Number of pathsin a grid with restrictions
$begingroup$
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
asked Nov 3 '15 at 7:58
Pls2Pls2
6713
$endgroup$
add a comment |
$begingroup$
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
asked Nov 3 '15 at 7:58
Pls2Pls2
6713
$endgroup$
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
add a comment |
$begingroup$
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
asked Nov 3 '15 at 7:58
Pls2Pls2
6713
$endgroup$
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
asked Nov 3 '15 at 7:58
Pls2Pls2
6713
asked Nov 3 '15 at 7:58
Pls2Pls2
6713
asked Nov 3 '15 at 7:58
Pls2Pls2
6713
asked Nov 3 '15 at 7:58
Pls2Pls2
6713
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
edited Dec 18 '18 at 4:20
Rohit Pandey
1,6331023
answered Nov 3 '15 at 16:21
Marcus MMarcus M
8,84911047
$endgroup$
add a comment |
$begingroup$
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.
answered Nov 3 '15 at 19:33
Michael EngenMichael Engen
393
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add a comment |
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$begingroup$
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.
edited Dec 18 '18 at 4:20
Rohit Pandey
1,6331023
answered Nov 3 '15 at 16:21
Marcus MMarcus M
8,84911047
$endgroup$
add a comment |
$begingroup$
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.
edited Dec 18 '18 at 4:20
Rohit Pandey
1,6331023
answered Nov 3 '15 at 16:21
Marcus MMarcus M
8,84911047
$endgroup$
add a comment |
$begingroup$
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.
edited Dec 18 '18 at 4:20
Rohit Pandey
1,6331023
answered Nov 3 '15 at 16:21
Marcus MMarcus M
8,84911047
$endgroup$
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.
edited Dec 18 '18 at 4:20
Rohit Pandey
1,6331023
answered Nov 3 '15 at 16:21
Marcus MMarcus M
8,84911047
edited Dec 18 '18 at 4:20
Rohit Pandey
1,6331023
edited Dec 18 '18 at 4:20
Rohit Pandey
1,6331023
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
answered Nov 3 '15 at 16:21
Marcus MMarcus M
8,84911047
answered Nov 3 '15 at 16:21
Marcus MMarcus M
8,84911047
8,84911047
add a comment |
add a comment |
$begingroup$
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.
answered Nov 3 '15 at 19:33
Michael EngenMichael Engen
393
$endgroup$
add a comment |
$begingroup$
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.
answered Nov 3 '15 at 19:33
Michael EngenMichael Engen
393
$endgroup$
add a comment |
$begingroup$
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.
answered Nov 3 '15 at 19:33
Michael EngenMichael Engen
393
$endgroup$
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.
answered Nov 3 '15 at 19:33
Michael EngenMichael Engen
393
answered Nov 3 '15 at 19:33
Michael EngenMichael Engen
393
answered Nov 3 '15 at 19:33
Michael EngenMichael Engen
393
answered Nov 3 '15 at 19:33
Michael EngenMichael Engen
393
393
add a comment |
add a comment |
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$begingroup$
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