# Dada una matriz NxN, ¿cómo podría encontrar todos los caminos posibles a una ubicación (i, i)?

Dado un `NxN` matrix, how could I find all possible paths to a location `(i,i)`. Navigation would be only `downwards` y hacia el `right`. The starting point would be `(0,0)`.

P.S. Not a homework.

preguntado el 31 de julio de 12 a las 14:07

Where's the starting point? (0,0)? -

@Zenzen Yes, assuming the starting point as (0,0) -

## 6 Respuestas

The easiest (for me) would be to do it recursively:

1) the stop condition would be arriving at (i,i)

2) you would try two next moves at each recursion level:

• can you go down
• can you go right

Assuming that your current position is (x,y): You cannot go right if your current "y" plus 1 would be bigger than "i" in the (i,i). You cannot go down if your current "x" plus 1 would be bigger than "i" in the (i,i).

You'd just have to remember the path in some variable or pass it downwards and print it when the stop condition occurs.

So assuming you start at (0,0) you remember this and then check if you can go right or down, if yes for both then you recursively call this same method for (0,1) and (1,0).

Respondido 31 Jul 12, 14:07

I had the same in mind but was thinking about how to store the path and didn't think about the idea of passing the current path down the recursion. Thanx a lot. - Rahul

?

To a location (i, i), i.e. one on the diagonal, there are exactly comb(i, 2 * i) paths (number of ways to select i elements from a set of 2 * i elements), each consisting of i movements rightward and i movements downward. Enumerating them is trivial.

Respondido 31 Jul 12, 14:07

Although you can use DP to solve this, this is just simle math:

Tienes que hacer `2i` steps, from which `i` must be steps to the right. The total combinations are `C(2i,i)`. Ver: computing the binomial coefficients.

Another similar problem is where you cannot cross the diagonal, called monotonic path, for which the interesting Número catalán sequence is the solution.

Respondido 31 Jul 12, 16:07

Look at the problem from a 45 degree angle. You'll see that you're trying to calculate the number of paths to a vertex on a triangular array. The solution is given by Triángulo de Pascal.

`2i choose i`

Respondido 31 Jul 12, 18:07

Solution 1 - Using dynamic programming :

Add up the solutions bottom up till (i,j) indices you want:

ans(i,i) = ans(i-1,j) + ans(i,j-1)

and use the same for position (i,i).

Solution 2 - Using simple permutation and combination

ans = (2i)!/ i!i!

(Since you have to choose any of the i steps towards the right and ji steps downwards starting from (0,0) and total are 2i steps).

Respondido 22 Abr '15, 17:04

It depends when you start from (0,0) or at a random location in the matrix.

If it is (0,0) ; use Depth First to go until the end of one branch then the others. Use Breadth First to iterate through all neigbouring nodes at a step. Check only down and right nodes.

If it is at a random location in the matrix, you need to trace all 4 neighbouring nodes.

Respondido 10 Feb 16, 13:02

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