Number of pathsin a grid with restrictions












2












$begingroup$


Please help, I am stuck with this example. Prove that the Catalan number $C_n$ equals the number of lattice paths from $(0,0)$
to $(2n, 0)$ using only upsteps $(1, 1)$ and downsteps $(1, -1)$ that never go above the
horizontal axis (so there are as many up steps as there are downsteps). (These
are sometimes called Dyck paths.) Thanks.






combinatorics catalan-numbers







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    If I were you I would try out the first several values of $n$ and see how they form the Catalan numbers, then try to understand and prove the relation $C_n=C_0C_{n-1}+C_1C_{n-2}+cdots+C_{n-1}C_0$.
    $endgroup$
    – Elliot G
    Nov 3 '15 at 8:18










  • $begingroup$
    Have you looked at Wikipedia?
    $endgroup$
    – Ross Millikan
    Dec 18 '18 at 5:29
















2












$begingroup$


Please help, I am stuck with this example. Prove that the Catalan number $C_n$ equals the number of lattice paths from $(0,0)$
to $(2n, 0)$ using only upsteps $(1, 1)$ and downsteps $(1, -1)$ that never go above the
horizontal axis (so there are as many up steps as there are downsteps). (These
are sometimes called Dyck paths.) Thanks.






combinatorics catalan-numbers







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    If I were you I would try out the first several values of $n$ and see how they form the Catalan numbers, then try to understand and prove the relation $C_n=C_0C_{n-1}+C_1C_{n-2}+cdots+C_{n-1}C_0$.
    $endgroup$
    – Elliot G
    Nov 3 '15 at 8:18










  • $begingroup$
    Have you looked at Wikipedia?
    $endgroup$
    – Ross Millikan
    Dec 18 '18 at 5:29














2












2








2


1



$begingroup$


Please help, I am stuck with this example. Prove that the Catalan number $C_n$ equals the number of lattice paths from $(0,0)$
to $(2n, 0)$ using only upsteps $(1, 1)$ and downsteps $(1, -1)$ that never go above the
horizontal axis (so there are as many up steps as there are downsteps). (These
are sometimes called Dyck paths.) Thanks.






combinatorics catalan-numbers







share|cite|improve this question









$endgroup$




Please help, I am stuck with this example. Prove that the Catalan number $C_n$ equals the number of lattice paths from $(0,0)$
to $(2n, 0)$ using only upsteps $(1, 1)$ and downsteps $(1, -1)$ that never go above the
horizontal axis (so there are as many up steps as there are downsteps). (These
are sometimes called Dyck paths.) Thanks.






combinatorics catalan-numbers




combinatorics catalan-numbers






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 3 '15 at 7:58









Pls2Pls2

6713




6713








  • 2




    $begingroup$
    If I were you I would try out the first several values of $n$ and see how they form the Catalan numbers, then try to understand and prove the relation $C_n=C_0C_{n-1}+C_1C_{n-2}+cdots+C_{n-1}C_0$.
    $endgroup$
    – Elliot G
    Nov 3 '15 at 8:18










  • $begingroup$
    Have you looked at Wikipedia?
    $endgroup$
    – Ross Millikan
    Dec 18 '18 at 5:29














  • 2




    $begingroup$
    If I were you I would try out the first several values of $n$ and see how they form the Catalan numbers, then try to understand and prove the relation $C_n=C_0C_{n-1}+C_1C_{n-2}+cdots+C_{n-1}C_0$.
    $endgroup$
    – Elliot G
    Nov 3 '15 at 8:18










  • $begingroup$
    Have you looked at Wikipedia?
    $endgroup$
    – Ross Millikan
    Dec 18 '18 at 5:29








2




2




$begingroup$
If I were you I would try out the first several values of $n$ and see how they form the Catalan numbers, then try to understand and prove the relation $C_n=C_0C_{n-1}+C_1C_{n-2}+cdots+C_{n-1}C_0$.
$endgroup$
– Elliot G
Nov 3 '15 at 8:18




$begingroup$
If I were you I would try out the first several values of $n$ and see how they form the Catalan numbers, then try to understand and prove the relation $C_n=C_0C_{n-1}+C_1C_{n-2}+cdots+C_{n-1}C_0$.
$endgroup$
– Elliot G
Nov 3 '15 at 8:18












$begingroup$
Have you looked at Wikipedia?
$endgroup$
– Ross Millikan
Dec 18 '18 at 5:29




$begingroup$
Have you looked at Wikipedia?
$endgroup$
– Ross Millikan
Dec 18 '18 at 5:29










2 Answers
2






active

oldest

votes


















2












$begingroup$

Note that these paths are the same as the lattice paths from $(0,0)$ to $(n,n)$ that stay below the diagonal line ${(x,x) : x in mathbb{R} }$. Let such paths be called "good paths" and let "bad paths" be lattice paths from $(0,0)$ to $(n,n)$ that cross the diagonal. Then



# good paths = # paths - # bad paths



The total number of lattice paths from $(0,0)$ to $(n,n)$ is $dbinom{2n}{n}$ since we have to take $2n$ steps, and we have to choose when to take the $n$ steps to the right.



To count the total number of bad paths, we do the following: every bad path crosses the main diagonal, implying that it touches the diagonal just above it. Specifically, every bad path must touch the line $L = {(x,x+1) : x in mathbb{R}}$. Given a bad path, break it up into two portions: the portion before the first time the path touches $L$, and the portion after. If we reflect the first portion over the line $L$, then we have a lattice path from $(-1,1)$ to $(n,n)$. This gives a bijection between bad paths and lattice paths from $(-1,1)$ to $(n,n)$. Since there are $dbinom{2n}{n+1}$ such lattice paths, there must be $dbinom{2n}{n+1}$ bad paths.



Putting this all together yields
$$binom{2n}{n} - binom{2n}{n-1} = C_n$$
total good paths.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    The typical method for counting Dyck paths (that I know) goes as follows:



    For every Dyck path $D$ from $(0,0)$ to $(2n,0)$, $D$ must begin with an upstep and eventually return to the x-axis with a downstep. Say $(2m,0)$ is the the first time $D$ returns to the x-axis. Then the sub-path of $D$ which goes from $(1,1)$ to $(2m-1,1)$ is a Dyck path (albeit shifted up) of length $m-1$, and the part of $D$ which goes from $(2m,0)$ to $(2n,0)$ is also a Dyck path (of length $n-m$).



    Every Dyck path admits a unique decomposition in this way. That is, each Dyck path of length $n$ yields an ordered pair of Dyck paths, the first of length $m-1$ and the second of length $n-m$. Here, $m$ may be any positive integer less than or equal to $n$.



    Too, every ordered pair of Dyck paths, the first of length $m-1$ and the second of length $n-m$ gives a unique Dyck path under the this construction.



    So $$C_n=sum_{m=1}^{n} C_{m-1}C_{n-m},$$ which gives the Catalan numbers.






    share|cite|improve this answer









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      2 Answers
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      2 Answers
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      2












      $begingroup$

      Note that these paths are the same as the lattice paths from $(0,0)$ to $(n,n)$ that stay below the diagonal line ${(x,x) : x in mathbb{R} }$. Let such paths be called "good paths" and let "bad paths" be lattice paths from $(0,0)$ to $(n,n)$ that cross the diagonal. Then



      # good paths = # paths - # bad paths



      The total number of lattice paths from $(0,0)$ to $(n,n)$ is $dbinom{2n}{n}$ since we have to take $2n$ steps, and we have to choose when to take the $n$ steps to the right.



      To count the total number of bad paths, we do the following: every bad path crosses the main diagonal, implying that it touches the diagonal just above it. Specifically, every bad path must touch the line $L = {(x,x+1) : x in mathbb{R}}$. Given a bad path, break it up into two portions: the portion before the first time the path touches $L$, and the portion after. If we reflect the first portion over the line $L$, then we have a lattice path from $(-1,1)$ to $(n,n)$. This gives a bijection between bad paths and lattice paths from $(-1,1)$ to $(n,n)$. Since there are $dbinom{2n}{n+1}$ such lattice paths, there must be $dbinom{2n}{n+1}$ bad paths.



      Putting this all together yields
      $$binom{2n}{n} - binom{2n}{n-1} = C_n$$
      total good paths.






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        Note that these paths are the same as the lattice paths from $(0,0)$ to $(n,n)$ that stay below the diagonal line ${(x,x) : x in mathbb{R} }$. Let such paths be called "good paths" and let "bad paths" be lattice paths from $(0,0)$ to $(n,n)$ that cross the diagonal. Then



        # good paths = # paths - # bad paths



        The total number of lattice paths from $(0,0)$ to $(n,n)$ is $dbinom{2n}{n}$ since we have to take $2n$ steps, and we have to choose when to take the $n$ steps to the right.



        To count the total number of bad paths, we do the following: every bad path crosses the main diagonal, implying that it touches the diagonal just above it. Specifically, every bad path must touch the line $L = {(x,x+1) : x in mathbb{R}}$. Given a bad path, break it up into two portions: the portion before the first time the path touches $L$, and the portion after. If we reflect the first portion over the line $L$, then we have a lattice path from $(-1,1)$ to $(n,n)$. This gives a bijection between bad paths and lattice paths from $(-1,1)$ to $(n,n)$. Since there are $dbinom{2n}{n+1}$ such lattice paths, there must be $dbinom{2n}{n+1}$ bad paths.



        Putting this all together yields
        $$binom{2n}{n} - binom{2n}{n-1} = C_n$$
        total good paths.






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          Note that these paths are the same as the lattice paths from $(0,0)$ to $(n,n)$ that stay below the diagonal line ${(x,x) : x in mathbb{R} }$. Let such paths be called "good paths" and let "bad paths" be lattice paths from $(0,0)$ to $(n,n)$ that cross the diagonal. Then



          # good paths = # paths - # bad paths



          The total number of lattice paths from $(0,0)$ to $(n,n)$ is $dbinom{2n}{n}$ since we have to take $2n$ steps, and we have to choose when to take the $n$ steps to the right.



          To count the total number of bad paths, we do the following: every bad path crosses the main diagonal, implying that it touches the diagonal just above it. Specifically, every bad path must touch the line $L = {(x,x+1) : x in mathbb{R}}$. Given a bad path, break it up into two portions: the portion before the first time the path touches $L$, and the portion after. If we reflect the first portion over the line $L$, then we have a lattice path from $(-1,1)$ to $(n,n)$. This gives a bijection between bad paths and lattice paths from $(-1,1)$ to $(n,n)$. Since there are $dbinom{2n}{n+1}$ such lattice paths, there must be $dbinom{2n}{n+1}$ bad paths.



          Putting this all together yields
          $$binom{2n}{n} - binom{2n}{n-1} = C_n$$
          total good paths.






          share|cite|improve this answer











          $endgroup$



          Note that these paths are the same as the lattice paths from $(0,0)$ to $(n,n)$ that stay below the diagonal line ${(x,x) : x in mathbb{R} }$. Let such paths be called "good paths" and let "bad paths" be lattice paths from $(0,0)$ to $(n,n)$ that cross the diagonal. Then



          # good paths = # paths - # bad paths



          The total number of lattice paths from $(0,0)$ to $(n,n)$ is $dbinom{2n}{n}$ since we have to take $2n$ steps, and we have to choose when to take the $n$ steps to the right.



          To count the total number of bad paths, we do the following: every bad path crosses the main diagonal, implying that it touches the diagonal just above it. Specifically, every bad path must touch the line $L = {(x,x+1) : x in mathbb{R}}$. Given a bad path, break it up into two portions: the portion before the first time the path touches $L$, and the portion after. If we reflect the first portion over the line $L$, then we have a lattice path from $(-1,1)$ to $(n,n)$. This gives a bijection between bad paths and lattice paths from $(-1,1)$ to $(n,n)$. Since there are $dbinom{2n}{n+1}$ such lattice paths, there must be $dbinom{2n}{n+1}$ bad paths.



          Putting this all together yields
          $$binom{2n}{n} - binom{2n}{n-1} = C_n$$
          total good paths.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 18 '18 at 4:20









          Rohit Pandey

          1,6331023




          1,6331023










          answered Nov 3 '15 at 16:21









          Marcus MMarcus M

          8,84911047




          8,84911047























              0












              $begingroup$

              The typical method for counting Dyck paths (that I know) goes as follows:



              For every Dyck path $D$ from $(0,0)$ to $(2n,0)$, $D$ must begin with an upstep and eventually return to the x-axis with a downstep. Say $(2m,0)$ is the the first time $D$ returns to the x-axis. Then the sub-path of $D$ which goes from $(1,1)$ to $(2m-1,1)$ is a Dyck path (albeit shifted up) of length $m-1$, and the part of $D$ which goes from $(2m,0)$ to $(2n,0)$ is also a Dyck path (of length $n-m$).



              Every Dyck path admits a unique decomposition in this way. That is, each Dyck path of length $n$ yields an ordered pair of Dyck paths, the first of length $m-1$ and the second of length $n-m$. Here, $m$ may be any positive integer less than or equal to $n$.



              Too, every ordered pair of Dyck paths, the first of length $m-1$ and the second of length $n-m$ gives a unique Dyck path under the this construction.



              So $$C_n=sum_{m=1}^{n} C_{m-1}C_{n-m},$$ which gives the Catalan numbers.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                The typical method for counting Dyck paths (that I know) goes as follows:



                For every Dyck path $D$ from $(0,0)$ to $(2n,0)$, $D$ must begin with an upstep and eventually return to the x-axis with a downstep. Say $(2m,0)$ is the the first time $D$ returns to the x-axis. Then the sub-path of $D$ which goes from $(1,1)$ to $(2m-1,1)$ is a Dyck path (albeit shifted up) of length $m-1$, and the part of $D$ which goes from $(2m,0)$ to $(2n,0)$ is also a Dyck path (of length $n-m$).



                Every Dyck path admits a unique decomposition in this way. That is, each Dyck path of length $n$ yields an ordered pair of Dyck paths, the first of length $m-1$ and the second of length $n-m$. Here, $m$ may be any positive integer less than or equal to $n$.



                Too, every ordered pair of Dyck paths, the first of length $m-1$ and the second of length $n-m$ gives a unique Dyck path under the this construction.



                So $$C_n=sum_{m=1}^{n} C_{m-1}C_{n-m},$$ which gives the Catalan numbers.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  The typical method for counting Dyck paths (that I know) goes as follows:



                  For every Dyck path $D$ from $(0,0)$ to $(2n,0)$, $D$ must begin with an upstep and eventually return to the x-axis with a downstep. Say $(2m,0)$ is the the first time $D$ returns to the x-axis. Then the sub-path of $D$ which goes from $(1,1)$ to $(2m-1,1)$ is a Dyck path (albeit shifted up) of length $m-1$, and the part of $D$ which goes from $(2m,0)$ to $(2n,0)$ is also a Dyck path (of length $n-m$).



                  Every Dyck path admits a unique decomposition in this way. That is, each Dyck path of length $n$ yields an ordered pair of Dyck paths, the first of length $m-1$ and the second of length $n-m$. Here, $m$ may be any positive integer less than or equal to $n$.



                  Too, every ordered pair of Dyck paths, the first of length $m-1$ and the second of length $n-m$ gives a unique Dyck path under the this construction.



                  So $$C_n=sum_{m=1}^{n} C_{m-1}C_{n-m},$$ which gives the Catalan numbers.






                  share|cite|improve this answer









                  $endgroup$



                  The typical method for counting Dyck paths (that I know) goes as follows:



                  For every Dyck path $D$ from $(0,0)$ to $(2n,0)$, $D$ must begin with an upstep and eventually return to the x-axis with a downstep. Say $(2m,0)$ is the the first time $D$ returns to the x-axis. Then the sub-path of $D$ which goes from $(1,1)$ to $(2m-1,1)$ is a Dyck path (albeit shifted up) of length $m-1$, and the part of $D$ which goes from $(2m,0)$ to $(2n,0)$ is also a Dyck path (of length $n-m$).



                  Every Dyck path admits a unique decomposition in this way. That is, each Dyck path of length $n$ yields an ordered pair of Dyck paths, the first of length $m-1$ and the second of length $n-m$. Here, $m$ may be any positive integer less than or equal to $n$.



                  Too, every ordered pair of Dyck paths, the first of length $m-1$ and the second of length $n-m$ gives a unique Dyck path under the this construction.



                  So $$C_n=sum_{m=1}^{n} C_{m-1}C_{n-m},$$ which gives the Catalan numbers.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 3 '15 at 19:33









                  Michael EngenMichael Engen

                  393




                  393






























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