13) / 
     / (34) (04)
    (12) (23) / 
     (45) (05)





    share|cite|improve this answer























    • I'll likely accept, after I absorb it better. Thanks, +1 for now.
      – coffeemath
      Jul 30 at 16:43






    • 2




      Here’s an excellent series of blog posts about this correspondence with tons of cool applications to binary search trees.
      – templatetypedef
      Jul 30 at 17:53














    up vote
    6
    down vote



    accepted










    There is a nice bijection between both objects and binary trees:



    • Given a triangulation on the polygon with vertices numbered $0$ to $n$, build a tree whose vertices are the edges of the figure, including the $n+1$ edges of the polygon and the $n-2$ internal edges. The root is the external edge connected vertex $0$ to vertex $1$. For each internal edge $e$, which borders two triangles $T_1$ and $T_2$, then supposing $T_1$ is further from the root edge $(0,1)$, then the right (left) child of $e$ is the edge of $T_1$ which is clockwise (anti-clockwise) adjacent to $e$.


    • Given a parenthesization of $x_1cdots x_n$, build a tree with one vertex for each intermediate result when performing all of the multiplications. The result $a$ has left and right children $b$ and $c$ if $a$ was computed as the product of $b$ and $c$ in that order.


    For example, with $(1,2)(3,(4,5))$, the tree is



     120
     / 
     / 60
     2 / 
     / 3 20
    1 2 / 
     4 5


    This tree corresponds to the triangultion



     1–––––0
     /| /| 
     / | / | 
    2 | / | 5
     | / | /
     |/ |/
     3–––––4


    whose tree is



     (01)
     / 
     / (03)
     (13) / 
     / (34) (04)
    (12) (23) / 
     (45) (05)





    share|cite|improve this answer























    • I'll likely accept, after I absorb it better. Thanks, +1 for now.
      – coffeemath
      Jul 30 at 16:43






    • 2




      Here’s an excellent series of blog posts about this correspondence with tons of cool applications to binary search trees.
      – templatetypedef
      Jul 30 at 17:53












    up vote
    6
    down vote



    accepted







    up vote
    6
    down vote



    accepted






    There is a nice bijection between both objects and binary trees:



    • Given a triangulation on the polygon with vertices numbered $0$ to $n$, build a tree whose vertices are the edges of the figure, including the $n+1$ edges of the polygon and the $n-2$ internal edges. The root is the external edge connected vertex $0$ to vertex $1$. For each internal edge $e$, which borders two triangles $T_1$ and $T_2$, then supposing $T_1$ is further from the root edge $(0,1)$, then the right (left) child of $e$ is the edge of $T_1$ which is clockwise (anti-clockwise) adjacent to $e$.


    • Given a parenthesization of $x_1cdots x_n$, build a tree with one vertex for each intermediate result when performing all of the multiplications. The result $a$ has left and right children $b$ and $c$ if $a$ was computed as the product of $b$ and $c$ in that order.


    For example, with $(1,2)(3,(4,5))$, the tree is



     120
     / 
     / 60
     2 / 
     / 3 20
    1 2 / 
     4 5


    This tree corresponds to the triangultion



     1–––––0
     /| /| 
     / | / | 
    2 | / | 5
     | / | /
     |/ |/
     3–––––4


    whose tree is



     (01)
     / 
     / (03)
     (13) / 
     / (34) (04)
    (12) (23) / 
     (45) (05)





    share|cite|improve this answer















    There is a nice bijection between both objects and binary trees:



    • Given a triangulation on the polygon with vertices numbered $0$ to $n$, build a tree whose vertices are the edges of the figure, including the $n+1$ edges of the polygon and the $n-2$ internal edges. The root is the external edge connected vertex $0$ to vertex $1$. For each internal edge $e$, which borders two triangles $T_1$ and $T_2$, then supposing $T_1$ is further from the root edge $(0,1)$, then the right (left) child of $e$ is the edge of $T_1$ which is clockwise (anti-clockwise) adjacent to $e$.


    • Given a parenthesization of $x_1cdots x_n$, build a tree with one vertex for each intermediate result when performing all of the multiplications. The result $a$ has left and right children $b$ and $c$ if $a$ was computed as the product of $b$ and $c$ in that order.


    For example, with $(1,2)(3,(4,5))$, the tree is



     120
     / 
     / 60
     2 / 
     / 3 20
    1 2 / 
     4 5


    This tree corresponds to the triangultion



     1–––––0
     /| /| 
     / | / | 
    2 | / | 5
     | / | /
     |/ |/
     3–––––4


    whose tree is



     (01)
     / 
     / (03)
     (13) / 
     / (34) (04)
    (12) (23) / 
     (45) (05)






    share|cite|improve this answer















    share|cite|improve this answer



    share|cite|improve this answer








    edited Jul 30 at 16:44


























    answered Jul 30 at 16:20









    Mike Earnest

    14.8k11644




    14.8k11644











    • I'll likely accept, after I absorb it better. Thanks, +1 for now.
      – coffeemath
      Jul 30 at 16:43






    • 2




      Here’s an excellent series of blog posts about this correspondence with tons of cool applications to binary search trees.
      – templatetypedef
      Jul 30 at 17:53
















    • I'll likely accept, after I absorb it better. Thanks, +1 for now.
      – coffeemath
      Jul 30 at 16:43






    • 2




      Here’s an excellent series of blog posts about this correspondence with tons of cool applications to binary search trees.
      – templatetypedef
      Jul 30 at 17:53















    I'll likely accept, after I absorb it better. Thanks, +1 for now.
    – coffeemath
    Jul 30 at 16:43




    I'll likely accept, after I absorb it better. Thanks, +1 for now.
    – coffeemath
    Jul 30 at 16:43




    2




    2




    Here’s an excellent series of blog posts about this correspondence with tons of cool applications to binary search trees.
    – templatetypedef
    Jul 30 at 17:53




    Here’s an excellent series of blog posts about this correspondence with tons of cool applications to binary search trees.
    – templatetypedef
    Jul 30 at 17:53










    up vote
    3
    down vote













    Here is a possible way, which makes use of the intrinsically similar recursive property of both numbers.



    I'm assuming the vertices of the $(n+1)$-gon are numbered counterclockwise from $1$ to $n+1$.



    Starting from a triangulation.



    In every triangulation, there is one triangle containing the edge $1to 2$. Call the third vertex $k$. The corresponding parenthesizing is going to start with $$(1 ldots k-2)(k-1 ldots n)$$
    Now "on the right" of the edge $2to k$, you have a triangulated $(k-1)$-gon. Rename its vertices from $1$ to $k-1$ starting from the lowest and iterate.



    Similarly "on the left" of the edge $1to k$ there is a triangulated $(n-k+3)$-gon. Rename its vertices from $k-1$ to $n+1$ starting from the lowest and iterate.



    Starting from a parenthesized product.



    • Consider the most external product, say $$(1 ldots k-2)(k-1 ldots n)$$

    • Draw a triangle $1 2 k$

    • Iterate after renumbering the vertices accordingly as before.





    share|cite|improve this answer



























      up vote
      3
      down vote













      Here is a possible way, which makes use of the intrinsically similar recursive property of both numbers.



      I'm assuming the vertices of the $(n+1)$-gon are numbered counterclockwise from $1$ to $n+1$.



      Starting from a triangulation.



      In every triangulation, there is one triangle containing the edge $1to 2$. Call the third vertex $k$. The corresponding parenthesizing is going to start with $$(1 ldots k-2)(k-1 ldots n)$$
      Now "on the right" of the edge $2to k$, you have a triangulated $(k-1)$-gon. Rename its vertices from $1$ to $k-1$ starting from the lowest and iterate.



      Similarly "on the left" of the edge $1to k$ there is a triangulated $(n-k+3)$-gon. Rename its vertices from $k-1$ to $n+1$ starting from the lowest and iterate.



      Starting from a parenthesized product.



      • Consider the most external product, say $$(1 ldots k-2)(k-1 ldots n)$$

      • Draw a triangle $1 2 k$

      • Iterate after renumbering the vertices accordingly as before.





      share|cite|improve this answer

























        up vote
        3
        down vote










        up vote
        3
        down vote









        Here is a possible way, which makes use of the intrinsically similar recursive property of both numbers.



        I'm assuming the vertices of the $(n+1)$-gon are numbered counterclockwise from $1$ to $n+1$.



        Starting from a triangulation.



        In every triangulation, there is one triangle containing the edge $1to 2$. Call the third vertex $k$. The corresponding parenthesizing is going to start with $$(1 ldots k-2)(k-1 ldots n)$$
        Now "on the right" of the edge $2to k$, you have a triangulated $(k-1)$-gon. Rename its vertices from $1$ to $k-1$ starting from the lowest and iterate.



        Similarly "on the left" of the edge $1to k$ there is a triangulated $(n-k+3)$-gon. Rename its vertices from $k-1$ to $n+1$ starting from the lowest and iterate.



        Starting from a parenthesized product.



        • Consider the most external product, say $$(1 ldots k-2)(k-1 ldots n)$$

        • Draw a triangle $1 2 k$

        • Iterate after renumbering the vertices accordingly as before.





        share|cite|improve this answer















        Here is a possible way, which makes use of the intrinsically similar recursive property of both numbers.



        I'm assuming the vertices of the $(n+1)$-gon are numbered counterclockwise from $1$ to $n+1$.



        Starting from a triangulation.



        In every triangulation, there is one triangle containing the edge $1to 2$. Call the third vertex $k$. The corresponding parenthesizing is going to start with $$(1 ldots k-2)(k-1 ldots n)$$
        Now "on the right" of the edge $2to k$, you have a triangulated $(k-1)$-gon. Rename its vertices from $1$ to $k-1$ starting from the lowest and iterate.



        Similarly "on the left" of the edge $1to k$ there is a triangulated $(n-k+3)$-gon. Rename its vertices from $k-1$ to $n+1$ starting from the lowest and iterate.



        Starting from a parenthesized product.



        • Consider the most external product, say $$(1 ldots k-2)(k-1 ldots n)$$

        • Draw a triangle $1 2 k$

        • Iterate after renumbering the vertices accordingly as before.






        share|cite|improve this answer















        share|cite|improve this answer



        share|cite|improve this answer








        edited Jul 30 at 21:44


























        answered Jul 30 at 16:35









        Arnaud Mortier

        18.1k21958




        18.1k21958






















             

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            Color the edges and diagonals of a regular polygon

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            Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne...
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            Relationship between determinant of matrix and determinant of adjoint?

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            Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/…...
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