Discrete Geometry and Extremal Graph Theory - IMSA
Discrete Geometry and Extremal Graph Theory - IMSA
Discrete Geometry and Extremal Graph Theory - IMSA
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<strong>Discrete</strong> <strong>Geometry</strong> <strong>and</strong> <strong>Extremal</strong> <strong>Graph</strong> <strong>Theory</strong><br />
Tony Liu, Illinois Mathematics & Science Academy<br />
under the direction of<br />
Dr. László Babai, University of Chicago<br />
Abstract<br />
In 1946, Paul Erdős proposed the following problem: what is the maximum number of unit<br />
distances among n points in the plane? Erdős established an upper bound of cn 3/2 <strong>and</strong> a lower<br />
bound that grows slightly faster than n.<br />
A graph is a collection of “vertices” <strong>and</strong> “edges” connecting some pairs of vertices. For example,<br />
in an airline route chart, airports are the vertices <strong>and</strong> direct connections correspond to<br />
the edges. We examine graphs without a 4-cycle, <strong>and</strong> show the maximum number of edges<br />
to be at most cn 3/2 . Replacing the 4-cycle by the “Θ-graph,” we obtain an upper bound of<br />
c ′ n 3/2 on the number of edges. Finally, we use this seemingly unrelated graph theoretic result<br />
to establish Erdős’ upper bound on the number of unit distances.<br />
Although we present the optimal solution, within a constant factor, to the graph theory<br />
problem, Erdős’ question on unit distances remains wide open after sixty years.<br />
A cycle (also known as a circuit) is a subset of the edge set of a graph G that forms a<br />
path such that the first node of the path corresponds to the last. A 4-cycle is merely a cycle<br />
with 4 vertices, as shown below.<br />
1
We begin by proving the following.<br />
Theorem 1. A graph G on n vertices without a 4-cycle contains at most cn 3/2 edges for<br />
some constant c > 0.<br />
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.<br />
Let E be the number of edges of G, so 2E = ∑ n<br />
i=1 d i. For a pair of vertices v 1 <strong>and</strong> v 2 , note<br />
that there can be at most one other vertex v 3 connected to both v 1 <strong>and</strong> v 2 . Otherwise, for<br />
if v 4 was also joined to v 1 <strong>and</strong> v 2 , we would have a 4-cycle with vertices v 1 , v 3 , v 2 , v 4 . Thus,<br />
more generally, for a vertex v i connected to d i other vertices, we count every pair of these d i<br />
vertices. If we repeat this count for all the vertices v 1 , v 2 , . . . , v n , note that our total count<br />
must less than the total number of pairs of vertices, ( n<br />
2)<br />
. This is because every pair of some<br />
d i vertices connected to v i can be uniquely associated with v i . In other words, we have the<br />
following inequality.<br />
( )<br />
d1<br />
+<br />
2<br />
(<br />
d2<br />
) ( ) (<br />
dn n<br />
+ · · · ≤ .<br />
2 2 2)<br />
We claim that ( d 1<br />
) (<br />
2 +<br />
d2<br />
) (<br />
2 + · · ·<br />
dn<br />
)<br />
2 ≥<br />
2E 2<br />
− E, so it will follow that<br />
n<br />
( )<br />
2E 2 n<br />
n − E ≤ n(n − 1)<br />
=<br />
2 2<br />
⇒ 4E 2 − 2En ≤ n 2 (n − 1).<br />
Thus, ( )<br />
2E − n 2<br />
2 ≤ n 2 (n − 1) + n2 ≤ 4 n3 , so we have 2E ≤ n 3/2 + n . Finally, we conclude<br />
2<br />
that E ∼ < n 3/2 (<strong>and</strong> actually, c ≈ 1 will suffice for sufficiently large n).<br />
2<br />
Now, we give two proofs of the inequality ( d 1<br />
) (<br />
2 +<br />
d2<br />
) (<br />
2 + · · · +<br />
dn<br />
)<br />
2 ≥<br />
2E 2<br />
− E. n<br />
• First Proof. By the Quadratic-Arithmetic Mean Inequality (or Cauchy-Schwartz), we<br />
have<br />
( )<br />
d1<br />
+<br />
2<br />
( )<br />
d2<br />
+ · · · +<br />
2<br />
( )<br />
dn<br />
= 1 2 2 (d2 1 + d 2 2 + · · · + d 2 n) − 1 2 (d 1 + d 2 + · · · + d n )<br />
≥ 1<br />
2n (2E)2 − E.<br />
• Second Proof. As a function, f(x) = ( )<br />
x<br />
2 =<br />
x(x−1)<br />
is convex, so by Jensen’s inequality,<br />
2<br />
we have ( ) ( ) ( ) 2E<br />
)<br />
d1 d2<br />
dn<br />
+ + · · · + ≥ n(<br />
n<br />
= 2E2<br />
2 2<br />
2 2 n − E.<br />
This concludes our proof. <br />
The “Θ-graph” is the union of three internally disjoint (simple) paths that have the same<br />
two end vertices, as shown below.<br />
2
We prove the following result using a similar technique.<br />
Theorem 2. A graph G on n vertices without a “Θ-graph” contains at most c ′ n 3/2 edges<br />
for some constant c ′ > 0.<br />
Proof of Theorem 2 (sketch). As in the proof of Theorem 1, let d 1 , d 2 , . . . , d n be the degrees<br />
of the vertices v 1 , v 2 , . . . , v n . For a vertex v i , we count the pairs of the d i vertices<br />
connected to v i . As we continue this count for all the vertices of G, note that each pair is<br />
counted at most twice, or else G must have a “Θ-graph.” In other words, we arrive at the<br />
inequality<br />
( ) ( ) ( ) (<br />
Create PDF with GO2PDF for free, if you d1 wish to d2 remove this line, dn click here n to buy Virtual PDF Printer<br />
2)<br />
2<br />
+<br />
2<br />
+ · · ·<br />
Thus, as in the proof of Theorem 1, we see c ′ ≈ √ 2c will work for sufficiently large n. <br />
Finally, we prove our main result, showing an unexpected connection between Erdős’ problem<br />
on unit distances <strong>and</strong> graph theoretic results.<br />
Theorem 3. Let f(n) be the maximum number of unit distances among n points in the<br />
plane. Then, f(n) ≤ c ′ n 3/2 .<br />
Proof of Theorem 3. Consider the unit distance graph of the n points defined as follows: two<br />
points are joined by a segment if <strong>and</strong> only if they are unit distance apart. Now, note that<br />
no “Θ-graph” can appear in such a graph, for let us consider the two end vertices. The two<br />
unit circles centered at these vertices must each pass through three common points, but two<br />
circles can intersect in at most two points, contradiction. Thus, by Theorem 2, the result<br />
follows. <br />
2<br />
≤ 2<br />
.<br />
3
Other topics I have explored with my mentor include the following problems.<br />
1. Let F n be the Fibonacci sequence defined by F 1 = F 2 = 1 <strong>and</strong> F n+2 = F n+1 + F n for<br />
n ≥ 1. Show that gcd(F m , F n ) = F gcd(m,n) .<br />
Proof. First we note the following (which are easily proven by induction)<br />
• gcd(F n+1 , F n ) = 1<br />
• F m+n = F m+1 F n + F m F n−1 = F m F n+1 + F m−1 F n<br />
• n | m ⇒ F n | F m<br />
For instance, the first two can be done with straightforward induction while the third<br />
follows from induction <strong>and</strong> setting m = (k − 1)n, where m = kn. Now, without loss of<br />
generality, m ≥ n <strong>and</strong> let m = nq + r where 0 ≤ r < n. We have<br />
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 )<br />
because F n | F nq <strong>and</strong> gcd(F nq+1 , F nq ) = 1. Now, we can write n = rq 1 + r 1 <strong>and</strong><br />
continue this procedure by writing r n = r n+1 q n+2 + r n+2 where r = r 0 , n = r −1 , <strong>and</strong><br />
0 ≤ r n+2 < r n+1 . Eventually, we will end up at r k = gcd(F m , F n ) <strong>and</strong> r k+1 = 0 for<br />
some k, at which point we have<br />
gcd(F m , F n ) = gcd(F rk , F rk+1 ) = gcd(F gcd(m,n) , 0) = F gcd(m,n)<br />
2. Show that ∏ p≤x p ≤ 4x where p is a prime.<br />
Proof. The base cases x ≤ 3 are easily verified. We may assume that x > 3 is<br />
an integer, as we will prove the result by induction. Note that for odd x, the even<br />
integer x + 1 clearly cannot be prime so we have<br />
∏<br />
p ≤ 4 x < 4 x+1<br />
p≤x+1<br />
p = ∏ p≤x<br />
so the result holds for x + 1 too. Now, for odd x = 2k + 1 where k > 1, note that<br />
primes k + 2 ≤ p ≤ 2k + 1 divide ( )<br />
2k+1<br />
k =<br />
(2k+1)!<br />
as they divide the numerator but<br />
k!(k+1)!<br />
not the denominator. Thus,<br />
∏<br />
( ) 2k + 1<br />
p ≤ = 1 (( ) ( ))<br />
2k + 1 2k + 1<br />
+<br />
< 1 k 2 k k + 1 2 (1 + 1)2k+1 = 4 k<br />
k+2≤p≤2k+1<br />
By the induction hypothesis, we have<br />
∏<br />
<strong>and</strong> multiplying these yields<br />
∏<br />
p≤k+1<br />
p<br />
p≤k+1<br />
∏<br />
k+2≤p≤2k+1<br />
p ≤ 4 k+1<br />
p =<br />
∏<br />
p≤2k+1<br />
p ≤ 4 2k+1<br />
4
3. Let R m (n) be the number of subsets of an n element set, with cardinality divisible by<br />
m. In other words, let<br />
⌊n/m⌋<br />
∑<br />
( n<br />
R m (n) =<br />
.<br />
mk)<br />
Show that R m (n) = 1 m<br />
k=0<br />
∑ m−1<br />
k=0 (1 + zk ) n where z is a primitive m-th root of unity.<br />
Proof. First, we note that 0 = z m − 1 = (z − 1)(z m−1 + z m−2 + · · · + 1) <strong>and</strong> z ≠ 1 so<br />
z m−1 + z m−2 + · · · + 1 = 0. Directly exp<strong>and</strong>ing, we have<br />
m−1<br />
1 ∑<br />
m<br />
k=0<br />
(1 + z k ) n = 1 m<br />
= 1 m<br />
m−1<br />
∑<br />
k=0<br />
n∑<br />
j=0<br />
n∑<br />
j=0<br />
( n<br />
j)<br />
z kj<br />
( m−1<br />
n ∑<br />
z<br />
j) kj<br />
k=0<br />
Now, note that ∑ m−1<br />
k=0 zkj equals m if m | j <strong>and</strong> 0 otherwise. Indeed, if m | j, then<br />
z j = 1 <strong>and</strong> the sum is clearly m. Otherwise, let g = gcd(m, j) <strong>and</strong> l = m g<br />
> 1 so<br />
m | lj. Then, z lj = 1 so z j is a primitive l-th root of unity. In particular, z lj − 1 = 0<br />
<strong>and</strong> z j ≠ 1 imply z j(l−1) + z j(l−2) + · · · + 1 = 0 <strong>and</strong> we have<br />
Thus, our sum becomes<br />
as claimed.<br />
R m (n) = 1 m<br />
4. Show that |R 3 (n) − 2n 3 | ≤ 2 3 .<br />
m−1<br />
∑<br />
k=0<br />
∑l−1<br />
z kj = gz kj = 0<br />
∑<br />
0≤j≤n,m|j<br />
k=0<br />
( n<br />
m =<br />
j)<br />
⌊n/m⌋<br />
∑<br />
k=0<br />
( n<br />
mk)<br />
Proof. Note that by the previous problem, we have R 3 (n) = 1 3 (2n +(1+ω) n +(1+ω 2 ) n )<br />
where ω 3 = 1 <strong>and</strong> ω ≠ 1. Thus, it suffices to prove that |(1 + ω) n + (1 + ω 2 ) n | ≤ 2.<br />
However, this follows immediately from the triangle inequality, as |1 + ω| = | − ω 2 | = 1<br />
<strong>and</strong> |1 + ω 2 | = | − ω| = 1, so<br />
|(1 + ω) n + (1 + ω 2 )| ≤ |(1 + ω) n | + |(1 + ω 2 ) n | = |(1 + ω)| n + |(1 + ω 2 )| n = 2.<br />
Comments. This can be generalized to a (much weaker) bound<br />
∣ R m(n) − 2n<br />
m ∣ ≤ (2 cos π m )n for m ≥ 4<br />
5
<strong>and</strong> the proof is pretty much analogous to the one above. By the formula in the previous<br />
problem, it suffices to prove | ∑ m−1<br />
k=1 (1 + zk ) n | ≤ (2 cos π m )n where z is a primitive<br />
m-th root of unity. Note that all the points 1 + z k must lie within the circle centered<br />
at the origin with radius 2 cos π . Thus, |(1 + m zk ) n | = |1 + z k | n ≤ 2 cos π m )n , <strong>and</strong> the<br />
result follows as before from the triangle inequality.<br />
5. A graph G is bipartite if <strong>and</strong> only if it contains no odd cycle.<br />
Proof: If G = G 1 ∪ G 2 is bipartite, where G 1 <strong>and</strong> G 2 are disjoint subgraphs of G,<br />
then it clearly contains no odd cycle. This is because two vertices are neighbors if <strong>and</strong><br />
only if they lie in different subsets, so a path must alternate between vertices in G 1<br />
<strong>and</strong> G 2 . If G has no odd cycles, let us pick a vertex v <strong>and</strong> color it red. We color all<br />
neighbors of v blue, <strong>and</strong> all uncolored neighbors of these blue vertices red, <strong>and</strong> so on.<br />
We might as well assume that G is connected, so all vertices have a color. We claim<br />
that setting G 1 <strong>and</strong> G 2 to contain all red <strong>and</strong> blue vertices, respectively, shows that<br />
G is bipartite. Assume otherwise, <strong>and</strong> say that v 1 , v 2 are neighbors of the same color.<br />
This means that there exist paths from v to v 1 <strong>and</strong> v 2 with lengths of the same parity.<br />
In other words, the path from v 1 to v 2 through v must have an even length. (If the<br />
paths from v 1 to v <strong>and</strong> v 2 to v first meet at v 3 , then we may replace v with v 3 <strong>and</strong> the<br />
proof continues.) Now, this path from v 1 to v 2 with the edge joining v 1 <strong>and</strong> v 2 creates<br />
an odd cycle, contradiction. Thus, G is bipartite if <strong>and</strong> only if it contains no odd cycle.<br />
6. Let G be a graph on n vertices without K k , a complete graph on k vertices. Then, G<br />
(k − 2)n2<br />
has at most edges. Moreover, there exists a graph with this number of edges.<br />
2(k − 1)<br />
Proof: I do admit to having seen this before, although I only vaguely remembered<br />
the ideas behind this proof of Turán’s Theorem.<br />
Let E(G) denote the number of edges of G. The proof proceeds by induction on<br />
n. Our base cases, for instance n = 1, 2, 3 are trivially true. Now, let G be a graph<br />
on n vertices without K k containing the maximal number of edges. Note that G must<br />
contain a K k−1 or else we could add edges, contradiction the maximality of E(G). Let<br />
H 1 be a K k−1 <strong>and</strong> let H 2 be G without H 1 . Note that E(H 1 ) = ( )<br />
k−1<br />
2 . By the induction<br />
hypothesis, we also have<br />
E(H 2 ) ≤<br />
(k − 2)(n − k + 1)2<br />
.<br />
2(k − 1)<br />
If v ∈ H 2 , then v is neighbors with at most k − 2 vertices in H 1 or else we have a K k .<br />
This implies that the number of edges with one vertex in each H 1 <strong>and</strong> H 2 is at most<br />
(n − k + 1)(k − 2), so<br />
( ) k − 1<br />
E(G) ≤ +<br />
2<br />
(k − 2)(n − k + 1)2<br />
2(k − 1)<br />
+ (n − k + 1)(k − 2) =<br />
(k − 2)n2<br />
2(k − 1) .<br />
Now, let n = (k −1)q +r where 0 ≤ r < k −1. Consider a partition G = G 1 ∪· · ·∪G k−1<br />
where r of the subgraphs have q + 1 vertices <strong>and</strong> the rest have q vertices. We join two<br />
6
vertices if <strong>and</strong> only if they lie in different G i . After a bit of computation, this shows<br />
that the bound above is essentially sharp.<br />
7. We prove a few results regarding even <strong>and</strong> odd permutations.<br />
Lemma 1: Every permutation σ ∈ S n can be written as a product of transpositions.<br />
Proof: Write σ = c 1 c 2 · · · c m where the c k are (possibly trivial) disjoint cycles. It<br />
suffices to show that any cycle can be written as a product of transpositions. Let<br />
c = (i 1 i 2 · · · i k ) be a cycle. We show that c can be written as a product of transpositions<br />
by induction on k. For k = 1, 2, the result is trivial. Now, note that<br />
c = (i 1 i 2 · · · i k ) = (i 1 i 2 · · · i k−1 )(i k−1 i k ) <strong>and</strong> (i 1 i 2 · · · i k−1 ) can be written as a product<br />
of transpositions by the induction hypothesis. The result follows.<br />
Lemma 2: Let τ 1 , τ 2 , . . . , τ m ∈ S n be transpositions. Then, the number of cycles<br />
in τ 1 τ 2 · · · τ m has the same parity as m + n.<br />
Proof: We first prove the following: if σ ∈ S n is a permutation <strong>and</strong> τ ∈ S n is a<br />
transposition, then the number of disjoint cycles in σ <strong>and</strong> στ have different parity.<br />
As before, write σ = c 1 c 2 · · · c m where the c k are (possibly trivial) disjoint cycles <strong>and</strong><br />
suppose τ = (ij). First we deal with the case where i <strong>and</strong> j are in the same cycle, say<br />
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<br />
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<br />
either case, the parity of the number of disjoint cycles changes.<br />
If i <strong>and</strong> j are in different cycles, say c t <strong>and</strong> c u respectively, we argue as follows. First,<br />
if c t = (i) <strong>and</strong> c u = (j), then στ is the product of all cycles c i <strong>and</strong> (ij) except for c t<br />
<strong>and</strong> c u . If c t = (i) <strong>and</strong> c u = (jj 1 · · · j v )) (or vice versa), then in στ, replace c u with<br />
(jij 1 · · · j v ) <strong>and</strong> remove (ij). If c t = (ii 1 · · · i s ) <strong>and</strong> c u = (jj 1 · · · j v ), then στ can be<br />
rewriten with c ′ t,u = (ij 1 · · · j v ji 1 · · · i s ) <strong>and</strong> (ij), c t , c u removed. Again, in either case,<br />
the parity changes.<br />
Now we prove Lemma 2 by induction on m, <strong>and</strong> our base cases are easily verified.<br />
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<br />
cycles in σ <strong>and</strong> στ m+1 have different parity, so this completes our induction.<br />
Comment: It immediately follows that a permutation σ can be written as a product<br />
of an even number of transpositions or an odd number of transpositions, but not<br />
both.<br />
7
References<br />
[1] I. Anderson, Combinatorics of Finite Sets, Oxford University Press, Oxford, Engl<strong>and</strong>,<br />
1989.<br />
[2] G. E. Andrews, Number <strong>Theory</strong>, W. B. Saunders Company, Philadelphia, 1971.<br />
[3] D. Reinhard, <strong>Graph</strong> <strong>Theory</strong>, Springer-Verlag, Heidelberg, July 2005.<br />
[4] L. Lovasz, Combinatorial Problems <strong>and</strong> Exercises, Akademiai Kiado, Budapest, Hungary,<br />
1993.<br />
8