Muhammad's Light Lattice

The resulting figure is a cube of side \­(2\­). Thus, only if two turned on light bulbs share an edge, they will explode. Painting the vertex of the cube with black and white, so that any two vertices of one edge have different color: 1. If a configuration includes one black and one white bulb on, there is no other bulb on and this two are opposite vertices of the cube. There are \­(4\­) ways to do this. 2. If all the turned on bulbs are the same color, there is no way that the bulbs explode, so there are \­({4 \choose 1}+{4 \choose 2}+{4 \choose 3}+{4 \choose 4}=15\­) configuration for each color. That gives a total of \­(30\­) configuration. 3. If all light bulbs are off, there is only \­(1\­) configuration.

So the total is \­(4+30+1=35\­)

It is best to imagine the bulbs placed on a cube of side length 2.In which the corners are the lattice points. We divide the problem into cases

First case:If all the lights are turned off there will obviously be no explosion.There is only one distinct way of doing this.

Second Case:If only one light is turned on,there will not be any explosion no matter which light we turn on. So there are eight possible ways of doing this.

Third Case: What if we turn on two lights?How many possible ways are there of placing these two lights so that they are more than 2 units away.Well it should be clear that as long as these two bulbs are not on the same edge of the cube they should not explode. For each bulb there are 4 bulbs that do not share its edge. Thus we multiply 8 by 4 giving us 32. But there are arrangements we counted twice,so we divide 32 by 2. Giving us 16 possible arrangements.

Fourth Case.Here we find the number of ways of arranging three coordinates such that any of the two points are not on the same edge.Building from each of the values from the third case we find 48 possible arrangements.But since we counted each permutation we have to divide 48 by $3!$ to count each distinct arrangement. This gives us 8 possible arrangements.

Fifth Case:The number of ways four bulbs can be arranged without exploding.This means finding the number of ways four bulbs can be turned on without no pair sharing an edge. We can see that for every three lattice points that satisfy the property there is only one other lattice point on the cube that satisfies the property.From the previous case we have 8*1 = 8 possible arrangements.But again there are repeats that are counted. This time they are four thus we have 8/4 = 2.We have 2 ways of turning on 4 bulbs without any explosion.

Sixth Case: We can see that from any arrangement from the previous case.If we add one or more bulbs two lighted bulbs will share an edge.From this we can conclude that there can not be more than four bulbs turned on simultaneously without an explosion.

To find our answer,we just add all the possible arrangements from the five cases 1+8+16+8+2 which gives 35.

Claim: If $S$ is a valid configuration, then $|S| \leq 4$ .

Proof: For a $x \in S$ , there are three neighbors in the lattice that are not allowed to belong to $S$ . Of the remaining $4$ points in the lattice, all of them cannot belong to $S$ since some two are neighbours. Thus at most $3$ more points other than $x$ can belong to $S$ . Thus $|S| \leq 4$ . $\blacksquare$

Lemma: If $x, -x \in S$ , then $|S|=2$

Proof: The three neighbours of $x$ and $-x$ are disjoint and thus any element other than $x$ and $-x$ is a neighbour to one of them. $\blacksquare$

Case 1: $|S| = 4$

For every lattice point $x$ , we make all possible sets, that contains $x$ , of four lattice points in which two co-ordinates differ for every pair. Since $|S| = 4$ , by the lemma, both $x$ and $-x$ cannot belong to $S$ . However, all the lattice points at a pairwise distance 4 from each other can belong to $S$ . There are only two such configs: $P = \{(1,-1,-1),(1,1,1),(-1,1,-1),(1,-1,-1)\}$ and $N = \{(-1,1,1),(-1,-1,-1),(1,-1,1),(-1,1,1)\}$ . Therefore there are $2$ configurations for $|S| = 4$ .

Case 2: $|S| = 3$

Again by the lemma, both $x$ and $-x$ cannot belong to $S$ . Further we can verify that $S$ cannot contain elements from both $P$ and $N$ . Thus there are $\binom{4}{3}$ ways of choosing a valid config from each P and N. Therefore there are $2\binom{4}{3}=8$ configurations for $|S| = 3$ .

Case 3: $|S| = 2$

A) If $x, -x \in S$ , there are four possible configs for every choice of totally opposite (antipodal?) pairs.

B) Suppose that both $x$ and $-x$ do not belong to $S$ . We can again verify that $S$ cannot contain elements from both $P$ and $N$ . Thus there are $\binom{4}{2}$ ways of choosing a valid config from each P and N. Therefore there are $2\binom{4}{2}=12$ configurations for this case.

Therefore there are totally $12 + 4 = 16$ configurations for $|S| = 2$ .

Case 4: $|S| = 1$

Simple. Every singleton from P and N is a valid config. Therefore there are totally $4 + 4 = 8$ configurations for $|S| = 1$ .

Case 5: $|S| = 0$

This can only mean $S$ is empty. Therefore there is only $1$ configuration for $|S| = 0$ .

Adding up all the above, we get that there are $2 + 8 + 16 + 8 + 1 = 35$ configurations in which the lights do not blow up.

The 8 positions of the light bulbs correspond a unit cube, 2 bulbs on on a same side of the cube = explosion!

No light on there is exactly 1 way

Only 1 light on the total ways are 8 for 8 positions

2 lights on there are $8 \choose 2$ minus 12 sides, i.e 12 cases of explosion, so 16 ways

3 lights, first one $8 \choose 1$ , if second light at the position in the other diagonal then no position else not neighbouring the first 2. So the second is chosen from 8 minus 1 (of the first) minus 3 (neighbouring the first) minus 1 (diagonal) = 3, eliminating 2 other neighbouring positions. So the third one has 4-2=2 choices left. Total ways = 8 3 2, each counted 3! times. So, 3 lights on, 8 3 2/3!=8 ways

4 lights on from 8 positions, obviously 2 ways

1+8+16+8+2=35

• If all light bulbs were turned off

• 8 C 0 = 1

• If a light bulb was turned on

• 8 C 1 = 8

• If 2 light bulbs were turned on

• 2*(4 C 2) = 12

• If 3 light bulbs were turned on

• 2*(4 C 3) = 8

• If 4 light bulbs were turned on

• 2*(4 C 4) = 2

• Then, to find the other possible configurations, imagine the position of the light bulbs. They were pretty much look like they were points of cube. So we can consider the diagonal position of light bulbs that the distance is greater than two units, and there are only 4 ways.

• So the total possible configurations of on/off light bulbs exist if no explosions occur were: 1 + 8 + 12 + 8 + 2 + 4 = 35

I see, of course, that many have answered this question. Yet I post a new solution, because it takes fewer reasoning steps and considers fewer separate cases than the other posted solutions.

Paint the bulbs red and blue, based on whether $xyz$ is positive or negative. Any two lightbulbs of the same color can be on simultaneously without causing an explosion.

Leaving all red bulbs off, we can turn on the blue lights in any desired configuration: $2^4 = 16$ possibilities. Likewise, with all blue bulbs off, we can turn on any combination of red lights. However, to prevent counting double the configuration with all lights off, subtract one. $31\ \text{possibilities with only red or only blue.}$

Now consider possible configurations with both red and blue bulbs on. Each red bulb that is on requires the three neighboring blue bulbs to be off, leaving only the blue bulb immediately across. Clearly, the only valid configurations with both red and blue are those with two "on" bulbs on a body diagonal. $4\ \text{possibilities with one red and one blue.}$

There are no other possibilities, so the answer is $31 + 4 = \boxed{35}$ .

It is clear that the points in the question form a cube with side length 2. All vertices of the cube which don't lie on the same edge have a length > 2. Hence, it implies that we need to find the number of configurations where the two vertices of the same edge aren't on.

Let us denote the points (-1,-1,-1) and (1,1,1) as A and H respectively.

Let the vertices which share an edge with A be B, C, D. Let the vertices which share an edge with H be E, F, G.

Note that B,C,D don't share an edge with each other and so is with E,F,G

We can denote (E,F,G) based on (B,C,D) as follows:

E doesn't share an edge with B, but does share an edge with C,D F doesn't share an edge with C, but does share an edge with B,D G doesn't share an edge with D, but does share an edge with B,C

Now the following cases arise:

Case I: Both A and H are off

This case is trivial -> All of B,C,D,E,F,G should not be on. Hence 1 combination.

Case II: A is on but H is off

B,C,D will not be on. Now each one of E,F,G can be on or non-on independently (because all vertices they share and edge with are off). Hence the number of cases that arise here are 2x2x2 = 8.

Case III: A is off but H is on

Similar argument as Case II ensues, just need to replace B,C,D with E,F,G.

Case IV: Both A and H are off

Let us now give complete freedom to be B,C,D in terms of them being off or on (as they don't interact with each other) and find the number of combinations of E,F,G in each such case.

IV.a All of B,C,D are on: All of E,F,G should be off. 1 Case

IV.b Exactly two of B,C,D are on: All of E,F,G should be off. 3 Cases

IV.c Exactly one of B,C,D is on: Two of E,F,G should be off, and the third one can be on or off. 3x2 = 6 cases

IV.d None of B,C,D are on: Each of E,F,G can be off or on: 2x2x2 = 8 cases

Hence, total combinations for Case IV is 1+3+6+8 = 18

Hence the answer is 1+8+8+18 = 35.

I think we need to divide it in many case such as 5 points so it will be exploded so we start at 4,the 3 points,and 2 points and one point

lattice point mean lights are on unit cube

now assume when no light on the total ways c(8,0)=1

when 1 light on the total ways c(8,1)=8

when 2 light on the total ways(on same like coordinates explosion will not explode)= 2c(8,2)=12

when 3 light on the total ways(on same like coordinates )=2C(4,3)=8

when all 4 light on on same like coordinates =2C(4,4)=2

now also 4 diagonals of cube(distance more than 2 )=4 ways 1+8+12+8+4+2=35