Lights-Out on Graphs

1
Lights-Out on Graphs

• Nadav Azaria

2
What is Lights-Out ?

• Lights-Out is a hand-held electronic game by
  Tiger electronics. It is played on a 55 keypad
  of lightable buttons.
• Other versions of the game exist.

3

• 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!

4
Lets Play!
5
From ToysRus to BGU
The game gave inspiration to several researches.
6
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
0,1V, where a 0 coordinate indicates that the
light at the corresponding vertex is off, while a
1 means that it is on.
7
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.

8
Back to the previews example
9
Question

• Given two configurations, decide whether it is
  possible to pass from one to the other by some
  sequence of switch toggles.

10
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
• CM(G)v,
• where M(G)v is the row of M(G) corresponding to v.

11

• (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
Press
12
Meaning.

• We can pass from C1 to C2 if and only if there
  exists an
• xÎ0,1V such that
• C1 M(G)x C2
• Or, equivalently
• M(G)x C2 - C1

13
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
14
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?
15
Oh Yes, and find algorithms for evaluating
light-deficiency for specific graph
types. Naturally they need to perform better then
O(n2.376).
16
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
17

• The light-deficiency d(G) of G is the dimension
  of
• the kernel of M(G).
• Thus, there exist 2V- d(G) reachable
  configurations.

18
Universal Configurations

• A universal configuration is a non-trivial
  configuration which is reachable for each graph.
• Theorem The all-on configuration is the only
  universal configuration.

19
Invariant Graphs (soon)
A rooted union of two rooted graphs
20
Invariant Graphs
An invariant graph I satisfy d(GÇ I)d(G), for
any rooted graph G. Ç - rooted union.
r
r
21
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.
22
Example
r
Actually, any light-transitive cycle with an
arbitrary vertex as the root is invariant.
23
S1
S2
S3
r
r
r
24
Meaning
G
G
The number of unreachable configuration in G
stays the same .
25
Introducing S5
r
A 0-combination
26
Now, what do I know about G?
G
The same 0-combination
Thus, G is not light-transitive!
27
Introducing S4
r
28

• Proposition Let G (V, E, r) be any rooted
  graph. Then

r
r
G
The number of 0-combination now is exactly half
29
The Trees 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
30
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

31
Status?
Add 1 to the final result
(Null, 0)
(Father, 1)
(Leaf, 0)
32
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
33
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)