6 lamp at the circumcircle

Question

In the circumcircle of a circle the exist $6$ lamp which form a reguler Hexagonal.The lamp only can be press switch. Note if the lamp is off and we press the switch thne the lamp is on and vice versa

There will be procedure: 1) press $3$ consecutive switch lamp. 2) press $3$ switch lamp where the lamp form an equilateral triangle. 3) press $2$ switch lamp where the lamp form an diameter of the circle.

Is the statement true: Whatever the first given configuration we can have all the lamp switch off.

My comment i think it is not true for $1,2,3,4,5$ lamp switch off at the first configuration, where the first configuration mean the condition before any switch was pressed

Check Me If I Wrong ( CMIIW )

Sounds like Galois permutation groups can be used in order to prove this. — barak manos, Jan 16 '15 at 08:51
In fact, you can check that the only configurations that are reachable from the all-off position are: those reachable in one move; those reachable by two distinct diameter moves; and the all-on position. — Greg Martin, Jan 16 '15 at 08:55

score 1 · Answer 1 · answered Jan 16 '15 at 14:38

There are $6$ choices of $3$ consecutive lamps to toggle at the first step, $2$ choices of an equilateral triangle at the second step, and 3 choices of a diameter at the third step, so there are at most $36$ different configurations which can be handled successfully. But there are $2^6 = 64$ possible initial configurations, so it's certainly not true that all initial configurations can be handled successfully.

As to the specific configuration you mention: consider combining the second and third steps. Exactly one of the two lamps in the diameter is also in the triangle, so the net effect of those two steps is to toggle a second sequence of three consecutive lamps. The overlap with the first sequence of three consecutive lamps can be zero (net effect: toggle every lamp), one (net effect: toggle four lamps), two (net effect: toggle two lamps), or three (net effect: nothing changes). So any starting position with an odd number of lamps which need toggling serves as a counterexample.

score 1 · Answer 2 · answered Jan 16 '15 at 14:58

You are correct. For any configuration, consider the number of 0-1-pairs, by which I mean diagonal pairs of lamps whose state differs (i.e. one is on, the other is off). You can show by induction that starting from the configuration you give, this number is always 1 or 2:

The first two moves change the state of one lamp in each pair, so a pair is a 0-1-pair afterwards iff it wasn't one before. So the number of 0-1-pairs changes from $i$ to $3-i$.

The third move leaves the set of 0-1-pairs unchanged.

6 lamp at the circumcircle

2 Answers2