Physically Incorrect

Here’s the solution to the spinning wheel puzzle.

Let’s use binary numbers to represent the wheel’s positions. For example, 1101 will represent the case in which buttons 1, 2 and 4 are pressed and button 3 is depressed.  Since the wheel spins randomly after each attempt, a key insight is to try and group the states into “groups”, the members of which can be obtained from each other by spinning the wheel. Let’s call these groups A, B, C, D and W (for Win), and use binary numbers to denote pressed/depressed states:

We’ll also denote binary numbers to denote button presses. For example, 1001 will mean pressing buttons 1 & 4 simultaneously, while 0111 will mean pressings buttons 2 3 & 4 simultaneously. So, for example, we’d might have:

So how do the various groups A, B, C, D respond to the different keystrokes? It seems we have a total of 16 key presses to investigate (0000, 0001, 0010, 0100, 1000, 0011, 0110, …, 1111). However, we can use the following insights to reduce the number of relevant options:

• The problem is symmetric - that both 0000 and 1111 are end states, and so, for example, starting at 0001 and pushing 1100 is just like starting at 1110 and pushing 0011. In short, investigating 1100 and 0011 is redundant. Picking just one is enough. This halves the number of  actions.
• The same rotational symmetry that renders the states 0011 and 0110 equivalent also renders pressing 0110 and 0011 equivalent.
• 0000 is uninteresting (does nothing).

To sum up, we only have 3 meaningful actions that need exploring: 0001, 0011 and 0101. How do the groups A, B, C, D fare under them? A bit of thought reveals the following observations:

• We can assume neither A, B, C or D become W. Although this is possible, this is precisely what we want - if this happens, we’ve won. We’ll have to always assume otherwise.
• Under 0001, states from A and C turn into states in B. States from B can become either states in A, C, or D.
• Under 0011, states from B turn into states from D. States from A and C turn into other states from A and C.
• Under 0101, states from D turn into winning states necessarily. States from B remain in B, states from A become states in C and states from C become states in A.

These observations are all summed up neatly in the following table:

Starting off we can be in either A, B, C, or D (or W, but then we’d just win automatically). How can we get to W for sure? Just follow the following algorithm:

• Denote the initial state by A/B/C/D.
• Press 0101. This will turn your state into C/B/A/W - I’ll omit the W and retain just the C/B/A.
• Press 0001: your state will become B/ACD/B, where by ACD I mean that if your state was previously B, it is now either A, C or D.
• Press 0101: your state will become D/AC/D.
• Press 0001: your state will become W/AC/W, or simply AC, omitting the possible winning states.
• Press 0001: your state will become B.
• Press 0011: your state will become D.
• Press 0101: the LED lights up.

In fact, you might win even earlier - this chain of events simply describes the ‘worst-case-scenario’, the greatest number of pushes you’ll need (7).