Three Switches, One Visit, and the State Light Cannot Carry

Three switches are mounted outside a closed room, and each one controls one of three bulbs inside. You may set the switches however you like and take as long as you want over it. Then you open the door exactly once, look, and must name which switch runs which bulb. The switches cannot be reached from inside.

Almost everyone reaches the same dead end, and then treats it as the answer rather than as information. The dead end is real: with light as the only readable state, one visit is provably not enough. The way out is to notice that a bulb has a second physical state you can read, and to count how many states you actually need.

How much has to be separated

A pairing is a one-to-one assignment of switches to bulbs, so there are

of them, and one visit produces one observation. For the puzzle to be solvable, different pairings must produce different observations. That is the whole requirement, and it turns the problem into arithmetic about how many observations a visit can produce.

Light alone cannot do it

Suppose the only thing you can see is which bulbs are lit. Your strategy is a choice of which switches to leave on, say a set $S$ of size $k$. What you see is the set of bulbs driven by $S$, and nothing else.

Here is the fatal part. That sight tells you which bulbs belong to $S$, and says nothing about the order inside $S$ or inside its complement. The pairings consistent with what you saw number

The minimum is 2, achieved at $k = 1$ and $k = 2$. Leave one switch on and you identify that switch and are left flipping a mental coin between the other two. Leave two on and you identify the third, with the same two-way ambiguity among the pair. Leave none or all three on and you learn nothing at all, with all 6 pairings still standing.

A crude count says the same thing faster. There are at most $\max_k \binom{3}{k} = 3$ different sights available from a single switch setting, and 3 is smaller than 6, so by the pigeonhole principle two pairings must look identical. This was checked exhaustively as well, over all 8 switch settings against all 6 pairings: no light-only strategy separates them.

The third state

A bulb that has been on is warm, and it stays warm for a while after the switch goes off. So a bulb can be in one of three conditions rather than two: lit, dark and warm, or dark and cold. That is a physical state of the bulb, readable at the moment you walk in, and the puzzle never said light was the only thing you could observe.

The protocol writes itself once you have three states and three bulbs to distribute them over:

Assigning three distinct states to three bulbs gives $3 \cdot 2 \cdot 1 = 3! = 6$ possible readings, exactly matching the number of pairings. The map from pairing to reading was verified to be one-to-one over all six pairings, which is the difference between a protocol that works and a protocol that usually works.

Nothing spare, and nothing wasted

Two states offer $2^3 = 8$conceivable labellings of the bulbs, which is more than 6, and yet the puzzle is impossible with two states. The gap between 8 and what is achievable is the point: only 3 of those 8 labellings can be produced by a single switch setting, because "on" and "off" are not assigned to bulbs freely, they are assigned to switches and then pushed through the unknown pairing.

Three states are enough because they can be handed out injectively. The general statement is the same counting argument with the numbers left open.

With $s$ states you may give each switch its own state only if $s \ge n$. If two switches share a state, the two bulbs they drive look the same and cannot be told apart. So the warmth trick handles three switches and fails at four, which was also brute-forced: every three-state protocol on four switches collides somewhere.

More visits, more states

Two natural relaxations both make the problem easy, and it is worth seeing why they are the same relaxation.

Allow two visits with light only. Each bulb now shows you a two-bit signature, on or off at each visit, so there are $2^2 = 4$ signatures available and $4 \ge 3$ is enough. Leave switch 1 on for the first visit, then switch 2 for the second, and read the pattern. In general $v$ visits with a two-state readout identify $n$ switches when $2^v \ge n$, which is the binary encoding of the switch index.

Alternatively keep one visit and add temperature levels. If you can distinguish hot, warm, cool and cold by giving switches different amounts of on-time, you have four states and four switches become possible. This is where the puzzle stops being mathematics: whether a real bulb gives you four reliably distinguishable temperatures is a question about the bulb, not about the counting. An LED may not give you two.

What the argument assumes

The impossibility result in equation (2) leans on the switches being unreachable from inside the room. If you could flip a switch while watching the bulbs, one visit would trivially be enough and there would be no puzzle. That assumption is what makes a visit a single observation rather than an interactive session, and it is the first thing to state clearly if you are the one posing the problem.

It also assumes the pairing is a genuine bijection: three switches, three bulbs, one each. Allow a switch to control two bulbs, or a bulb to answer to two switches, and the space of configurations is no longer of size $3!$, and the whole count in equation (1) has to be redone.

Sources and further reading

• The counting bound in the crude form — Pigeonhole principle
• What a protocol has to be for the puzzle to be solved — Bijection
• The quantity in equation (1) — Factorial

Both halves were checked by exhaustive enumeration rather than by argument alone. All 8 switch subsets were run against all 6 pairings, giving a worst case of 2 surviving candidates for the six useful subsets and 6 for the two useless ones, with no light-only strategy separating everything. The heat protocol produces 6 distinct readings, each using every state exactly once. The four-switch boundary was brute-forced separately.