python - Code to find number of route-combinations in a N by N grid -

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i have write code has count number of different combinations start upper left of n n grid, , end in lower right corner. can go down , right!

so (if give each of vertices number 1 - (n+1)^2), how many different combinations there 1 - (n+1)^2. can add 1 or add n+1.

an example: 2x2 grid:

1 -- 2 -- 3 |    |    | 4 -- 5 -- 6 |    |    | 7 -- 8 -- 9

and combinations be:

[1-2-3-6-9] [1-2-5-6-9] [1-2-5-8-9] [1-4-5-6-9] [1-4-5-8-9] [1-4-7-8-9]

now, how should write code? appreciated!
in advance

ah. see trying on 1 of eulerproject problems? point manage figure out , learn new things along way! tricking cheating. oh well

anyways, consider grid of m rows , n columns (we not need assume grid square). counting upper-left , starting @ zero, denote intersection/node in i-th row , j-th column n_(i,j).

thus, upper-left node n_(0,0), bottom-left n_(m,0) , bottom-right n_(m,n). clearly, number of paths n_(0,0) node along far right or far top of grid 1 (since may proceed down or right).

now, consider how many paths there n_(1,1). must first travel through n_(0,1) or n_(1,0). yields 2 paths n_(1,1). can continue process. in order determine total number of paths node n_(i,j), need sum total number of paths n_(i,j−1) , n_(i−1,j). process understood graphically in following diagram, each new integer represents number of paths node.

enter image description here

if code in python get:

def route_num(cube_size):     l = [1] * cube_size     in range(cube_size):         j in range(i):             l[j] = l[j]+l[j-1]         l[i] = 2 * l[i - 1]     return l[cube_size - 1]

print route_num(20) give 137846528820.

source: http://code.jasonbhill.com/


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