Discrete Geometry and Extremal Graph Theory - IMSA

Discrete Geometry and Extremal Graph Theory
Tony Liu, Illinois Mathematics & Science Academy
under the direction of
Dr. László Babai, University of Chicago
Abstract
In 1946, Paul Erdős proposed the following problem: what is the maximum number of unit
distances among n points in the plane? Erdős established an upper bound of cn 3/2 and a lower
bound that grows slightly faster than n.
A graph is a collection of “vertices” and “edges” connecting some pairs of vertices. For example,
in an airline route chart, airports are the vertices and direct connections correspond to
the edges. We examine graphs without a 4-cycle, and show the maximum number of edges
to be at most cn 3/2 . Replacing the 4-cycle by the “Θ-graph,” we obtain an upper bound of
c ′ n 3/2 on the number of edges. Finally, we use this seemingly unrelated graph theoretic result
to establish Erdős’ upper bound on the number of unit distances.
Although we present the optimal solution, within a constant factor, to the graph theory
problem, Erdős’ question on unit distances remains wide open after sixty years.
A cycle (also known as a circuit) is a subset of the edge set of a graph G that forms a
path such that the first node of the path corresponds to the last. A 4-cycle is merely a cycle
with 4 vertices, as shown below.
1

We begin by proving the following.
Theorem 1. A graph G on n vertices without a 4-cycle contains at most cn 3/2 edges for
some constant c > 0.
Proof of Theorem 1. Let d 1 , d 2 , . . . , d n be the degrees of the vertices v 1 , v 2 , . . . , v n of G.
Let E be the number of edges of G, so 2E = ∑ n
i=1 d i. For a pair of vertices v 1 and v 2 , note
that there can be at most one other vertex v 3 connected to both v 1 and v 2 . Otherwise, for
if v 4 was also joined to v 1 and v 2 , we would have a 4-cycle with vertices v 1 , v 3 , v 2 , v 4 . Thus,
more generally, for a vertex v i connected to d i other vertices, we count every pair of these d i
vertices. If we repeat this count for all the vertices v 1 , v 2 , . . . , v n , note that our total count
must less than the total number of pairs of vertices, ( n
2)
. This is because every pair of some
d i vertices connected to v i can be uniquely associated with v i . In other words, we have the
following inequality.
( )
d1
+
2
(
d2
) ( ) (
dn n
+ · · · ≤ .
2 2 2)
We claim that ( d 1
) (
2 +
d2
) (
2 + · · ·
dn
)
2 ≥
2E 2
− E, so it will follow that
n
( )
2E 2 n
n − E ≤ n(n − 1)
=
2 2
⇒ 4E 2 − 2En ≤ n 2 (n − 1).
Thus, ( )
2E − n 2
2 ≤ n 2 (n − 1) + n2 ≤ 4 n3 , so we have 2E ≤ n 3/2 + n . Finally, we conclude
2
that E ∼ < n 3/2 (and actually, c ≈ 1 will suffice for sufficiently large n).
2
Now, we give two proofs of the inequality ( d 1
) (
2 +
d2
) (
2 + · · · +
dn
)
2 ≥
2E 2
− E. n
• First Proof. By the Quadratic-Arithmetic Mean Inequality (or Cauchy-Schwartz), we
have
( )
d1
+
2
( )
d2
+ · · · +
2
( )
dn
= 1 2 2 (d2 1 + d 2 2 + · · · + d 2 n) − 1 2 (d 1 + d 2 + · · · + d n )
≥ 1
2n (2E)2 − E.
• Second Proof. As a function, f(x) = ( )
x
2 =
x(x−1)
is convex, so by Jensen’s inequality,
2
we have ( ) ( ) ( ) 2E
)
d1 d2
dn
+ + · · · + ≥ n(
n
= 2E2
2 2
2 2 n − E.
This concludes our proof.
The “Θ-graph” is the union of three internally disjoint (simple) paths that have the same
two end vertices, as shown below.
2

We prove the following result using a similar technique.
Theorem 2. A graph G on n vertices without a “Θ-graph” contains at most c ′ n 3/2 edges
for some constant c ′ > 0.
Proof of Theorem 2 (sketch). As in the proof of Theorem 1, let d 1 , d 2 , . . . , d n be the degrees
of the vertices v 1 , v 2 , . . . , v n . For a vertex v i , we count the pairs of the d i vertices
connected to v i . As we continue this count for all the vertices of G, note that each pair is
counted at most twice, or else G must have a “Θ-graph.” In other words, we arrive at the
inequality
( ) ( ) ( ) (
Create PDF with GO2PDF for free, if you d1 wish to d2 remove this line, dn click here n to buy Virtual PDF Printer
2)
2
+
2
+ · · ·
Thus, as in the proof of Theorem 1, we see c ′ ≈ √ 2c will work for sufficiently large n.
Finally, we prove our main result, showing an unexpected connection between Erdős’ problem
on unit distances and graph theoretic results.
Theorem 3. Let f(n) be the maximum number of unit distances among n points in the
plane. Then, f(n) ≤ c ′ n 3/2 .
Proof of Theorem 3. Consider the unit distance graph of the n points defined as follows: two
points are joined by a segment if and only if they are unit distance apart. Now, note that
no “Θ-graph” can appear in such a graph, for let us consider the two end vertices. The two
unit circles centered at these vertices must each pass through three common points, but two
circles can intersect in at most two points, contradiction. Thus, by Theorem 2, the result
follows.
2
≤ 2
.
3

Other topics I have explored with my mentor include the following problems.
1. Let F n be the Fibonacci sequence defined by F 1 = F 2 = 1 and F n+2 = F n+1 + F n for
n ≥ 1. Show that gcd(F m , F n ) = F gcd(m,n) .
Proof. First we note the following (which are easily proven by induction)
• gcd(F n+1 , F n ) = 1
• F m+n = F m+1 F n + F m F n−1 = F m F n+1 + F m−1 F n
• n | m ⇒ F n | F m
For instance, the first two can be done with straightforward induction while the third
follows from induction and setting m = (k − 1)n, where m = kn. Now, without loss of
generality, m ≥ n and let m = nq + r where 0 ≤ r < n. We have
gcd(F m , F n ) = gcd(F nq+r , F n ) = gcd(F nq+1 F r + F nq F r+1 , F n ) = gcd(F r , F n )
because F n | F nq and gcd(F nq+1 , F nq ) = 1. Now, we can write n = rq 1 + r 1 and
continue this procedure by writing r n = r n+1 q n+2 + r n+2 where r = r 0 , n = r −1 , and
0 ≤ r n+2 < r n+1 . Eventually, we will end up at r k = gcd(F m , F n ) and r k+1 = 0 for
some k, at which point we have
gcd(F m , F n ) = gcd(F rk , F rk+1 ) = gcd(F gcd(m,n) , 0) = F gcd(m,n)
2. Show that ∏ p≤x p ≤ 4x where p is a prime.
Proof. The base cases x ≤ 3 are easily verified. We may assume that x > 3 is
an integer, as we will prove the result by induction. Note that for odd x, the even
integer x + 1 clearly cannot be prime so we have
∏
p ≤ 4 x < 4 x+1
p≤x+1
p = ∏ p≤x
so the result holds for x + 1 too. Now, for odd x = 2k + 1 where k > 1, note that
primes k + 2 ≤ p ≤ 2k + 1 divide ( )
2k+1
k =
(2k+1)!
as they divide the numerator but
k!(k+1)!
not the denominator. Thus,
∏
( ) 2k + 1
p ≤ = 1 (( ) ( ))
2k + 1 2k + 1
+
< 1 k 2 k k + 1 2 (1 + 1)2k+1 = 4 k
k+2≤p≤2k+1
By the induction hypothesis, we have
∏
and multiplying these yields
∏
p≤k+1
p
p≤k+1
∏
k+2≤p≤2k+1
p ≤ 4 k+1
p =
∏
p≤2k+1
p ≤ 4 2k+1
4

3. Let R m (n) be the number of subsets of an n element set, with cardinality divisible by
m. In other words, let
⌊n/m⌋
∑
( n
R m (n) =
.
mk)
Show that R m (n) = 1 m
k=0
∑ m−1
k=0 (1 + zk ) n where z is a primitive m-th root of unity.
Proof. First, we note that 0 = z m − 1 = (z − 1)(z m−1 + z m−2 + · · · + 1) and z ≠ 1 so
z m−1 + z m−2 + · · · + 1 = 0. Directly expanding, we have
m−1
1 ∑
m
k=0
(1 + z k ) n = 1 m
= 1 m
m−1
∑
k=0
n∑
j=0
n∑
j=0
( n
j)
z kj
( m−1
n ∑
z
j) kj
k=0
Now, note that ∑ m−1
k=0 zkj equals m if m | j and 0 otherwise. Indeed, if m | j, then
z j = 1 and the sum is clearly m. Otherwise, let g = gcd(m, j) and l = m g
> 1 so
m | lj. Then, z lj = 1 so z j is a primitive l-th root of unity. In particular, z lj − 1 = 0
and z j ≠ 1 imply z j(l−1) + z j(l−2) + · · · + 1 = 0 and we have
Thus, our sum becomes
as claimed.
R m (n) = 1 m
4. Show that |R 3 (n) − 2n 3 | ≤ 2 3 .
m−1
∑
k=0
∑l−1
z kj = gz kj = 0
∑
0≤j≤n,m|j
k=0
( n
m =
j)
⌊n/m⌋
∑
k=0
( n
mk)
Proof. Note that by the previous problem, we have R 3 (n) = 1 3 (2n +(1+ω) n +(1+ω 2 ) n )
where ω 3 = 1 and ω ≠ 1. Thus, it suffices to prove that |(1 + ω) n + (1 + ω 2 ) n | ≤ 2.
However, this follows immediately from the triangle inequality, as |1 + ω| = | − ω 2 | = 1
and |1 + ω 2 | = | − ω| = 1, so
|(1 + ω) n + (1 + ω 2 )| ≤ |(1 + ω) n | + |(1 + ω 2 ) n | = |(1 + ω)| n + |(1 + ω 2 )| n = 2.
Comments. This can be generalized to a (much weaker) bound
∣ R m(n) − 2n
m ∣ ≤ (2 cos π m )n for m ≥ 4
5

and the proof is pretty much analogous to the one above. By the formula in the previous
problem, it suffices to prove | ∑ m−1
k=1 (1 + zk ) n | ≤ (2 cos π m )n where z is a primitive
m-th root of unity. Note that all the points 1 + z k must lie within the circle centered
at the origin with radius 2 cos π . Thus, |(1 + m zk ) n | = |1 + z k | n ≤ 2 cos π m )n , and the
result follows as before from the triangle inequality.
5. A graph G is bipartite if and only if it contains no odd cycle.
Proof: If G = G 1 ∪ G 2 is bipartite, where G 1 and G 2 are disjoint subgraphs of G,
then it clearly contains no odd cycle. This is because two vertices are neighbors if and
only if they lie in different subsets, so a path must alternate between vertices in G 1
and G 2 . If G has no odd cycles, let us pick a vertex v and color it red. We color all
neighbors of v blue, and all uncolored neighbors of these blue vertices red, and so on.
We might as well assume that G is connected, so all vertices have a color. We claim
that setting G 1 and G 2 to contain all red and blue vertices, respectively, shows that
G is bipartite. Assume otherwise, and say that v 1 , v 2 are neighbors of the same color.
This means that there exist paths from v to v 1 and v 2 with lengths of the same parity.
In other words, the path from v 1 to v 2 through v must have an even length. (If the
paths from v 1 to v and v 2 to v first meet at v 3 , then we may replace v with v 3 and the
proof continues.) Now, this path from v 1 to v 2 with the edge joining v 1 and v 2 creates
an odd cycle, contradiction. Thus, G is bipartite if and only if it contains no odd cycle.
6. Let G be a graph on n vertices without K k , a complete graph on k vertices. Then, G
(k − 2)n2
has at most edges. Moreover, there exists a graph with this number of edges.
2(k − 1)
Proof: I do admit to having seen this before, although I only vaguely remembered
the ideas behind this proof of Turán’s Theorem.
Let E(G) denote the number of edges of G. The proof proceeds by induction on
n. Our base cases, for instance n = 1, 2, 3 are trivially true. Now, let G be a graph
on n vertices without K k containing the maximal number of edges. Note that G must
contain a K k−1 or else we could add edges, contradiction the maximality of E(G). Let
H 1 be a K k−1 and let H 2 be G without H 1 . Note that E(H 1 ) = ( )
k−1
2 . By the induction
hypothesis, we also have
E(H 2 ) ≤
(k − 2)(n − k + 1)2
.
2(k − 1)
If v ∈ H 2 , then v is neighbors with at most k − 2 vertices in H 1 or else we have a K k .
This implies that the number of edges with one vertex in each H 1 and H 2 is at most
(n − k + 1)(k − 2), so
( ) k − 1
E(G) ≤ +
2
(k − 2)(n − k + 1)2
2(k − 1)
+ (n − k + 1)(k − 2) =
(k − 2)n2
2(k − 1) .
Now, let n = (k −1)q +r where 0 ≤ r < k −1. Consider a partition G = G 1 ∪· · ·∪G k−1
where r of the subgraphs have q + 1 vertices and the rest have q vertices. We join two
6

vertices if and only if they lie in different G i . After a bit of computation, this shows
that the bound above is essentially sharp.
7. We prove a few results regarding even and odd permutations.
Lemma 1: Every permutation σ ∈ S n can be written as a product of transpositions.
Proof: Write σ = c 1 c 2 · · · c m where the c k are (possibly trivial) disjoint cycles. It
suffices to show that any cycle can be written as a product of transpositions. Let
c = (i 1 i 2 · · · i k ) be a cycle. We show that c can be written as a product of transpositions
by induction on k. For k = 1, 2, the result is trivial. Now, note that
c = (i 1 i 2 · · · i k ) = (i 1 i 2 · · · i k−1 )(i k−1 i k ) and (i 1 i 2 · · · i k−1 ) can be written as a product
of transpositions by the induction hypothesis. The result follows.
Lemma 2: Let τ 1 , τ 2 , . . . , τ m ∈ S n be transpositions. Then, the number of cycles
in τ 1 τ 2 · · · τ m has the same parity as m + n.
Proof: We first prove the following: if σ ∈ S n is a permutation and τ ∈ S n is a
transposition, then the number of disjoint cycles in σ and στ have different parity.
As before, write σ = c 1 c 2 · · · c m where the c k are (possibly trivial) disjoint cycles and
suppose τ = (ij). First we deal with the case where i and j are in the same cycle, say
c t . If σ(i) = j (or vice versa), then c t = (ij) or (iji 1 · · · i s ). If c t = (ij), then we may
replace c t with (i)(j) in στ. If c t = (iji 1 · · · i s ), then we replace c t with (ii 1 · · · i s )(j) In
either case, the parity of the number of disjoint cycles changes.
If i and j are in different cycles, say c t and c u respectively, we argue as follows. First,
if c t = (i) and c u = (j), then στ is the product of all cycles c i and (ij) except for c t
and c u . If c t = (i) and c u = (jj 1 · · · j v )) (or vice versa), then in στ, replace c u with
(jij 1 · · · j v ) and remove (ij). If c t = (ii 1 · · · i s ) and c u = (jj 1 · · · j v ), then στ can be
rewriten with c ′ t,u = (ij 1 · · · j v ji 1 · · · i s ) and (ij), c t , c u removed. Again, in either case,
the parity changes.
Now we prove Lemma 2 by induction on m, and our base cases are easily verified.
We have τ 1 τ 2 · · · τ m+1 = στ m+1 , where we set σ = τ 1 τ 2 · · · τ m . The number of disjoint
cycles in σ and στ m+1 have different parity, so this completes our induction.
Comment: It immediately follows that a permutation σ can be written as a product
of an even number of transpositions or an odd number of transpositions, but not
both.
7

References
[1] I. Anderson, Combinatorics of Finite Sets, Oxford University Press, Oxford, England,
1989.
[2] G. E. Andrews, Number Theory, W. B. Saunders Company, Philadelphia, 1971.
[3] D. Reinhard, Graph Theory, Springer-Verlag, Heidelberg, July 2005.
[4] L. Lovasz, Combinatorial Problems and Exercises, Akademiai Kiado, Budapest, Hungary,
1993.
8