Download Theorem.

The game L(d ,1) - labeling
problem of graphs
game
chromatic
number
Given a graph G and a set X of colors.
Consider t he two - person game played on
G:
Alice and Bob alternate turns. A move
consisting of selecting an uncolored vertex
v of G and assigning it a color from X
distince from the color assigned previously
to neighbors of v. Alice wins if all the vertices
are successful ly labeled, else Bob wins.
game chromatic number of G
least cardinalit y of a color set X for which
Alice has a winning strategy
 (G) : game chromatic number of G in
1
g
the game Alice plays first
 (G) : game chromatic number of G in
2
g
the game Bob plays first
Question.
 (C4 )  ?
1
g
Alice
1
Alice
?
2
Bob
 (C4 )  2
1
g
Since  (G)  (G)  1, we also have
i
g
 (C4 )  3
1
g
 (C4 )  3
1
g
Question.
 (C4 )  ?
2
g
Alice
1
1
Bob
Alice
2
2
Bob
 (C4 )  2
2
g
Since  (G)   (G), we also have
i
g
 (C4 )  2
2
g
 (C4 )  2
2
g
Theorem. (Faigle et al.)
The game chromatic number of the class of
forests is 4.
Theorem. (Kierstead and Trotter)
The game chromatic number of the class of
planar graph is between 8 and 33.
Theorem. (Zhu)
The game chromatic number of the class of
planar graph is between 8 and 17.
Theorem. (Zhu)
The game chromatic number of the class of
outerplana r graph is between 6 and 7.
L(p,q)-labeling
L( p, q) - labeling of G
a function f from V (G ) to nonnegativ e
integers satisfying
(1) | f (u)  f (v ) | p if d (u, v )  1
(2) | f (u)  f (v ) | q if d (u, v )  2
k - L( p, q) - labeling of G
an L( p, q) - labeling such that no label is
greater th an k
L( p, q) - labeling number of G
 p ,q (G ) : smallest number k , such that G
has a k - L( p, q) - labeling
(when q  1, we use  p (G ) to replace  p ,q (G ))
 (C4 )  ?
Question.
Since
0
23
4
?1
((C
4
f is a 3 - L
2,14 ) -labeling
of C4
 (C4 )  4
the four verti ces are labeled by 0,1,2,3
one vertex is labeled by 2
 (C4 )  4
Theorem. (Griggs and Yeh)
The L(2,1) - labeling problem is NP - complete
for general graphs.
Conjecture. (Griggs and Yeh)
For any graph G with (G)  2,  (G)   (G).
2
Theorem. (Gonçalves)
For any graph G,  (G )   (G )  (G )  2.
2
game
L(p,q)-labeling
Given a graph G and nonnegativ e integers
k , p, q, p  q  0. Consider t he two - person
game played on G :
Alice and Bob alternate turns. A move
consisting of selecting an unlabeled vertex
v of G and label it with a number a in
{0,1,2,..., k} satisfying the conditions that
(1) If u is labeled by b and d (u, v )  1, then
|b-a|  p.
(2) If u is labeled by b and d (u, v )  2, then
|b-a|  q.
Alice wins if all the vertices are successful ly
labeled, else Bob wins.
game L( p, q) - labeling number of G
smallest k , so that Alice has a winning strategy
~ p ,q
 1 (G ) :
Alice plays first
~ p ,q
 2 (G ) :
Bob plays first
~2
Question.
 1 (C4 )  ?
Alice
a  2
a
0
Bob
b
a
a+3
2
b
a  1 or 6
1 (C 4 ) 
6
5
4
Bob
Alice
~ 2
Alice can ~complete
this game using
2
1 (in
C 4{)0
the numbers
,1,62,...,6}
~ 2
Question.
Alice
2 (C4 )  ?
a  b  1
a
2
b
Bob
~ 2

5
4this game using
2 (C 4 ) 
Alice cancomplete
Note. the numbers in {0,1,2,...,5}
~ 2
~ 2
1 (G)  2 (G) dose not hold in general!
Lemma. For any graph G ,
~ d
i (G )  d (G ) (i  1,2)
Lemma. If G is a graph with maximum
degree , then
~ d
i (G )  (  2d  2)
for i  1,2 and d  2.
Theorem.
 ( K n )  2n  2
0
2
4
6
Example.
Alice
Bob plays first
Alice
30
~ 2
12
12 ( K 22)  23
Bob
Example.
Alice
Bob plays
playsfirst
first
Alice
Alice
Bob
32
54
10
~ 2
12 ( K 3 )  45
Bob
Example.
Alice plays first
Alice
2
0
Alice
Bob
5
7
Bob
~ 2
1 ( K 4 )  7
b  3
Alice
a  3
a
b
Alice
Bob
5
?
Bob
~ 2
1 ( K 4 )  6
~ 2
1 ( K 4 )  7
Bob plays first
Alice
5
Bob
1
~ 2
1 ( K 3 )  4
~ 2
~ 2
2 ( K 4 )  1 ( K 3 )  3  7
Alice
Bob plays first
a  0 or b  2
a Alice
2
b  4
Bob
b
c
~ 2
2 ( K 4 )  7
~ 2
2 ( K 4 )  7
Bob
Question.
~ 2
1 ( K n )  ?
~ 2
2 ( K n )  ?
Observation 1.
~ 2
~ 2
1 ( K n )  2 ( K n 2  K 1 )  4
Alice
2
Bob
b
0
Bob
1
5
0
1
2
3
4
5
6
7 …
Observation 2.
~ 2
~ 2
2 ( K n )  1 ( K n1 )  3
2
5
Bob
Alice
1
0
1
2
3
4
5
6
7 …
Theorem.
 7n  9 
1 ( K n )  

 3 
~ 2
 7n  7 
2 ( K n )  

3


~ 2
0
1
2
3
4
3 vertices, 7 numbers
5
6
 7n 3
Question.
How to prove this theorem?
(Use induction, we already know that
~ 2
~ 2
~ 2
~ 2
1 ( K n )  2 ( K n 2  K 1 )  4,
2 ( K n )  1 ( K n1 )  3)
Idea.
Alice’s strategy
If v is still unlabeled, choose a smallest
i , so that all the numbers in {i , i  1, i  2}
can be used to label v. Labels v with i  2.
If no such i exists, choose a number j
that can be used to label v. Labels v with
j.
Idea.
Bob’s strategy
If v is still unlabeled, choose a smallest
i , so that all the numbers in {i , i  1} can
be used to label v. Labels v with i  1.
If no such i exists, choose a number j
that can be used to label v. Labels v with
j.
~ 2
1 ( K14 )
Example.
2nd
4th
6th
3rd
0 1 2 3 4 5 6 7 8 9 10 11 12
13 14 15 16 17 18 19 20 21 22 …
7th
5th
~ 2
1st
1 ( K11  K1 )
2nd
4th
0 1 2
3 4 5
6
7
8
9
3rd
10 11 12 13 14 15 …
5th
1st
Theorem. For all n  1, r  0,
~ d
1 ( K n  rK 1 )
and
n  3
 (4d  1) 

[(
n

1
)
mod
3
]
d

3


~ d
2 ( K n  rK 1 )
 n  1
 (4d  1) 

2
d

[(
n

2
)
mod
3
]
d

3


Theorem. For all d , m, n, m  n  1,
d ( K m ,n )  d  m  n  2
11 ( K 7,5 )  ?
Example.
0
17
18
19
20
21
1
2
3
4
5
6
11 ( K 7,5 )  7  5  (11  1)  1  21
~ 11
Question.
1 ( K 7 , 5 )  ?
Y
0
17
Alice
Bob
X
can' t use the numbers in {7,...,16}  {18,...,27}
~ 11
guess :  1 ( K 7 ,5 )  11 ( K 7 , 5 )  11
~ 11
1 ( K 7 , 5 )  31
Bob’s strategy
a  16
15
a
proof.
Alice
27
4
Bob
Alice
Bob
a
15
16
10
21
a  17
14
~ 11
less
than
5
numbers
can
be
used
to
less than 
71numbers
( K 7 , 5 ) can
31be used to
label
these
vertices
label these vertices
Y
X
~ 11
proof.
Alice
1 ( K 7 ,5 )  32
Alice’s strategy
c
4
30
28
b  11  7, if 11  b  14,
c Alice

 b  11  7, if 18  b  21
16
29
Alice
Bob
Alice
Bob
17
11
12
15
b
Y
X
(11 
14, or
bb10
22
game18using
 b  the
21)
Alice can complete this
allallnumbers
the
the
thenumbers
numbers
numbers
{can
0,29
132
,be
2,...,
,}3used
,4
32
}Bob
}can
can
to labels
bebe a
in {0in,0in
1,{,1228
,...,
if
used
usedlabel
toXlabel
label
these
these
these
vertices
vertices
vertices
vertex
into
at
the
second
step
~ 11
1 ( K 7 ,5 )  32
Alice’s strategy
proof.
Alice
16
2
1
b
Bob Y
(0  b(
, or
29528
27
(43
bb 
))b  32)
Bob Alice
20
c
19
22
10
X
b  11  7, if 0  b  3,
Alice can completec this
 game using the
b,,29
,11

74
, }}
ifcan
29be
be
b  32.
11 inin

~
allallthe
the
numbers
numbers
{
28
{
0
1
2
,...,
,
3
,
32
can
the numbers 0,1 can be used to
numbers in1{0,(1K
,27,...,
32
}
if
Bob
labels
a
)

32
,5
used
used
to
to label
label
these
these
vertices
vertices
label
these
vertices
vertex in Y at the second step
Theorem. For all d , m, n, d  m  n  2,
~ d
1 ( K m ,n )  2d  m  n  2   m ,n
(where  m,n  [( m  n) mod 2])
Thanks!
Note.
It is not true that if Alice(resp. Bob) plays
first, and at some step, he can move twice,
then the smallest number needed to2 complete
~
the game is less than or equal to 1 (G ) (resp.
~ 2
~ 2
2 (G )). For example, 1 (2C4  K1 )  5,
but if at the first step, Alice need to move
twice, then the smallest number needed to
complete this game is 6({0,1,2,…,6}).