Finding distinct paths for an NxN chessboard. Is there a pattern?

Question

I decided to come up with a puzzle for my friends at lunch today, and it seems to be much more complicated then I would have ever imagined. Here's the puzzle:

A rook is located at the top left corner of a $4\times4$ chessboard, as shown (image 1).

At any given point, the rook only has two legal moves:

Move down 1 square
Move right 1 square

The rook cannot move diagonally. The rook cannot move up or left.

Here's the puzzle: how many distinct paths can the rook take to get to the bottom right corner (Point A)?

I quickly realized that counting all the paths would work, but it would be a pain to keep track of all of them. So I spent a lot of time confused, but after a while, I found a pretty elegant solution (image 2).

All of the paths (that the rook can take) must pass through exactly one of the following points: $B, C, D$, or $E$ (see image). And so tracking all of the paths gets much easier – you just have to add up the valid paths that pass through each letter.

Only 1 path can pass through B – since the rook has to hug the wall to get to Point B. And by symmetry, only 1 path can pass through E. Point C gets a bit trickier. From the rook's starting position, the rook has 3 ways to get to C – and from C, the rook has $3$ ways to get to $A$. So we multiply $3\times 3$, so we have 9 paths through C. And by symmetry, we also have 9 paths through D.

So in total, for a 4x4 board, we have 20 valid paths from the top left corner to A.

Now, there's also the case of the $2\times2$ and $3\times3$ chessboards, with the same setup (rook in top left, A in bottom right). Finding the valid paths for a 2x2 board is trivial – there are only 2 valid paths. And you can pretty easily count the valid paths for a 3x3 board: there are 6 valid paths. (I'm gonna disregard a $1\times1$ board, because there isn't really a "path".)

So for a square chessboard of dimensions $N\times N$, the rook has a certain number of valid paths:

N	Valid Paths
2	2
3	6
4	20

Now, you could just use the method I described earlier for finding the number of valid paths for a NxN board. But that's long and time-consuming, especially as N increases. So I had ChatGPT write me a bit of Python code (which it did):

def count_paths(N):
    dp = [[0 for _ in range(N)] for _ in range(N)]
    dp[0][0] = 1
    for i in range(N):
        dp[i][0] = 1
        dp[0][i] = 1
    for i in range(1, N):
        for j in range(1, N):
            dp[i][j] = dp[i-1][j] + dp[i][j-1]
    return dp[N-1][N-1]

Now I'm not very good at Python, so I don't really know what ChatGPT is doing – but when I asked it to return 'count_paths(4)', it gave me the correct answer of 20. And it gave me the correct answer for 'count_paths(3)' [which is 6] and 'count_paths(2)' [which is 2] – so I guess the code works.

Also, I didn't want to manually input 'count_paths(N)' and manually put those entries into a table. So I slapped the code into a Python compiler, and added this line to the code that ChatGPT wrote:

for i in range(2, 20):
    print(i, count_paths(i), round(count_paths(i)/count_paths(i - 1), 7))

So what's going on here? Well, for 2 ≤ N ≤ 19, I made the program output N, the number of valid paths for an NxN board, and the ratio of count_paths(N) / count_paths(N - 1) [rounded to 7 decimal places]. Here's the output:

N	Valid Paths	Ratio
2	2	2.0
3	6	3.0
4	20	3.3333333
5	70	3.5
6	252	3.6
7	924	3.6666667
8	3432	3.7142857
9	12870	3.75
10	48620	3.7777778
11	184756	3.8
12	705432	3.8181818
13	2704156	3.8333333
14	10400600	3.8461538
15	40116600	3.8571429
16	155117520	3.8666667
17	601080390	3.875
18	2333606220	3.8823529
19	9075135300	3.8888889

So as you can see, the valid paths for an NxN board gets really big, really fast. That's what happens whenever you have an exponential function like this.

But what's the pattern here?

The ratio doesn't follow any common patterns. Not the Fibonacci sequence, not a typical pattern, not an algebraic rule. Nothing. Maybe ChatGPT hallucinated and created incorrect code?

Notice that the ratio gradually increases – but it doesn't seem to increase beyond 4. And this holds true if you change 'for i in range(2, 20)' to 'for i in range(2, 100)'. The ratio seems to asymptote at 4, which is super strange.

What is going on here? Why does the ratio asymptote at 4? Why does this seemingly simple question so complicated?

Would appreciate any help. – waffles2124

You are looking for shortest paths. Every path has to go 3 times down and 3 times right. Therefore there are $6!/(3!)^2$ = 20paths. The same applies to $N\times N$. — Student, May 10 '23 at 19:45
Instead of ChatGPT you should have tried OEIS: https://oeis.org/ — JBL, May 10 '23 at 20:10
ChatGPT is implementing the fact that the number of ways to get to a square is equal to the number of ways to get to the square above it plus the number of ways to get to the square left of it. In fact, if you fill in the squares this way, you get Pascal's triangle, meaning the number of paths through a $m\times n$ board is $\binom{n+m-2}{n-1}$. — eyeballfrog, May 10 '23 at 20:15
JBL – looked up the first seven numbers on OEIS and found a sequence on "central binomial coefficients", which matches up with David Suh's answer. — waffles2124, May 10 '23 at 20:24

score 1 · Accepted Answer · answered May 10 '23 at 19:50

Since we can only move down or right, we denote a single "path" in terms of the moves e.g. $DDDRRR$ to move down 3 times then right 3 times. Since we cannot "undo" any movements, any path in a $N\times N$ grid is just a permutation of $N-1$ down movements and $N-1$ right movements, thus the number of paths is just the distinct number of permutations of $DDD\dots DRRR\dots R$. It is easy to verify that this is in fact

$$ \frac{[2(N-1)]!}{[(N-1)!]^2}. $$ We can empirically confirm this by using the $N=19$ case you provide:

$$ \frac{(2\cdot 18)!}{(18!)^2} = 9075135300. $$

Never thought about factorials. Great explanation, thank you! — waffles2124, May 10 '23 at 20:22

Finding distinct paths for an NxN chessboard. Is there a pattern?

1 Answers1