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I'm writing a video game, and I'm trying to find an efficient way of calculating this. The goal is to count the number of paths of length $n$ that a character can take, where the character can move left, right, or up. The caveat is that the character cannot go back to the same position in a path.

I've come up with a mediocre brute-force method the count the paths, but any ideas on an efficient way to calculate this?

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  • $\begingroup$ Are some cells of the grid forbidden ? $\endgroup$ – Vince Jun 11 at 17:47
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This is a question in combinatorics, and can be calculated in a closed formula.

The key settings are:

  • "Down" is not allowed
  • Visiting previously visited square is not allowed

From the two requirements, we can draw the following conclusions:

  • Up is always a valid move (since we never went down, going up is essentially revealing a new square)
  • Left is not valid after Right, but is valid after Up (revealing new row) or after Left (which is essentially unvisited)
  • Right is not valid after Left, but is valid after Up (revealing new row) or after Right (which is essentially unvisited)

Denote $U$, $L$ and $R$ for Up, Left and Right respectively. We can now represent a path in a string like so: $P = (U,U,U,U,L)$

The question is now: How many valid paths strings of length $n$ are there?


Let $T(n)$ be the number of valid strings of length $n$

  • If the first move is $U$ then the remaining strings are $T(n-1)$

  • If the first move is $L$ then the remaining strings are those starting with $L$ or $U$

  • If the first move is $R$ then the remaining strings are those starting with $R$ or $U$

Looking at strings where the first move is either $L$ or $R$: the remaining strings are: strings starting with $U$ (twice), starting with $L$, or starting with $R$. (simple summation of all the options in 2nd and 3rd bullet).

Note that all the strings starting with $U$ + all the strings starting with $L$ + all the strings starting with $R$ is exactly $T(n-1)$ since the first move is already set.

We are left with one more instance of "all the strings starting with $U$" = $T(n-2)$ (again, first move is set)

Which results the following recursive formula: $$T(n) = 2T(n-1)+T(n-2)$$ where:

$$T(1) = 3, \space T(2) = 7 $$


Calculations omitted, the closed formula for the relation above is:

$$T(n)= \frac{(1+\sqrt{2})^{n+1}}{2} + \frac{(1-\sqrt{2})^{n+1}}{2} $$

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Obervations :

  • As you only allow for up and left/right moves, when you leave a row (by going up), you can't come back to it.
  • As you cannot revisit the same node in a given path, when you go right on a row, you can't go left afterwards.

I think this can lead to a direct formula for paths of length $n$, but you can first try to compute the paths with $k$ moves up, and then sum for $k$ from 0 to $n$ (and you get a recurrence relationship here)

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