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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.**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.**Invariant Graphs (soon)**A rooted union of two rooted graphs:**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 Actually, any light-transitive cycle with an arbitrary vertex as the root is invariant.**S1**S2 S3 r r r**Meaning…**G’ G The number of unreachable configuration in G stays the same .**Introducing S5…**r A 0-combination**Now, what do I know about G’?**G’ The same 0-combination Thus, G’ is not light-transitive!**r**r Proposition:Let G = (V, E, r) be any rooted graph. Then G The number of 0-combination now is exactly half**The Tree’s Theorem**• Any tree T with at least three vertices a rooted union of some rooted tree T’ and one of the four rooted trees S1,S2,S3,S5. Proof: S2 S3 S1 S5**The Algorithm**The Goal: • Given a tree, evaluate its light-deficiency. The Means: • Assign a status to each vertex. The status of a vertex determined by the status of its sons. A status is pair consist of an integer and one of the following: • Null • Leaf • Father**Status?**Add 1 to the final result (Null, 0) (Father, 1) (Leaf, 0)**Ha,ha My status is (Null, 9)**What is my status? The leaves … The fathers… The nulls… 1 2 0 3 0 0 0 1 Remember : 3 9**See it in action…**(Null, 2) (Father, 1) (Null, 0) (Father, 0) (Father, 0) (Leaf, 0) (Leaf, 0) (Null,1) (Leaf, 0) (Leaf, 0) (Father, 0) (Father, 0) (Leaf, 0) (Leaf, 0)