Distance Coloring
Alexa Sharp
Cornell University, Ithaca, NY 14853
[email protected]
Abstract. Given a graph G = (V, E), a (d, k)-coloring is a function from
the vertices V to colors {1, 2, . . . , k} such that any two vertices within
distance d of each other are assigned different colors. We determine the
complexity of the (d, k)-coloring problem for all d and k, and enumerate
some interesting properties of (d, k)-colorable graphs. Our main result is
the discovery of a dichotomy between polynomial and NP-hard instances:
for fixed d ≥ 2, the distance coloring problem is polynomial time for
k ≤ b 3d
c and NP-hard for k > b 3d
c.
2
2
Key words: graph coloring, power graphs, complexity threshold
1
Introduction
The classic k-coloring problem tries to assign a color from 1 to k to each vertex
in a graph such that no two adjacent vertices share the same color [1]. The kcoloring problem, along with many variations and generalizations, is well-studied
in both computer science and mathematics [2–6]. Its applications range from
frequency assignment [7] to circuit board testing [8], among others.
The distance (d, k)-coloring problem is a generalization of k-coloring that
tries to assign a color from 1 to k to each vertex such that no two vertices within
distance d of each other share the same color. Clearly, k-coloring is a special case
of (d, k)-coloring with d = 1. Conversely, (d, k)-coloring a graph G is equivalent
to k-coloring Gd , the dth power graph of G. (The graph Gd has the same vertex
set as G and an edge between two vertices if and only if they are within distance d
of each other in G.) In this way (d, k)-coloring is also a special case of k-coloring.
Although k-coloring is NP-complete for k ≥ 3 and polynomial-time for k ≤ 2
[1], the complexity of (d, k)-coloring is not so straightforward. It is NP-complete
for large k [9, 10]; other special cases have been studied in depth, such as cubic
graphs [11], planar graphs [12], and trees [13, 14]. However, there are many values
of d and k for which (d, k)-coloring is polynomial-time; the dichotomy between
polynomial and NP-hard instances is the subject of this paper. The results are
not only of theoretical interest, but also of practical use in applications with
underlying structures that fit the power graph model. For example, we may
want to assign k frequencies to a set of radio stations, but require that any two
stations within distance d of each other use different frequencies.
The main result is significant, as it classifies the complexity of all instances
of (d, k)-coloring for d ≥ 2: determining whether a graph is (d, k)-colorable is
2
Alexa Sharp
3d
polynomial-time for k ≤ b 3d
2 c, but NP-hard for k > b 2 c. Moreover, (d, k)coloring on trees is solvable in polynomial time.
This paper presents many opportunities for future work; further improvements on these results, as well as their implications to generalized graph decompositions and complexity are explored in [15].
2
Preliminaries
Given a graph G = (V, E) and integers d ≥ 1 and k ≥ d + 1, the goal of (d, k)
coloring is to find an assignment of k colors to the vertices of G such that no two
vertices within distance d of each other share the same color. More precisely, we
want a function f : V → {1, 2, . . . , k} such that d(u, v) ≤ d ⇒ f (u) 6= f (v). We
say G is (d, k)-colorable if such a function exists, and call the function itself a
(d, k)-coloring. The standard k-coloring problem is the special case when d = 1.
Observe that each connected component of G can be colored independently;
moreover, any component of size less than or equal to k can be (d, k)-colored by
assigning a distinct color to each vertex. Thus in this paper we assume that G
is connected and has at least k + 1 vertices.
Definition 1. For a vertex v and integer r, let Grv ⊆ G be the subgraph of
radius r around v, that is, the subgraph induced by {w | d(w, v) ≤ r}.
Definition 2. For a subgraph G0 ⊆ G, the diameter of G0 , denoted diam(G0 ),
is the maximum shortest path distance between any two vertices of G0 , that is,
diam(G0 ) = max{d(u, v) | u, v ∈ G0 }.
Definition 3. Given a connected graph G = (V, E), the set V 0 ⊆ V is a cutset
if the subgraph induced by V \ V 0 is no longer connected.
Definition 4. Given a graph G of size at least k + 1, a graph G0 = (V 0 , E 0 ) ⊆ G
is a forbidden (d, k) subgraph if diam(G0 ) ≤ min{d, |V 0 | − (k − d) − 1}.
2.1
Properties of (d, k)-colorable Graphs
Theorem 1. If G contains a forbidden (d, k) subgraph then it is not (d, k)colorable.
Proof. Suppose G0 = (V 0 , E 0 ) is a subgraph of G with diam(G0 ) ≤ min{d, |V 0 | −
(k −d)−1}. Let G00 be a connected graph induced by V 0 and max{0, k +1−|V 0 |}
vertices of V \V 0 ; G00 has ≥ k+1 vertices and diameter at most d, which precludes
a (d, k)-coloring.
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Corollary 1. If there is a vertex v ∈ G such that |Grv | ≥ (k − d) + 2r + 1 for
any 1 ≤ r ≤ b d2 c then G cannot be (d, k)-colored.
Proof. The diameter of Grv is at most 2r ≤ d. But 2r = (k − d + 2r + 1) − (k −
d) − 1 ≤ |Grv | − (k − d) − 1 and Theorem 1 completes the proof.
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Distance Coloring
3
v
v
(a)
(b)
v
(c)
Fig. 1. Examples of Corollary 1 for (6, 9)-coloring. The graphs of 1(a), 1(b) cannot be
(6, 9)-colored because |G3v | = 10; the graph of 1(c) is (6, 9)-colorable because |Grv | ≤
2r + 3 for all r ≤ 3.
k−d
Theorem 2. Given a (d, k)-colorable graph G with k ≤ b 3d
is a strict
2 c, if Gv
k−d
subgraph of G for some v ∈ G, then Gv
contains a cutset of size ≤ 2 discon.
necting v from G \ Gk−d
v
is connected to v through
Proof. First observe that any vertex y ∈ G \ Gk−d
v
at least one vertex of each Giv \ Gi−1
for 1 ≤ i ≤ k − d. We will show there is
v
i
i−1
1 ≤ i ≤ k − d such that |Giv \ Gi−1
v | ≤ 2. Then the removal of Gv \ Gv , of at
most two vertices, would disconnect y from v.
By way of contradiction, suppose |Giv \ Gi−1
v | ≥ 3 for all 1 ≤ i ≤ k − d. Then
0
| ≥ 2(k − d) + (k − d) + 1, and G is
|Gv | = 1 implies |Giv | ≥ 3i + 1. Hence |Gk−d
v
not (d, k)-colorable by Corollary 1, a contradiction.
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Lemma 1. Let P be a path of length 0 ≤ p ≤ b d2 c, with left and right endpoints
vL and vR , respectively, in a (d, k)-colorable graph G, k ≤ b 3d
2 c. Suppose there
exist disjoint subgraphs PL , PR in G \ P such that PL ∪ {vL }, PR ∪ {vR } are
connected and |PL |, |PR | ≥ b d2 c; let O be the non-P vertices connected to P in
the graph G \ (PL ∪ PR ). Then |O| ≤ k − d − 1.
Proof. See Appendix for the proof and an illustration of the notation.
3
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Negative Complexity Results
We know that (1, k)-coloring is NP-hard for k ≥ 3; we now use the results of
Section 2 to show that for d ≥ 2, (d, k)-coloring is NP-hard for k > b 3d
2 c.
Theorem 3. The (d, k)-coloring problem is NP-hard for d ≥ 2, k > b 3d
2 c.
4
Alexa Sharp
Theorem 3 follows via a reduction from (1, k)-coloring (k-COL). Given a graph
G that we wish to k-color, we construct a graph G0 such that G is k-colorable
if and only if G0 is (d, k)-colorable. The building block of this reduction is a
triangle gadget G∆ , which is shown in Figure 2.
z
z
! d2 "
! d2 "
! d2 "
! d2 "
! d2 "
y
x
k − " 3d
2 #−1
(a)
! d2 "
y
x
k − " 3d
2 #−1
(b)
Fig. 2. 2(a) and 2(b) show Theorem 3’s G∆ gadget for odd and even d, respectively.
Lemma 2. If G∆ is (d, k)-colorable then x, y and z have the same color.
Proof. The k − 1 vertices in G∆ \ {x, y, z} are within distance d of each other,
and use k − 1 distinct colors. The vertices x, y and z are within distance d of all
these k − 1 colors, but distance d + 1 from each other. If G∆ is (d, k)-colorable
then x, y and z are colored the same.
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Reduction from k-COL. Given an instance G = (V, E) of k-COL, create the
graph G0 = (V 0 , E 0 ) as follows. For each vertex u ∈ V , create gadget Gu ⊆ G0 by
concatenating deg(u) · 2k 4 copies of G∆ , overlapping the x and y vertices, always
leaving the z vertices open. Every 2k 4 th z vertex is reserved for use as follows:
for each edge e = (u, v) ∈ G, create an edge (ue , ve ) ∈ G0 where ue and ve are
reserved z vertices of Gu and Gv , respectively; an example of this reduction is
shown
in Figure 3. Note that G0 is polysize, as |G∆ | = k + 2 and it is copied
P
4
u∈V deg(u) · 2k times.
Lemma 3. If G0 is (d, k)-colorable then ue , uf ∈ G0 have the same color for all
edges e, f incident to u ∈ G.
Proof. By Lemma 2 we know that Gu ’s x, y and z vertices must be the same
color. The vertices ue and uf are simply z vertices, and the result follows.
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Lemma 4. If G0 is (d, k)-colorable then for edge e = (u, v) ∈ G, ue , ve ∈ G0 are
different colors.
Proof. If e = (u, v) ∈ G then (ue , ve ) ∈ G0 . Thus d(ue , ve ) ≤ d and ue , ve are
different colors in G0 .
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Distance Coloring
v
e
ugt
f
w
(a)
ve
ue
2k 4 − 1
wf
uf
2k 4 − 1
5
tg
ug
(b)
Fig. 3. 3(a) An instance G of k-COL. 3(b) A subgraph of the graph G0 constructed
from G in Theorem 3.
Lemma 5. If G0 is (d, k)-colorable then G is k-colorable.
Proof. If G0 has a (d, k)-coloring C then create a k-coloring D of G by setting
D(u) = C(ue ) for any e incident to u. Since C(ue ) = C(uf ) for all e, f incident
to u by Lemma 3, D is well-defined. Moreover, D(u) 6= D(v) for (u, v) ∈ G by
Lemma 4, and D uses at most k colors.
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Lemma 6. If G is k-colorable then G0 is (d, k)-colorable.
The proof of Lemma 6 requires some additional framework. Consider the colors
of the vertices of G∆ \ {x, y, z}, labeled as in figure 4(a). If we use the coloring of
this G∆ to color an adjacent G∆ as shown in Figures 4(b) or 4(c) then we will
have defined a symmetry group based on the colors of G∆ \ {x, y, z}; the base
set is A = {a1 , a2 , . . . , ak−1 } and we have two elements σ, π : A → A, where σ is
the shift operator (a1 a2 · · · ak−1 ) and τ is the adjacent transposition operator
(a1 a2 )(a3 )(a4 ) · · · (ak−1 ).
Lemma 7. Applying either the shift σ or the transposition τ to the colors of
G∆ yields a valid coloring.
Proof. The vertices of the same color in adjacent G∆ s are at least distance d + 1
apart.
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Lemma 8. Any adjacent transposition π = (aj aj+1 ) can be generated from k
compositions of σ and τ .
Proof. Use the shift σ j − 1 times, transpose with τ once, then shift again
(k − 1) − (j − 1) times. The only elements transposed are j and j + 1; the
remaining elements shift k − 1 times and hence are unchanged.
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Lemma 9. Any transposition π = (ai aj ) can be generated from ≤ 2k compositions of adjacent transpositions.
Proof. Without loss of generality, suppose i < j. Then
(ai aj ) = (aj−1 aj )(aj−2 aj−1 ) · · · (ai ai+1 )(ai+1 ai+2 ) · · · (aj−2 aj−1 )(aj−1 aj ),
which uses 2(j − i) − 1 ≤ 2k transpositions, as required.
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6
Alexa Sharp
z
a2
a4
a2! d "
2
a2! d "+1 a
a3
x
ak−1
a1
d+1
2
a5
y
ak−" d #+1
2
ad+2 ad+3 ak−" d #
2
(a)
z
z
shift σ
a2
a1
a4
a2! d "
a3
2
a2! d "+1 a
x
a3
ak−1
a1
d+1
2
a5
a5
a2! d "+1 ak−" d #+1
y
ak−" d #+1
a2
a4
2
2
2
y
ak−" d # ak−1
2
a2! d "
2
ad+1ad+2 ad+3
ad+2 ad+3 ak−" d #
2
(b)
z
z
adjacent transposition τ
a2
a1
a4
a2! d "
a4
a2! d "
2
2
a2! d "+1 a
2
x
a3
ak−1
a1
d+1
a5
a2! d "+1 a
2
y
ak−" d #+1
a3
ak−1
a2
d+1
a5
y
ak−" d #+1
2
2
ad+2 ad+3 ak−" d #
ad+2 ad+3 ak−" d #
2
2
(c)
Fig. 4. 4(a) The set A for the basis of the permutation group for odd d. The same
labeling can be used for even d, ignoring the center vertex. 4(b) The shift operator σ.
4(c) The adjacent transposition operator τ .
Lemma 10. Any cycle π = (π1 π2 · · · πp ) can be generated from ≤ p compositions of transpositions.
Proof. Note that (π1 π2 · · · πp ) = (π1 πp )(π1 πp−1 ) · · · (π1 π3 )(π1 π2 ).
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Lemma 11. Any permutation on k elements can be generated from ≤ 2k 4 compositions of the shift σ and the adjacent transposition τ .
Distance Coloring
7
Proof. First note that any permutation on A can be generated by at most k
cycles of length k. By Lemmas 8-10 we have that
k cycles of length k are generated by ≤ k 2 transpositions,
which are generated by ≤ 2k 3 adjacent transpositions,
which are generated by ≤ 2k 4 compositions of σ and τ.
Thus any permutation is generated by ≤ 2k 4 compositions of σ and τ .
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Proof of Lemma 6. Given a k-coloring D of G we show how to (d, k)-color the
gadgets of G0 . First, for each vertex u ∈ G, color all of Gu ’s x, y and z vertices
with the color D(u). This is feasible because these vertices are distance d + 1
apart, and their color is different from all neighboring gadgets’ x, y, z colors.
Next, for each vertex u ∈ G, color the G∆ ’s directly connected to Gv for some
v 6= u ∈ G. Now the remaining uncolored G∆ gadgets in Gu are chains of 2k 4 − 1
G∆ s between two colored G∆ s. We know from Lemma 11 that there is some
chain of at most 2k 4 compositions of σ and τ that lead from any one coloring
of G∆ to the next, and each composition corresponds to a valid coloring of each
G∆ . Thus we know there is a sequence of colorings getting us from one G∆ to
the one distance 2k 4 away, and G0 can thus be (d, k) colored.
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Proof of Theorem 3. The polynomial reduction constructs a graph G0 that is
(d, k)-colorable if and only if G is k-colorable, by Lemmas 5 and 6. Therefore
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(d, k)-coloring is NP-complete for d ≥ 2 and k ≥ b 3d
2 c + 1.
4
Positive Complexity Results
Thus (d, k)-coloring is NP-hard for d ≥ 2, k > b 3d
2 c; we now show that the
remaining parameters lead to polynomial-time solutions.
4.1
(d, d + 1)-coloring
Theorem 4. The (d, d + 1)-coloring problem is polynomial-time solvable.
Proof. A graph is (1, 2)-colorable if and only if it is bipartite. For d ≥ 2, if G
contains a vertex of degree 3 or greater, then it is not (d, d + 1)-colorable by
Corollary 1. Otherwise, G is a path or cycle. If it is a cycle but its length is not
a multiple of (d + 1) then it is not (d, d + 1)-colorable. Otherwise G is a path
or a cycle whose length is a multiple of (d + 1). In either case, cycle through
the colors 1, 2, . . . , (d + 1), which ensures that vertices within distance d of each
other are colored differently.
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4.2
(d, k)-coloring for k ≤ b 3d
c
2
We describe an algorithm polynomial in |G| that either produces a (d, k)-coloring
of G or declares that no such coloring exists. The algorithm finds a bounded treewidth tree decomposition of G [16], then colors the graph using known coloring
algorithms for bounded tree-width graphs [17, 18].
8
Alexa Sharp
Definition 5. A tree decomposition of G = (V, E) is a triple (T, F, X) consisting of an undirected tree (T, F ) and a map X : T → 2V associating a subset
Xi ⊆ V with each i ∈ T such that
S
I. V = i∈T Xi ;
II. for all edges (u, v) ∈ E, there exists i ∈ T such that u, v ∈ Xi ;
III. if j lies on the path between i and k in (T, F ), then Xi ∩ Xk ⊆ Xj .
Definition 6. The width of (T, F, X) is maxi {|Xi | − 1}.
Definition 7. A path decomposition (T, F, X) is a tree decomposition in which
the graph (T, F ) is a simple path.
We first try to compute a path decomposition of the graph G. Let P be a
simple path of length ≥ d + 1 in G, and let s be a center vertex; if no such path
exists then diam(G) ≤ d and G is not (d, k)-colorable. Otherwise, perform a
breadth-first search on G starting from s to get level sets L0 , L1 , . . . , Lm , where
Li consists of the set of vertices of distance i from the root s. The level graph
H is the graph consisting of vertices V and directed edges from Li to Li+1 ,
ignoring edges between vertices on the same level. We take Lj = ∅ for j > m.
For 0 ≤ i ≤ max{0, m − d}, let
def
Xi =
i+d
[
Lj .
(1)
j=i
Lemma 12. If G is (d, k)-colorable, then |Xi | ≤ 5d.
R
R
Proof. Label the right side of P as s = sR
0 , s1 , . . . , smR and the left side of P
L L
L
R L
as s = s0 , s1 , . . . , smL such that si , si ∈ Li (where mL , mR ≥ d d2 e because
`(P ) ≥ d + 1). For 0 ≤ i ≤ m, let TiR , TiL be the subgraphs of the level graph
L
R L
rooted at sR
i , si , respectively, consisting of all vertices reachable from si , si by
L
R
R
L
a (directed) path in H. Let T = H \T1 , T = H \T1 . Note that H = T L ∪T R .
See Figure 5.
To obtain the result for |Xi |, we bound |Xi ∩ T R | and |Xi ∩ T L | separately,
showing |Xi ∩ T R |, |Xi ∩ T L | ≤ b 5d
2 c. The arguments are identical except for
notation, so without loss of generality we argue only the former.
For |Xi ∩ T R |, let j = min{mR , i + d} − d d2 e and ` = max{0, i − (k − d)} (see
Figure 6). Then j ≥ 0 and ` ≤ i ≤ mR − d d2 e, and
|Xi ∩ T R | = |Xi ∩ TiR | + |Xi ∩ (T R \ TiR )|
= |Xi ∩ TjR | + |Xi ∩ (TiR \ TjR )| + |Xi ∩ (T R \ TiR )|
(2)
(3)
≤ |Xi ∩ TjR | + |Xi ∩ (TiR \ TjR )| + |Xi ∩ (T`R \ TiR )|,
(4)
where the last inequality follows because any vertex v ∈ (Xi ∩ T R ) \ T`R is at
least distance i − ` + 1 ≥ (k − d) from its connection point on P , which has
b d2 c and d d2 e vertices of P on either side of it; this forms a forbidden subgraph,
therefore no such v exists since G is (d, k)-colorable.
Distance Coloring
9
sR
mR
sR
sR
1
2
TR
L
sR
0 = s = s0
TL
sL
sL
1
2
sL
mL
Fig. 5. Notation used in Lemma 12.
sR
!
sR
i
sR
j
d
! "
2
sR
min{mR ,i+d}
Xi ∩ T R
Fig. 6. The above inequalities shown pictorially. The light gray triangle is Xi ∩ TjR .
The dark gray triangles are Xi ∩ (TiR \ TjR ). The small white triangle is Xi ∩ (T`R \ TiR ).
For |Xi ∩ TjR |, note that there are b d2 c vertices of P \ TjR within b d2 c of sR
j .
d
R
Either G is not (d, k)-colorable or there are ≤ k − b 2 c ≤ d vertices in Tj within
R
R
distance d d2 e of sR
j ; since Tj ∩ Xi is precisely this set of vertices, |Tj ∩ Xi | ≤ d.
For |Xi ∩ (TiR \ TjR )|, consider the interval of length p = j − i ≤ b d2 c from sR
i
d
R
R
to sR
.
There
are
at
least
b
c
vertices
in
P
\
T
and
P
∩
T
,
respectively,
thus
j
i
j
2
by Lemma 1, either G is not (d, k)-colorable or |(TiR \ TjR ) \ P | ≤ k − d − 1, and
|Xi ∩ (TiR \ TjR )| ≤ k − d − 1 + p + 1 ≤ k − d d2 e ≤ d.
d
R
For |Xi ∩ (T`R \ TiR )| consider the interval sR
` to si of length p = i − ` ≤ b 2 c.
d
There are ≥ b 2 c vertices in P \ T`R and P ∩ TiR , respectively, and either G is not
(d, k)-colorable, or |T`R \ TiR \ P | ≤ k − d − 1, and |Xi ∩ (T`R \ TiR )| ≤ b d2 c.
Thus |Xi ∩ T R | ≤ b 5d
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2 c, and |Xi ∩ H| = |Xi | ≤ 5d, as required.
Theorem 5. For fixed constants d, k with k ≤ b 3d
2 c, there is an O(n) algorithm
for finding a (d, k)-coloring of G if one exists.
Proof. We use the algorithm of Section 4.2 to either find a path decomposition
(T, F, X) of G of width at most 5d or determine that G is not (d, k)-colorable.
10
Alexa Sharp
Assume the former. Then (T, F, X) is also a path decomposition of the dth power
graph Gd . The algorithms of [17, 18] can determine the (1, k)-colorability of Gd
in linear time, and a (1, k)-coloring of Gd is precisely a (d, k)-coloring of G. u
t
5
Coloring on Trees
Although it is already known that (d, k)-coloring trees is polynomial-time solvable [13], the results of this paper lead to a nice combinatorial method to achieve
the same result. This does not require d or k to be fixed.
Algorithm to Find a (d, k)-coloring for a Tree T .
1. If d = 1 and k ≥ 2 then we (1, 2)-color T , a bipartite graph.
2. Otherwise, perform a breadth-first search on T , starting from a leaf node s
to get level sets L0 , L1 , . . . , Lm , where Li consists of the set of vertices of
distance i from the root s.
3. For i = 0, 1, 2, . . . , m, color the vertices of Li in some arbitrary order by
assigning v ∈ Li the lowest color not yet used in Gdv .
4. If we need more than k colors then T is not (d, k)-colorable, otherwise return
the assigned coloring.
Lemma 13. The algorithm returns a coloring of T if and only if T is (d, k)colorable.
Proof. [⇒] Consider a coloring found by the algorithm; it uses at most k colors.
Moreover, no two vertices within distance d of each other are assigned the same
color. Indeed, consider any two vertices u and v such that d(u, v) ≤ d, and
suppose u preceded v in the algorithm. Then when v is considered, u’s color is
excluded from consideration; consequently, the algorithm will not assign v the
color that it used for u.
[⇐] Suppose T is (d, k)-colorable. Consider one of the vertices v, and suppose
there are t vertices already colored within distance d of v. Then these t vertices
along with v are within distance d of each other by properties of the BFS tree,
and hence t+1 ≤ k necessarily. It follows that this set of t vertices uses ≤ t ≤ k−1
colors, and so there is a color that can be assigned to v without exceeding the
allotted k colors.
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This work was supported by NSF grant CCF-0635028. Any views and conclusions expressed herein are those of the author and do not necessarily represent the official
policies or endorsements of the National Science Foundation or the United States government.
Acknowledgments. I would like to thank Dexter Kozen for his many useful comments and insights, as well as the anonymous reviewers for their helpful suggestions.
Distance Coloring
11
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Appendix
Lemma 14. Let P be a path of length 0 ≤ p ≤ b d2 c, with left and right endpoints
vL and vR , respectively, in a (d, k)-colorable graph G, k ≤ b 3d
c. Suppose there exist
2
12
Alexa Sharp
disjoint subgraphs PL , PR in G \ P such that PL ∪ {vL }, PR ∪ {vR } are connected and
|PL |, |PR | ≥ b d2 c; let O be the non-P vertices connected to P in the graph G \ (PL ∪ PR ).
Then |O| ≤ k − d − 1.
Proof. Label P as vL = v0 , v1 , . . . , vp = vR . Let O0 , O1 , . . . , Op be a partition of O
such that the induced subgraph Oi ∪ {vi } is connected for all 0 ≤ i ≤ p. See Figure 7.
O1
O2
O
PL
PR
vL = v0 v1
v2
v3
vR = vp
Fig. 7. Notation for Lemma 14.
bdc
Note that |Oi | ≤ k − d − 1 ≤ b d2 c − 1 for all i ≥ 0 otherwise Gvi2
forbidden subgraph and G not (d, k)-colorable by Theorem 1. Let
would be a
d
d
d
dL = max{2b c − max{|Oi | + i, b c}, max{|Oi | − i}} ≥ b c − p .
i≥0
i≥0
2
2
2
(5)
Consider the subgraph G0 consisting of P , the dL ≤ b d2 c closest vertices of PL to vL
and the 2b d2 c−p−dL ≤ b d2 c closest vertices of PR to vR ; G0 consists of 2b d2 c+1 vertices
within distance 2b d2 c of each other. The vertices of O are within dL + maxj≥0 {|Oj | + j}
of the G0 ∩ PL vertices, where
d
d
dL + max{|Oj | + j} = max{2b c − max{|Oi | + i, b c} + max{|Oj | + j},
j≥0
i≥0
j≥0
2
2
max{|Oi | − i} + max{|Oj | + j}}
i≥0
j≥0
d
≤ max{2b c, max {2|Oi |, |Oi | + |Oj | + p}}
2 i≥0,j6=i
d
≤ max{2b c, |O| + p} .
2
(6)
(7)
(8)
(9)
Similarly, the vertices of O are within 2b d2 c − p − dL + maxj≥0 {|Oj | + p − j} of the
G0 ∩ PR vertices, where
d
d
d
d
2b c − dL + max{|Oj | − j} = 2b c − max{2b c − max{|Oi | + i, b c}
j≥0
i≥0
2
2
2
2
− max{|Oj | − j}, 0}
j≥0
(10)
(11)
d
≤ 2b c .
(12)
2
Lastly, the vertices of O are at most distance p + |O| from each other. Thus we have
2b d2 c + 1 + |O| vertices within distance ≤ max{2b d2 c, |O| + p} of each other, and hence
|O| ≤ k − d − 1 otherwise there is a forbidden subgraph.
t
u