Thomson's lamp

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Thomson's lamp is a puzzle that is a variation on Zeno's paradoxes. It was devised by philosopher James F. Thomson, who also coined the term supertask.

Time	Lamp state transition
0:00.000	turn on
1:00.000	turn off
1:30.000	turn on
1:45.000	turn off
1:52.500	turn on
...	...
2:00.000	?

Consider a lamp with a toggle switch. Flicking the switch once turns the lamp on. Another flick will turn the lamp off. Now suppose a being able to perform the following task: starting a timer, he turns the lamp on. At the end of one minute, he turns it off. At the end of another half minute, he turns it on again. At the end of another quarter of a minute, he turns it off. At the next eighth of a minute, he turns it on again, and he continues thus, flicking the switch each time after waiting exactly one-half the time he waited before flicking it previously. The sum of all these progressively smaller times is exactly two minutes.

The following questions are then considered:

Is the lamp on or off after exactly two minutes?
Is the lamp switch on or off after exactly two minutes?
Would it make any difference if the lamp had started out being on, instead of off?

1 Contrast with Zeno's Paradoxes
2 Discussion
3 Mathematical proof of absence of a solution
4 See also
5 Reference

[edit] Contrast with Zeno's Paradoxes

Two notable features of contrast between Thomson's Lamp and Zeno's Paradoxes is that in the case of the lamp, the focus is on two discrete positions and there is a pause between them. Several proposed solutions to Zeno's Paradoxes fail if there is a pause before each movement in the series.

[edit] Discussion

The status of the lamp and the switch is known for all times strictly less than two minutes. However the question does not state how the sequence finishes, and so the status of the switch at exactly two minutes is indeterminate. Though acceptance of this indeterminacy is resolution enough for some, problems do continue to present themselves under the intuitive assumption that one should be able to determine the status of the lamp and the switch at any time given full knowledge of all previous statuses and actions taken.

One response is that one must consider how much time is spent moving the switch. Questions of lamp physics aside, one can simplify the problem to flipping a single bit of information to either a 0 or 1 state. If the flip takes any constant positive amount of time, then an infinite number of flips would take forever. So the only way this paradox will reach the 2 minute mark, under the assumption of constant flip time, is if the flip is not a delaying factor — essentially, if the flip takes zero amount of time. Yet if one can change the state of a bit instantly, then what does the question of a bit's state at a certain time mean? One could turn it off and on again without any time passing. One could even turn it off and on an infinite number of times. This response, however, does not deal with the case where successive flips take less and less time, so that the entire supertask can be performed in the given two minutes.

One possible solution to this problem, at least in the physical world, is provided by special relativity, i.e. the existence of a speed limit. That is, no one and nothing would be able to flick the switch infinitely fast, as would be required at the end of the sequence. There is a limit (the speed of light) to how quickly we can flip the switch.

[edit] Mathematical proof of absence of a solution

If the flipping of the switch may be taken as being instantaneous, then the question is mathematically equivalent to determining the value of Grandi's series as n tends to infinity:

$\sum_{i=0}^n{(-1)^i}$

For even values of n, the series sums to 1; for odd values, it sums to 0. In other words, as n takes the values of each of the non-negative integers 0, 1, 2, 3, ... in turn, the series generates the sequence {0, 1, 0, 1, 0, 1, ...}, representing the changing state of the lamp.

The sequence does not converge as n tends to infinity, and neither does the series. This demonstrates that the original problem does not have a solution.

Another way of illustrating this problem is to let the series look like this:

$S = 1 - 1 + 1 - 1 + 1 - 1 + \cdots$

To find if the light is on or off after an infinite number of switchings all we have to do is find the infinite sum of the series. However the series can be rearranged as:

$S = 1 - (1 - 1 + 1 - 1 + 1 - 1 + \cdots)$

Nothing odd about this, but notice that the unending series in the brackets is exactly the same as the original series S. This means S = 1 - S which implies S = ½. The lamp is neither off or on, but in a state that is like an average of the two.