Solutions

Problem Set 4
MATH 778P, Fall 2008, Cooper
Expiration: Wednesday, December 10
You are awarded up to 15 points for each problem, 5 points for submitting
solutions in L ATEX, and 5 points per solution that is used for the answer key.
1. Let G = (V, E) be the graph whose vertices are all 7 n vectors of length n
over Z7, in which two vertices are adjacent iff they differ in precisely one
coordinate. Let U ⊂ V be a set of 7 n−1 vertices of G, and let W be the
set of all vertices of G whose distance from U exceeds (c + 2) √ n, where
c > 0 is a constant. Prove that |W | ≤ 7 n e −c2 /2 .
Solution (Mark Walters): Fix c > 0 and define ɛ = e −c2 /2 . Consider
{0, 1, 2, 3, 4, 5, 6} n as the underlying probability space, where all points are
equally likely. For y ∈ {0, 1, 2, 3, 4, 5, 6} n set
X(y) = min ρ(x, y)
x∈U
where ρ is defined to be the Hamming metric on {0, 1, 2, 3, 4, 5, 6} n . Let
X0, X1, . . . , Xn = X be the martingale given by exposing one coordinate
of {0, 1, 2, 3, 4, 5, 6} n at a time. The Lipschitz condition holds for X: If
y, y ′ differ in just one coordinate then |X(y) − X(y ′ )| ≤ 1. Thus, with
µ = E[X], Azuma’s inequality yields
But
so µ ≤ 2 √ n. Thus
and
Pr[X < µ − 2 √ n] < e −22 /2 1
= ,
e2 Pr[X > µ + c √ n] < e −c2 /2 = ɛ.
Pr[X = 0] = |U|7 −n = 1
7
Pr[X > (c + 2) √ n] < ɛ
1
> ,
e2 |B(U, (c + 2) √ n)| = 7 n Pr[X ≤ (c + 2) √ n] ≥ 7 n (1 − ɛ).
It follows that |W | ≤ 7 n − 7 n (1 − ɛ) = 7 n ɛ.
1

2. Let σ be a random permutation of [n], and fix a permutation τ of [k]. Recall
that, for a set I ∈ � � [n]
k , we say that “the pattern τ occurs in σ on I” if σ|I
is order-isomorphic to τ. Denote by Xτ (σ) the number of sets I ∈ � � [n]
k
for which τ occurs in σ on I. Show that
��
���
Pr X τ (σ) − 1
k! ·
� �� � �� �
n ��� 2 n
−(1 + o(1))ɛ n
≥ ɛ < exp
k k
2k4 �
.
Solution (Aaron Dutle):
Note that E[Xτ (σ)] = 1
� � n
k! k
Consider the function Xτ (σ) and the martingale given by at step i revealing
the location of the number i in σ. Note that given two partially
revealed permutations, the next revealed number could on one extreme
be completely unusable in finding τ in σ, or on the other, be usable as
any of the k symbols of τ and with any choice of k − 1 elements from the
remaining n − 1. Thus our martingale satisfies
� �
n − 1
|Xi+1 − Xi| ≤ k
k − 1
So if we define a new martingale Y by Yi = Xi
k( n−1 then the martingale
k−1)
Y0, Y1, . . . Yn is actually Lipschitz. Set µ = E[Y ] = Y0, and Y = Yn.
Now we note that
��
���
Pr Xτ (σ) − 1
� �� � ��
n ��� n
≥ ɛ
k! k k
��
����
Xτ (σ)
= Pr
k � � �
1 n
k! k � n−1 −
k−1 k � �
� � � n
� k � n−1 � ≥ ɛ
�
k−1 k � �
� n−1
k−1
�
= Pr |Y − µ| ≥ ɛ n
k2 �
�
= Pr
by Azuma’s inequality.
Noting that
�
2 exp − ɛ2n 2k4 �
|Y − µ| ≥ ɛ√ n
k 2
� �
√
n < 2 exp − ɛ2n 2k4 �
�
= exp ln(2) − ɛ2n 2k4 � � �
= exp − 1 − 2k4 ln(2)
ɛ2 � 2 ɛ n
n 2k4 �
and that k is fixed, this is
� 2 −(1 + o(1))ɛ n
exp
2k4 �
,
the desired bound.

3. Show that, with high probability, G(n, 1/2) contains a path of length n(1/2+
o(1)).
Solution (Wei-Tian Li): Here we apply the Theorem 8.5.1. Set p =
[(log n)/n]ω(n), where ω(n) → ∞ arbitrary slowly. Then in G(n, p) almost
always deg(x) ∼ (n − 1)p for all vertices x. Let ω(n) = n/2 log n,
then p = 1
n
2 and hence δ(G) > 2 with high probability. By the fact that if
δ(G) > k, then there exists a path of length k in G, so with high probability
G(n, 1/2) contains a path of length n(1/2 + o(1)).
4. Prove that, for every ɛ > 0 there is some n0 = n0(ɛ) so that for every n > n0,
there is a graph on n vertices containing every graph on k ≤ (2 − ɛ) log 2 n
vertices as an induced subgraph.
Solution (Xing Peng): Here, we want to use the general form of Janson’s
inequality. The statement is following.
Let I be the index set and {Bi}i∈I be a collection of events. Set XBi
be the indicator variable of Bi and X = X . Set ∼ be the dependence
i∈IBi
graph. If µ = E[X], ∆ = �
�
Pr[Bi ∧ Bj] and δ = max Pr[Bj]. Then
Pr[ �
i∈I
i∼j
i
i∼j
Bi] ≤ exp(−min{ µ2 µ µ
, ,
8∆ 2 6δ }).
For any natural number n, take a graph G(n, 1
2 ). For a fixed graph H
on k = (2 − ɛ) log2 n vertices, Let the index set be � � [n]
k and BS be the
event that H is isomorphic to the subgraph induced by some vertex set
S ∈ � � [n]
k . Then
Pr[BS] = 1
� � n
k , µ =
2) 2) ,
and
∆ =
� � k−1
n �
k
i=2
� n−k
k−i
2 (k
�� k
i
�
2 2(k 2)−( i , δ =
2)
2 (k
k−1 �
i=2
� n−k
k−i
Then 8∆
µ 2 = 8 �k−1 i=2 g(i), where g(i) is defined in Section 4.5. By the proof
of Theorem 4.5.1, we have 8∆
µ 2 < k
µ2 n
n and then 8∆ > k . Simply estimation,
µ n
2 > k , here we need k = (2 − ɛ) log2 n and we can not omit ɛ. Moreover,
µ n
6δ > k . Hence,
Pr[ �
Bs] ≤ e −n/k .
S∈( [n]
k )
For this H, let AH be the event that H is not a induced subgraph of
G(n, 1
2 ) and An be the event that there is some graph on k vertices which
is not a induced subgraph of G(n, 1
2 ). Then
Pr[AH] < e −n/k .
2 (k
�� k
i
2)
�
.

and
Pr[An] < 2 (k2)
−n/k k
e < 2 2 /2 −n/k k −n/k 1
e < n e <
nα Here, α > 1 and n sufficiently large.
The Borel-Cantelli Lemma applies, alway all An fail for n sufficiently large.
Thus, there is some n0 so that for every n > n0, there is a graph on n
vertices containing every graph on k vertices as an induced subgraph.
5. Find a threshold function for the property: G(n, p) contains at least n/6
pairwise vertex disjoint triangles.
Solution (Aaron Dutle):
Claim that r(n) = n −2/3 is a threshold function for G(n, p) to contain n/6
vertex disjoint triangles.
Suppose that p(n) ≪ n −2/3 .
If H is the graph made of n/6 vertex disjoint triangles, note that for a
particular choice of the of 3-sets of vertices of the triangles, the probability
that H occurs with these as the triangles in G(n, p) is p n/2 , since n/2 is
the number of edges of H.
If we consider making this choice by choosing three vertices for the first
triangle, then choosing three for the second, then the third, etc. until we
have n/6 of them, noting that there are then (n/6)! ways that we can
choose the same set, we see there are
� � n
3,3,...,3,n/2
(n/6)!
such possible choices. So if X denotes the number of times H occurs in
the graph, the expectation of X is
E[X] =
� n �
3,3,...,3,n/2
(n/6)!
p n/2
Rewriting this in an equivalent form, and then estimating using � �
n n
e <
n! < nn we have
E[X] =
≤
n!
6n/6 � � � �
n n p
2 ! 6 ! n/2
n n
6n/6 � �
n n/2 � � p
n n/6
2e 6e
n/2
= n n/3 2 n/2 e 2n/3 p n/2

Since p ≪ n−2/3 , eventually p <
is
< n n/3 2 n/2 e 2n/3
�
1
8e4/3n2/3 1
8e 4/3 n 2/3 after which the quantity above
� n/2
= 2 −n = o(1)
Hence X = 0 almost surely, proving one direction of the threshold function.
For the opposite direction, suppose that p(n) ≫ n −2/3 . To make the proof
go easier, note that if p(n) > ln(n)n −2/3 we can randomly remove edges
from our G(n, p) to reduce to the case p(n) = ln(n)n −2/3 . So we can
assume that p(n) ≤ ln(n)n −2/3 .
Consider a subset of the vertices of G(n, p) of size n/2. For each set of
three vertices in this set, we let Bi be the event that the three vertices are
a triangle. The expected number of triangles is then
� �
n/2
µ = E[X] = p
3
3
We’re going to use Janson’s inequality, so we compute ∆ = �
i∼j Pr[Bi ∧
Bj]. We choose a triangle, choose an edge of it, and choose a third vertex
to count all of the i ∼ j. The probability of each of these is p 5 , giving
� �
n/2
∆ = 3(n/2 − 3)p
3
5 .
When applying Janson’s inequality, we get Pr �� �
−µ+∆/2
Bi < e
Since this is the probability that a single choice of n/2 of the vertices
contains no triangle, and the probability for any other choice of an n/2
subset would be the same, conclude that � � �� � n
n/2 Pr Bi is the number of
n/2 subsets of the vertices that have no triangle. Estimating this, we find
<
� �
n
�� �
Pr Bi
n/2
� �n/2 ne
e
n/2
−µ+∆/2
< e n e −µ+∆/2 = e n−µ+∆/2
We want this quantity to go to 0 as n → ∞, so we need that the exponent
is going to −∞.
Since we know p < ln(n)n −2/3 , we can approximate
n − µ + ∆/2

� �
n/2
= n − p
3
3 � �
n/2
+ 3(n/2 − 3)p
3
5
< n − n3
216e 3 p3 + n4
64 p5
≤ n − n3
10 4 p3 + (ln(n))5
n 1/3
Since p ≫ n −2/3 , eventually p > 10 5/3 n −2/3 , so this quantity is
which clearly goes to −∞.
Thus � � �� � n
n/2 Pr Bi = o(1).
< n − 10n + (ln(n))5
n 1/3
= −9n + (ln(n))5
n 1/3
So with high probability every n/2 sized subset of G(n, p) has a triangle.
To find n/6 vertex disjoint triangles, we take them greedily, choosing an
n/2 sized subset, taking its triangle, choosing another n/2 subset avoiding
our chosen triangle, and so on. With high probability, we only run out of
n/2 subsets to choose when our triangles use n/2 of the vertices, which is
when we have n/6 triangles chosen.
Thus n −2/3 is a threshold function for G(n, p) having n/6 vertex disjoint
triangles.
6. Let G = (V, E) be a graph with chromatic number χ(G) = 1000. Let
U ⊂ V be a random subset of V chosen uniformly among all 2 |V | subsets
of V . Let H = G[U] be the induced subgraph of G on U. Prove that
Pr[χ(H) ≤ 400] < 1
100 .
Solution (Yiting Yang): Let L(U) be the chromatic number of G[U].
Let X0, X1, . . . , Xm be the martingale on L by exposing a color class each
time. Xm(U) = L(U).
Let U and U ′ be two subsets of V which differ only in one color class. Then
|L(U ′ ) − L(U)| ≤ 1, which implies L satisfied the Lipschitz Condition. By
Theorem 7.4.2, we have
where µ = E[L].
P r[L < µ − λ √ m] < e −λ2 /2 ,
Recalling E[L] ≥ 500 in Homework 2, we have

P r[L[U] < 400] = P r[L[U] < 500 − √ 10 √ 1000]
≤ P r[L[U] < µ − √ 10 √ m] < e −(√ 10) 2 /2
= e −5 < 1
100 .
7. Show that there exists a binary string S of length at most c2n log n/n so
that, for every w ∈ {0, 1} n , there is a length n subword of S identical to
w except in at most one coordinate. (We say that w1w2 · · · wk−1wk is a
subword of S1S2 · · · SN−1SN if there is some j so that Sj+i = wi for all
1 ≤ i ≤ k.) Hint: Take a random string. How many w’s are missed? Tack
them onto the end of the string.
Solution (Mark Walters): Let w ∈ {0, 1} n and randomly select a
string S of length 2n log n
n . Let Ai be the event that the length n subword
beginning at the ith coordinate of S is identical to w except in at most
one coordinate. Note that
Pr[Ai] =
n + 1
2 n
because there are n choices for the “free” coordinate and of course one
additional subword that didn’t require a “free” coordinate. Also, note
that
µ = �
Pr[Ai] = (|S| − n + 1)
i∈I
n + 1
2 n
where I = {1, 2, . . . , |S| − n + 1}. Now, in the event that Ai and Aj are
dependent events for i < j, we need to assess the amount of overlap, i.e.,
the value of j − i, in order to calculate ∆. It’s obvious that the smallest
overlap is 1 while the largest overlap is n − 1. If j − i = k then we get
Pr[Ai ∧ Aj] ≤ n2 + k 2 + (2n − 1)(1 − k)
2 2n−k
because there are 2n − k coordinates involved and we have to allow that
at least one of them is a “free” coordinate. So, we get
∆ = �
Pr[Ai ∧ Aj] =
i∼j
|S|−n+1 �
|S|
i=1
i=1
k=1
n−1 �
2
k=1
Pr[Ai ∧ Aj]
� n−1 � n
≤ 2
2 + k2 + (2n − 1)(1 − k)
22n−k |S|
�
<
i=1
�
2 k
2 −2n+2 n 3
n−1
k=1

|S|
�
= 2 −2n+2 n 3 (2 n − 1)
i=1
|S|
�
< 2 −n+2 n 3
i=1
= |S|2 −n+2 n 3
Now plugging into The Janson Inequality we get
� �
�
Pr ≤ exp � − (|S| − n + 1) n2 −n+1 + � |S|2 −n+2� /2 �
i∈I
Ai
= exp � (−|S|n + n 2 − n)2 −n+1 + |S|2 −n+1�
= exp � |S|2 −n+1 (1 − n) + (n 2 − n)2 −n+1�
� n 2 log n
= exp
n
�
2 log n
= exp
n − 2 log n + n2 − n
2n−1 �
−2 log n
→ e
= n −2
2 −n+1 (1 − n) + (n 2 − n)2 −n+1
In other words, given a word w ∈ {0, 1} n , the probability that there does
not exist a length n subword of S identical to w except in at most one
coordinate is at most n −2 . Since there are 2 n possible words w ∈ {0, 1} n ,
we need to tack on 2 n n −2 words of length n. In other words, we need to
tack on 2n
n many coordinates to S to form a new string ¯ S, which has fewer
than 2 2n log n
n
coordinates. Since we are not even doubling the size of S,
the size of ¯ S is c2 n log n/n and now for each w ∈ {0, 1} n there exists a
length n subword of ¯ S identical to w except in at most one coordinate.
8. Let G = G(n, p) with p = [(3 log n)/n2 ] 1/3 . Show that, with high probability,
every pair of vertices in G is connected by a path of length 3. (Note
that a path of length three has four vertices.)
Solution (Bill Kay): First, we calculate the probability that a given
pair of vertices has a path of length 3 between them. If � � n
2 is o(P r[a
given pair of vertices do not have a path of length 3) , we can apply the
union bound. Pick two vertices at random, call them x0 and x3. For
counting purposes, it is convenient to think of this path as going from
x0, x1, x2, x3. Now, there are n − 2 ways to pick x1 and then n − 3
ways to pick x2. Also, each of the edges has to be in the graph. If X
is the random variable that counts the number of paths between x0 and
x3, E[X] = µ = (n − 2)(n − 3)p3 . To calculate ∆, simply notice that two
paths can only intersect in one place; If they intersect in the edge between
x1, x2 then the other edges are fixed. We first count the number of ways
�

two paths could intersect in the edge (x0, x1), and by symmetry we get
how many ways two paths could intersect in the edge (x2, x3). First, there
are n − 2 ways to choose the common (x1). Then, there are � � n−3
2 ways
to pick the middle edges. The last two edges are now fixed. Each edge
has to be in, so the calculation gives ∆ = 2(n − 2) � � n−3 5
2 p . We now apply
Jansen’s inequality. The probability that none of the paths are in the
graph is bounded above by e −(n−2)(n−3)p3 +(n−2)( n−3
2 )p 5
. Plugging in the
proposed value for p, a little work gives:
e −(n−2)(n−3)p3 +(n−2)( n−3
2 )p 5
= e −(n−2)(n−3)([(3 log n)/n2 ] 1/3 ) 3 +(n−2)( n−3
2 )([(3 log n)/n 2 ] 1/3 ) 5
≤ e −n2 ([(3 log n)/n 2 ])+n 3 ([(3 log n)/n 2 ] 5/3 )
= e
3 log n
−3 log n+ 1
n 3
−3 log n+o(1)
= e
Clearly, � � n − log n
2 = o(e ), so the union bound can be applied as desired.