1. We have n 2 2 lamps L L in a row, each of them being either on or off. Every
ID: 3210045 • Letter: 1
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1. We have n 2 2 lamps L L in a row, each of them being either on or off. Every second we simultaneously modify the state of each lamp as follows: _ if the lamp L, and its neighbours (only one neighbour for i-l or i-n, two neighbours for other i) are in the same state, then L is switched off - otherwise, Li is switched on Initially all the lamps are off except the leftmost one which is on. (a) Prove that there are infinitely many integers n for which all the lamps will eventually be off. (b) Prove that there are infinitely many integers n for which the lamps will never be all off.Explanation / Answer
Total n lamps are there and they are noted as L1......Ln and
We have L>=2 lamps in a row.
Each of them being on (or) off.
If they on means take binary state as 1 and otherwise if off means 1 state.
Initially all lamps are off except left most first one is on.
If state of neibhours is same as that ones state,then that particular lamp is off means it has 0 state.
Other wise it is 1 state that means it is oon.
Let us take lamps in row with left most one is on and remaining all are off.
10000000000.........---(1)
n>=2 lamps means we have to see from third one except left most one.
(a) Now let us assume all lamps are off.As all have same states with their neighbors.So,infinitely many n integers are there for all lamps eventually be off like 00,000,0000,00000,00000....etc.
(b).Here already given that Initially left most one should on So,all lamps never be off.
From thisit i proved that infinitely many n integers are there for which lamps will never be all off.Examples 10,100,1000,10000,1000000. ...etc.
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