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 cn3/2 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 cn3/2 . Replacing the 4-cycle by the “Θ-graph,” we obtain an upper bound of
c0 n3/2 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 cn3/2 edges for
some constant c > 0.
Proof of Theorem 1. Let d1 , d2 , . . . , dn be the
P degrees of the vertices v1 , v2 , . . . , vn of G.
Let E be the number of edges of G, so 2E = ni=1 di . For a pair of vertices v1 and v2 , note
that there can be at most one other vertex v3 connected to both v1 and v2 . Otherwise, for
if v4 was also joined to v1 and v2 , we would have a 4-cycle with vertices v1 , v3 , v2 , v4 . Thus,
more generally, for a vertex vi connected to di other vertices, we count every pair of these di
vertices. If we repeat this count for all the vertices v1 ,v2 , . . . , vn , note that our total count
must less than the total number of pairs of vertices, n2 . This is because every pair of some
di vertices connected to vi can be uniquely associated with vi . In other words, we have the
following inequality.
d1
d2
dn
n
+
+ ···
≤
.
2
2
2
2
2
We claim that d21 + d22 + · · · d2n ≥ 2En − E, so it will follow that
2E 2
n
n(n − 1)
−E ≤
=
2
n
2
⇒ 4E 2 − 2En ≤ n2 (n − 1).
2
2
Thus, 2E − n2 ≤ n2 (n − 1) + n4 ≤ n3 , so we have 2E ≤ n3/2 + n2 . Finally, we conclude
<
that E ∼ n3/2 (and actually, c ≈ 12 will suffice for sufficiently large n).
Now, we give two proofs of the inequality
d1
2
+
d2
2
+ ··· +
dn
2
≥
2E 2
n
− E.
• First Proof. By the Quadratic-Arithmetic Mean Inequality (or Cauchy-Schwartz), we
have
1
d1
d2
dn
1
+
+ ··· +
= (d21 + d22 + · · · + d2n ) − (d1 + d2 + · · · + dn )
2
2
2
2
2
1
(2E)2 − E.
2n
• Second Proof. As a function, f (x) = x2 = x(x−1)
is convex, so by Jensen’s inequality,
2
we have
2E d1
d2
dn
2E 2
+
+ ··· +
− E.
≥n n =
2
2
2
2
n
≥
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 c0 n3/2 edges
for some constant c0 > 0.
Proof of Theorem 2 (sketch). As in the proof of Theorem 1, let d1 , d2 , . . . , dn be the degrees of the vertices v1 , v2 , . . . , vn . For a vertex vi , we count the pairs of the di vertices
connected to vi . 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
d1 wish todremove
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2
Create PDF with GO2PDF for free, if you
this line,
+
+ ···
≤2
.
2
2
2
2
√
Thus, as in the proof of Theorem 1, we see c0 ≈ 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) ≤ c0 n3/2 .
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. 3
Other topics I have explored with my mentor include the following problems.
1. Let Fn be the Fibonacci sequence defined by F1 = F2 = 1 and Fn+2 = Fn+1 + Fn for
n ≥ 1. Show that gcd(Fm , Fn ) = Fgcd(m,n) .
Proof. First we note the following (which are easily proven by induction)
• gcd(Fn+1 , Fn ) = 1
• Fm+n = Fm+1 Fn + Fm Fn−1 = Fm Fn+1 + Fm−1 Fn
• n | m ⇒ Fn | Fm
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(Fm , Fn ) = gcd(Fnq+r , Fn ) = gcd(Fnq+1 Fr + Fnq Fr+1 , Fn ) = gcd(Fr , Fn )
because Fn | Fnq and gcd(Fnq+1 , Fnq ) = 1. Now, we can write n = rq1 + r1 and
continue this procedure by writing rn = rn+1 qn+2 + rn+2 where r = r0 , n = r−1 , and
0 ≤ rn+2 < rn+1 . Eventually, we will end up at rk = gcd(Fm , Fn ) and rk+1 = 0 for
some k, at which point we have
gcd(Fm , Fn ) = gcd(Frk , Frk+1 ) = gcd(Fgcd(m,n) , 0) = Fgcd(m,n)
Q
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
Y
Y
p=
p ≤ 4x < 4x+1
p≤x+1
p≤x
so the result holds for x + 1 too. Now,for odd x = 2k + 1 where k > 1, note that
(2k+1)!
as they divide the numerator but
primes k + 2 ≤ p ≤ 2k + 1 divide 2k+1
= k!(k+1)!
k
not the denominator. Thus,
Y
2k + 1
1
2k + 1
2k + 1
1
p≤
=
+
< (1 + 1)2k+1 = 4k
k
2
k
k+1
2
k+2≤p≤2k+1
By the induction hypothesis, we have
Y
p ≤ 4k+1
p≤k+1
and multiplying these yields
Y
p≤k+1
p
Y
p=
k+2≤p≤2k+1
4
Y
p≤2k+1
p ≤ 42k+1
3. Let Rm (n) be the number of subsets of an n element set, with cardinality divisible by
m. In other words, let
bn/mc X
n
.
Rm (n) =
mk
k=0
Show that Rm (n) =
1
m
Pm−1
k=0
(1 + z k )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 6= 1 so
z m−1 + z m−2 + · · · + 1 = 0. Directly expanding, we have
m−1
m−1 n 1 X
1 X X n kj
k n
(1 + z ) =
z
m k=0
m k=0 j=0 j
n m−1
1 X n X kj
=
z
m j=0 j k=0
P
kj
equals m if m | j and 0 otherwise. Indeed, if m | j, then
Now, note that m−1
k=0 z
j
z = 1 and the sum is clearly m. Otherwise, let g = gcd(m, j) and ` = mg > 1 so
m | `j. Then, z `j = 1 so z j is a primitive `-th root of unity. In particular, z `j − 1 = 0
and z j 6= 1 imply z j(`−1) + z j(`−2) + · · · + 1 = 0 and we have
m−1
X
z
kj
=
`−1
X
gz kj = 0
k=0
k=0
Thus, our sum becomes
1
Rm (n) =
m
X
0≤j≤n,m|j
bn/mc
X n n
m
=
j
mk
k=0
as claimed.
4. Show that |R3 (n) −
2n
|
3
≤ 23 .
Proof. Note that by the previous problem, we have R3 (n) = 31 (2n +(1+ω)n +(1+ω 2 )n )
where ω 3 = 1 and ω 6= 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
n
2
Rm (n) − ≤ (2 cos π )n for m ≥ 4
m
m
5
and the proof is pretty much analogous
above. By the formula in the previPm−1to the kone
π n
n
) where z is a primitive
ous problem, it suffices to prove | k=1 (1 + z ) | ≤ (2 cos m
k
m-th root of unity. Note that all the points 1 + z must lie within the circle centered
π
π n
at the origin with radius 2 cos m
. Thus, |(1 + z k )n | = |1 + z k |n ≤ 2 cos m
) , 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 = G1 ∪ G2 is bipartite, where G1 and G2 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 G1
and G2 . 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 G1 and G2 to contain all red and blue vertices, respectively, shows that
G is bipartite. Assume otherwise, and say that v1 , v2 are neighbors of the same color.
This means that there exist paths from v to v1 and v2 with lengths of the same parity.
In other words, the path from v1 to v2 through v must have an even length. (If the
paths from v1 to v and v2 to v first meet at v3 , then we may replace v with v3 and the
proof continues.) Now, this path from v1 to v2 with the edge joining v1 and v2 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 Kk , a complete graph on k vertices. Then, G
(k − 2)n2
edges. Moreover, there exists a graph with this number of edges.
has at most
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 Kk containing the maximal number of edges. Note that G must
contain a Kk−1 or else we could add edges, contradiction the maximality
of E(G). Let
k−1
H1 be a Kk−1 and let H2 be G without H1 . Note that E(H1 ) = 2 . By the induction
hypothesis, we also have
E(H2 ) ≤
(k − 2)(n − k + 1)2
.
2(k − 1)
If v ∈ H2 , then v is neighbors with at most k − 2 vertices in H1 or else we have a Kk .
This implies that the number of edges with one vertex in each H1 and H2 is at most
(n − k + 1)(k − 2), so
k−1
(k − 2)(n − k + 1)2
(k − 2)n2
E(G) ≤
+
+ (n − k + 1)(k − 2) =
.
2
2(k − 1)
2(k − 1)
Now, let n = (k −1)q +r where 0 ≤ r < k −1. Consider a partition G = G1 ∪· · ·∪Gk−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 Gi . 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 ∈ Sn can be written as a product of transpositions.
Proof: Write σ = c1 c2 · · · cm where the ck are (possibly trivial) disjoint cycles. It
suffices to show that any cycle can be written as a product of transpositions. Let
c = (i1 i2 · · · ik ) 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 = (i1 i2 · · · ik ) = (i1 i2 · · · ik−1 )(ik−1 ik ) and (i1 i2 · · · ik−1 ) can be written as a product of transpositions by the induction hypothesis. The result follows.
Lemma 2: Let τ1 , τ2 , . . . , τm ∈ Sn be transpositions. Then, the number of cycles
in τ1 τ2 · · · τm has the same parity as m + n.
Proof: We first prove the following: if σ ∈ Sn is a permutation and τ ∈ Sn is a
transposition, then the number of disjoint cycles in σ and στ have different parity.
As before, write σ = c1 c2 · · · cm where the ck 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
ct . If σ(i) = j (or vice versa), then ct = (ij) or (iji1 · · · is ). If ct = (ij), then we may
replace ct with (i)(j) in στ . If ct = (iji1 · · · is ), then we replace ct with (ii1 · · · is )(j) In
either case, the parity of the number of disjoint cycles changes.
If i and j are in different cycles, say ct and cu respectively, we argue as follows. First,
if ct = (i) and cu = (j), then στ is the product of all cycles ci and (ij) except for ct
and cu . If ct = (i) and cu = (jj1 · · · jv )) (or vice versa), then in στ , replace cu with
(jij1 · · · jv ) and remove (ij). If ct = (ii1 · · · is ) and cu = (jj1 · · · jv ), then στ can be
rewriten with c0t,u = (ij1 · · · jv ji1 · · · is ) and (ij), ct , cu 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 τ1 τ2 · · · τm+1 = στm+1 , where we set σ = τ1 τ2 · · · τm . The number of disjoint
cycles in σ and στm+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.
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