# Bulbs in circle puzzle

On a circle there are 2014 light bulbs, 2 are ON, and 2012 are OFF.

You can choose any bulb and change the neighbor’s state i.e. * ON to OFF* or

*.*

**OFF to ON**Doing so,

What is the maximum number of bulbs we can turn on?

**ANSWER– **All the 2014 can be turned on

**EXPLANATION-
**

Let’s assume initial state of circuit as {**0,0,0,1,1,0,0,0,0…..,0,0**} where 1 = Switched ON & 0 = Switched OFF

Choose second last bulb from bulb which has state =1, then 2 bulb will be lighted on either sides

next state becomes {**1,0,1,1,1,1,0,1,0,0,0…..0,0**}

So, at each step we are lighting 4 bulbs, since 2012 is multiple of 4, there will come a stage where all bulbs will be ON.

I think it depends on the number of lightbulbs and the order the two lighted bulbs are in.

4 cases

1. If the two lighted bulbs are next to each other and the total number of lightbulbs is an odd multiple of 2 (odd # * 2), then we can turn on all bulbs.

2. If the two lighted bulbs are next to each other and the total number of lightbulbs is an even multiple of 2 (even # * 2), then we can turn on all but 2, **unless the total number of bulbs is 2, which doesn’t make sense here as a scenario anyways.

3. If the two lighted bulbs have a one bulb gap between them, and the total number of lightbulbs is an odd multiple of 2 (odd # *2), then we can turn on all but 2.

4. If the two lighted bulbs have a one bulb gap between them, and the total number of lightbulbs is an even multiple of 2 (even # * 2), then we can turn on all bulbs.

The two cases relevant here are case #2 and case #4. Try this for the simplest cases: 4 light bulbs and 6 light bulbs, and you’ll quickly see that these conclusions should hold true for larger sets.

One. Note that it says “change the neighbor’s state” NOT “neighbors’ stateS”. It does not say that changing state from 0-1 has an impact on the next neighbor to iterate.

### Your Answer

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