University of South Carolina Scholar Commons eses and Dissertations 2015 e Packing Chromatic Number of Random d- regular Graphs Ann Wells Cliſton University of South Carolina Follow this and additional works at: hps://scholarcommons.sc.edu/etd Part of the Mathematics Commons is Open Access esis is brought to you by Scholar Commons. It has been accepted for inclusion in eses and Dissertations by an authorized administrator of Scholar Commons. For more information, please contact [email protected]. Recommended Citation Cliſton, A. W.(2015). e Packing Chromatic Number of Random d-regular Graphs. (Master's thesis). Retrieved from hps://scholarcommons.sc.edu/etd/3697
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University of South CarolinaScholar Commons
Theses and Dissertations
2015
The Packing Chromatic Number of Random d-regular GraphsAnn Wells CliftonUniversity of South Carolina
Follow this and additional works at: https://scholarcommons.sc.edu/etd
Part of the Mathematics Commons
This Open Access Thesis is brought to you by Scholar Commons. It has been accepted for inclusion in Theses and Dissertations by an authorizedadministrator of Scholar Commons. For more information, please contact [email protected].
Recommended CitationClifton, A. W.(2015). The Packing Chromatic Number of Random d-regular Graphs. (Master's thesis). Retrieved fromhttps://scholarcommons.sc.edu/etd/3697
Now, observe that |V2i∩Ni(u)| ≤ 1 for i = 1, 2, . . . , bk2c and |V2i−1∩Ni−1(uv)| ≤ 1
for i = 1, 2, . . . , dk2e. Also note that ∪uNi(u) and ∪uvNi(uv) cover all the vertices of
G evenly.
15
Thus,
|V2i| ≤n
|Ni(u)| = n(d− 2)d(d− 1)i − 2 for 1 ≤ i ≤
⌊k
2
⌋
and
|V2i−1| ≤n
|Ni(uv)| = n(d− 2)2d(d− 1)i − 2 for 1 ≤ i ≤
⌈k
2
⌉.
As the series φi(d) ..= ∑∞i=1
d−2d(d−1)i−2 and νi(d) ..= ∑∞
i=1d−2
2d(d−1)i−2 converge and are
decreasing functions of d, we have φi(d) + νi(d) < 1 for all d ≥ 4. Hence,
n =k∑i=1|Vi|
=d k2 e∑i=1|V2i−1|+
b k2 c∑i=1|V2i|
≤d k2 e∑i=1
n(d− 2)2d(d− 1)i − 2 +
b k2 c∑i=1
n(d− 2)d(d− 1)i − 2
< n(φi(d) + νi(d))
< n,
a contradiction.
Corollary 4.1. For any d ≥ 4 and any integer k, there exists a d-regular graph G
with χρ(G) ≥ k.
In the configuration model, the induced graph of G on Ni(u) is a tree with high
probability but we must account for the possibility of overlaps. We refer to the case
in which two subsets, say vi and vj, contain vertices which are matched to vertices of
a third subset, vk, as an overlap. Note that two overlaps occur with probability less
than ε. Let D = diam(Gn,d). Define fi(d) ..= d(d−1)i−2(d−2) and gi(d) ..= 2d(d−1)i−2
(d−2) .
Lemma 4.1. Let u be a vertex of Gn,d and let Ni(u) denote the set of vertices of
distance at most i from u in Gn,d where 1 ≤ i ≤ (1 − o(1))D/2. For fixed u, with
probability 1−O( 1n), |Ni(u)| = fi(d), with probability O( 1
n), |Ni(u)| = fi(d)− 1, and
with probability O( 1n2 ), |Ni(u)| ≤ fi(d)− 2.
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Proof. First observe that
P (|Ni(u)| = fi(d)) =(
1− d− 1nd− 1
)(1− 2d− 3
nd− 3
)· · · (1− ξ)
≥ 1− d− 1nd− 1 −
2d− 3nd− 3 −
3d− 5nd− 5 − · · · − ξ
≥ 1− cid
nd
= 1− cin
for ξ ..= (2fi(d)− 1)(d− 1) + 1nd− (2fi(d)− 1) and ci = O(f 2
i (d)). Since i ≤ (1− o(1))D/2, we have
fi(d) = o(√n) and ci = o(n). As
P (∃ u s.t. |Ni(u)| = fi(d)− 2) ≤ ncin2 ≤
cin
= o(1),
we need only consider the case in which there is one overlap. Hence,
P (|Ni(u)| ≤ fi(d)− 1) ≤(cin
)2.
The proof of Lemma 4.2 follows a similar argument.
Lemma 4.2. Let uv be an edge of Gn,d and let Ni(uv) denote the set of vertices of
distance at most i from u or v in Gn,d, where 1 ≤ i ≤ (1−o(1))D/2. With probability
1− o(1), for all u, v, |Ni(uv)| ∈ gi(d), gi(d)− 1.
Next we consider the case when i ∈ [(1− o(1))D/2, 3D/4].
Lemma 4.3. Let u be a vertex of Gn,d and let Ni(u) denote the set of vertices of
distance at most i from u in Gn,d where (1− o(1))D/2 ≤ i ≤ 3D/4. With probability
1-o(1), for all u, fi(d)− 4 ≤ |Ni(u)| ≤ fi(d).
Proof. Let Av1,...,vα be the event that an overlap occurs at the subsets v1, . . . , vα.
Then,
P(Av1,...,vα) ≤(
d− 1nd− 1− 2fi(d)
)α.
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Thus,
P(|Ni(u)| = fi(d)− α) ≤ P (∪Av1,...,vα)
≤∑
v1,...,vα
P(Av1,...,vα)
≤ (fi(d))α(
d− 1nd− 1− 2fi(d)
)α≤ c · n
34αn−α
≤ c1
nα/4
= o( 1n
),
since i ≤ 3D/4 we have fi(d) ≤ n34 . Thus, for all u,
P(∃ u such that |Ni(u)| ≤ fi(d)− α) ≤ no(1/n) = o(1).
Hence, choosing α = 5, we have with probability 1−o(1) for all u, |Ni(u)| ≥ fi(d)−4 ≥12fi(d).
Again, a similar argument proves the following Lemma.
Lemma 4.4. Let uv be an edge of Gn,d and let Ni(uv) denote the set of vertices of
distance at most i from u or v in Gn,d, where (1 − o(1))D/2 ≤ i ≤ 3D/4. With
probability 1− o(1), for all u, v, gi(d)− 4 ≤ |Ni(uv)| ≤ gi(d).
We may now prove the main result.
Proof. By Theorem 4.2, we may choose ε > 0 small enough so that ∑∞i=11
fi(d) +∑∞i=1
1gi(d) ≤ 1− ε. Now, there exists an M such that
∞∑i=M
1fi(d) +
∞∑i=1
1gi(d) <
ε
6 .
By Theorem 3.1, we may choose n0 such that for n ≥ n0, D = (1 + o(1)) logd−1(n) ≥
4M .
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Recall Gi = (V,Ei) where Ei = (u, v) : d(u, v) ≤ i in G.
Let i ∈ [3D/4 + 1, D]. Note that as α(Gin,d) ≤ α(Gj
n,d) for i ≥ j, where α is the
independence number, we have that |Vi| ≤ α(Gin,d) ≤ α(Gj
n,d) ≤ 2fi(d)n for i ≥ j.
Thus, ∑Di=3/4D |Vi| ≤
∑3/4Dj=D/2
2fj(d)n <
ε6n.
When i > D observe that |Vi| ≤ 1.
Thus,
n =k∑i=1|Vi|
=(1−o(1))D/2∑
i=1|Vi|+
3D/4∑i=(1+o(1))D/2+1
|Vi|+D∑
i=3D/4+1|Vi|+
∑i>D
|Vi|
< (1− o(1/n))(1+o(1))D/2∑
i=1
(1
fi(d) + 1gi(d)
)
+o(1/n)(1+o(1))D/2∑
i=1
(1
fi(d)− 1 + 1gi(d)− 1
)
+3D/4∑
i=(1−o(1))D/2
(1
fi(d)− 5 + 1gi(d)− 5
)+
3/4D∑j=D/2
2fj(d)n+
∑i>D
|Vi|
< (1− ε)n+ ε
6n+ ε
6n+ ε
6n+∑i>D
|Vi|
≤(
1− ε
2
)n+
∑i>D
|Vi|
Thus, ∑i>D |Vi| ≥ ε2n and hence, with high probability, χρ(Gn,d) ≥ cdn for some
constant cd.
In the future, we would like to determine how the packing chromatic number of
random cubic graphs behaves.
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