All the Way to the Top - A problem left for readers

We can code the state of each input as $0$ for "off" and $1$ for "on". A particular combination is then represented in binary by a number in the range $\left[ 0,2^n-1 \right]$ , where there are $n$ input boxes ( $5$ in the case of this problem).

Since there are $2^n$ states, and we start in a given state, the best we can possibly do is $2^n-1$ clicks; but is this possible?

Let's start with a simpler case: $n=2$ . The four states are $00,01,10,11$ ; but this ordering doesn't work (at one point we change two bits). But reordering to $00 \to 01 \to 11 \to 10$ does work.

If we think about the two bits as coordinates, the four points form the vertices of a square. "Clicking" moves us along an edge of this square.

So we need to find a path around a square that visits every vertex exactly once by travelling along edges.

This is easy for $n=2$ ; the path we found above has this shape: $\sqsupset$

Now let's try $n=3$ . Here the points we have to visit form a cube; again, clicking corresponds to moving along an edge. This is a bit harder to picture (but worth trying), but we can solve this with the $n=2$ solution. Split the cube into two parallel squares; the path we'll take will visit the four vertices of the first square, then jump to the second square, and visit those four vertices. The path we get is $000 \to 001 \to 011 \to 010 \to 110 \to 100 \to 101 \to 111$

Note that, as required, we only click one box each time. Also, there are various points we can make a choice about the path; but we only have to find one solution.

We can continue this idea for larger $n$ ; a $4$ -d hypercube can be thought of as two parallel cubes; a $5$ -d hypercube is two parallel $4$ -d hypercubes.

An explicit example for $n=5$ using this approach is below: $\begin{aligned} &00000 \to 00001 \to 00011 \to 00010 \to 00110 \to 00100 \to 00101 \to 00111 \\ \to &01111 \to 01101 \to 01100 \to 01110 \to 01010 \to 01011 \to 01001 \to 01000 \\ \to &10000 \to 10001 \to 10011 \to 10010 \to 10110 \to 10100 \to 10101 \to 10111 \\ \to &11111 \to 11101 \to 11000 \to 11110 \to 11010 \to 11011 \to 11001 \to 11000 \end{aligned}$

(each row above corresponds to visiting the vertices of a cubic cell of the $5$ -d hypercube)

Since Alice need to check all the 32 possibilities, which are all UNIQUE, Alice would have to either turn an input on or off. Since the each click makes a new combinations of inputs, Alice need 2^5=32 clicks to check every single possibility. But Alice has to subtract one off of 32 because the problem automatically checks one of the possible inputs, which is when all the inputs are off. Therefore, Alice needs 32-1 = 31 clicks to check all possibilities.