Closed-form for finding the number of paths on an nxm board
$begingroup$
Let a board $B$ be of size $n times m $ squares where $n, m in mathbb{Z}$ and $n, m ge 1$. Starting from the upper-left square, $B_{1,1}$, find the number of paths to the bottom-right square, $B_{n,m}$, by going right or down at any given square.
For example, let $B$ be $2times 2$, then the total number of paths is two: $(B_{1,1}, B_{1,2}, B_{2,2}), (B_{1,1},B_{2,1},B_{2,2})$.
One can calculate the total number of paths for an $ntimes m$ board by representing the board as a matrix where the top row are 1s and the left-most column are 1s, then the number of paths to get to $B_{n,m} = B_{n-1,m} + B_{n,m-1}$. Going further, the total paths can be expressed as the closed-form $frac{(n-1+m-1)!}{(n-1)!(m-1)!}$.
When the problem is extended to allow movements of right, right-down, and down at any given square, the closed-form above breaks down. As an example for a board $B$ of size $2times 2$, the total number of paths is three now: $(B_{1,1}, B_{1,2}, B_{2,2}), (B_{1,1}, B_{2,2}), (B_{1,1},B_{2,1},B_{2,2})$. One could still calculate the total paths doing the matrix method above, but I'm interested in whether a closed-form exist?
combinatorics combinations
asked Jan 11 at 7:26
Miket25Miket25
1106
$endgroup$
add a comment |
$begingroup$
Let a board $B$ be of size $n times m $ squares where $n, m in mathbb{Z}$ and $n, m ge 1$. Starting from the upper-left square, $B_{1,1}$, find the number of paths to the bottom-right square, $B_{n,m}$, by going right or down at any given square.
For example, let $B$ be $2times 2$, then the total number of paths is two: $(B_{1,1}, B_{1,2}, B_{2,2}), (B_{1,1},B_{2,1},B_{2,2})$.
One can calculate the total number of paths for an $ntimes m$ board by representing the board as a matrix where the top row are 1s and the left-most column are 1s, then the number of paths to get to $B_{n,m} = B_{n-1,m} + B_{n,m-1}$. Going further, the total paths can be expressed as the closed-form $frac{(n-1+m-1)!}{(n-1)!(m-1)!}$.
When the problem is extended to allow movements of right, right-down, and down at any given square, the closed-form above breaks down. As an example for a board $B$ of size $2times 2$, the total number of paths is three now: $(B_{1,1}, B_{1,2}, B_{2,2}), (B_{1,1}, B_{2,2}), (B_{1,1},B_{2,1},B_{2,2})$. One could still calculate the total paths doing the matrix method above, but I'm interested in whether a closed-form exist?
combinatorics combinations
asked Jan 11 at 7:26
Miket25Miket25
1106
$endgroup$
1
$begingroup$
See en.wikipedia.org/wiki/Delannoy_number . Using keyword "Delannoy" brings interesting tracks on this site.
$endgroup$
– Jean Marie
Jan 11 at 10:05
add a comment |
$begingroup$
Let a board $B$ be of size $n times m $ squares where $n, m in mathbb{Z}$ and $n, m ge 1$. Starting from the upper-left square, $B_{1,1}$, find the number of paths to the bottom-right square, $B_{n,m}$, by going right or down at any given square.
For example, let $B$ be $2times 2$, then the total number of paths is two: $(B_{1,1}, B_{1,2}, B_{2,2}), (B_{1,1},B_{2,1},B_{2,2})$.
One can calculate the total number of paths for an $ntimes m$ board by representing the board as a matrix where the top row are 1s and the left-most column are 1s, then the number of paths to get to $B_{n,m} = B_{n-1,m} + B_{n,m-1}$. Going further, the total paths can be expressed as the closed-form $frac{(n-1+m-1)!}{(n-1)!(m-1)!}$.
When the problem is extended to allow movements of right, right-down, and down at any given square, the closed-form above breaks down. As an example for a board $B$ of size $2times 2$, the total number of paths is three now: $(B_{1,1}, B_{1,2}, B_{2,2}), (B_{1,1}, B_{2,2}), (B_{1,1},B_{2,1},B_{2,2})$. One could still calculate the total paths doing the matrix method above, but I'm interested in whether a closed-form exist?
combinatorics combinations
asked Jan 11 at 7:26
Miket25Miket25
1106
$endgroup$
Let a board $B$ be of size $n times m $ squares where $n, m in mathbb{Z}$ and $n, m ge 1$. Starting from the upper-left square, $B_{1,1}$, find the number of paths to the bottom-right square, $B_{n,m}$, by going right or down at any given square.
For example, let $B$ be $2times 2$, then the total number of paths is two: $(B_{1,1}, B_{1,2}, B_{2,2}), (B_{1,1},B_{2,1},B_{2,2})$.
One can calculate the total number of paths for an $ntimes m$ board by representing the board as a matrix where the top row are 1s and the left-most column are 1s, then the number of paths to get to $B_{n,m} = B_{n-1,m} + B_{n,m-1}$. Going further, the total paths can be expressed as the closed-form $frac{(n-1+m-1)!}{(n-1)!(m-1)!}$.
When the problem is extended to allow movements of right, right-down, and down at any given square, the closed-form above breaks down. As an example for a board $B$ of size $2times 2$, the total number of paths is three now: $(B_{1,1}, B_{1,2}, B_{2,2}), (B_{1,1}, B_{2,2}), (B_{1,1},B_{2,1},B_{2,2})$. One could still calculate the total paths doing the matrix method above, but I'm interested in whether a closed-form exist?
combinatorics combinations
combinatorics combinations
asked Jan 11 at 7:26
Miket25Miket25
1106
asked Jan 11 at 7:26
Miket25Miket25
1106
edited Jan 11 at 7:35
Miket25
asked Jan 11 at 7:26
Miket25Miket25
1106
asked Jan 11 at 7:26
Miket25Miket25
1106
asked Jan 11 at 7:26
Miket25Miket25
1106
1106
1
$begingroup$
See en.wikipedia.org/wiki/Delannoy_number . Using keyword "Delannoy" brings interesting tracks on this site.
$endgroup$
– Jean Marie
Jan 11 at 10:05
add a comment |
1
$begingroup$
See en.wikipedia.org/wiki/Delannoy_number . Using keyword "Delannoy" brings interesting tracks on this site.
$endgroup$
– Jean Marie
Jan 11 at 10:05
1
1
$begingroup$
See en.wikipedia.org/wiki/Delannoy_number . Using keyword "Delannoy" brings interesting tracks on this site.
$endgroup$
– Jean Marie
Jan 11 at 10:05
$begingroup$
See en.wikipedia.org/wiki/Delannoy_number . Using keyword "Delannoy" brings interesting tracks on this site.
$endgroup$
– Jean Marie
Jan 11 at 10:05
add a comment |
1 Answer
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$begingroup$
Let $t$ be the number of diagonal moves. Then there are $n-1-t$ horizontal moves and $m-1-t$ vertical moves, for a total of $binom{n+m-2-t}{t,n-1-t,m-1-t}$ paths (this is a multinomial coefficient). Therefore the total number of paths is
$$
sum_{t=0}^{min(n,m)-1} binom{n+m-2-t}{t,n-1-t,m-1-t}.
$$
I'm not sure this can be simplified any further.
answered Jan 11 at 8:58
Yuval FilmusYuval Filmus
48.7k472145
$endgroup$
add a comment |
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$begingroup$
Let $t$ be the number of diagonal moves. Then there are $n-1-t$ horizontal moves and $m-1-t$ vertical moves, for a total of $binom{n+m-2-t}{t,n-1-t,m-1-t}$ paths (this is a multinomial coefficient). Therefore the total number of paths is
$$
sum_{t=0}^{min(n,m)-1} binom{n+m-2-t}{t,n-1-t,m-1-t}.
$$
I'm not sure this can be simplified any further.
answered Jan 11 at 8:58
Yuval FilmusYuval Filmus
48.7k472145
$endgroup$
add a comment |
$begingroup$
Let $t$ be the number of diagonal moves. Then there are $n-1-t$ horizontal moves and $m-1-t$ vertical moves, for a total of $binom{n+m-2-t}{t,n-1-t,m-1-t}$ paths (this is a multinomial coefficient). Therefore the total number of paths is
$$
sum_{t=0}^{min(n,m)-1} binom{n+m-2-t}{t,n-1-t,m-1-t}.
$$
I'm not sure this can be simplified any further.
answered Jan 11 at 8:58
Yuval FilmusYuval Filmus
48.7k472145
$endgroup$
add a comment |
$begingroup$
Let $t$ be the number of diagonal moves. Then there are $n-1-t$ horizontal moves and $m-1-t$ vertical moves, for a total of $binom{n+m-2-t}{t,n-1-t,m-1-t}$ paths (this is a multinomial coefficient). Therefore the total number of paths is
$$
sum_{t=0}^{min(n,m)-1} binom{n+m-2-t}{t,n-1-t,m-1-t}.
$$
I'm not sure this can be simplified any further.
answered Jan 11 at 8:58
Yuval FilmusYuval Filmus
48.7k472145
$endgroup$
Let $t$ be the number of diagonal moves. Then there are $n-1-t$ horizontal moves and $m-1-t$ vertical moves, for a total of $binom{n+m-2-t}{t,n-1-t,m-1-t}$ paths (this is a multinomial coefficient). Therefore the total number of paths is
$$
sum_{t=0}^{min(n,m)-1} binom{n+m-2-t}{t,n-1-t,m-1-t}.
$$
I'm not sure this can be simplified any further.
answered Jan 11 at 8:58
Yuval FilmusYuval Filmus
48.7k472145
answered Jan 11 at 8:58
Yuval FilmusYuval Filmus
48.7k472145
answered Jan 11 at 8:58
Yuval FilmusYuval Filmus
48.7k472145
answered Jan 11 at 8:58
Yuval FilmusYuval Filmus
48.7k472145
48.7k472145
add a comment |
add a comment |
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$begingroup$
See en.wikipedia.org/wiki/Delannoy_number . Using keyword "Delannoy" brings interesting tracks on this site.
$endgroup$
– Jean Marie
Jan 11 at 10:05