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## Lights-Out on Graphs

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**Lights-Out on Graphs**Nadav Azaria**What is “Lights-Out” ?**• “Lights-Out” is a hand-held electronic game by Tiger electronics. It is played on a 5‰5keypad of lightable buttons. • Other versions of the game exist.**On Start:**• Some random buttons are lit. Object: • To turn all the lights out on the keypad. The difficulty is that each time you press a lit or an unlit button, it not only changes that button, but also all adjacent buttons!**From ToysRus to BGU**The game gave inspiration to several researches. Some of the work done, will be presented today.**Lights-Out on Arbitrary Graphs**Let G=(V,E) be a given graph. Suppose that at each vertex there is a light bulb and a switch. Toggling the switch at a vertex, we flip the light at this vertex and all its neighbors - those that were off are turned on and vice versa. A configuration of the system is a point of {0,1}V, where a 0 coordinate indicates that the light at the corresponding vertex is off, while a 1 means that it is on.**Did you notice that…**While solving the game: • There is no point pressing the same button more than once. • The order in which you press the buttons has no effect on the final configuration. Thus: A solution may be identified with a subset of V.**Question**• Given two configurations, decide whether it is possible to pass from one to the other by some sequence of switch toggles.**Answer**• Let M(G) be the neighborhood matrix of G. • If C is some configuration and we press some vertex v, the resulting configuration is C+M(G)v, where M(G)v is the row of M(G) corresponding to v.**Press**(0,1,0,0) (0,1,1,1) (0,0,1,1) (0,0,1,1) (1,1,0,0) (1,1,1,1) Press**Meaning….**We can pass from C1 to C2 if and only if there exists an xÎ{0,1}V such that C1 + M(G)x= C2 Or, equivalently: M(G)x= C2 - C1**Conclusions**We can now always assume starting with the all-off configuration and only ask which configurations can be reached. AND All configurations can be reached M(G) is non- singular over Z2**The End ?!**We are interested in: We are interested in: Which graphs have the property that one can pass from any configuration to any other? Which graphs have the property that one can pass from any configuration to any other?**Oh Yes, and find algorithms for evaluating light-deficiency**for specific graph types. Naturally they need to perform better then O(n2.376).**Definitions**A 0-combination is a non-zero vector in Ker(M(G)). For example, in the graph , (1,1) is a 0- combination. A graph is light-transitive if each configuration can be reached. v1 v2**Thelight-deficiencyd(G) of G is the dimension of**the kernel of M(G). Thus, there exist 2|V|- d(G) reachable configurations.**Universal Configurations**• A universal configuration is a non-trivial configuration which is reachable for each graph. • TheoremThe all-on configuration is the only universal configuration.**Proof**By considering the complete graph, we get the “only” part of the theorem.**We use induction on the number of vertices for the other**direction. n=1 Now Let V={v1,v2,…,vn}. By the induction hypothesis we get: For each vertex vithere exist a set SviÎV\{vi} such that after pressing all switches in Svi, all vertices in V\{vi} are on .**If n is even:**For each vertex v, press all vertices in Sv. Now, every vertex changes its state n-1 times meaning it is on. If n is odd: Let v be a vertex with an even degree. Let C be the set of all neighbors of v and v itself. For any u in V\C, press all vertices in Su.**Now, every vertex except the ones in C is on. Now by**pressing on v we arrive to the all-on configuration.**Lights Out on Circles**For nº0(mod3) we have 0-combinations. Example (press the following vertices): • 1,2,4,5,...,n-2,n-1 • 2,3,5,6,...,n-1,n Meaning light-deficiency is at least 2.**Remark - Note that if we press all the vertices**we arrive at the all-on configuration. In case nº1,2(mod3), we press vertices: 2,5,8,… Now all lights are on except one. Using the remark above we get a configuration where only one light bulb is on - meaning we can arrive at any configuration.**m‰n Grids**Definition p0(g)=1, p1(g)=g, pn(g)=gpn-1(g)+pn-2(g) Theorem.The Lights Out game has a unique solution iff pm(g) and pn(g+1) are relatively prime. SO**Here is how..**קצת סימונים... Let: • Am+ denote the m‰m tridiagonal matrix. • Am be Am++I. • C be an m‰n matrix representing a configuration. Then C has a unique solution iff the equation AmX+XAn+=C has a unique solution X in M(m,n,Z2).**It is known that:**The equation AX+XB=C has a unique solution iff the characteristic polynomials of A and B are relatively prime. And for similar reasons to the ones we saw in Paths det(Am-gI)=pm(g) thus det(Am-g+I)=pm(g+1).**Invariant Graphs (soon)**A rooted union of two rooted graphs G2 G1 Is X root of G1 and G2**Invariant Graphs**An invariant graphI satisfy d(GÇ I)=d(G), for any rooted graph G. Ç - rooted union. r r**Q. When does a light-transitive rooted graph is invariant?**A. A light-transitive rooted graph is invariant •The configuration which all lights if off except for the root, must be lit using the root itself.**Example:**r r r r r