You dig a hole in Minecraft straight down $n$ blocks, how many blocks must you dig to get out?

Question

I was thinking on this question and came to the solution $3(x-2)$ (where $x$ is the depth of the hole and $x\ge3$) which I am fairly confident in. Although, I wanted to prove it for any hole that is $n$ blocks deep, could anyone illustrate how I could formally do this using something like induction?

Answer 1

You are correct, and you can prove this by induction. Let's do it!

Let $a_n$ be this sequence, and let $A$ be the subset of elements in $\{ 3, 4, ... \}$ for which the statement $a_n = 3 (n - 2)$ holds.

Base Case

We first show that it holds for $n = 3$. In this case, you need to dig $2$ blocks in the column directly in front of the hole, and $1$ block in the next column. The total is $3$, which is equal to $3 (3 - 2) = 3 (1)$.

Inductive Step

Suppose the statement is true for $n$; we show it is true for $n + 1$. Since we have already shown it to be true for $n = 3$, we will assume that $n \ge 4$. In this case, one must again dig $2$ blocks in the column in front of the hole. But in order to continue moving upwards (assuming no other paths, of course), an additional block above those $2$ must be removed. Now, the player moves to the new column and her elevation increases by $1$. The situation at this point is the same as if the player had started in a hole with depth $n$ (you should convince yourself of why this is true). Therefore, the total number should be $a_{n + 1} = 3 + a_n = 3 + 3 (n - 2) = 3 n - 3 = 3 ((n + 1) - 2)$, as desired.

Therefore, we have shown that $A = \{ 3, 4, ... \}$; i.e., $a_n = 3 (n - 2)$ holds for all $n \ge 3$.

It might be worth extending the sequence to include $0, 1, 2$. For $n = 0$, $a_n = 0$ obviously. For $n = 1$, $a_n = 0$ too, since you mentioned that the player can simply jump out of $1$-high holes. For $n = 2$, $a_n = 1$. It turns out to be this sequence, for what it's worth.