Light Bulbs in a Circle

[Winkler] Light bulbs placed on a circle are numbered 1 through N, all initially on. At time t, you examine bulb number t, and if it's on, you change the state of bulb t+1 (mod N); i.e., you turn it off if it's on, and on if it's off. If bulb t is off you do nothing. Prove that if you continue around and around the circle in this manner, eventually all the bulbs will again be on.

The applet allows you to experiment with the problem. Orange circles impersonate the on state of the bulbs, the red ones the off state. The green arrow points to the bulb under examination. The blue arrow points to the bulb whose state is going to be affected.

The applet generalizes the problem in two ways. First of all, the look ahead may be an integer other than 1; i.e., the examination of bulb t may affect bulb t+L, L > 0, where L is the "look ahead" parameter, 1 originally. Secondly, instead of examining the bulbs in their natural order, you may skip a chosen number (S) of bulbs.

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at, download and install Java VM and enjoy the applet.

What if applet does not run?



  1. P. Winkler, Mathematical Puzzles: A Connoisseur's Collection, A K Peters, 2004, pp. 81-82 (Light bulbs in a circle)

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Copyright © 1996-2018 Alexander Bogomolny

In the original problem, N is arbitrary, L = 1, S = 0. Except when S is a multiple of N -- S = 0 (mod N), Winkler's solution works without change. The solution is based on two observations:

  1. The process is reversible: a state of affairs at time t is uniquely determined by that at time t+1.

  2. Since the number of possible states is finite, the process is bound to enter a loop by the Pigeonhole principle.

The generalization does not require special treatment, but makes it obvious that there is one special case, viz., the state wherein all the bulbs are off. Except when S = 0 (mod N), this state is unreachable:

  1. Starting with all the bulbs on, it is impossible to reach a dead end combination where all the bulbs are off.

Indeed, to turn off an apparently last bulb, one would have to be examining a bulb that is on. This logic does not apply when S = 0 (mod N), because then the bulb to be turned off is exactly the bulb being examined.

Now for the proof.

Because the number of states is finite, state repetition is bound to happen. Assume at time t1 a state is repeated for the first time. Assume also, that the repeated state has occurred at time t0 < t1. We want to show that t0 = 0. If it were not so, we could reverse one step and verify that the states at times t1-1 and t0-1 are also the same, which would contradict our assumption that a repetition occurred for the first time at time t1.

Related material

  • Six integers out of 10: Pigeonhole Principle
  • Pigeonhole Principle (Same sum)
  • Pigeonhole in a Matrix
  • Pigeonhole in Calendar
  • A nice puzzle modeled on the Petersen graph
  • Proizvolov's identity in a game format
  • Pigeonhole with Disjoint Intervals
  • All antichains
  • Euclid via Pigeonhole
  • Light Bulbs in a Circle (an Interactive Gizmo)
  • Seven integers under 127 and their Ratios

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    Copyright © 1996-2018 Alexander Bogomolny