Math778P Homework 3
due Oct. 5, Friday lecture
Choose any 5 problems to solve.
1. Use alteration method to prove the Ramsey number
t 2
R(4, t) ≥ c
ln t
for some constant c > 0.
Proof by Edward Boehnlein: By theorem 3.3.2, we have for any
p ∈ [0, 1] and n ∈ N,
n
n
n (4)
(1 − p)( t )
R(4, t) > n −
p2 −
t
4
4
n
In particular, if n4 p(2) < n4 and nt (1 − p)( t ) < n4 for some n then
1
R(4, t) > n2 . If we set p = n1/2
, then we obtain
6
n (4)
n
1
n4 1
n
p2 =
<
=
3
1/2
4
4
24 n
24
n
Further,
t2 /2 ne t 2
n
n
1
ne t − t1/2
(
)
t
(1 − p)
<
1 − 1/2
<
e 2n
t
t
t
n
So we want to force
ne t
−
e
t
t2
2n1/2
<
n
2
or equivalently, force
−t2
< ln n − ln 2
2n1/2
or, when disregarding terms of relatively small magnitudes,
t ln n + t − t ln t +
−t ln t +
−t2
2n1/2
−t2
2n1/2
< 0
(1)
< t ln t
(2)
n1/2 >
n >
1
t
2 ln t
2
t
1
4 2 ln t
(3)
(4)
So taking n >
1
4
t
2 ln t
2
implies
R(4, t) >
n
1
>
2
8
t
2 ln t
2
2. Prove that every 3-uniform hypergraph with n vertices and m ≥ n/3
edges contains an independent set of size at least
2n3/2
√ √ .
3 3 m
Proof by Cliff Gaddy: Let S be a random set of vertices of a 3uniform hypergraph G, with P r(v ∈ S) = p. Let X = |S| and Y be
the number of edges in G|S . Let e = {i, j, k} be an edge of G and Ye
3
be the indicator
P function of the3 event that e ∈ G|S . So E[Ye ] = p ,
and E[Y ] = e∈G E[Ye ] = mp . Also E[X] = np. By linearity of
expectation, E[X − Y ] = np − mp3 . So if p is between 0 and 1, then
there exists a specific S in which X − Y ≥ np − mp3 . So for each
edge in G|S , we choose one vertex and remove it from S. Then we
are guaranteed to have an independent set, since a vertex has been
removed from each edge. Also we have removed Y vertices from S,
and so our new set has X − Y vertices ≥ np − mp3 . So there exists
an independent set of size at least np − mp3 . So now we choose p
to maximize np − mp3 . Taking the derivative with
respect to p and
√
n
2
setting this to 0, we get n − 3mp = 0 ⇒ p = √3m . This is less than
or equal √
to 1 since m ≥ n/3. So we have an independent set of size at
n
mn3/2
2n3/2
√
√
) − (3m)
least n( 3m
3/2 = 33/2 m .
3. Show that there is a finite n0 such that any directed graph on n ≥
1
n0 vertices in which each outdegree is at least log2 n − 10
log2 log2 n
contains an even simple directed cycle.
Proof by Danny Rorabaugh: Take digraph G on n vertices and m
1
edges where each vertex has outdegree of at least log2 n− 10
log2 log2 n.
Remove any edges necessary in order that every edge has outdegree
1
of exactly dlog2 n − 10
log2 log2 ne. Define the hypergraph H = HG
on vertex set V = V (G) with edgeset F = {N (u) | u ∈ V } where
2
N (u) = {v ∈ V | u → v} ∪ {u}. Thus H is an s-uniform hypergraph
with
1
n
s = log2 n −
+ 1.
log2 log2 n + 1 = log2
10
(log2 n)1/10
Suppose H has property B: there is a two-coloring of H such that
no edge is monochromatic. Then take such coloring and find an even
directed cycle as follows: pick any vertex v0 ∈ V ; for vi , let vi+1 be
any vertex in N (vi ) with a different color then vi ; since the graph is
finite, continue until you have a cycle, which must be even, since we
are alternating between two colors. Therefore, for G to have a directed
even cycle, is suffices for HG to have property B. We have that
r 1/2 r
m(r) ≥ c
2
ln r
for some universal constant c. Since H has n edges, it suffices to have
s 1/2 s
n≥c
2 .
ln s
For sufficiently large n,


!1/2 1
n
1/2
log2
+1
log2 n − 10 log2 log2 n + 1
s
(log2 n)1/10

c
2
2s
= c
1
ln s
ln log2 n − 10 log2 log2 n + 1


!1/2 1
n
log
log2 n − 10 log2 log2 n
2
(log2 n)1/10

2
≥ c
ln log2 n
!
1
log2 log2 n)1/2
(log2 n − 10
n
= c
(ln log2 n)1/2
(log2 n)1/10
!
( 12 log2 n)1/2
n
≥ c
(∗)
((log2 n)1/5 )1/2 (log2 n)1/10
!
r
1
=
c(log2 n)3/10 n
2
≥ n,
where the 21 and 15 which appear in step (∗) work for large enough n
since log2 log2 n = o(log2 n) and ln log2 n = o((log2 n)1/5 ).
3
4. Show that there is a positive constant c such
P that the following holds.
For any n reals a1 , a2 , . . ., an satisfying ni=1 a2i = 1, if 1 , . . . , n ) is
a {−1, 1}- random vector obtained by choosing each i randomly and
independently with uniform distribution to be either −1 or 1, then
" n
#
X
Pr i ai ≤ 1 ≥ c.
i=1
Proof by Heather Smith: We break this problem into two cases.
Pn
2
Case 1: If all ai ≤ √12 then since
i=1 ai = 1, there must exist
P
P
1 ≤ j ≤ n such that 14 ≤ ji=1 a2i ≤ 34 . Therefore 14 ≤ ni=j+1 a2i ≤ 34 .
P
P
Let X = ji=1 i a2i and Y = ni=j+1 i a2i .
P
Observe E(X) = ji=1 E(i )a2i = 0 and likewise E(Y ) = 0. Also,



!2 
j
j
j
X
X
X
X
3
E(X 2 ) = E 
i ai  = E  (i ai )2 +
i k ai ak  =
a2i ≤ .
4
i=1
i=1
i=1
i6=k
Therefore, P
Varr(X) = E(X 2 ) − (E(X))2 = E(X 2 ) ≤
Varr(Y ) = ni=j+1 a2i ≤ 34 .
3
4
and similarly
By Chebyshev’s inequality,
Pr(|X| ≥ 1) ≤
3
4
⇒
Pr(|X| ≤ 1) ≥
1
4
1
3
⇒
Pr(|Y | ≤ 1) ≥
4
4
And since the distribution is symmetric,
Pr(|Y | ≥ 1) ≤
Pr(−1 ≤ X ≤ 0) = Pr(0 ≤ X ≤ 1) ≥
1
8
Pr(−1 ≤ Y ≤ 0) = Pr(0 ≤ Y ≤ 1) ≥
1
8
Now consider the following:
#
" n
X
i ai ≤ 1 = Pr(|X + Y | ≤ 1) ≥ Pr(−1 ≤ X ≤ 0 and 0 ≤ Y ≤ 1)
Pr i=1
= Pr(−1 ≤ X ≤ 0)Pr(0 ≤ Y ≤ 1) ≥
4
1
=c
64
P
Case 2: Suppose ai > √12 for some i. Say a1 > √12 . Since ni=1 a2i = 1,
P
a1 ≤ 1. Let X = ni=2 i ai and Y = 1 a1 . Observe E(X) = 0 and
Varr(X) = 1 − a21 . By Chebyshev’s inequality,
Pr(|X| ≥ 1) ≤ Varr(X) = 1 − a21 ≤
1
2
Therefore Pr(|X| ≤ 1) ≥ 21 and since the distribution is symmetric,
Pr(−1 ≤ X ≤ 0) ≥ 14 . For Y , since a1 is fixed, Pr(0 ≤ Y ≤ 1) = 12 .
Therefore
1
Pr(|X+Y | ≤ 1) ≥ Pr(−1 ≤ X ≤ 0 and 0 ≤ Y ≤ 1) = Pr(−1 ≤ X ≤ 0)Pr(0 ≤ Y ≤ 1) ≥ .
8
No matter the choice of a1 , . . . , an ,
" n
#
X
1
1 1
Pr ,
=
= c.
i ai ≤ 1 ≥ min
64 8
64
i=1
5. Let X be a random variable with expectation E(X) = 0 and variance
σ 2 . Prove that for all λ > 0,
Pr[X ≥ λ] ≤
σ2
.
σ 2 + λ2
Proof by Talor Short: Let λ > 0, then
Pr(X ≥ λ) = Pr(λX ≥ λ2 )
= Pr(λX + σ 2 ≥ λ2 + σ 2 ).
Now by Markov’s inequality, linearity of expectation and using E(X) =
0,
E(λX + σ 2 )
λ2 + σ 2
λE(X) + E(σ 2 )
=
λ2 + σ 2
2
σ
= 2
λ + σ2
Pr(λX + σ 2 ≥ λ2 + σ 2 ) ≤
as desired.
5
6. Prove that for every X of at least 4k 2 distinct residue classes modulo
a prime p, there is an integer a such that the set {ax mod p : x ∈ X}
intersects every interval in {0, 1, . . . , p − 1} of length at least p/k.
Proof by Travis Johnston: First, partition Zp into 2k intervals of
length p/2k. Let S1 , . . . , S2k be the intervals. Note that any interval
of length p/k strictly contains (at least) one of the intervals Si . It
suffices to find an a ∈ Zp such that aX intersects every Si . Note that
one might create intervals as: S1 = {0, 1, ..., p/2k −1}, etc in sequence.
One might just as well shift all the intervals (uniformly) by −b. Thus,
it suffices to show that we can find a, b ∈ Zp such that aX +b intersects
every Si –the b being the amount we shift the intervals. Choose a and
b uniformly and independently at random from Zp .
For x ∈ X and z ∈ Zp define
Ixz
(
1 if ax + b = z
=
0 otherwise
We want to investigate the random variable Xi = |Si ∩ aX + b|. Note
that P(Ixz = 1) = p1 since for any choice of b there is exactly one choice
of a satisfying ax + b = z when x 6= 0. If x = 0 then a can be anything,
but there is exactly one choice for b, namely b = z. It then follows
that
XX
E(Xi ) = E(
Ixz )
x∈X z∈Si
=
XX
E(Ixz )
x∈X z∈Si
=
XX1
p
x∈X z∈Si
= |X| · |Si | ·
=
|X|
2k
6
1
p
Now, we compute the variance:
Var(Xi ) = E(Xi2 ) − E(Xi )2

2 
XX
|X|2
= E 
Ixz   −
4k 2
x∈X z∈Si
=
X X X X
E(Ix1 z1 Ix2 z2 ) −
x1 ∈X x2 ∈X z1 ∈Si z2 ∈Si
=
XX
X
2
E(Ixz
)+
E(Ix1 z Ix2 z )
x1 6=x2 ∈X z∈Si
x∈X z∈Si
+
X
|X|2
4k 2
X
X
E(Ixz1 Ixz2 ) +
x∈X z1 6=z2 ∈Si
X
X
E(Ix1 z1 Ix2 z2 ) −
x1 6=x2 ∈X z1 6=z2 ∈Si
Note that
Ix1 z Ix2 z
(
1 if ax1 + b = z and ax2 + b = z
=
0 otherwise
If x1 6= x2 then this only occurs when a = 0 and b = z. So P(Ix1 z Ix2 z =
1) = p12 if x1 6= x2 .
Similarly,
Ixz1 Ixz2
(
1
=
0
if ax + b = z1 and ax + b = z2
otherwise
If z1 6= z2 then this never occurs. So P(Ixz1 Ixz2 = 1) = 0 if z1 6= z2 .
Finally,
Ix1 z1 Ix2 z2
(
1 if ax1 + b = z1 and ax2 + b = z2
=
0 otherwise
If x1 6= x2 and z1 6= z2 then there is a unique intersection point of the
two lines, i.e. a unique a and b satisfying the two equations. Hence
P(Ix1 z1 Ix2 z2 = 1) = p12 when x1 6= x2 and z1 6= z2 .
7
|X|2
4k 2
Continuing our computation of the variance we have:
X X
X
X 1
X X 1
|X|2
+
0
+
−
p2
p2
4k 2
x∈X z1 6=z2 ∈Si
x∈X z∈Si
x1 6=x2 ∈X z1 6=z2 ∈Si
x1 6=x2 z∈Si
|X|
|X|
|Si |
1
|X|2
|X|
1
=
2
· 2−
+2
|Si | · 2 + 2
2k
p
2
2
p
4k 2
2
|X|
1
|X|2
1
=
+ (|X|2 − |X|)|Si | · 2 + (|X|2 − |X|)(|Si |2 − |Si |) · 2 −
2k
p
p
4k 2
|X|
1
|X|2
=
+ (|X|2 − |X|)|Si |2 · 2 −
2k
p
4k 2
2
2
|X| |X| − |X| |X|
+
−
=
2k
4k 2
4k 2
|X| |X|
=
− 2
2k
4k
Var(Xi ) =
XX
2
E(Ixz
)+
We say that any given choice of a and b is bad if there is an interval
Si such that |Si ∩ aX + b| = Xi = 0. It follows that
P(a and b are bad) ≤
≤
2k
X
i=1
2k
X
i=1
= 2k
P(Xi = 0)
Var(Xi )
E(Xi )2
|X|
|X|
2k − 4k2
|X|2
4k2
!
4k 2
2k
−
|X| |X|
2k
<1
≤1−
|X|
=
Thus, there exists a choice of a and b such that every interval Si intersects aX +b. Since every interval of length p/k contains (properly) one
of the Si intervals, we have that every interval of length p/k intersects
aX.
8