Source: TSTST-2021-P7
Category: Counting

Problem:
Let $M$ be a finite set of lattice points and $n$ be a positive integer. A *mine-avoiding path* is a path of lattice points with length $n$, beginning at $(0,0)$ and ending at a point on the line $x+y=n$, that does not contain any point in $M$. Prove that if there exists a mine-avoiding path, then there exist at least $2^{\,n-|M|}$ mine-avoiding paths.

Solution:
We prove the statement by induction on $n$. We use $n=0$ as a base case, where there is exactly $1$ path of length $0$, and the statement follows since $1\ge 2^{-|M|}$. 

For the inductive step, let $n>0$. Since a path of length $n$ beginning at $(0,0)$ and ending on the line $x+y=n$ must consist entirely of steps of the form $(1,0)$ and $(0,1)$, any mine-avoiding path must first pass through either $(1,0)$ or $(0,1)$. By assumption, there exists at least one mine-avoiding path. We consider two cases:

**Case 1:** There exist mine-avoiding paths passing through $(1,0)$ and mine-avoiding paths passing through $(0,1)$.
By the inductive hypothesis, there are at least $2^{n-1-|M|}$ mine-avoiding paths of length $n-1$ starting from $(1,0)$, and at least $2^{n-1-|M|}$ mine-avoiding paths of length $n-1$ starting from $(0,1)$. Then the total number of mine-avoiding paths from $(0,0)$ is at least
$$ 2^{n-1-|M|} + 2^{n-1-|M|} = 2^{n-|M|}. $$

**Case 2:** Only one of $(1,0)$ and $(0,1)$ is on a mine-avoiding path.
Without loss of generality, suppose no mine-avoiding path passes through $(0,1)$. Then some element of $M$ must be of the form $(0,k)$ for $1\le k\le n$ in order to block the path that moves entirely along the $y$-axis. This element can be ignored for any mine-avoiding path passing through $(1,0)$, since such a path can never return to the $y$-axis. Thus, for paths starting from $(1,0)$, the number of relevant mines is at most $|M|-1$. By the inductive hypothesis, there are at least
$$ 2^{n-1-(|M|-1)} = 2^{n-|M|} $$
mine-avoiding paths starting from $(1,0)$. 

This completes the induction step, which solves the problem.


Grading guidelines:
(Partial) 1. Observed that any length-$n$ path from $(0,0)$ to the line $x+y=n$ uses only steps $(1,0)$ and $(0,1)$, hence every mine-avoiding path must pass through exactly one of $(1,0)$ or $(0,1)$ as its first step. 2. Set up a correct induction on $n$ (including a valid base case) and reduced the counting of mine-avoiding paths of length $n$ to counting mine-avoiding paths of length $n-1$ starting from $(1,0)$ and/or from $(0,1)$. 3. Handled correctly the ‘both first steps possible’ case: showed that if mine-avoiding paths exist through both $(1,0)$ and $(0,1)$ then the counts add and give at least $2^{n-|M|}$. (Almost) 1. Presented the full induction with the correct case split, but in Case 2 did not fully justify the claim that if no mine-avoiding path goes through $(0,1)$ then $M$ must contain some point $(0,k)$ (or the symmetric statement), or did not justify why this mine can be discarded for paths starting with $(1,0)$. 2. Carried out the argument with the correct mine-removal idea in Case 2 (reducing the number of relevant mines by $1$), but made a localized exponent/index mistake (e.g. $2^{n-1-|M|}$ vs $2^{n-(|M|-1)}$) while the structure is otherwise complete. 3. Solution is almost complete, but made minor mistakes which are not negligible.