Theorem 1 (Simple Containers Theorem). Let G be a simple, r-graph of average degree d and of order n. Let 0 < δ < 1. If d is large enough, then there
exists a collection of sets C ⊆ P(V (G)) satisfying:
(i) for every independent set I ⊆ [n], there exists C ∈ C with I ⊆ C,
(ii) e(G[C]) < δe(G) for every C ∈ C,
1
(iii) |C| 6 2βn where β = ( d1 ) 2r−1 .
1
Degree Measure
Definition. Let G be a hypergraph. For v ∈ V (G) we define
d(v) = |{e ∈ E(G) : v ∈ e}|.
Definition (Degree Measure). Let G be a r-graph of order n and average degree
d. Let S ⊆ V (G). The degree measure µ(S) of S is defined by,
µ(S) =
1 X
d(v).
nd
v∈S
Note that µ is probability measure, so particularly additive and µ(V (G)) = 1.
The degree measure has several useful properties, which are stated below.
• First property tell us that µ(S) small implies e(G[S]) small.
e(G[S]) 6
1X
nd
d(v) =
µ(S) = µ(S)e(G)
r
r
v∈S
e(G[S]) 6 µ(S)e(G)
(1)
• Let e(A, B) denote number of edges between sets A, B ⊆ V (G). We have
X
(r − 1)ndµ(V (G)\S) = (r − 1)
d(v) > (r − 1)e(V (G)\S, S) >
v ∈S
/
>
X
d(v) − re(G[S]).
v∈S
That is,
(r − 1)(1 − µ(S)) = (r − 1)µ(V \S) > µ(S) −
r
e(G[S]).
nd
So finally,
nd
e(G[S]) >
(rµ(S) − r + 1) =
r
1
1
µ(S) − 1 +
nd.
r
(2)
• If I is an independent set in G. From (2) we get,
1
0 = e(G[I]) > µ(I) − 1 +
nd
⇔
r
1−
1
> µ(I).
r
• If G is d-regular, inequality e(G[S]) < δe(G) with (2) implies:
nd
1
d|S|
1
δ
= δe(G) > e(G[S]) > µ(S) − 1 +
nd =
−1+
nd.
r
r
nd
r
What gives us
|S| <
2
1−
1 δ
+
r
r
n.
Proof of Theorem 1
For a proof of Theorem 1, we will need following lemma.
Lemma 1. Let G be a simple, r-graph of average degree d and of order n. If d
is large enough, then there exists a collection of sets C ⊆ P(V (G)) satisfying:
(i) for every independent set I ⊆ [n], there exists C ∈ C with I ⊆ C,
(ii) µ(C) < (1 −
1
4r 2 )
for every C ∈ C,
1
(iii) |C| 6 2αn where α = c(r)( d1 ) 2r−2 log d.
Statement of the lemma is quite similar to the statement of the Theorem 1,
condition (i) is exactly the same. However (ii) is expressed in terms of degree
measure. We can suspect that it will be used to bound number of edges in G[C]
using property (1) of µ. In (iii) cardinality of the container family is smaller
than in the Theorem 1, provided d sufficiently large.
Definition. Let G be a r-graph. For a given 0 6 j 6 r−1 and sets R, S ⊆ V (G)
we define Γj (R, S), by
R
S
Γj (R, S) = v ∈ V (G) : ∃ e = {v} ∪ f ∪ g, f ∈
,g ∈
.
j
r−j−1
e∈E(G)
Proof. Let V = V (G), E = E(G). For a given 0 6 j 6 r − 1 and sets R, S,
T ⊆ V (G) we define Cj (R, S, T ), by
V \(Γj (R, S)\T )
if µ(Γj (R, S)\T ) > 4r12 ,
Cj (R, S, T ) =
.
∅
otherwise.
Since 1 = µ(V \(Γj (R, S)\T )) + µ(Γj (R, S)\T ), if Cj (R, S, T ) would be a container condition (ii) will be satisfied.
2
Let
1
u= √
3r
6r
d
1
2r−2
√
q = 15ru = 5 3r
and
6r
d
1
2r−2
.
By the definition q is small if d is large. Now we will define a family C which
will be candidate for a containers family,
C = {Cj (R, S, T ) : 0 6 j 6 r − 1, |R|, |S|, |T | 6 qn}.
Then
7qn
3 7 3qn en
n
e
d
n
√
6
6
=
6
qn
qn
qn
5 3r 6r
1 7qn
7qn/(2r−2)
)
6 d 2r−2
= 2log2 (d
.
|C| 6 r(qn)3
We will now focus on the exponent of 2,
7nq/(2r−2)
log2 (d
1
2r−2
√
1 log d
6r
1 log d
= 7 · 5 3r
=
) = 7qn
·
2r − 2 log 2
d
2r − 2 log 2
√
1
2r−2
7 · 5 3r(6r)1/(2r−2)
1
=n·
·
log d.
(2r − 2) log 2
d
{z
}
|
c(r)
So
1 1
α = c(r)( ) 2r−2 log d.
d
We will use probabilistic method, to show that the probability that C is not
satisfying conditions (i),(ii),(iii) is smaller that 1. For a given independent set
I we will define sets R, S, T , each not bigger than qn such that for some j,
Cj (R, S, T ) is a container for I.
Fix an independent set I, For a subset A ⊆ V with I ⊆ A and for 0 6 j 6 r
we define
Ej (A) = {e ∈ E : e ⊆ A, |e ∩ A| > j}.
|C| 6 2αn ,
Now, let P (j) be a statement:
1
j
nduj
I ⊆ A ∧ µ(A) > 1 −
+ 2 ⇒ (|Ej (A)| >
∀
2r 2r
2r
A⊆V
1
nd
P (0):
I ⊆ A ∧ µ(A) > 1 −
⇒ |E0 (A)| >
. From property (2) of
∀
2r
2r
A⊆V
degree measure we have
1
nd
|E0 (A)| = e(G[A]) > µ(A) − 1 +
nd =
.
r
2r
Hence the statement P (0) is true.
3
P (r):
∀ (I ⊆ A ∧ µ(A) > 1) ⇒ |Er (A)| >
A⊆V
contain any edges, 0 = |Er (A)| false.
ndur
2r
ndur
. So A = V . Since I cannot
2r
> 0. Hence the statement P (r) is
Therefore there exists 0 6 j 6 r − 1 such that the statement P (j) is true
and P (j + 1) is false. Fix a set A witnessing falsity of P (j + 1). Such set exists
j
1
1
since µ(A) > 1 − 2r
+ j+1
2r 2 > 1 − 2r + 2r 2 . Thus
I ⊆ A ∧ µ(A) > 1 −
nduj+1
j+1
1
∧
|E
(A)|
<
+
.
j+1
2r
2r2
2r
Let p = (6r/duj )1/(r−1) . Since u < 1 and 0 6 j 6 r − 1, we have
p=
6r
duj
= 3r
1
r−1
6
6r
d
1
r−1
1
= 3r
u
1
√
3r
6r
d
!2
1
2r−2
·
1
=
u
u2
q
= 3ru = .
u
5
Let R ⊆ I and S ⊆ A\I be random sets, where each vertex (if I and
A\I respectively) is included independently with probability p. By Markov’s
inequality
E(|R|)
|I|p
np
1
Pr(|R| > qn) 6
=
6
6 ,
qn
qn
qn
5
and
Pr(|S| > qn) 6
E(|S|)
|A\I|p
np
1
=
6
6 .
qn
qn
qn
5
Now, let T = Γj (R, S) ∩ I. Provided µ(Γj (R, S)\T ) > 1/4r2 , Cj (R, S, T ) is
a container for I. Indeed, if v ∈ I, then v ∈
/ Γj (R, S)\T , by definition of T . So
clearly v ∈ V \(Γj (R, S)\T ) = Cj (R, S, T ). To complete the proof we have to
show that with the adequate probabilities |T | > qn and µ(Γj (R, S)\T ) < 1/4r2
holds.
Let v ∈ I. If v ∈ Γj (R, S) then, by definition v ∈ T . This would happen
if
S
there exists e ∈ E(G) such that e = {v} ∪ f ∪ g, f ∈ Rj , g ∈ r−j−1
. Since
|e∩I| = j+1 we have that e ∈ Ej+1 (A). There are j+1 partitions e = {v}∪f ∪g,
A\I
v ∈ I, f ∈ Ij , g ∈ r−j−1
, for each such partition, the probability that both
S
f ∈ Rj and g ∈ r−j−1
is pr−1 .
Observation. Let us consider fixed R and S. Let Y denote pairs
(v, e) Ssuch
that v ∈ I and e ∈ Ej+1 (A) with e = {v} ∪ f ∪ g, v ∈ I, f ∈ Rj , g ∈ r−j−1
.
Then
nduj+1
1
|T | 6 |Y | 6 (j + 1)|Ej+1 (A)| < (j + 1)
6 nduj+1 .
2r
2
4
Taking expectations, by observation we obtain
E(|T |) 6 E(|Y |) <
1
qn
1
6r
nduj+1 pr−1 = nduj+1 j = 3run =
.
2
2
du
5
Applying Markov’s inequality
Pr(|T | > qn) 6
E(|T |)
qn
1
=
= .
qn
5qn
5
Now we will show Pr(µ(Γj (R, S)\T ) < 1/4r2 ) 6 51 . For v ∈ A, let
Fj (v) = {e ∈ E : v ∈ e, e ∈ Ej (A), e ∈
/ Ej+1 (A)} = {e : v ∈ e ⊆ A, |e ∩ I| = j},
and
D=
1 j
v ∈ A\I : |Fj (v)| > du (1 − u) .
2r
Note that I ⊆ A\D. Consider an edge e ∈ Ej (A\D) and e ∈
/ Ej+1 (A). Then
e ⊆ A\D and |e ∩ I| = j. Since j < r there is v ∈ (A\D)\I and v ∈ e. Note
that Ej (A\D) ⊆ Ej (A), so e ∈ Fj (v). By the definition of D and facts that
1
v∈
/ D and v ∈
/ I we have |Fj (v)| < 2r
duj (1 − u). Therefore
|Ej (A\D)\Ej+1 (A)| < |A\D|
By the choice of A, |Ej+1 (A)| <
1 j
1
du (1 − u) 6 nduj (1 − u).
2r
2r
1
j+1
,
2r ndu
hence
|Ej (A\D)| 6 |Ej (A\D)\Ej+1 (A)| + |Ej+1 (A)| <
1
1
1
< nduj (1 − u) + nduj+1 = nduj .
2r
2r
2r
However, we know that the statement P (j) is true, so it has to be µ(A\D) <
1
1
1 − 2r
+ 2rj 2 . On the other hand A was chosen such that µ(A) > 1 − 2r
+ j+1
2r 2 .
Hence
1−
1
j
1
j+1
+ 2 > µ(A\D) = µ(A) − µ(D) > 1 −
+
− µ(D),
2r 2r
2r
2r2
and so µ(D) > 2r12 .
By the definition of D there is D ∩ I = ∅, in particular D ∩ T = ∅. Let D∗ =
D\Γj (R, S). Then D\D∗ = D ∩ Γj (R, S) ⊆ Γj (R, S)\T . So µ(Γj (R, S)\T ) >
µ(D\D∗ ) = µ(D) − µ(D∗ ).
2
1
duj (1 − u) > 5r
duj (since u is small).
Now, let v ∈ D, then |Fj (v)| > 2r
I
Each e ∈ Fj (v) has a partition e = f ∪ g with f ∈ j , g ∈ A\I
r−j where v ∈ g
since v ∈
/ I. For every edge the probability that f ⊆ R and g\{v} ⊆ S is pr−1 .
Moreover these events are independent over all e ∈ Fj (v), because G is simple.
Hence
5
!
R
R
=
,g ∈
Pr(v ∈ D ) 6 Pr
∀ e 6= {v} ∪ f ∪ g, v ∈ D, f ∈
j
r−j−1
e∈Fj (v)


 2

12
1
r−1 |Fj (v)|
j r−1
= (1 − p )
6 exp − du p
= exp −
<
.
 5r | {z }
5
10
∗
=6r
P
P
1
1
Now, µ(D∗ ) = nd
v∈D ∗ d(v) = nd
v∈D d(v)Iv , where Iv is an indicator
1
∗
of event v ∈ D . Since E(Iv ) = Pr(v ∈ D∗ ) < 10
,
E(µ(D∗ )) =
1 X
1
d(v)E(Iv ) < µ(D) .
nd
10
v∈D
Applying Markov’s inequality, we get
µ(D)
2µ(D)
1
2E(µ(D∗ ))
∗
Pr µ(D ) >
<
= .
6
2
µ(D)
10µ(D)
5
Therefore
µ(D)
1
> 2.
2
4r
Finally, the probability that Cj (R, S, T ) is not the container for I is equal
µ(Γj (R, S)\T ) > µ(D) − µ(D∗ ) >
to
1
Pr |R| > qn ∨ |S| > qn ∨ |T | > qn ∨ µ(Γj (R, S)\T ) > 2 6
4r
6 Pr(|R| > qn) + Pr(|S| > qn) + Pr(|R| > qn) + Pr(µ(Γj (R, S)\T ) >
64·
1
)6
4r2
1
< 1.
5
Proof of Theorem 1. The proof strategy is as follows. We will iteratively apply
lemma 1, starting with G. If we will get a container, say C, with at least
δe(G) edges, we will apply the lemma 1 to G[C], and so on to each of the new
containers, until we obtain a collection of containers each with less than δe(G)
edges.
If C is a container obtained after applying lemma 1 to G, and e(G[C]) >
δe(G), if dC stands for athe average degree of G[C], we have
dC =
1 X
1
1
d(v) =
e(G[C]) >
δe(G) = δd.
|C|r
|C|
|C|
v∈C
So we can apply lemma 1 to G[C] with dC > δd, we will obtain a collection
C 0 of containers for G[C] with
6
(i) for every independent set I ⊆ V (C), there exists C 0 ∈ C 0 with I ⊆ C 0 ,
(ii) µ(C 0 ) < 1 −
0
1
4r 2
for every C 0 ∈ C 0 ,
1
(iii) |C 0 | 6 2α n where α0 = c(r)( d1C ) 2r−2 log dC .
Let I ∈ I(G) and I ⊆ C, it is equivalent to I ∈ I(G[C]). Hence every
independent set contained in C will be also contained in some member C 0 of C 0 .
Now, by (1) we have
1 2
) e(G).
4r2
One can see that each application of the lemma decreases the fraction of edges
by 1 − 1/4r2 . Note that new containers are smaller concerning the number of
edges, they do not necessarily have much less vertices. Let m be the number of
levels of iteration. From above observation we ant to have (1 − 1/4r2 )m < δ.
Note that
1
log δ
.
(1 − 1/4r2 )m < δ ⇔ m log 1 − 2 < log δ ⇔ m >
4r
log 1 − 4r12
log δ
So we need at most m = log 1− 1
+ 1 < −c log δ levels of iteration, where
( 4r2 )
c > 0 is some positive constant.
By iterative applications of the lemma the fraction of edges in each container
decreases, however the cardinality of our family of containers can grow rapidly.
We must assure that it will be not bigger than 2βn , as stated in Theorem 1. Let
αmax denote maximum over all applications. Then
1
1
2r−2
2r−2
1
1
log dmax < c(r)
log d.
αmax = c(r)
dmax
δm d
e(G[C 0 ]) 6 µ(C 0 )e(G[C]) 6 µ(C 0 )µ(C)e(G) 6 (1 −
Since dmax = δ a d, for some 0 6 a 6 m. We want to show m · αmax < β. Let
1/δ < log d, we have
1
2r−2
1
1
1
m · αmax < m · c(r)
log d · d 2r−1 =
m
β
δ d
1
1
2r−2
(2r−1)(2r−2)
1
1
= c(r)m
log d <
δm
d
c
1
2r−2
log δ1 (2r−1)(2r−2)
1
1
1
< c(r)c log ·
log d <
δ
δ
d
1
(2r−1)(2r−2)
c
1
log log d
2r−2
< c(r)c log log d · (log d)
log d <
d
C3
1
d→∞
< C1 (log d)C2 log log d
−−−→ 0.
d
Hence, m · αmax < β for d sufficiently large.
7
3
List chromatic number
Theorem. Let G be a simple d-regular r-graph. Then, as d → ∞,
χL (G) > (1 + o(1))
1 log d
.
2r − 2 log r
1 log d
Note: Originally (2012), χL (G) > (1 + o(1)) 4r
log 2r .
Proof. First, since G is simple, d-regular, r-graph, we can apply Theorem 1 to G.
Then we have a collection C of containers of size not greater than 2βn . Moreover,
since G is d-regular, for every container C we have |C| < (1 − 1/r + δ/r)n where
0 < δ < 1.
Let c = (1 − δ)/r, k = d1/(2r−2) , with this values we can apply C-compability
theorem (theorem 2 from 20.10) to a family C. We obtain collection of lists L,
log k
.
each of size l, which are not C-compatible, where as k → ∞, l > (1+o(1)) log(1/c)
Hence,
χL (G) > (1 + o(1))
log d1/(2r−2)
1 log d
log k
= (1 + o(1))
= (1 + o(1))
.
log(1/c)
log(r/(1 − δ))
2r − 2 log r
Since δ was arbitrary.
8