96
Discrete Mathematics
REU 2004. Info:
http://people.cs.uchicago.edu/�laci/reu04.
Instructor: L�aszl�o Babai
8/13/04
2
Contents
1 7
1.1 Points in the plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.2 Games . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.3 Ramsey Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
2 11
2.1 Games continued . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.2 Hamilton cycles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
3 15
3.1 Graph theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
3.2 Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
3.3 Graphs without short cycles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
4 19
4.1 Graph theory continued . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
4.2 Puzzles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
4.3 Set game and Szemer�edi's theorem . . . . . . . . . . . . . . . . . . . . . . . . . 20
5 23
5.1 Density versions of coloring theorems . . . . . . . . . . . . . . . . . . . . . . . . 23
5.2 Hypergraphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
5.3 Steiner Triple Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
5.4 Projective planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
5.5 Galois Planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
5.6 Internet and library resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
3
4 CONTENTS
6 31
6.1 North-South Game . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
6.2 Number of unit distances in the plane < n32 . . . . . . . . . . . . . . . . . . . . 31
6.3 Max distance occurs � n times . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
6.4 Fitting 1020 + 1 squares on a�1010 + 0:1
�� �1010 + 0:1�square . . . . . . . . . 32
6.5 Tiling a� b rectangle by rectangles with at least one integer dimension . . . . 32
7 35
7.1 Fishermen's clubs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
7.2 Ramsey Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
8 39
8.1 Steiner Triple Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
9 43
9.1 Extremal Set Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
9.1.1 Eventown . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
9.1.2 Oddtown . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
10 47
10.1 Cauchy-Hilbert Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
10.2 Cauchy's Functional Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
10.3 Number Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
10.4 Ramsey Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
11 51
11.1 Totally unimodular matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
11.2 Latin squares . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
11.3 Counting perfect matchings in a bipartite graph; the permanent . . . . . . . . . 52
11.4 Doubly Stochastic Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
11.5 Back to Latin squares . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
11.6 Orthogonal Latin Squares . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
CONTENTS 5
12 57
12.1 Constructive Proofs of Negative Results in Ramsey Theory . . . . . . . . . . . 57
12.2 Bipartite Ramsey Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
12.3 Hadamard Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
13 61
13.1 The Gale-Berlekamp Switching game . . . . . . . . . . . . . . . . . . . . . . . . 61
13.2 The RAY-CHAUDHURI - WILSON theorem . . . . . . . . . . . . . . . . . . . 63
14 65
14.1 Points in general position . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
14.2 Pattern in proofs of inequalities using the Linear Algebra Method . . . . . . . . 65
14.3 Additional Exercises in Probability Theory . . . . . . . . . . . . . . . . . . . . 67
15 69
15.1 Bases for Vector Spaces of Polynomials . . . . . . . . . . . . . . . . . . . . . . . 69
15.2 Projective Representation of a Graph . . . . . . . . . . . . . . . . . . . . . . . . 70
15.3 The Number of Zero-Patterns of a Sequence of Polynomials . . . . . . . . . . . 71
16 73
16.1 Random Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
16.2 Independence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
16.3 Conditional Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
17 77
17.1 Puzzles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
17.2 Statistical Independence vs Linear Independence . . . . . . . . . . . . . . . . . 77
17.3 Algorithmic Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
17.4 Algebraic Coding Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
18 81
18.1 Projective dimension of a graph, continued . . . . . . . . . . . . . . . . . . . . 81
18.2 Theorem of Milnor-Thom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
6 CONTENTS
19 85
19.1 Matrix rigidity (Valiant 1978) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
20 89
20.1 Matrix Rigidity continued . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
21 91
21.1 Linear Programming . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
21.2 Shannon Capacity of a graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
21.2.1 Lov�asz's theta function (Lov�asz's capacity) . . . . . . . . . . . . . . . . 94
22 95
22.1 0; 1-measures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
22.2 Sign-Rigidity (Paturi-Simon) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
Chapter 11st day, Monday 6/21/04 (Scribe: Charilaos Skiadas, Eric Patterson, and Travis Schedler)
1.1 Points in the plane
Exercise 1.1 (Sylvester). Given n points in the plane, not all on the same line, show thatthere is a line in the plane that passes through exactly two of them. What if we switch therole of points and lines, i.e. given n lines in the plane such that not all of them pass throughthe same point, is it true that there is a point that belongs to exactly two of them? What ifin addition, no two of the lines are parallel?
Exercise 1.2 (Erd}os-E. Klein). Show that there is a constant c, independent of n, suchthat given n points in the plane, the number of pairs of points with distance 1 is less thancn3=2.
Note: It is known that this number is in fact less that cn5=4 (for possibly a di�erentconstant c).
OPEN QUESTION: Is it less than cn1:1?
1.2 Games
Exercise 1.3. Let n be even, and consider an n � n chessboard, with two opposite cornersremoved. Show that it cannot be covered with dominoes. (A domino covers two neighboringsquares, and dominoes don't overlap.) (An \Aha" problem.)
Exercise 1.4 (John Conway). Consider an in�nite chessboard: that means, a chessboardwith in�nitely many squares in all directions. We will draw one line which cuts the chessboardinto to in�nite halves, the North and the South half. This could be viewed as the coordinateplane where the integer lattice (i.e. points (a; b) for a and b integers) are the corners of squares,and the x-axis is the equator separating the board into the North and South parts. The Northside has some army of pieces, all identical, with at most one on each square, and the South
7
8 CHAPTER 1.
side is empty. North's goal is to take its pieces and invade as deep as possible into South'sterritory. To do this, the only move allowed is to take two of North's pieces which are adjacent(possibly diagonally adjacent) and have one jump (or step) over the other piece, thereby killingthe second piece, ending up across on the other side. In other words, two adjacent pieces canbe replaced with a third piece which makes a line of three adjacent pieces with the �rst two.Prove that North cannot invade too far into South's territory: say, prove that North cannotadvance beyond 100 steps south of the equator. (Note that North may have begun with anin�nite number of pieces.) (Note: 8 is the maximum depth.)
Exercise 1.5 (Paul Erd}os). a) Take n+ 1 numbers between 1 and 2n, inclusive. (assignedto Laj�os P�osa at P�osa's age of 12). Prove that two of them are relatively prime. (an \aha"problem)b) Again take n+1 numbers between 1 and 2n, inclusive. Prove that one is a factor of another.(an \ahaaa" problem)
Exercise 1.6. Take a rectangular pool table, and a large collection of pennies, all round ofthe same radius (assume that the pool table �ts at least one penny). Suppose two peopletake turns, each turn consisting of placing another penny on the pool table so that it does notoverlap with any other penny and does not dangle o� the side. The �rst player who cannot �ta penny on the board loses. Prove that the �rst player has a winning strategy.
Divisor Game. Fix an integer n > 1: Player 1 starts the game by picking a divisor of n:Player 2 picks another divisor of n that is not a divisor of the divisor picked by Player 1. Playalternates between the two players with each player choosing a divisor of n that is not a divisorof any divisor already picked. Eventually, one player is forced to choose n; and that playerloses. A winning strategy for Player 1 was found in class for n = 30; 12: For primes p; q; andr; we generalized winning strategies for n = pqr; pk; pkq:
Exercise 1.7. Find a winning strategy for Player 1 when n = p2q2:
Exercise 1.8. Prove that Player 1 has a winning strategy for all n � 2: (Hint: Prove theexistence of a winning strategy. No explicit strategy is known that works for all n:)
On an n� n chessboard, we can \infect" any initial con�guration of cells. Then the infectionspreads: a cell becomes infected if at least two adjacent cells are infected (diagonal neighborsdo not count). How few cells can we initially infect so that the whole board becomes infected?Note that if we infect the n cells on a diagonal, the whole board becomes infected.
Exercise 1.9. Show that in order to get the entire board infected, we need at least n initiallyinfected cells.
1.3 Ramsey Numbers
Ramsey Game. Player 1 has a blue pencil, and Player 2 has a red pencil. Given n pointsin the plane, no three on a line, Player 1 starts by connecting a pair of points with a blue
1.3. RAMSEY NUMBERS 9
line. Player 2 connects a di�erent pair with a red line. The players continue taking turns byassigning their color to an uncolored pair of points. Player 1 loses if a blue triangle is formed,that is three of the n points are connected to each other with blue lines. Similarly, Player 2loses if a red triangle is formed.
For the game n = 5; it is possible that no player loses: color the �ve sides of a regularpentagon blue, and color the �ve diagonals red. In this way, no triangles are made. We willsee that this is not possible in the n = 6 game, but �rst we introduce some notation.
De�nition 1.10 (Erd}os-Rado Arrow Notation). The notation n! (k; `) means that forany red and blue coloring of the
�n2
�pairs of n points, there is either a subset of k points all of
whose pairs are colored red or a subset of ` points all of whose pairs are colored blue.
We saw above that 5 6! (3; 3) and we prove
Theorem 1.11.6! (3; 3)
Proof. Select one point, call it 1: Since 1 is paired with �ve other points, at least three ofthese pairings are the same color, say blue. Denote these three other numbers by 2; 3; and 4: Ifa pair among f2; 3; 4g is also blue, say f2; 4g; then f1; 2; 4g is a blue triangle. If no pair amongf2; 3; 4g is blue, then all three pairs are red, so f2; 3; 4g is a red triangle.�
Notice that the arrow notation is more general than the Ramsey game because the arrownotation does not put any restrictions on the proportion of blue or red pairs. In particular, wefound a coloring in the case n = 6 so that there are 9 red pairs but no red triangles (the threehouses, three utilities example).
Exercise 1.12. Show 10! (3; 4):
10 CHAPTER 1.
Chapter 22nd day, Tuesday 6/22/04 (Scribe: Charilaos Skiadas, Eric Patterson, and Travis Schedler)
2.1 Games continued
Exercise 2.1 (Dominoes). Prove: if we remove two opposite corners from the chessboard,the board cannot be covered by dominoes. (Each domino covers two neighboring cells of thechessboard.) Look for an \Ah-ha" proof: clear, convincing, no cases to distinguish.
Exercise 2.2 (Triominoes). Remove a corner from a 101� 101 chessboard. Prove that therest cannot be covered by triominoes. A triomino is like a domino except it consists of threesquares in a row; each cell can cover one cell on a chessboard. Each triomino can either \stand"or \lie." Look for an \Ah-ha" proof.
Exercise 2.3 (Knight's trail). Consider a knight moving around on a 4� 4 chessboard. Welet the knight start at any cell of our choosing, and we wish to guide it through 15 moves soit never steps on a previously visited cell. So, after the 15 moves, the knight will have visitedeach cell. Prove that this is impossible. Find an \Ah-ha" proof.
Graph of knight moves on a 4� 4 chessboard.
Hint: Show that from any trail of the knight's moves, if you delete 4 cells, the trail splits intono more than 5 connected parts. (A set S of cells is connected from the knight's point ofview if the knight can move from any cell of S to any other cell of S without leaving S.)
11
12 CHAPTER 2.
Exercise 2.4. A mouse �nds a 3 � 3 � 3 chunk of cheese, cut into 27 blocks (cubes), andwishes to eat one block per day, always moving from a block to an adjacent block (a blockthat touches the previous block along a face). Moreover, the mouse wants to leave the centercube last. Prove that this is impossible. Find two \Ah-ha!" proofs; one along the lines of thesolution of knight's trail problem, the other inspired by the solution of the dominoes problem.
Exercise 2.5. 96 kids wait for us to split an 8 � 12 chocolate bar along the grooves into 96small rectangles. It is up to us in what order we do the splitting; we can start, for instance, bybreaking the 7 long grooves and then split each of the 8 long (1 � 12) pieces; or we can startwith the short grooves, or halve the bar each time, or any other way. The one thing we arenot permitted to do is stack the pieces. At any one time, we have to pick up one piece andbreak it into two.
Each break takes us 1 second. Find the fastest method. (This is another \Ah-ha" problem.)
Exercise 2.6. n teams play a straight-elimination tournament; there are no ties. How manymatches do they need to play before the winner is declared? (An \Ah-ha" problem again.)
� � �
Study the handout for the basic de�nitions involving graphs, including complete graphs, (com-plete) bi(multi-)partite graphs, (induced) subgraphs, the degree of a vertex, complement of agraph and Hamilton cycles. In particular, recall that an isomorphism from the graph G = (V;E)to the graph H = (W;F ) is a bijection f : V ! W between the vertices that preserves theadjacency relation, i.e. for any two vertices x; y 2 V , x; y are adjacent in G i� (if and only if)f(x); f(y) are adjacent in H. G is said to be isomorphic to H if there exists an isomorphismfrom G to H.Exercise 2.7. Show that isomorphism of graphs is a transitive relation.
Exercise 2.8. Show that the number of functions f : A! B is jBjjAj.
� � �
We de�ned decision tree and the related concepts of root, node, leaves, parent,and child. For example, ipping n coins can be made into a decision tree with 2n leaves.Similarly, we can make a decision tree from the moves in a chess game, but the tree is morecomplicated. For example, the leaves will not all be at the same level since di�erent games endafter di�erent numbers of moves. Furthermore, the possible moves at each node are dependenton the preceding moves or the history. Although the decision tree for a chess game is fartoo large to store in a computer (it has more con�gurations than the number of atoms in theearth), such trees can on principle be analyzed in a �nite time for a winning strategy.
First, assign a value of B, W, or D to the leaves of the tree if the outcome is a black win, awhite win, or a draw, respectively. The nodes above the leaves can be assigned a value of B,
2.2. HAMILTON CYCLES 13
W, or D if there is an optimal strategy for black to win, white to win, or a draw. Suppose thenode x is a white move. Then the value of x is W if 9 W child of x; B if all children are B,and D if there is no W child and 9 D child.
De�nition 2.9. A strategy function for white is a function from the set of histories precedinga white move to the set of possible next moves. A winning strategy function for white makesmoves to nodes with value W.
If a winning strategy exists for white, then the root must have value W, and there is a Wchild all of whose children are W. Each of these children must have a W child all of whosechildren are W, etc.
Theorem 2.10. If a �nite game only has win or lose outcomes, then one of the players has a
winning strategy.
This theorem can be used to solve the Divisor Game problem. (Hint: Proof by contradic-tion.)
Exercise 2.11. Exercise 6:1:5 p. 44 of the notes.
2.2 Hamilton cycles
De�nition 2.12. AHamilton cycle is a subgraph that is an n-cycle in a graph on n vertices.A graph is Hamiltonian if it has a Hamilton cycle.
Exercise 2.13. The k � ` grid is Hamiltonian if and only if k` is even and k; ` � 2:
Method 1. Make the vertices of the grid into the cells of a chess board. Hint: How manysteps do you move to get back to your starting color? How many steps do you move to get allthe cells?
Method 2. Sam's hint: Use the lemma:
Lemma 2.14. If G is Hamiltonian then removing k vertices splits G into at most k connected
components.
De�nition 2.15. A graph G is bipartite if its vertices can be colored red and blue such thatadjacent vertices have di�erent colors.
Observe: if G is bipartite and Hamiltonian, then the two parts are equal. Exercise 2.13 isa consequence of this.
Method 3. Alex's hint: What goes up, must come down. What moves left, must returnto the right.
14 CHAPTER 2.
Exercise 2.16 (Dirac). Prove: if every vertex in a graph G of n vertices has degree � n=2,then G is Hamiltonian (i.e. it contains a Hamiltonian cycle).
Exercise 2.17. Let G be a graph with n vertices and m edges. If m � �n�12 � + 2 then G isHamiltonian.
De�nition 2.18. A graph G is bipartite if the set of vertices V partitions into two disjointsubsets, V1 and V2 (i.e. so that V1 [ V2 = V and V1 \ V2 = ;) such that no vertex of V1 isadjacent to any other vertex of V1, and no vertex of V2 is adjacent to any other vertex of V2.In other words, all edges of G connect one vertex from V1 and one vertex from V2.
In other words, a graph is bipartite i� it is a subgraph of a complete bipartite graph (cf.p. 44 of the notes).
Exercise 2.19. Prove that a graph G is bipartite i� (if and only if) G has no odd cycle.
Exercise 2.20. Exercise 6.1.10, p. 45 of the notes.
Exercise 2.21. Exercise 6.1.11, p. 45 of the notes.
Chapter 3
3rd day, Wednesday 6/23/04 (Scribe: Justin Sinz and Travis Schedler)
3.1 Graph theory
Exercise 3.1. Let R be an equivalence relation on a set . Prove that we can obtain from Ra (unique) partition of (such that the partition induces the original equivalence relation inthe natural way).
Exercise 3.2. Recall: Every connected graph contains a spanning tree. Prove this by in-duction on the number of edges (i.e., complete the proof given in class). Recall that if weremove an edge from a cycle in a connected graph, the graph remains connected, and that thisdecreases the (�nite) number of cycles (cycles have no repeated vertices).
Exercise 3.3. (A)
(a) In every tree, there exists a vertex that is common to all the longest paths (i.e., paths ofgreatest length).
(b) In every tree, if the length of a longest path is odd then there is an edge common to allthe longest paths.
(c)� Disprove (a) for connected graphs.
Exercise 3.4. (B)(Helly-type Theorem) Let R1; : : : ; Rk be subtrees of a tree, and suppose that for all i; j wehave V (Ri) \ V (Rj) 6= ;. Then V (R1) \ : : : \ V (Rk) 6= ;:
Exercise 3.5. (C)Let G be a connected graph. Let P1; P2 be longest paths in G. Prove that V (P1)\V (P2) 6= ;:
Note: (C) and (B) imply (A).
15
16 CHAPTER 3.
De�nition 3.6. A legal coloring of a graph G is an assignment of a \color" to each vertexsuch that adjacent vertices have di�erent \colors".The chromatic number �(G) of G is de�ned to be the minimum number of colors necessaryto color G legally.
Exercise 3.7. The graph G is bipartite if and only if �(G) � 2.
Exercise 3.8. Construct a graph G such that K3 6� G and �(G) = 4:Hint: 11 vertices, 5-fold symmetry.
3.2 Inequalities
For a set of n positive real numbers, x1; : : : ; xn we can de�ne several means:
(1) the arithmetic mean A(x1; : : : ; xn) =x1 + x2 + � � �+ xn
n;
(2) the quadratic mean Q(x1; : : : ; xn) =
rx21 + � � �+ x2n
n=qA(x21; : : : ; x
2n);
(3) the geometric mean G(x1; : : : ; xn) = npx1x2 � � �xn;
(4) the harmonic mean H(x1; : : : ; xn) =n
1x1
+ � � �+ 1xn
=1
A( 1x1; : : : ; 1
xn).
Exercise 3.9. Prove, for any x1; : : : ; xn, the inequalities
Q(x1; : : : ; xn) � A(x1; : : : ; xn) � G(x1; : : : ; xn) � H(x1; : : : ; xn): (3.1)
Hint: Q � A is easy; A � G is di�cult; G � H follows immediately from A � G.
Exercise 3.10 (JENSEN'S INEQUALITY). There is a generalization of all of the aboveinequalities: the discrete version of Jensen's inequality. Let I � R be a �nite or in�nite intervaland let f : I ! R be a function. If f satis�es the inequality
(8x; y 2 I)f
�x+ y
2
�� f(x) + f(y)
2; (3.2)
then
f
�x1 + : : :+ xn
n
�� f(x1) + : : :+ f(xn)
n; (3.3)
i.e.f(A(x1; : : : ; xn)) � A(f(x1); : : : ; f(xn)): (3.4)
Similarly, if:
(8x; y 2 I)f
�x+ y
2
�� f(x) + f(y)
2; (3.5)
thenf(A(x1; : : : ; xn)) � A(f(x1); : : : ; f(xn)): (3.6)
3.3. GRAPHS WITHOUT SHORT CYCLES 17
Note: A continuous function satisfying inequality (3.2) is concave and a continuous func-tion satisfying inequality (3.5) is convex. However, not every function satisfying either (oreven both) inequalities is continuous. (Prove!�)
Exercise 3.11. Prove Jensen's inequality. Hint: First show inductively that the inequality istrue for n = 2k (that is, by induction on k). Then, show that the inequality for 2k implies theinequality for any n < 2k as well, thus proving the inequality for all n.
Exercise 3.12. Deduce Exercise 3.9 from Exercise 3.11.
3.3 Graphs without short cycles
We proved in class the
Theorem 3.13 (K}ov�ari-S�os-Tur�an). (cf. Exercise 6.1.22 in the text) There exists c > 0such that, if G is a graph with m edges and n vertices, and G has no 4-cycles, i.e. G 6� C4,
then m � cn3=2. In other words, m = O(n3=2).
In fact we showed that, as m tends to in�nity, m . 12n
3=2.
The following exercise strengthens this result:
Exercise 3.14. Show that, in the above situation of G with m edges, n vertices, s.t. G 6� C4,that
m � n3=2 + n
2(3.7)
Hint: use the following inequality from the proof of the theorem:
n� 1
2��2m
n
2
�(3.8)
The following exercise shows the tightness of the bound:
Exercise 3.15. Show that there exists c0 > 0 such that for in�nitely many choices of n, thereexists a graph G 6� C4 with m edges and n vertices, such that m > c0n3=2.
Exercise 3.16. Show that (8k)(9G 6� K3); (�(G) = k). In words, there exist graphs which donot contain triangles which have arbitrarily high chromatic number.
Exercise 3.17. Show that (8k; `)(9G such that G has no odd cycles of length � `; and �(G) =k). In words, show that for any odd ` � 1, there exist graphs with arbitrarily high chromaticnumber which do not contain any odd cycles of length 3; 5; : : : ; `.
Remark 3.18. Exercise 3.17 remains valid if we drop the word "odd":
18 CHAPTER 3.
Theorem 3.19 (Erd}os 1960). (8k; `)(9G such that G has no cycles of length � `; and �(G) =k). In words, for any ` � 1, there exist graphs with arbitrarily high chromatic number which
do not contain any cycles of length 3; 4; 5; : : : ; `.
Erd}os proved this in 1960 using the probabilistic method but did not exhibit any speci�cexamples of such graphs. No explicit construction was known of such graphs until 30 yearslater, around 1990.
Chapter 44th day, Thursday 6/24/04 (Scribe: Eric Patterson and Travis Schedler)
4.1 Graph theory continued
Exercise 4.1. If G is connected with n � 2 vertices, then there exists a vertex v such thatG n v is connected (the latter means the subgraph we get by throwing out the vertex v and alledges incident with v). Note that we cannot pick just any vertex, as evidenced by the case ofthrowing out the center of a star (which leaves a highly disconnected graph).
De�nition 4.2. A graph G is k-edge-connected between vertices x and y if there are kedge-disjoint paths between x and y:
De�nition 4.3. A graph G is k-vertex-connected between vertices x and y if there are kvertex-disjoint paths between x and y:
Suppose we found 57-edge-disjoint paths between x and y; and we think 57 is optimal.How can we demonstrate that 58 such paths cannot be found?
Ben's Conjecture: If there are no more than 57 edge-disjoint paths from x to y; then thereis a partition V = A _[B such that x 2 A; y 2 B; and the maximum number of edges betweenA and B is less than or equal to 57:
De�nition 4.4. An (x; y)-cut of a graph G = (V;E) is a partition V = A _[B such that x 2 Aand y 2 B:
Theorem 4.5 (Menger's Theorem). The maximum number of edge-disjoint paths between
x and y equals the minimum number of edges between the two parts of an (x; y)-cut.
4.2 Puzzles
Exercise 4.6. If we have a 1010 � 1010-size square that we want to cover with 1� 1-squares,evidently we can �t 1020 squares. Show that if we have a (1010 + 0:1) � (1010 + 0:1)-squarethen we can �t 1020 + 1 squares.
19
20 CHAPTER 4.
Exercise 4.7. Suppose we have a big rectangle which is a � b in size, and we tile it withvarious size rectangles an � bn (situated parallel to the sides of the original rectangle), wherefor each n, at least one of an or bn is an integer. Show that either a and b is an integer.
4.3 Set game and Szemer�edi's theorem
Set Game. The Set game is played with a deck of cards. Each card has four properties withthree possible values: a shape (oval,diamond,squiggle), a color (red, green, purple), a shading(blank, solid, shaded), and a number (1,2,3). Each card is distinct, so there are 34 = 81 cards.Twelve cards are placed face up on a table. The objective is to �nd a \set" among the twelvecards. A \set" is formed by three cards so that each property is either the same on all cardsor di�erent on all cards. The �rst player to �nd a \set" gets that \set," after which three newcards from the deck are placed face up to replace them. The player who �nds the most \sets"wins.
We can generalize the Set game so that there are n properties on each card, and eachproperty has three possible values. In this n-dimensional Set game we have 3n cards. We areinterested in how many cards we can draw from the deck without �nding a \set." Let �(n)denote the maximum number.
Let Z3 be the set f0; 1; 2g with addition mod 3; so 2+2 = 1 for example. Let Zn3 be the setf(x1; : : : ; xn) : xi 2 Z3g: For x = (x1; : : : ; xn) and y = (y1; : : : ; yn) in Z
n3 ; we de�ne addition by
x+y = (x1+y1; : : : ; xn+yn): In the Set game (n = 4), the cards correspond to the quadruplesin Z43; for instance [oval, green, blank, 3] $ [1; 1; 0; 2] :
Observation 4.8. Three distinct elements x; y; z 2 Z43 form a set if and only if x+ y+ z = 0:
Similarly, we can work in Zn3 for the n-dimensional Set game. So an alternative de�nitionfor �(n) is the maximum number of elements of Zn3 without a solution to x+ y + z = 0 wherex; y; z are distinct.
De�nition 4.9. A function f : N! R is supermultiplicative if f(n+m) � f(n) � f(m):
Exercise 4.10. Prove: if f is supermultiplicative, then L = limn!1 npf(n) exists, and L �
npf(n) for all n:
Exercise 4.11. Prove �(n) is supermultiplicative.
Observe that 2n � �(n) � 3n; so 2 � L � 3 by the previous two exercises, where L =limn!1 n
p�(n):
Exercise 4.12. Prove: �(4) � 20: Infer: L � 4p20 � 2:115:
Exercise 4.13. Prove: �(2) = 4: Infer �(4) � 36:
4.3. SET GAME AND SZEMER�EDI'S THEOREM 21
Exercise 4.14. Prove: �(3) = 9: Infer �(4) � 27:
Exercise 4.15. Prove: �(4) � 24:
Exercise 4.16. � Prove �(4) = 20:
Conjecture 4.17 (Open Problem). limn!1 np�(n) = 3:
Exercise 4.18. Color the integers red and blue. Is there a set of 3 red or 3 blue integers thatform an arithmetic progression? Find a coloring of the positive integers such that there is noarithmetic sequence of 3 red or 3 blue integers in the �rst n numbers. What is the largest nfor which you can �nd an example?
Theorem 4.19 (Van der Waerden, 1928). If we color the positive integers with k colors,
then for all ` there is an `-term arithmetic progression in one color.
Theorem 4.20 (Szemer�edi, 1975). For all � > 0 and `; there is an N such that if A �f1; : : : ; Ng and jAj � �N;then A contains an `-term arithmetic progression.
Exercise 4.21. Prove that Szemer�edi's Theorem implies van der Waerden's Theorem.
History:
� Erd}os and Tur�an conjectured this result in 1936.
� In 1952, K. F. Roth found an analytic proof of the result for ` = 3: This result, in additionto another major result, were part of his Fields Medal citation in 1958.
� In the late 1960s, E. Szemer�edi found a combinatorial proof for ` = 3:
� Later, E. Szemer�edi found a combinatorial proof for ` = 4:
� In 1975, E. Szemer�edi found a combinatorial proof of the whole conjecture.
� Subsequently, Furstenberg found an analytic proof of Szemer�edi's Theorem. His methodsled to further generalizations. For example:
Theorem 4.22. �(n) = o(3n):
On April 8, 2004, a major breakthrough was announced in a very old problem that hasfascinated mathematicians for ages:
Theorem 4.23 (Green-Tao, 2004). For all `; there is an `-term arithmetic progression
among prime numbers.
Ben Green and Terence Tao announced this result in their paper \The Primes Contain Arbi-trarily Long Arithmetic Progressions." The proof uses Szemer�edi's Theorem and the ideas ofFurstenberg's proof. It is posted on arXiv.
22 CHAPTER 4.
Chapter 55th day, Friday 6/25/04 (Scribe: Justin Sinz and Charilaos Skiadas)
5.1 Density versions of coloring theorems
Notation: [k] = f1; 2; : : : ; kg; [k]n = f(x1; : : : ; xn) : xi 2 [k]g.De�nition 5.1. A subset L � [k]n with k elements is called a combinatorial line if we canarrange the elements in an order so that in every coordinate either all the entries are the sameor they form the sequence 1; 2 : : : ; k, in that order.
Note: A combinatorial line is a \SET" in the game \SET." The converse is not true.(why?)
Theorem 5.2 (Hales-Jewett, 1963). For all k; ` there exists an n0 such that for all n � n0and any k-coloring of [`]n, there exists a combinatorial line in one color.
Note: This theorem implies Van der Waerden's theorem:
Theorem 5.3 (Van der Waerden, 1927). For any k; ` there exists N0 such that for all
N � N0, if we color [N ] with k colors, then there is an `-term arithmetic progression in one
color.
The idea is to let N0 = `n0 , and to associate with every number in [N ] the digits in itsexpansion in the base-` number system. Then any combinatorial line would correspond to anarithmetic progression. (But not conversely. Why?)
Recall Szemer�edi's Theorem, which is the \density version" of van der Waerden's Theorem:
Theorem 5.4 (Szemer�edi, 1975). For all � > 0 and ` there exists N0 such that for all
N � N0, if A � f1; : : : ; Ng and jAj � �N then A contains an `-term arithmetic progression.
The density version of the Hales-Jewett Theorem is:
23
24 CHAPTER 5.
Theorem 5.5 (Furstenberg-Katznelson). For all � > 0 and all ` there exists n0 such that
for all n � n0, if A � [`]n and jAj � �`n, then A contains a combinatorial line.
Exercise 5.6. Use the Furstenberg-Katznelson Theorem to prove that �(n) = o(3n) in thenotation of the fourth set of notes to this course. (Recall that we used �(n) to denote themaximum number of cards in the n-dimensional \SET" game without a \SET.")
5.2 Hypergraphs
De�nition 5.7. (a) A hypergraph is a pair H = (V;E), where V is a set of vertices andE is a set of edges, an edge simply being a subset of V .
(b) A hypergraph is called t-uniform, if every edge has t vertices.
Note:
(a) A graph is a 2-uniform hypergraph.
(b) If a hypergraph on n vertices is t-uniform, then jEj � �nt�.(c) Given a set V of vertices, jV j = n, the number of t-uniform hypergraphs on V is 2(
n
t).
The number of t-uniform hypergraphs with m-edges is
��nt
�m
�.
(d) The \SET" game is based on a 3-uniform hypergraph with 81 vertices, where the edgesare the \SETs."
Exercise 5.8. Count the total number of \SETs" in the n-dimensional \SET" game. (Self-check: for n = 1, your formula should give 1.) What number do you get for n = 4 (the actual\SET" game)?
Extremal set theory is concerned with problems of the following type: determine (or esti-mate) the maximum number of edges of a hypergraph satisfying certain conditions. We haveseen some examples of this already in graph theory, for instance the Mandel-Tur�an Theoremand the K}ov�ari{S�os-Tur�an Theorem. There are many more that deal with hypergraphs. Ruzsaand Szemer�edi, in examining the combinatorial roots of Roth's Theorem on 3-term arithmeticprogressions, were lead to the following natural problem in extremal set theorey:
Question: What is the maximum number of edges of a 3-uniform hypergraph satisfying thefollowing two conditions:
(C1) No two edges share more than one point.
5.3. STEINER TRIPLE SYSTEMS 25
(C2) There are no triangles, i. e., there is no set of three edges that intersect pairwise but havean empty intersection.
Note that the maximum possible number of edges in a 3-uniform hypergraph is�n3
� � n3
6 ,the total number of triples of vertices. Condition (C1) already forces a lower (uadratic) rateof growth:
Exercise 5.9. Prove: if a 3-uniform hypergraph on n vertices satis�es condition (C1) then ithas at most n(n� 1)=6 edges.
Hint. For every pair of vertices there is at most one edge containing them; and every edgecontains 3 pairs of vertices.
When we add condition C2, the rate of growth of the number of edges drops even further,below quadratic:
Lemma 5.10 (Ruzsa-Szemer�edi). The maximum number m(n) of edges in a 3-uniform
hypergraph satisfying conditions (C1) and (C2) is m(n) = o(n2), i. e., limn!1m(n)=n2 = 0.
We will see next time that this implies Roth's theorem (i. e., Szemer�edi's theorem for ` = 3):
Roth's Theorem. Given � > 0 there exists N0 such that for all N � N0, if jAj � [N ] andjAj � �N , then A contains a 3-term arithmetic progression.
Exercise 5.11. Assuming that we have proved Roth's theorem for sets A consisting only ofodd numbers, and prove Roth's theorem for all sets.
5.3 Steiner Triple Systems
De�nition 5.12. A Steiner Triple System (STS) is a hypergraph H (in which the edges arecalled lines) such that
(a) every line consists of exactly 3 points(=vertices) (i. e., H is 3-uniform);
(b) every pair of (distinct) points is contained in a unique line.
In other words, a STS satis�es condition (C1) above and is maximal with that property:The number of edges is exactly equal to the upper bound obtained above: jEj = n(n � 1)=6,where n is the number of vertices. In particular we must have that n(n� 1) is divisible by 6.
Exercise 5.13. With n as above, prove that n is odd.
Exercise 5.14. Deduce from the preceding two exercises that the number of points of a STSis congruent to 1 or 3 mod 6.
26 CHAPTER 5.
This necessary condition on n also turns out to be su�cient.
We construct a STS F on 7 vertices (with 7 edges) as follows: Take the vertices of a(nequilateral) triangle along with the bisectors of the edges and the (in)center of the triangle.Let the edges of the STS consist of the following:
(a) the 3 vertices on each edge of the triangle (for a total of 3 edges).
(b) the 3 vertices on each angle bisector.
(c) the 3 vertices on the (natural) inscribed circle.
This con�guration is called the \Fano Plane" see Figure 5.1.
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0
1 2
34
5
6
ab
c
d
e
f
g
Figure 5.1: The Fano Plane
The incidence matrix for the Fano plane is as follows (the rows correspond to lines, columnsto points, \1" indicates incidence):
0 1 2 3 4 5 6
a 1 1 0 1 0 0 0b 0 1 1 0 1 0 0c 0 0 1 1 0 1 0d 0 0 0 1 1 0 1e 1 0 0 0 1 1 0f 0 1 0 0 0 1 1g 1 0 1 0 0 0 1
5.4. PROJECTIVE PLANES 27
Exercise 5.15. Prove that the Fano plane has 168 automorphisms (line-preserving permuta-tions).
The group of automorphisms of the Fano plane is a very important group: it is the secondsmallest �nite simple group. \Simple groups" to \groups" are like \atoms" to \chemicalcompounds," so according to this analogy, the automorphism group of the Fano plane is the\Helium of group theory."
Exercise 5.16. Construct STSs for n = 13 and for n = 3k and n = 2k � 1 for all k � 1.
Exercise 5.17. Given STSs H1 and H2 with n1 and n2 points, respectively, construct a STSwith n = n1n2 points.
Exercise 5.18. Given a STS with k points, construct (explicitly) STSs on 3k; 3k � 2; and3k � 6 points.
Exercise 5.19. From Exercises 5.16 and 5.18 above, deduce that if n is congruent to 1 or 3mod 6 then there exists a STS on n points.
5.4 Projective planes
De�nition 5.20. An incidence geometry is a set P of \points," a set L of \lines," and anincidence relation I � P � L.
Notation 5.21. If (p; `) 2 I then we say that p is incident with ` and we write p � `. If(p; `) =2 I, we write p
�`.
De�nition 5.22. The dual of the incidence geometry (P;L; I) is the incidence geometry(L;P; I�1) (we switch the roles of points and lines; the same pairs remain incident).
De�nition 5.23. A projective plane is an incidence geometry satisfying the following threeaxioms:
Axiom 1. (8p1 6= p2 2 P )(9!` 2 L)(p1 � ` and p2 � `).
Axiom 2. (8`1 6= `2 2 L)(9!p 2 P )(p � `1 and p � `2).
Axiom 3. (Non-degeneracy) 9p1; p2; p3; p4 2 P such that no three are on the same line.
Exercise 5.24. Prove that the dual of a projective plane is a projective plane. (Note the dualof Axiom 1 is Axiom 2 and vice versa. State the dual of Axiom 3 and prove that it followsfrom Axioms 1{3.)
28 CHAPTER 5.
Unless expressly stated otherwise, all projective planes considered here will be �nite. Forp 2 P , let deg(p), the degree of p, be the number of lines incident with p. For ` 2 L, letrk(`), the rank of `, be the number of points incident with `.
The following observation is an immediate consequence of Axioms 1 and 2 and does notrequire Axiom 3.
Exercise 5.25. If p�`, then deg(p) = rk(`).
Exercise 5.26. In a projective plane, 8p1; p2 2 P , 9` 2 L, p1�` and p2
�`.
The proof of this requires some care.
The following is immediate from the preceding two exercises.
Exercise 5.27. In a projective plane, all points have the same degree.
Use the fact that the dual of a projective plane is a projective plane (Ex: 5.24) to infer:
Exercise 5.28. In a projective plane, all lines have the same rank.
Exercise 5.29. In a projective plane, the degree of every point and the rank of each line isthe same.
In other words, projective planes are regular and uniform, and their degree and rank areequal.
For reasons of tradition, we denote this common value by n+ 1. Every point of the planeis thus incident with n+ 1 lines and every line is incident with n+ 1 points. The number n iscalled the order of the projective plane.
Proposition 5.30. jP j = jLj = n2 + n+ 1.
The smallest projective plane is the Fano plane, which has order n = 2 (see Figure 5.1).
5.5 Galois Planes
A class of projective planes called Galois planes is constructed as follows. Let F be a �nite�eld of order q. Let F 3 be the 3-dimensional space over F . We de�ne the inner product overF 3 in the usual way: for u = (�1; �2; �3) and v = (�1; �2; �3) we set u �v = �1�1+�2�2+�3�3.We say that u and v are perpendicular if u � v = 0.
Let us say that two nonzero vectors u; v 2 F 3 are equivalent if u = �v for some � 2 F .
Let S be the set of equivalence classes on F 3�0. Note that each equivalence class has q�1elements and therefore the number of equivalence classes is (q3 � 1)=(q � 1) = q2 + q + 1.
5.5. GALOIS PLANES 29
Set P = L = S and let us say that p 2 P and ` 2 L are incident if u � v = 0 where u 2 p(u is a vector in the equivalence class p) and v 2 `. The coordinates of u are called homoge-neous coordinates of p (they are not unique{every point has q � 1 triples of homogeneouscoordinates); similarly, the coordinates of v are called homogeneous coordinates of `.
Exercise 5.31. Prove that this de�nition gives a projective plane of order q. It is called aGalois plane after �Evariste Galois (1811{1832), the discoverer of �nite �elds and of modernalgebra.
Exercise 5.32. Prove that the Fano plane is a Galois plane (necessarily over the �eld of order2) by assigning homogeneous coordinates to the points and lines of the Fano plane.
A set of points is collinear if there is a line with which all of them are incident.
We say that four points are in general position if no three of them are collinear.
A collineation is a transformation of the projective plane consisting of a permutation ofthe points and a permutation of the lines which preserves incidence.
Theorem 5.33 (Fundamental Theorem of Projective Geometry.). If a1; : : : ; a4 and
b1; : : : ; b4 are two quadruples of points in general position in a Galois plane then there exists a
collineation ' such that (8i)('(ai) = bi).
Exercise 5.34. (Prove!) Use Theorem 5.33 to show that the Fano plane has 168 collineations.
Exercise 5.35. Consider the projective plane � = PG(2; F ) over the �eld F . for i = 1; 2; 3;let pi = (ai; bi; ci) be three points in �, given by homogeneous coordinates (ai; bi; ci 2 F , notall zero). Prove: the three points are collinear (lie on a line) if and only if the 3�3 determinantjai bi cij is zero.Exercise 5.36. A projective plane �1 = (P1; L1; I1) is a subplane of the projective plane�2 = (P2; L2; I2) if P1 � P2, L1 � L2, and the incidence relation I1 is the restriction of I2 toP1 �L1. Prove: if �1 is a proper subplane of �2 then n1 �
pn2 (where ni is the order of �i).
Exercise 5.37. Let P (n) be the number of projective planes of order n. Prove: P (n) <(ne)(n+1)3 . Hint. �rst prove that
P (n) �� �n2+n+1
n+1
�n2 + n+ 1
�:
Exercise 5.38. Let us consider the Galois plane PG(2; 5) (over the �eld of 5 elements).
1. How many points does this plane have, and what is the number of points per line?
2. Points are given by \homogeneous coordinates." Determine whether or not the pointsgiven by a = [1; 4; 0], b = [3; 2; 2], and c = [4; 1; 2] are collinear (belong to the same line).(Coordinates are mod 5.) Prove your answer.
30 CHAPTER 5.
� � �
Puzzle 5.39. Find two in�nite sets of nonnegative integers, A;B, such that every non-negativeinteger can be written in a unique way as a+ b, with a 2 A and b 2 B.
Note:
(a) We necessarily have that A \B = f0g.(b) If A is allowed to be �nite, then we could take, for our favorite number d > 0, A =
f0; : : : ; d� 1g and B = dZ. The hard part is to make both A and B in�nite.
5.6 Internet and library resources
Regarding arithmetic progressions of prime number, please check Ivars Peterson's \MathTrek" article, \Progressive Primes," at
http://www.maa.org/mathland/mathtrek 04 26 04.html
and the new article by Ben Green and Terence Tao: \The primes contain arbitrarily longarithmetic progressions." Peterson's article includes a link to the Green { Tao paper.
Regarding the card game \SET," consult the article \The card game SET" by B. L.Davis and Diane Maclagan at
http://math.stanford.edu/�maclagan/papers/set.pdf and some of the references
mentioned in that paper, especially
Roy Meshulam: ``On subsets of finite abelian groups with no 3-term
arithmetic progressions," J. Comb. Theory-A 71 (1995), 168--172.
Chapter 66th day, Monday 6/28/04 (Scribe: Eric Patterson and Travis Schedler)
6.1 North-South Game
Exercise 6.1. Assign positive real numbers to the cells in the grid so that
(a) the sum of the North squares is �nite,
(b) a move never increases the sum of occupied squares, and
(c) there is a cell (in the South) with a number greater than the sum of all the cells in theNorth.
(Conclusion: that square cannot be reached.)
6.2 Number of unit distances in the plane < n3
2
Exercise 6.2. First prove geometrically that a unit distance graph does not contain a K2;3:Then prove the K}ov�ari-S�os-Tur�an theorem for K2;3:
Exercise 6.3. In R4; �nd n points with bn24 c unit distances between them (for all n).
Exercise 6.4. Find a method of shifting a grid and doubling points to get n points withcn log n unit distances.
Exercise 6.5. What numbers can be written as a2 + b2 in many ways? In particular, what isthe smallest number n that you can write as a2 + b2 in at least 200 ways?
Exercise 6.6. Use the previous exercise to show that there is a dn such that the distance dnoccurs n
1+ clog logn times in the
pn�p
n grid. This is the best known result for this problem.
Exercise 6.7. Prove (log x)100 = o (x) : Show nc
log logn > log n: (Thus, the best known resultis in fact better than cn log n) .
31
32 CHAPTER 6.
6.3 Max distance occurs � n times
Exercise 6.8. If each vertex in a graph has degree � 2; then the number of edges is � n:
Let S be a set of points in the plane. Let G(S) = (S;E) be the \max distance graph"de�ned by joining two points if they are at maximum distance.
Exercise 6.9. Prove: in a max distance graph, if deg(x) � 3; then x has a neighbor of degreeone (a \dangling neighbor").
Exercise 6.10. Prove: if G has the property in the previous exercise, hereditarily to allinduced subgraphs, then m � n:
Exercise 6.11. Use the preceding exercises to conclude that the max distance occurs at mostn times.
6.4 Fitting 1020+1 squares on a�1010 + 0:1
��
�1010 + 0:1
�square
Exercise 6.12. Let the upper�1010 � 106
�� �1010� rectangle be region A; and let the rest ofthe square be region B: So B is a
�106 + 0:1
���1010 + 0:1�rectangle. Tile region A by aligned
squares (1020 � 1016 of them, no surprise here). B is a long stripe. Squeeze in the square hereby tilting the columns. You may �nd a number better than 106 for the width of this stripe.
Exercise 6.13. In an (n + :01) � (n + :01) square, we can put in n2 + cn� squares for someconstants c and �: Compute as large an � as you can get.
6.5 Tiling a�b rectangle by rectangles with at least one integerdimension
Let R be an a � b rectangle, where a and b are positive real numbers. Suppose that we cantile R by \aligned" rectangles (the sides of the rectangles are parallel to the sides of R) suchthat each raligned rectangle has at least one side with integer length. Show that at least oneside of R has integer length. Assume one corner of R is at the origin, and the sides of R arealigned with the axes.
There is a measure theory solution and a graph theory solution. This hint applies to themeasure theory solution. Assign weights (positive or negative) to each rectangle in R2 so thatif a rectangle has an integer side, the weight is zero. The weights should be additive (if arectangle is cut up into rectangles, the weights should add up).
This can be done continuously or discretely. For a rectangle S; we can de�ne its weight asZ ZSsin(2�x) sin(2�y)dx dy:
6.5. TILINGA�B RECTANGLE BY RECTANGLESWITH AT LEAST ONE INTEGER DIMENSION33
To de�ne the weight discretely, tile R2 on the half integers with a chess board coloring. If aregion is white, its weight is its area. If a region is black, its weight is the negative of its area.
Exercise 6.14. Prove that the weights have the following properties:
(a) if a side of an aligned rectangle is an integer, then the measure is zero, and
(b) if the measure of an aligned rectangle is zero and one corner is the origin, then a side isan integer.
Exercise 6.15. Use the previous exercise to prove that at least one side of R has integerlength.
34 CHAPTER 6.
Chapter 7
7th day, Wednesday 6/30/04 (Scribe: Justin Sinz and Charilaos Skiadas)
Puzzle 7.1. (a) A 3-way lamp (red,yellow,green) is connected to n 3-way switches. Eachcon�guration of the switches corresponds to a state of the lamp. In other words we havea function f : [3]n ! [3]. It is known that if we change all the switches, then the lampwill switch also. Show that all but one of the switches are dummies. In other words, thefunction f depends on only one of the variables: one switch determines the state of thelamp.
(b) Show that this is not necessarily true if we have in�nitely many switches. In fact, we canarrange it so that the lamp does not change whenever we change �nitely many switches.(But it does change if we change all of them.)
Puzzle 7.2. A hundred prisoners are on their way to a prison island. They know a lot aboutthe workings of the prison. First of all, the moment they arrive they will be taken to individualcells, never to see or talk to each other again. Also, there is a special room that they are takento, where they are being interrogated by the warden. They are taken there one at a time, butin unknown order. They have no way of knowing who else has entered the room before them.They know that each one of them will be called to the room in�nitely many times, but theydon't know in what order. In the room there is a two-way switch (up,down), which they getto see and are allowed to change if they want to. Nobody other than the prisoners will evertouch the switch. There is also a red button, which has an amazing property: If one of theprisoners presses it and by that time all the prisoners have been in this room at least once,then the game is over and they all go free. If one of them hasn't, then they all spend the restof their lives in prison. (By the way, the prisoners are immortal, so this will be a really longtime.) The prisoners have a lot of time before they get to the prison to make their plans. Finda way for them to get out.
There are two versions to this problem. In the easier version, the prisoners know the initialposition of the switch. In the other they don't. Figure out a strategy for each problem.
35
36 CHAPTER 7.
7.1 Fishermen's clubs
In a small �shing town by a lake, the n �shermen of the town are in the habit of forming asmany clubs as they can. To control the situation a bit the state has formed a rule that any twoclubs must share exactly one member. The question is raised, how many clubs can there be.It is easy to come up with n clubs. For instance we could have one really passionate person,who has formed n � 1 clubs, one with each other �sherman. (So all these clubs so far haveexactly 2 members each.) Then all the other �shermen, having gotten tired of him, decide toform a club, for a total of n clubs. Or maybe he was faster than them and formed a club byhimself. Notice though that not both of those last clubs can be formed, because they don'tshare a member. Another way to form n clubs is if n = k2 + k + 1 and there is a projectiveplane of order k; the points correspond to the �shermen and the lines to the clubs. The naturalquestion then is whether we could have more than n clubs. The surprising answer is given bythe following theorem:
Theorem 7.3 (Erd}os-deBruijn, 1949). Given a set with n elements, there cannot be more
than n subsets with the property that any two of them share exactly one member.
A reasonable next step would be to ask: What if we allow them to have exactly k membersin common? The answer remains the same:
Theorem 7.4 (Fisher-R.C.Bose-Majumdar). Given k � 1 and given a set with n el-
ements, there cannot be more than n subsets with the property that any two of them share
exactly k members.
The proof is not hard but is quite ingenious: it invokes an unexpected tool, linear algebra.The method, introduced by R.C. Bose, has since produced volumes of startling results.
De�nition 7.5. (a) The characteristic vector (incidence vector) of a subset A � [n] isthe vector (x1; : : : ; xn) in R
n, where xi = 1 if i 2 A and xi = 0 if i 62 A.
(b) If A1; : : : ; Ak are subsets of [n], the incidence matrix of this set-system (hypergraph)is the k � n matrix whose i-th row is the characteristic vector of Ai.
Exercise 7.6 (R.C. Bose, 1949). Let k � 1 and let A1; : : : ; Ak � [n]. Show that if (8i 6=j)(jAi \Aj j = k), the characteristic vectors of the Ai are linearly independent.
Since there cannot be more than n linearly independent vectors in Rn (\Fundamental Factof Linear Algebra"), this proves Theorem 7.4.
(Note: Bose proved this for uniform hypergraphs; Majumdar extended the result to thenonuniform case.)
7.2. RAMSEY THEORY 37
7.2 Ramsey Theory
Notation: If A is a set then
�A
k
�denotes the set of all k-subsets (subsets of size k) of A. So
if jAj = n then
������A
k
������ =�n
k
�. Recall that [n] = 1; : : : ; n:
De�nition 7.7 (Erd}os-Rado arrow notation). (n)rk ! (s1; : : : ; sk) if for any partition of�[n]
r
�= A1 _[ : : : _[Ak there exists i, 1 � i � k and H � [n] such that jHj � si and
�Hr
� � Ai.
Absence of a superscript on (n) indicates that r = 2. The subscript is redundant and oftenomitted.
Theorem 7.8 (Ramsey). For all r; k; s1; : : : ; sk there exists n such that (n)rk ! (s1; : : : ; sk).
Notation: Rr(s1; : : : ; sk) denotes the smallest such n and is called a Ramsey number. Ifr = 2 then r is omitted.
Exercise 7.9. Prove that 17! (3; 3; 3).
We have 6! (3; 3) and 17! (3; 3; 3). In general, dn!ee ! (3; : : : ; 3) (n parts, i. e., (dn!ee)2n !(3; : : : ; 3):
Here are some known Ramsey numbers.
6! (3; 3); 59 (3; 3), so R(3; 3) = 6.
10! (3; 4); 99 (3; 4), so R(3; 4) = 10.
17! (3; 3; 3); 169 (3; 3; 3), so R(3; 3; 3) = 17.
The R(5; 5) is known to be at least 43 and this is conjectured to be the correct value.However, in order to prove that it is 43 (that is, for every graph G on 43 vertices either G orits complement contains a K5), we would either have to be quite clever or perform a computer
search. A (naive) computer search would involve examining each of the 2(432 ) = 2902 2-colorings
of the complete graph on 43 vertices. 2903 is rather large (larger than the number of particlesin the known universe), so a computer search is unfeasible.
Exercise 7.10. Prove that Ramsey's theorem implies the weak form of Szekeres' theorem (i.e.,it gives a \Happy Ending").
Theorem 7.11 (Erd}os-Szekeres, 1935). R(k; `) � �k+`�2k�1�, i.e.,
�k+`k
�! (k + 1; `+ 1).
From the above statement, it follows that R(k; k) < 4k. Using the probabilistic method, Erd}osproved that 2k=2 < R(k; k).
More information about Ramsey numbers can be found in [CG98], the dynamic survey ofStanislaw Radziszowski at http://www.combinatorics.org/ and at web-pages of StanislawRadziszowski (http://www.cs.rit.edu/~spr/homepage.html) and Brendan McKay(http://cs.anu.edu.au/~bdm/).
38 CHAPTER 7.
[CG98] Fan Chung, Ron Graham: Erd}os on graphs. His legacy of unsolved problems. A. K.Peters, Ltd., Wellesley, MA, 1998.
Chapter 8
8th day, Friday 7/2/04 (Scribe: Eric Patterson and Travis Schedler)
8.1 Steiner Triple Systems
Exercise 8.1 (Puzzle). Let A1 : : : An be a regular n-gon inscribed in the unit circle. Provethat A1A2 A1A3 � � � A1An = n. Here XY is the length of the segment from X to Y . Hint: C.
Recall [De�nition 5.11] that a Steiner Triple System (STS) is a 3-uniform hypergraph [Def-inition 5.7(a,b)] such that for any pair of points (vertices) there is exactly one line passingthrough them.
Recall Exercise 5.12: Prove that in a STS, the number of vertices n is odd.
Recall Exercise 5.18: If n � 1; 3 (mod 6), then there exists a STS on n points. (note thatExercise 5.13 is the converse of this statement). To get started on this, make sure to proveand use the
Lemma 8.2. For any k, we get a STS on 3k points from the k-dimensional SET game (i.e. with
k characteristics).
The above lemma is part of Exercise 5.15.
De�nition 8.3. The �nite �eld F3 = f0; 1; 2g is de�ned by taking arithmetic modulo 3. Forany k we can consider the k-dimensional F3-vector space, F
k3 = f(�1; : : : ; �k) : � 2 F3g.
De�nition 8.4. More generally, for any p we can de�ne Fp = f0; 1; 2; : : : ; p � 1g, the �nite�eld of order p, by taking arithmetic modulo p. We can also de�ne the k-dimensional Fp-vector space, denoted by Fkp.
De�nition 8.5. For any vector space F k (where F is any �eld such as Fp;C;Q;Q[p2], etc.),
we de�ne a linear subspace to be a nonempty subset which is closed under addition andmultiplication by scalars (elements of F ). We de�ne an a�ne subspace to be any translate of
39
40 CHAPTER 8.
a linear subspace (i.e. a linear subspace with origin moved to any given vector). (In particular,any linear subspace is also a�ne.) For any v1; : : : ; vj 2 F k, we can consider the subspace
Span(v1; : : : ; vs) =
(sX
i=1
�ivi : �i 2 F
): (8.1)
We can also consider the a�ne span,
A�(v1 : : : ; vs) = fsX
i=1
�ivi : �i 2 F;sX
i=1
�i = 1g: (8.2)
Note that the linear (resp. a�ne) span is the smallest linear (resp. a�ne) subspace containingthe given vectors. Also note that the span of v1; : : : ; vs is the same as the a�ne span ofv1; : : : ; vs; 0. Intuitively, the a�ne span is the smallest hyperplane of any dimension (notnecessarily passing through the origin) containing the given vectors, and the linear span is thesmallest hyperplane of any dimension which passes through 0 and contains the given vectors.
Observation 8.6. Note that if T is an a�ne subspace, and v 2 T , then T = U + v where Uis a linear subspace. Also, U is independent of the choice of v.
De�nition 8.7. If T is an a�ne subspace T = v + U where U is a linear subspace, then wede�ne dim(T ) = dim(U) =the maximum number of linearly independent vectors in U .
Now we will consider some enumerative geometry over �nite �elds Fp.
Exercise 8.8. First of all, show that jFkpj = pk.
Exercise 8.9. Prove that, for all x 6= y 2 Fkp, there exists a unique line through x; y, namely
A�(x; y). (A line is by de�nition a set fv + twgt2Fp , for a �xed choice of v; w 2 Fkp.)Exercise 8.10. Show that the total number of lines in Fkp is�pk
2
��p2
� =pk(pk � 1)=2
p(p� 1)=2=pk(pk � 1)
p(p� 1): (8.3)
Exercise 8.11. Show that the total number of planes (i.e. two-dimensional a�ne subspaces)in Fkp is
pk(pk � 1)(pk � p)
p2(p2 � 1)(p2 � p): (8.4)
De�nition 8.12. For any k and p, we de�ne AG(k; p) (\a�ne geometry") to be the set Fkpwith its a�ne lines, planes, etc.
Exercise 8.13. Verify that the \set" cards form the space AG(4; 3), where the SETs are a�nelines. (cf. Lecture 5, 6/25/04)
8.1. STEINER TRIPLE SYSTEMS 41
Exercise 8.14. Verify that the SET game AG(k; 3) gives an STS for n = 3k.
For the relevant de�nition of projective space and the Fundamental Theorem of ProjectiveGeometry, see the Linear Algebra Methods in Combinatorics handout, pages 49 and 50.
Theorem 8.15 (Galois). A �nite �eld of order n exists i� n = pk:
Consequence 8.16. If n = pk; then there is a projective plane of order n:
Puzzle 8.17 (Euler's 36 O�cers Problem). Given n2 o�cers with n ranks and n divisionssuch that no o�cers have the same rank and division, put the o�cers in a n� n grid so thatneither rank nor division occurs twice in any row or column. It can be done for any primepower p but not for 6:
Exercise 8.18. Prove that the impossibility of Euler's 36 o�cers problem implies the nonex-istence of a projective plane of order 6:
42 CHAPTER 8.
Chapter 99th day, Wednesday 7/7/04 (Scribe: Ivona Bez�akov�a)
9.1 Extremal Set Theory
Notation: [n] = f1; : : : ; ng. The incidence vector of a set A � [n] is de�ned as �A 2 Fn2where (�A)i = 1 if and only if i 2 A. Standard inner product of two vectors a; b 2 Fn isde�ned as a � b =Pn
i=1 aibi. Notice that �A � �B = jA \Bj and �A � �A = jAj.Theorem 9.1 (Fisher's Inequality). Let A1; : : : ; Am � [n], and a � 1. If jAi \Aj j = a for
every i 6= j, then m � n.
In 1949 R.C. Bose proved Fisher's Inequality by showing linear independence of a well-chosen set of vectors over a well-chosen �eld, establishing the \linear algebra method" incombinatorics. The proof is based on the following lemma:
Lemma 9.2. The incidence vectors of the Ai are linearly independent.
If the size of one of the sets is a, e.g. jAij = a, then Aj � Ai for every j and the systemforms a sun ower (no two sets intersect outside the common intersection of all sets).
9.1.1 Eventown
Suppose there is a town of n citizens and there are m clubs A1; : : : ; Am � [n]. The lawmakerstend to create new laws and curiously examine the highest possible number of clubs under thecurrent set of rules.
Rules in the Eventown:
(0) The clubs have to be distinct.
(1) Each club has even number of members.
43
44 CHAPTER 9.
(2) jAi \Aj j is even for every i; j.
Under rule (0), there could be total 2n clubs. Under rules (0) and (1), the total numberof clubs drops to 2n�1. If all three rules need to be satis�ed, the \married couples solution"provides 2n=2 clubs. This solution is maximal, i. e., the solution cannot be extended, addinganother club would violate one of the rules. Is this solution also maximum, i. e., there is nosolution with higher number of clubs?
Exercise 9.3. In Eventown, m � 2bn=2c. (The \married couples solution" is maximum.)Hint. Prove that the statement follows from Exercise 9.5.
Exercise 9.4. Find another maximum solution which contains three clubs A1; A2; A3 suchthat jA1 \A2 \A3j is odd. (So not all maximum solutions are isomorphic.)
Exercise 9.5. In Eventown every maximal set of clubs is maximum.
9.1.2 Oddtown
Rules in Oddtown:
(1) jAij is odd for every i
(2) jAi \Aj j is even for every i 6= j
Exercise 9.6. The number of ways to create a system of n clubs in Oddtown is � 2n2=8.
Theorem 9.7 (Oddtown Theorem, Berlekamp). In Oddtown, m � n.
Lemma 9.8. Under Oddtown rules the incidence vectors of the clubs are linearly independent.
Exercise 9.9. Prove the previous lemma over the following �elds: (a) over Q, (b) over F2, (c)over R.
De�nition 9.10. For A � [n], the incidence vector (or characteristic vector) of A is thevector �A = (�1; : : : ; �n) where
�i =n 1 if i 2 A0 if i 62 A
:
Let B be a matrix with rows �Ai. Part (a) implies that B has a full rank over Q. Notice
that rank is invariant under extension of the �eld (Gaussian elimination process keeps allcoe�cients in the original �eld), therefore part (c) is a consequence of part (a). Notice thatF2 is not a sub�eld of Q. However, part (a) follows from (b) and the following exercise.
Exercise 9.11. Let A be a (0; 1)-matrix, i. e., its entries are from f0; 1g. Let rkp(A) denotethe rank of A over Fp and let rk0(A) be its rank over Q. Prove: rkp(A) � rk0(A).
9.1. EXTREMAL SET THEORY 45
De�nition 9.12. Vectors a; b 2 Fn are perpendicular, denoted a?b, if a �b = 0. For S � Fn,the set S? = fx j (8y 2 S)(x?y)g is called S-perp. A vector v 2 Fn is called isotropic ifv?v. A set U � Fn is totally isotropic if U?U .Exercise 9.13. Prove S? = (Span S)?.
Exercise 9.14. Let U � Fn. Prove: dim(U) + dim(U?) = n.
Let U � Fn2 be a maximal set of clubs in Eventown. Then U � U? and therefore Span(U) �Span(U)?. Since U is maximal, we can conclude that U = Span(U). By previous exercise,
dim(U) � n=2. Therefore jUj � jFn=22 j = 2n=2, i. e., the married couples solution is optimal.
Exercise 9.15. For what p does there exist a nonzero isotropic vector in F2p? (Answer isappealing.)
Exercise 9.16. Prove that there exists an n-dimensional totally isotropic subspace over C2n.
Question: How many sets can have pairwise intersection of size 0 or 1? If we take all setsof size at most 2, we get a set system of
�n2
�+ n+ 1 sets.
Exercise+ 9.17. Let A1; : : : ; Am � [n] be such that jAi \ Aj j � 1 for i 6= j. Prove: m ��n2
�+ n+ 1
Hint. Use linear algebra method. The trick lies in �nding a good set of vectors in dimension�n2
�+ n+ 1.
46 CHAPTER 9.
Chapter 10
10th day, Friday 7/9/04 (Scribe: Ivona Bez�akov�a)
10.1 Cauchy-Hilbert Matrix
The Hilbert matrix is de�ned as:266664
11
12
13 : : : 1
n12
13
14 : : : 1
n+1
......
1n
1n+1
1n+2 : : : 1
2n�1
377775
The Hilbert matrix is a special case of a Cauchy matrix:2666664
1�1��1
1�1��2
1�1��3 : : : 1
�1��n1
�2��11
�2��21
�2��3 : : : 1�2��n
......
1�n��1
1�n��2
1�n��3 : : : 1
�n��n
3777775
Exercise 10.1. Prove: If all �i, �i are distinct then the Cauchy-Hilbert matrix has full rank.
10.2 Cauchy's Functional Equation
Let f : R! R be a real function.
De�nition 10.2. Cauchy's functional equation speci�es:
(8x; y)(f(x+ y) = f(x) + f(y)):
47
48 CHAPTER 10.
Which functions are solutions to Cauchy's functional equation (CFE)? Clearly, f(x) = cxis a solution.
We will make several observations: if f satis�es CFE then
1. f(0) = 0. (Why?)
2. Let c := f(1). Then (8x 2 Z)(f(x) = cx).
3. f (p=q) = c � p=q for p; q 2 Z. (Why?)
Therefore (8x 2 Q)(f(x) = cx).
Exercise 10.3. If f is continuous then (8x)(f(x) = cx).
Exercise 10.4. If f is a solution of CFE and it is bounded in an interval then (8x)(f(x) = cx).
This is a known fact from linear algebra:
De�nition 10.5. Let F be a �eld and U;W vector spaces over F. A function ' : V ! W iscalled linear if (8x; y 2 V )('(x+ y) = '(x) + '(y)) and (8x 2 V )(8� 2 F)('(�x) = �'(x)).
Lemma 10.6. Let F be a �eld and V;W be two vector spaces over F. Suppose B � V be a
basis and ' : B ! W be an arbitary function. Then ' can be uniquely extended to a linear
function ' : V !W .
Hamel basis H is a basis of R over Q (its existence is implied by Zorn's Lemma). Itscardinality is continuum, denoted as c. CFE translates to �nding all Q-linear (linear over Q)functions f : R! R. Using previous lemma we can de�ne the function f by assigning a valuef(h) to every vector h in the Hamel basis.
Corollary 10.7. There are solutions to CFE not of the form f(x) = cx. In fact in �1; �2; : : :are linearly independent over Q then f(�1); f(�2); : : : can be independently prescribed.
10.3 Number Theory
Exercise 10.8. Pick two positive integers x; y at random, what is the probability that x; y arerelatively prime? (Hint. What does \integer chosen at random" mean? We want all integersto be chosen with the same probability. First make sense out of the problem statement. Youwill de�ne the probability as a limit. Assume the limit exists. Then use
P1n=1
1n2
= �2
6 .)
A little diversion:P1
n=11n2
= �2
6 was stated by Fermat and proved by Leonhard Euler.Other examples of Fermat's theorems that Euler worked on include Fermat's Last Theoremstating that xn + yn = zn, n � 3 implies xyz = 0 (recently proved by Andrew Wiles), and thefollowing remarkable result:
10.4. RAMSEY THEORY 49
Exercise 10.9. Let p > 2 be a prime. Prove that p = a2 + b2 for a; b 2 Z if and only if p � 1mod 4.
Exercise 10.10. Prove that Fp[p�1] = fa+ bi j a; b 2 Fpg (\complex numbers mod p") is a
�eld if and only if p � �1 mod 4.
Exercise 10.11. Let faig1i=0 be de�ned as a0 = 1, an+1 =p2an. Prove that (8n)(an < 2).
Exercise 10.12. Prove that an ! 2 for the ai from the previous exercise.
Exercise 10.13. Find the maximum r s.t. rrr:::
is bounded.
Exercise 10.14. Find two in�nite sets A;B � N = f0; 1; 2; : : : g such that for every n 2 Nthere exist unique a 2 A and b 2 B such that a + b = n. (Hint. Look at the digits of theintegers. Recall Hales-Jewett's implies the Van der Waerden's Theorem.)
10.4 Ramsey Theory
By Erd}os-Szekeres N =�k+`�2
k�1�! (k; `), i.e. if G is a graph with N vertices then G contains
a Kk or G contains a K`. Tur�an's example, a graph of n2 vertices consisting of n independentcopies of Kn, shows that n
2 6! (n + 1; n + 1). The following theorem states a much strongerresult.
Theorem 10.15 (Erd}os). 2n=2 6! (n+ 1; n+ 1)
In other words, N 6! (2 log2N + 1; 2 log2N + 1). The proof is based on the \probabilisticmethod."
Corollary 10.16. If 2�nk
� � 2(k
2) then n 6! (k; k).
Exercise 10.17. Prove:�nk
� � nk
k! .
We know that 4n ! (n + 1; n + 1) (why?), i.e. N ! (1=2 logN; 1=2 logN). By the aboveN 6! (2 logN +1; 2 logN +1). There is a multiplicative gap of 4 between the lower bound andthe upper bound.
Open problems.
(1) Shrink this gap: either proveN ! ��12 + "
�N;�12 + "
�N�or proveN 6! ((2� ")N; (2� ")N).
(2) Prove: (9c)(8" > 0)(9n0)(8n > n0)�n! ((c� ")n; (c� ")n) and n 6! ((c+ ")n; (c+ ")n)
�
50 CHAPTER 10.
Chapter 1111th day, Tuesday 7/13/04 (Scribe: Ivona Bez�akov�a)
11.1 Totally unimodular matrices
De�nition 11.1. A matrix is totally unimodular if the determinant of every square sub-matrix is 0, 1, or �1.De�nition 11.2. Let G be a digraph with n vertices and m edges. The incidence matrixA of G is a n �m matrix with rows corresponding to vertices and columns corresponding toedges s.t.
A[v; e] =
8<:
0 if v is not incident to e;1 if e starts at v;�1 if e ends at v:
Exercise 11.3. Prove: The incidence matrix of a digraph is totally unimodular.
11.2 Latin squares
De�nition 11.4. Let S be an n-element set of symbols. A n� n matrix with entries from Sis called a Latin square if each symbol appears exactly once in each row and in each column.L(n) denotes the number of n� n Latin squares.
Theorem 11.5. For every " > 0 there exists n0 such that for every n > n0
n(1�")n2 � L(n) � nn
2
Rephrasing the theorem we get logL(n) � n2 log n.
We say that two Latin squares A and B are equivalent, denoted A � B, if A is ob-tained from B by a sequence of the following: (1) permuting rows, (2) permuting columns, (3)permuting symbols, (4) permuting the roles of rows, column indices, and symbols.
An example of type (4) equivalence would replace entry a in row b, column c by entry c inrow a, column b.
51
52 CHAPTER 11.
Exercise 11.6. The number of Latin squares equivalent to a given Latin square is � 6(n!)3.
Exercise 11.7. Let ~L(n) be the number of inequivalent Latin squares of order n. Provelog ~L(n) � logL(n).
It is easy to prove the upper bound from Theorem 11.5, since L(n) < (n!)n < nn2(why?).
The lower bound is related to counting perfect matchings in a bipartite graph.
11.3 Counting perfect matchings in a bipartite graph; the per-manent
A graph G is bipartite if its vertex set can be partitioned into V1 and V2 so that every edgein G has one endpoint in V1 and the other in V2. A matching M � E(G) is a set of edgessuch that no two edges in M share an endpoint. A matching is perfect if it contains jV (G)j=2edges.
De�nition 11.8. Let A = (�i;j)n�n be a matrix. The permanent of A is de�ned as
perA =X�2Sn
nYi=1
�i;�(i);
where Sn is the set of all permutations of [n] = f1; : : : ; ng.Let G be a bipartite graph with partitions V1 and V2 where V1 = [k1] � f1g and V2 =
[k2]� f2g. The incidence matrix of G is a k1 � k2 matrix M = (�i;j) de�ned by
�i;j =
�1 if (i; 1) � (j; 2), i. e., (i; 1) is adjacent to (j; 2);0 otherwise:
Theorem 11.9. Let G be a bipartite graph with jV1j = jV2 and let M be its incidence matrix.
Then the number of perfect matchings of G is per(M).
Theorem 11.10. A regular bipartite graph of degree r with n+n vertices has > (r=e)n perfect
matchings.
11.4 Doubly Stochastic Matrices
De�nition 11.11. A n� n matrix is called stochastic if all its entries are nonnegative andevery row sums to 1. Matrix A is doubly stochastic if both A and AT are stochastic, i. e.,the entries are nonnegative and every row and every column sums to 1.
11.4. DOUBLY STOCHASTIC MATRICES 53
Let I be the identity matrix and J be the all-ones matrix.
I =
0BBB@
1 0 : : : 00 1 : : : 0...
. . ....
0 0 : : : 1
1CCCA J =
0BBB@
1 1 : : : 11 1 : : : 1...
. . ....
1 1 : : : 1
1CCCA
Example 11.12. I and 1nJ are doubly stochastic.
Exercise 11.13. n! > (n=e)n. (Hint: ex =P1
n=0 xn=n!)
Clearly, per(I) = 1 and per( 1nJ) =n!nn > 1
en .
Comment on how not to prove this inequality: Notice that the Stirling's formula
n! ��ne
�np2�n
implies the inequality n! > (ne )n only for su�ciently large n, while the above method proves
it for every n. A more precise Stirling's formula
n! =�ne
�np2�n
�1 +
#n12n
�;
where j#nj � 1, would imply the inequality for every n.
Exercise 11.14. If A is doubly stochastic then per(A) � 1. The equality holds exactly whenA is a permutation matrix (i. e., A has exactly one 1 in each row and column, otherwise zeros.)
Theorem 11.15 (The Permanent Inequality). If A is doubly stochastic then
perA � per
�1
nJ
�=
n!
nn:
The theorem was conjectured by van der Waerden and was known as van der Waerden'sPermanent Conjecture. It was proved independently by Egorichev and Falikman in 1980.
Exercise 11.16. LetM be the incidence matrix of a regular bipartite graph of degree r. Provethat 1
rM is doubly stochastic.
We show that the permanent inequality implies Theorem 11.10. Notice that per(�A) =�n per(A). Let M be the incidence matrix of the bipartite graph from Theorem 11.10. Byapplying the permanent inequality and Exercise 11.13 we get
per(M) = rn per
�1
rM
�> rn
1
en:
54 CHAPTER 11.
11.5 Back to Latin squares
De�nition 11.17. For k � n, a k � n Latin rectangle is a k � n matrix of n symbols suchthat every symbol appears at most once in each row and in each column. L(k; n) denotes thenumber of k � n Latin rectangles.
Obviously, L(n; k) � (n!)k and L(n; 1) = n!.
Exercise 11.18. Prove: L(n; 2) � (n!)2
e .
Theorem 11.19. Let k < n. Every k � n Latin rectangle can be extended to a (k + 1) � nLatin rectangle.
Theorem 11.20. Every regular bipartite graph of degree � 1 has a perfect matching.
Exercise 11.21. Prove Theorem 11.19 using Theorem 11.20.
Suppose we have k � n Latin rectangle L. De�ne a bipartite graph G on n+ n vertices asfollows. An edge goes from (i; 1) to (j; 2) if the i-th column in L does not contain the symbolj. This way we obtain a regular bipartite graph (of degree n � k). The perfect matchings ofG correspond to all possible extensions of L to a (k + 1)� n Latin rectangle.
Therefore L(n; k + 1) � L(n; k)(n�ke )n, implying
L(n) = L(n; n) ��ne
�n�n� 1
e
�n�n� 2
e
�n
: : :
�1
e
�n
=(n!)n
en2:
Exercise 11.22. Prove: n ln(n!)� n2 � n ln(n!); therefore logL(n) � n2 log n.
11.6 Orthogonal Latin Squares
De�nition 11.23. Let A and B be two n� n Latin squares. A and B are orthogonal if theset f(ai;j ; bi;j) j i; j 2 [n]g contains all the n2 distinct pairs, i. e., no pair (ai;j ; bi;j) is repeated.
Orthogonal Latin squares are related to Euler's \36 o�cers" problem: There are 36o�cers, 6 divisions and 6 ranks. The task is to assign a division and a rank to every o�cer sothat no two o�cers are assigned the same combination of division and rank; and arrange theo�cers in 6� 6 array such that each division is represented in each row and each column andeach rank is represented in each row and each column.
Exercise 11.24. Let p be a prime. Prove that there exist p � 1 pairwise orthogonal Latinsquares of order p.
Exercise 11.25. Prove previous exercise for p a prime power.
11.6. ORTHOGONAL LATIN SQUARES 55
Theorem 11.26 (Tarry'1900). Euler's 36 o�cers problem does not have a solution, i. e.,
there does not exist a pair of orthogonal 6� 6 Latin squares. (Proof: tedious.)
Exercise 11.27. Prove that for every n there are at most n � 1 pairwise orthogonal n � nLatin squares.
Exercise 11.28. Show that the following are equivalent: (a) there exists a set of n�1 pairwiseorthogonal n� n Latin squares, (b) there exists a projective plane of order n.
Exercise 11.29. Prove:�nk
� � (nk )k.
Exercise 11.30. Prove:�nk
�< ( enk )
k. Find an elegant solution, like the proof of n! > (ne )n.
Exercise 11.31. Prove: ( enk )k >
�nk
�+� nk�1�+ � � �+ �n0�.
Exercise 11.32. In Rn �nd cn2 points with only two pairwise distances.
The vertices of a regular pentagon are an example of a set in R2 with only two pairwisedistances.
Exercise 11.33. Prove: Any six points in the plane span at least three di�erent distances.
56 CHAPTER 11.
Chapter 1212th day, Thursday 7/15/04 (Scribe: Ivona Bez�akov�a)
12.1 Constructive Proofs of Negative Results in Ramsey The-ory
Recall that we discussed the following results:
� 4n ! (n+ 1; n+ 1), proved by Erd}os-Szekeres,
� 2n=2 6! (n+ 1; n+ 1), showed by Erd}os using a probabilistic proof of existence.
So far we have seen only one constructive proof of a negative result, the trivial exampleobserved by Tur�an showing that n2 6! (n+ 1; n+ 1).
Theorem 12.1 (Zsigmond Nagy, 1973).�n3
� 6! (n+1; n+1) can be proved constructively.
Nagy's construction de�nes the graph as follows. The vertices are all the 3-element subsetsof [n]. Sets A and B are adjacent if jA\Bj = 1. The Erd}os-deBruijn Theorem, a special case ofFisher's inequality for sets with intersection size 1, implies that Nagy's graph does not containa clique of size n + 1. By the Oddtown Theorem the graph does not contain an anticlique ofsize n+ 1.
The best known explicit construction is by Frankl and Wilson.
Theorem 12.2 (Frankl - Wilson). Let " > 0. For su�ciently large n one can construct a
graph of at most n(1�")lnn
4 ln lnn vertices with no clique or anticlique of size n+ 1.
Let p be a prime. The vertices in the Frankl - Wilson graph are all the subsets of [2p2 � 1]of size p2 � 1. Sets A and B are adjacent if jA \Bj � �1 mod p.
Claim 12.3. The Frankl - Wilson example proves that�2p2 � 1
p2 � 1
�6!�2p2 � 1
p� 1
�+ 1:
57
58 CHAPTER 12.
The proof of the claim is based on two theorems on extremal set theory:
Theorem 12.4 (Ray-Chaudhuri - Wilson, 1975). Fix k and let l1 < � � � < ls < k. If
A1; : : : ; Am � [n] are sets of size k such that jAi \ Aj j 2 fl1; : : : ; lsg for every i 6= j, thenm � �ns�.Exercise 12.5. Prove that the Ray-Chaudhuri - Wilson Theorem is tight, i.e. �nd
�ns
�sets
with s di�erent intersection sizes.
Frankl and Wilson generalized the Ray-Chaudhuri - Wilson Theorem:
Theorem 12.6 (Frankl - Wilson, 1981). Let p be a prime and let k; l1; : : : ; ls 2 Zp be suchthat k 6� l1; : : : ; ls mod p. If A1; : : : ; Am 2 [n] are sets of size k such that jAi\Aj j 2 fl1; : : : ; lsgmod p for every i 6= j, then m � �ns�.
A clique in the Frankl - Wilson graph corresponds to a set system A1; : : : ; Am, jAij = p2�1for every i, such that jAi \Aj j 2 fp� 1; 2p� 1; : : : ; p(p� 1)� 1g. Thus the Ray-Chaudhuri -Wilson Theorem implies that m � �2p2�1p�1
�.
An anticlique corresponds to a set system B1; : : : ; Bm � [2p2 � 1], jBij = p2 � 1 � �1mod p for every i, such that jBi \ Bj j 2 f0; 1; : : : ; p � 2g mod p. By the Frankl - Wilson
Theorem, m � �2p2�1p�1�.
12.2 Bipartite Ramsey Theory
We de�ne a bipartite version of the Erd}os-Rado arrow.
De�nition 12.7. We say that a (b; c) if every bipartite graph G with a vertices contains abipartite clique Kb;b or the complement G contains a bipartite clique Kc;c.
Exercise 12.8. Prove: 4n (n+ 1; n+ 1).
Exercise 12.9. Prove: 2n=2 6 (n+ 1; n+ 1).Hint. Probabilistic proof of existence.
12.3 Hadamard Matrices
Theorem 12.10 (Hadamard's Inequality). Let A 2Mn(R), i.e. A is a n� n matrix over
R. Then
j det(A)j �nYi=1
kaik;
where ai is the vector in the i-th row of A and kaik =qPn
j=1 a2i;j is its norm. The equality
holds if and only if there exists a zero row or if the rows are pairwise orthogonal.
12.3. HADAMARD MATRICES 59
De�nition 12.11. An Hadamard matrix is a �1-matrix with all rows orthogonal.
The Sylvester matrices are 2k � 2k matrices de�ned by the following matrix recurrence.
H0 = (1)
Hk+1 =
�Hk Hk
Hk �Hk
�for k > 0
Exercise 12.12. Prove: Hk is an Hadamard matrix.
De�nition 12.13. n is an Hadamard number if there exists a n� n Hadamard matrix.
Exercise 12.14. If n is an Hadamard number then n � 2 or n is divisible by 4.
Exercise+ 12.15. If p � �1 mod 4 is a prime power, then p+ 1 is an Hadamard number.Hint. Quadratic residues.
Exercise 12.16. If H is an Hadamard matrix then HT is also an Hadamard matrix.Hint. Examine HHT .
Exercise 12.17. Construct a matrix with all rows orthogonal such that the columns are notorthogonal.
De�nition 12.18. Let M = (mi;j) be a n � n matrix and let I; J � [n]. A rectangleis a submatrix of M corresponding to rows de�ned by I and columns de�ned by J . Thediscrepancy of the rectangle is de�ned as
discI;J(M) =
������X
i2I;j2Jmi;j
������ :Theorem 12.19 (Lindsay's Inequality). Let a = jIj and b = jJ j. If H is an Hadamard
matrix then
discI;J(H) �pnab
Note 12.20 (Back to bipartite Ramsey numbers.). Let M be an incidence matrix of abipartite graph, with zeros replaced by �1. The rectangle corresponding to a bipartite cliqueor anticlique of size t has discrepancy t2. By Lindsay's Inequality, t2 � t
pn, so t � p
n.Therefore
n2 6 (n+ 1; n+ 1):
The proof of Lindsay's inequality is based on the following facts:
Theorem 12.21 (Cauchy-Schwarz Inequality). Let a;b 2 Rn. Then
ja � bj � kakkbk;where � denotes the standard inner product.
60 CHAPTER 12.
De�nition 12.22. A matrix K is orthogonal if KT = K�1.
Lemma 12.23. If a n�n matrix K is orthogonal, then kKxk = kxk for every vector x 2 Fn.
Proof: [of Lindsay's Inequality] Let eS 2 Rn be the characteristic vector of the set S � [n].Then X
i2I;j2Jmi;j = e�IMeJ
If H is Hadamard then 1pnH is orthogonal. By Cauchy-Schwarz and the above lemma,
discI;J(H) = je�IHeJ j � keIkkHeJk =papnb
Exercise 12.24. Prove: If a and b are Hadamard numbers then ab is a Hadamard number.
Conjecture 12.25. If n is divisible by 4 then n is a Hadamard number.
De�nition 12.26. The upper density of a set A � N is
lim supn!1
jA(n)=nj;
where A(n) = fx 2 A jx � ng. The lower density is
lim infn!1 jA(n)=nj:
We say that A has density if is both the lower and the upper density.
Exercise 12.27. Construct a set with lower density 0 and upper density 1.
Exercise 12.28. The upper density of Hadamard numbers is � 1=4.
The Conjecture 12.25 would imply that the density of Hadamard numbers is 1=4.
OPEN PROBLEM: Is the (upper) density of Hadamard numbers positive? The Sylvestermatrices example shows that there are in�nitely many Hadamard matrices. The quadraticresidue Hadamard matrices give asymptotic density 1
2 lnn . However, the density of this set isstill 0.
Chapter 13
13th day, Tuesday 7/20/04 (Scribe: Charilaos Skiadas)
13.1 The Gale-Berlekamp Switching game
In this game there are 2 players and they play on a n�n checkerboard. There are exactly twomoves in this game. In the �rst move, player 1 assigns a sign (�1) to every square. Then, thesecond player switches the signs in any rows and columns he wants. After he �nishes, theysum all numbers in the �nal con�guration and look at the absolute value of the result. Thisis the amount the 1st player pays to the 2nd player. So naturally, the 1st player wants tominimize this number, while the 2nd player wants to maximize it. We want to know what thevalue of this game is, at least asymptotically in n. The value is the amount that the player 2is guaranteed to receive, or equivalently the largest amount that the player 1 will have to pay,assuming both players follow optimal strategies. This value is certainly is no more than n2.Player 2 can easily achieve n in the following way: Make everything in the �rst row a plus,by switching columns. Then, if the sum is a, we could switch the �rst row to get sum a� 2n.Since maxfjaj; ja� 2njg � n, we see that the value of the game is at least n. The question is:Can he achieve more?
Let us �rst make some observations: First of all, the order in which player 2 makes theswitches does not matter, so we can assume that he switches the columns �rst, and thenswitches the rows. After he has switched the columns, then his best strategy is to switch therows with negative sum, so that all the rows will have nonnegative sum. In particular, hisrow-switching moves are completely determined (except for the rows with zero sum, whereswitching makes no di�erence). If player 2 could arrange it so that many row sums are large,then he could improve his win.
So our goal should be to switch the columns so as to maximize the absolute value of thesum on an \average" row. Our goal is to achieve approximately c
pn gain per row. So this
would give c0n3=2 total gain. The answer is to randomly decide whether to switch a column ornot. The main observation is that a random sequence of n pluses and minuses has expectedabsolute sum �(
pn):
61
62 CHAPTER 13.
Assume for simplicity that n is even. We see that the probability
P (n coin ips, k heads) =
�nk
�2n
:
Asymptotically, �nn2
�2n
=
n!(n2)!2
2n= : : : �
r2
�
1pn=
cpn;
were we used Stirling's formula to estimate the factorials. So
P (#heads withinn
2� k) =
� nn2�k�+ � � �+ � n
n2+k
�2n
:
There are precise estimates for this sum, but for our purposes, a rough estimate is enough.Since the middle binomial coe�cient is the largest one, we see that the above sum is less than
(2k + 1)�nn2
�2n
� (2k + 1)
r2
�
1pn:
As long as (2k+1)q
2�
1pn� 1
2 , i. e., as long as k < cpn, where c :=
p�
4p2, the above probability
is less than 12 . Then P (#heads is within n
2 � cpn) is asymptotically at most 1
2 .
(Note: The Central Limit Theorem, says that asymptotically, the binomial coe�cients �ton a bell curve, and can hence obtain a better value for the constant above.)
So if in our game the 2nd player randomly switches the columns, then we expect at leastroughly half of the rows to have absolute sum greater than c
pn.
This means that on average, at least n2 of the rows have this advantage. But this means
that it is possible for player 2 to achieve that advantage, which would give him overall at least12cn
3=2 total gain.
The question now arises: Can we recommend a strategy for player 1 to force the total to beno more than that order? One could argue in a similar way with a random choice, but in thiscase there actually is a deterministic strategy. The idea is to use Hadamard matrices. Recallthat a Hadamard matrix is a matrix with entries only �1, where the rows are orthogonal. As aconsequence, all columns are orthogonal. We know that such matrices exist when n = 2k andwhen n = p+ 1, where p is a prime and p � �1 mod 4. If H is a Hadamard matrix, then Hprovides a good strategy for player 1: Any switch of rows or columns gives again a Hadamardmatrix. Then Lindsay's lemma tells us that a a� b submatrix has sum of entries less than orequal to
pnab in absolute value. For a = b = n, we get that the total sum of the entries in a
Hadamard matrix is no more than n3=2. Hence, in this case player 2 can't achieve a total gainof more than n3=2.
If n is not itself a Hadamard number, just take a Hadamard number ` larger than it,no bigger than 2n. (Doable, since between n and 2n there is always a power of 2.) Then
13.2. THE RAY-CHAUDHURI - WILSON THEOREM 63
take a Hadamard matrix of size ` � `, and choose your matrix to be any n � n submatrix ofthat. Then any row and column changes that player 2 performs, we can assume that they areperformed to the whole matrix. By Lindsay's lemma we then get that the sum is no morethan
p`n2 <
p2n3=2. (Note: If we use Hadamard matrices of size p + 1, we can actually get
1+ " as the constant instead ofp2.) Notice that there is a gap between the constants we have
obtained in the deterministic upper bound and the probabilistic lower bound for the value.While this gap can be reduced, the instructor believes it has not been closed: the lower bounddi�ers from the upper bound by cn3=2.
13.2 The RAY-CHAUDHURI - WILSON theorem
See separate handout.
64 CHAPTER 13.
Chapter 1414th day, Thursday 7/22/04 (Scribe: Ivona Bez�akov�a)
14.1 Points in general position
Exercise 14.1. Find a curve f : R ! Rn mapping any n distinct points into n points in\general position". In other words, if t1; : : : ; tn 2 R are distinct, then f(t1); : : : ; f(tn) arelinearly independent.
Recall the de�nition of the Vandermonde determinant and the corresponding closedform expression
Vn(t1; : : : ; tn) :=
���������
1 t1 t21 : : : tn�11
1 t2 t22 : : : tn�12...
...1 tn t2n : : : tn�1n
���������=Yi6=j
(ti � tj)
Then a solution to the above exercise is given by f(t) = (f0(t); : : : ; fn�1(t)) = (1; t; t2; : : : ; tn�1).
14.2 Pattern in proofs of inequalities using the Linear AlgebraMethod
The pattern can be summarized as follows. We have objects a1; : : : ; am 2 , where is someabstract domain. We want to obtain an upper-bound on m. We de�ne functions f1; : : : ; fm :!W , where W is a vector space, satisfying the \diagonal condition"
fi(aj) =
� 6= 0 if i = j0 if i 6= j
The set of functions ! W forms a vector space in W. The diagonal condition implieslinear independence of f1; : : : ; fm. If in addition we �nd another set of functions g1; : : : ; gk
65
66 CHAPTER 14.
such that f1; : : : ; fm 2 Span(g1; : : : ; gk) then m � k follows by the Fundamental Fact of LinearAlgebra. A weaker condition su�ces for this.
Claim 14.2. Suppose the fi satisfy \triangular condition":
fi(aj) =
� 6= 0 if i = j0 if i > j
Then the fi are linearly independent. (Note that we know nothing about fi(aj) when i < j.)
Exercise 14.3. Prove Claim 14.2.
Claim 14.4. If W = F and the matrix (fi(aj))mi;j=1 is nonsigular then the fi are linearly
independent.
Exercise 14.5. Prove Claim 14.4.
Exercise 14.6. Assuming W = F, prove that the following statement is false: If the fi arelinearly independent then the matrix (fi(aj))
mi;j=1 is nonsigular.
The following theorem is a nonuniform version of the Ray-Chaudhuri { Wilson's Theorem(the sets do not have to be equal size.)
Theorem 14.7 (Frankl-Wilson). Let A1; : : : ; Am � [n] be such that (8i 6= j)(jAi \ Aj j 2f`1; : : : ; `sg). Then
m ��n
s
�+
�n
s� 1
�+ � � �+
�n
0
�
The proof is a modi�cation of the proof of the uniform case (all the Ai are of the samesize.) We cannot use the same functions fi(x) to prove the linear independence. The originalfi were de�ned as
fi(x) =sY
t=1
(vi � x� `t);
where x; vi 2 Rn, vi is the incidence vector of Ai. Notice that fi(vi) 6= 0 only if jAij 62f`1; : : : ; `sg. Therefore we de�ne gi(x) as follows
gi(x) =Y
t: `t<jAij(vi � x� `t)
Then the gi satisfy
gi(vj) =
� 6= 0 if i = j0 if jAi \Aj j < jAij, i.e. Aj 6� Ai
If we reorder the Ai so that Aj 6� Ai for i > j, then the condition on gi is a triangularcondition. Therefore we could conclude that the gi are linearly independent. Such an orderingexists: it su�ces to order the Ai by their size | let jA1j � jA2j � � � � � jAmj. From this pointon the proof follows along the lines of the uniform case.
14.3. ADDITIONAL EXERCISES IN PROBABILITY THEORY 67
14.3 Additional Exercises in Probability Theory
Exercise 14.8. A poker hand is a set of �ve cards. We say that the poker hand has a pairif there are two cards of the same kind (two Kings, or two 9s, for example). Compute theprobability that a poker hand has at least one pair.
De�nition 14.9. Two random variables are uncorrelated if E(XY ) = E(X)E(Y ).
Exercise 14.10. Let c and d be constants. Prove: If X and Y are uncorrelated then X + cand Y + d are also uncorrelated.
Exercise 14.11. Suppose X1; : : : ; Xk are non-zero pairwise uncorrelated random variableswith E(Xi) = 0. Prove: X1; : : : ; Xk are linearly independent (over R).
Exercise 14.12. If X1; : : : ; Xm are random variables with E(Xi) = 0 then there exist pair-wise uncorrelated random variables Y1; : : : ; Yk with E(Yi) = 0 such that Span(X1; : : : ; Xm) =Span(Y1; : : : ; Yk).
The process of �nding Y1; : : : ; Yk is called factor analysis in statistics. The Yi are calledfactors of the Xi.
Exercise 14.13. Prove: If there exist k independent non-constant random variables over aprobability space , then jj � 2k.
Exercise 14.14.
(a) Prove: If X1; : : : ; Xk are non-constant pairwise independent random variables over ,then jj � k + 1.
(b) Prove that this inequality is tight for k = 2` � 1.
68 CHAPTER 14.
Chapter 1515th day, Friday 7/23/04 (Scribe: Eric Patterson)
15.1 Bases for Vector Spaces of Polynomials
Suppose there are n cookies to distribute among k children. How many ways are there todistribute the cookies? First, consider the number of ways to distribute the cookies so thateach child gets at least one cookie. If we suppose the i-th child gets xi cookies, then we cansolve the problem by looking for a solution to the equation
Pki=1 xi = n such that the unknowns
are integers xi � 1, i 2 f1; : : : ; ng:If the children are in order and we put the cookies in an order, all we have to do is insert
partitions among n objects. The �rst set of cookies ends at the �rst inserted partition, andthis set goes to the �rst child. The second set of cookies begins at the �rst inserted partitionand ends at the second inserted partition, and this set goes to the second child, and so on. Forn cookies, there are n� 1 slots between cookies, and for k children, there are k � 1 partitionsthat we must insert, so we get
�n�1k�1�ways to distribute the cookies.
Suppose now that we want to allow the possibility that some children might not get acookie. Then we are trying to solve
Pki=1 yi = n with yi � 0: By giving one \dummy" cookie
to each child before we distribute the cookies, we can reduce this to the previous problem.Instead of n cookies, however, we need n+ k cookies, so then we get
�n+k�1k�1
�=�n+k�1
n
�ways
to distribute the cookies.
Now, suppose we look at homogeneous polynomials in r variables of degree d. They forma vector space with a basis of all monic monomials of degree d in the r variables x1; : : : ; xr.To compute the dimension of this space, we need to count those monomials.
Denote the number of such monomials by H(r; k;F), where F is the �eld over which we aretaking these polynomials. This number is the same as the number of solutions in the secondcookie problem, but now we have degree d instead of n cookies, which we must divide betweenr variables instead of k children. Thus, we get
�d+r�1r�1
�monomials of degree d in r variables.
Now let P (r; d;F) be the space of polynomials of degree � d in x1; : : : ; xr. The conditionhere is that we are looking for ai � 0 such that
Pri=1 ai � d. We know how many ways we
69
70 CHAPTER 15.
can do this for each j � d; so the total dimension is just the sum. To computePd
j=0
�j+r�1r�1
�;
we can use Pascal's triangle. In Pascal's triangle, the sum of two adjacent entries is the entrybetween them on the next line. If we look at the entries
�j+r�1r�1
�in Pascal's triangle, we can
use the preceding fact to calculatePd
j=0
�j+r�1r�1
�=�r+d
r
�:
As an alternative method, we can use a \dummy" child xr+1 to get the remaining cookiesnot distributed. That is, if we have xa11 � � �xarr such that
Pri=1 ai � d; the we can insert
xd�Pr
i=1 air+1 to get a monomial in r + 1 variables of degree d: Conversely, given a monomial of
degree d in r+1 variables we can replace the r+1st variable by 1 to get a monomial of degreed in r variables. This directly gives
�r+dr
�as P (r; d;F):
15.2 Projective Representation of a Graph
De�nition 15.1. A projective representation of a graph G consists of the followinginformation. Let W be a vector space over a �eld F. To every vertex x in G; we assign asubspace U(x) �W in such a manner, that U(x) \ U(y) 6= f0g i� x � y.
We want to minimize the dimension of the spaceW where a graph has such a representation.
De�nition 15.2. The projective dimension of a graph G over a �eld F is
pdimF(G) = minfdimW : a projective representation of G in W existsg:
For example, for the empty graph �Kn, pdimF( �Kn) = 0. For the complete graph Kn,pdimF(Kn) = 1 by having each vertex correspond to the whole space.
For a cycle, say a 5-cycle, we produce a projective representation in F4: Assign nonzerovectors to each edge, and take the subspace corresponding to a vertex to be the subspacespanned by the vectors assigned to the incident edges. Since the subspaces associated toadjacent vertices will share a nonzero vector, the subspaces will have nonzero intersection. Toprevent subspaces corresponding to nonadjacent vertices from having nonzero intersection, weneed that any four of the vectors assigned to the edges be linearly independent, i.e., we needthe �ve vectors u1; : : : ; u5 assigned to the edges to be in general position.
We know we can produce as many points in general position in our space as the numberof elements in our �eld. So pdimF(Cn) � 4 if jFj � n.
Now look at a perfect matching graph with n vertices n2K2. If the �eld F has enough
elements (at least n2 � 1), its projective dimension is 2: send each pair to a di�erent line in F2:
Theorem 15.3. Over any �eld, every graph has a projective representation
Proof: Consider a vector space with basis a set of vectors in bijection with the set of edgeson the graph. Then assign each vertex to the subspace spanned by the vectors corresponding
15.3. THE NUMBER OF ZERO-PATTERNS OF A SEQUENCE OF POLYNOMIALS 71
to the incident edges. Actually we can do better: if � = maximum degree, then dimW = 2�su�ces because all we need is, from a vector space of that dimension, to pick the vectors sothat they are in general position. This we can do as long as jFj � m, where m is the numberof edges.
So, in fact we have shown that:
Theorem 15.4. pdimFG � 2�:
In particular, for a bipartite graph, we get that its projective dimension is less than orequal to n.
Exercise 15.5. Improve this bound by a lot (at least to o(n)).
Exercise 15.6. Give a logarithmic upper bound on the complement of a perfect matching,i.e., show dim
�n2K2
� � O(log n):
Now, can we guarantee a lower bound on the projective dimension? (This question isrelated to sub-linear space complexity).
Given a machine with a big read-only table with n bits, and small read-write tables, giveexample of something that can't be computed in sub-linear space (o(n)).
The best lower bounds we have are log(n). We can prove though that almost all graphshave at least
pn lower bound.
OPEN PROBLEM(Concept introduced by Pudl�ak-R�odl): Construct an explicit familyof graphs with projective dimension greater than nc, where c > 0:
15.3 The Number of Zero-Patterns of a Sequence of Polyno-mials
See separate handout.
72 CHAPTER 15.
Chapter 1616th day, Friday 7/30/04 (Scribe: Charilaos Skiadas)
16.1 Random Variables
Recall the de�nition of a random variable: Given a probability space (; P ), where is thesample space and P the probability distribution over the sample space, a random variable
is simply a function X : ! R. The expected value of X is E(X) =Xx2
X(x)P (x), i. e., it
is a weighted average of the values of X. Notice that
E(X) =Xx2
X(x)P (x) =Xy2R
yP (fX = yg):
Given an event A � , its probability is
P (A) =Xx2A
P (x):
The indicator variable of event A is the function �A : ! f0; 1g de�ned by
�A(x) =n 1 if x 2 A0 if x 62 A
:
16.2 Independence
The expected value of �A is E(�A) = P (A). We say that two events A;B are indepen-dent, if P (A \ B) = P (A)P (B). Three events A;B;C are independent, if P (A \ B \ C) =P (A)P (B)P (C) and they are pairwise independent. In general A1; : : : ; Ak are independent, iffor every I � [k] we have that
P�\i2I
Ai
�=Yi2I
P (Ai):
73
74 CHAPTER 16.
The random variables X1; : : : ; Xk are said to be independent, if for any a1; : : : ; ak
P�X1 = a1 and : : : and Xk = ak
�=
kYi=1
P (Ai):
Exercise 16.1. If X1; : : : ; Xk are independent random variables, then for any I � [k] thesub-collection (Xi : i 2 I) is also independent.
Exercise 16.2. The events A1; : : : ; Ak are independent if and only if their correspondingindicator variables are independent.
Notice: � �A = 1 � �A, �A\B = �A�B, and these two also give us: �A[B = 1 � �A[B =1� � �A\ �B = 1� � �A� �B = : : : = �A + �B � �A�B.
Exercise 16.3. If X1; : : : ; Xk are independent, then E� kYi=1
Xi
�=
kYi=1
E(Xi).
Let's see what this means for the indicator variables: E� kYi=1
�Ai
�=
kYi=1
E(�Ai) = E(�\Ai
),
in other words P� kTi=1
Ai
�=
kQi=1
P (Ai). So this is true almost by de�nition for indicator
variables.
Exercise 16.4. If X1 : : : ; Xk are random variables, then there exist polynomials in k variables,f1; : : : ; fm such that for each i, Yi := fi(X1; : : : ; Xk) is an indicator variable, the correspondingevents are disjoint, and (8j)(Xi 2 span(Y1; : : : ; Ym)).
Exercise 16.5. If X1; X2; X3; X4 are independent random variables, thenqX2
3 +1
X24+1
, eX1 ,
cos(X2) are independent.
In general, if we start with a number of random variables and split them in groups, andfor each group we create a new random variable by using any function of the variables in thatgroup, then those resulting random variables are independent.
16.3 Conditional Probability
For B 6= ; and any A, we de�ne the conditional probability P (AjB) as P (A \ B)=P (B). Itis easy to see that if A;B are independent and B is nonempty, then P (AjB) = P (A). Theadvantage of using the notion of independence instead of this last equality is that it doesn'trequire us to exclude the case B = ; and it shows clearly that the notion of independence issymmetric.
16.3. CONDITIONAL PROBABILITY 75
De�ne the conditional expectation of a variable X given the event B as: E(XjB) =Xx2B
X(x)P (fxgjB) =Xy
yP (fX = xgjB)
Given X;Y two random variables, what should E(XjY ) mean? It would have to be arandom variable: Z := E(XjY ). It is de�ned by (z 2 ) 7! Z(z) = E (XjY = Y (z)). Let's seewhat this does when X;Y are independent. Then
Z(z) =X
fxjY (x)=Y (z)g
X(x)P (x)
P (fY = Y (z)g) :
This is equal toXt
tP (fX = tgjfY = Y (z)g). If X and Y are independent, then it is further
equal toXt
tP (fX = tg) = E(X). So if X and Y are independent, then Z is going to be just
a number, the expected value of X.
Exercise 16.6. Show E(XjX) = X
We say that X;Y are uncorrelated, if E(XY ) = E(X)E(Y ). We know by Exercise 16.3that if X;Y are independent then they are uncorrelated.
Let us provide a counterexample for the converse: Let = fa; b; cg and P be the uniformprobability distribution. Let X take values f1; 0;�1g at fa; b; cg respectively, and let Y =X2, so it takes values f1; 0; 1g at fa; b; cg respectively. Then we have that XY = X andE(X) = 0, so they are uncorrelated. However, P (fX = 0g) = P (fY = 0g) = 1
3 , andP (XY ) 6= P (X)P (Y ).
Exercise 16.7. If jj = 2, then uncorrelated random variables are also independent.
Exercise 16.8. If X1; : : : ; Xk are independent and not constant, then jj = n � 2k.
Indeed, each variable can take at least two values. For each choice of a value ai for everyvariable Xi, we get a set fX1 = a1; : : : ; Xk = akg with positive probability, hence nonempty.There are at least 2k such sets, and they are all disjoint.
In particular, to get many independent events, we need a large sample space.
De�nition: A is called a trivial event, if A = ; or A = . So for nontrivial events wealways have: 0 < P (A) < 1.
Corollary 16.9. If A1; : : : ; Ak are independent non-trivial events, then n � 2k
Proof: use their indicator variables. �
If we only require pairwise independence, then how small can the size of the sample spacebe?
76 CHAPTER 16.
Theorem 16.10. If X1; : : : ; Xm are pairwise independent and non-constant, then m � n� 1.
Proof: Recall that if X;Y are independent, then X+c and Y +d are independent, so withoutloss of generality we can assume that (8i)(E(Xi) = 0). We claim that under this condition,and if the Xi are pairwise uncorrelated, the X1; : : : ; Xm are linearly independent over R (asfunctions from to R.) Recall that R denotes the space of functions ! R.
Since the dimension of this space is equal to jj, we get our result, since our functions liein the kernel of the non-zero functional E (the hyperplane consisting of the random variableswith expected value 0). Another way to argue this last step is to add the function X0 = 1.Then X0; : : : ; Xm are linearly independent. (They are still uncorrelated)
Now, the uncorrelated condition tells us that E(XiXj) = 0 i� i 6= j, since E(X2) > 0unless X is zero. This shows by the standard argument that the Xi are linearly independent.
As a consequence, if A1; : : : ; Am are nontrivial events, then m + 1 � n. This is actuallytight for in�nitely many values of n: Suppose n = 2k, and let = Fk2. Then any subspaceof dimension k � 1 gives an event with probability 1
2 . So for each u 2 Fk2 n f0g, P (u?) = 12 .
We need to know that these events are pairwise independent. If u1 6= u2, then their span hasdimension 2 since they are linearly independent (they can't be parallel), so P (u?1 \ u?2 ) = 1
4 .
More generally, let q be a prime power, = Fkq , and let u1; : : : ; um be elements in . Notice
that P (u?i ) =1q if ui 6= 0.
Exercise 16.11. Show that u?1 ; : : : ; u?m are independent events i� u1; : : : ; um are linearly
independent vectors.
Exercise 16.12. Find out how this example over Fk2 relates to the Sylvester matrix.
Exercise 16.13. Find n� 1 pairwise independent events of probability 12 over a sample spac
of size n for every Hadamard number n.
Exercise 16.14. Show that if there exist n� 1 pairwise independent events of probability 12
over a uniform probability space of size n, then n is a Hadamard number.
Exercise 16.15. Show that for in�nitely many n there exist n2 3-wise independent non-trivial
events over a sample space of size n.
Exercise 16.16. Show that if X1; : : : ; Xm are 4-wise independent nontrivial random variables,then
�m2
� � n.
Chapter 1717th day, Monday 8/2/04 (Scribe: Charilaos Skiadas and Eric Patterson)
17.1 Puzzles
The Monty Hall Paradox.
On a game show, there are three closed doors. There is a car behind one door. Behind each ofthe other two doors, there is a goat. You select one of the doors, and the game master opens adi�erent door, behind which is a goat. Then you are o�ered a choice to stay with your choiceor to switch. The chance that we picked the car initially is 1
3 ; so the chance that we get the carif we stay with our choice is 1
3 : Thus, the strategy that switches gives us a 23 chance of picking
the car.
The 2 Envelope Paradox
You get two envelopes. Each of them contains some money. One of the envelopes has twice asmuch money as the other. You are allowed to open one of the envelopes, see how much is in it,and then choose which envelope to pick. No matter what is in the envelope that you opened,you would expect that the other envelope would give you a larger expected amount of money.But you do not need to open the envelope to determine that. By this reasoning, we shouldkeep switching back and forth between the two envelopes to increase the amount of money weexpect to get.
Moral: be careful with the notion of expectation and probability space.
17.2 Statistical Independence vs Linear Independence
Problem 17.1. Can we create m 3-wise independent non-trivial events, such that n = jj =2m? (m = 2k)
For pairwise independent events, we know that n � m+1. Following the ideas in the pairwisecase, de�ne := F`q and suppose v1; : : : ; vt are vectors in : We have shown that v?1 ; : : : ; v
?t
77
78 CHAPTER 17.
are independent events i� the vi are linearly independent. If vi 6= 0; P (v?i ) =1q because the
number of elements of a hyperplane is q`�1: Now U := v?1 \ � � � \ v?t = span(v1; : : : ; vt)? had
dimension ` � dim(U) = ` � r where r is the rank of (v1; : : : ; vt). So the probability of U is1qr . If the events are independent, then this has to be equal to 1
qt : Hence r = t; and the vi arelinearly independent. Conversely, if the vi are linearly independent, then r = t and the eventsare independent.
To construct 3-wise independent events, we would need to construct vectors that are 3-wiselinearly independent. In F`2, we need to �nd 2`�1 3-wise independent vectors, so n = 2` andm = n
2 . Take an a�ne hyperplane not passing through 0 (a shift of a subspace). For example,we could put 1 in the �rst coordinate and either 0 or 1 in the other coordinates for a total of2`�1 elements of :
Claim: These vectors are 3-wise linearly independent.
Taking any three of the vectors, we would need to show that any nontrivial linear combinationcannot be 0. Since the only elements of F2 are 0 and 1; nontrivial combinations are sums ofone, two, or all three of the vectors. This means (i) no one of them is 0; (ii) no two of them addup to 0; and (iii) no three of them add up to 0: The vectors we chose begin with coordinate 1,so they are not zero. If v + v0 = 0, then v = �v0 = v0, but we assumed that we did not taketwo identical vectors. Any three of them add to a vector with �rst coordinate 1, so the sum isnot equal to 0. If m is not a power of 2, we can still get that n < 4m by taking the smallest `such that m < 2`:
17.3 Algorithmic Application
A Boolean function is a function f0; 1gn ! f0; 1g. A Boolean formula is a formula com-posed of literals (variables and their negations) using AND and OR operations. For instance,x1_(x2 ^ x1 ^ x2). A disjunction is an OR of literals. A CNF (conjunctive normal form)is an AND of clauses, each of which is a disjunction.
Exercise 17.2. Every Boolean formula can be represented as a CNF.
A 3-CNF formula is a formula in which every clause has 3 literals. To evaluate a Booleanformula on some substitution of values of the variables, recall that an OR of two variables iszero if and only if both variables are zero, and an AND of two variables is one if and only ifboth variables are one. An assignment of values to the variables in a Boolean formula satis�esthe formula if the substitution of the values returns 1:
Theorem 17.3. Satis�ability of 3-CNF formulae is NP-complete.
Claim: For a 3-CNF fomula, there always exists a substitution that satis�es at least 7=8 of theclauses.
17.4. ALGEBRAIC CODING THEORY 79
Let n be the number of variables and m be the number of clauses.
Hint 1: Flip coins for the value of each variable; that is, make a random substitution. If eachset of values is an element of ; then jj = 2n.
Hint 2: Find the expected number of satis�ed clauses.
LetX be the number of satis�ed clauses, soX is a random variable from the uniform probabilityspace :We can write X =
Pmi=1 #i, where #i is the indicator of the event that the ith clause is
satis�ed. Then the expected value is the sum of the expected values of the #i: Since the #i areindicator variables, E(#i) equals the probability that the clause Ci is satis�ed. By the rules forevaluating Boolean formulae, the probability is 7
8 that a disjunction of 3 literals with randomvariables is satis�ed. Therefore, the expected number of satis�ed clauses is
Pmi=1
78 = 7m
8 .Hence there exists an outcome x such that X(x) � 7
8m. So there exists a substitution thatsatis�es at least 7
8 of the clauses.
If we want to �nd such a substitution deterministically in polynomial time, we should noticethat we only used the fact that the variables occurring in a clause are 3-wise independent. Hencewe can replace our space of outcomes with a space of size less than 4n by the construction for�nding 3-wise independent vectors above. This gives us a tiny fraction of all the substitutions,but the expected number of satis�ed clauses is the same. Now we can simply try everythingin this space, which will �nish in quadratic time.
Moral: It is worth �ghting for small sample space.
Recall the following exercise: if X1; : : : ; Xm are 4-wise independent nonconstant random vari-ables, then n = jj � �m2 �.Proof: Without loss of generality, we can assume that the expected value of each randomvariable is 0. The space of random variables has dimension dimR = n. We want to construct�m2
�random variables that will be linearly independent in this space. Look at the pairwise
products of the Xi: �
Exercise 17.4. Prove that the�m2
�products XiXj for i � j are linearly independent.
17.4 Algebraic Coding Theory
Question: what is the maximum number of k-wise linearly independent vectors in F`q?
A codeword is a sequence (�1; : : : ; �n) 2 F`q: When you transmit the codeword, there issome noise, that is, some of the entries in the codeword might change value. If we have agiven set of codewords from F`q; we want to be able to distinguish them from one another evenafter some reasonable amount of noise interference. The Hamming distance between twocodewords is the number of substitutions that must be made to change one into the other. Ifthe Hamming distance is large among the codewords, the codewords will still be distinguishableafter some interference. The code space is a subspace U � F`q. We want dimU to be large(i.e., lots of codewords) and the Hamming distance between codewords to be at least something.
80 CHAPTER 17.
Exercise 17.5. Establish what this something is and relate it to the k above, i.e., establish aconnection between the two data above.
Chapter 1818th day, Wednesday 8/4/04 (Scribe: Charilaos Skiadas)
18.1 Projective dimension of a graph, continued
De�nition 18.1. A projective representation of a graph G over a �eld F is a vector spaceW over F and an assignment of subspaces to vertices i 7! Ui � W such that i � j (adjacentvertices) , Ui \ Uj 6= f0g.
Projective dimension of a graph: pdimF(G), G a graph, F a �eld, de�ned as the smallestpossible dimension of a projective representation of G. We showed that if jFj � m, m thenumber of edges, then the projective dimension is � 2 degmax, twice the maximum degree ofthe graph. We did it by associating to every edge of a graph a vector in a vector space ofdimension 2 degmax in such a way, that the vectors are in general position. Then we assignto each vertex the subspace spanned by the vectors corresponding to edges attached to thatvertex. If two vertices are adjacent, then the spaces corresponding to them both contain thevector of the edge joining them. If we take the vectors to be in general position in a spaceof dimension 2 degmax, then any 2 degmax of them will be linearly independent. So we get arepresentation. Recall that the vectors of the form f(1; �; �2; : : : ; �k�1) : � 2 Fg are in generalposition in Fk.
Theorem 18.2. For any �eld F, almost all graphs have projective dimension at least cq
nlogn .
\Almost all" means the lower bound is true for all but o(2(n
2)) graphs on a given set ofn vertices, so the probability that a random graph on n vertices does not have this propertytends to 0 as n!1.
It is an open question whether almost all graphs have large projective dimension over every�eld. Large here means greater than nc, where c > 0 is a constant. Not even (log n)1+" isknown.
For the proof, we will need the theorem on zero-patterns. Recall that given m polynomialsf = (f1; : : : ; fm) of degree � d in n variables over the �eld F, then by substituing � 2 Fn
81
82 CHAPTER 18.
in them, we get an element in Fm. Replacing nonzero entries with a star gives us a \zeropattern," i .e., a string in f0; �gm. Let Z(f) be the number of zero-patterns for f .
Theorem 18.3 (RBG(2000)). If d � 2, then Z(f) � �mdn
�<�emdn
�n.
The bound�emdn
�nholds for d = 1 also. We will use this to estimate the projective
dimension of a random graph.
Lemma 18.4. If pdimF(G) = k and jFj is large enough, then there exists a projective rep-
resentation over F of dimension 2k such that for every i, the dimension of every subspace
representing a vertex has dimension k.
Proof: Let Ui correspond to the vertex i. Each of them has dimension at most k. Addnew vectors that are linearly independent (do it in a large space) such that each space hasdimension k. Then the dimension of this space is � k + nk. Now each Ui has dimension k,and the intersections are still the same.
Exercise 18.5. The new Ui represent G.
We need to make a random projection to a 2k dimensional space. Suppose W is a spaceover F, F not too small, and let U1; : : : ; Un �W , dimUi = k and let S be a space dimS = 2k.Let � :W ! S be a random map (i.e. a random 2k � dimW matrix). Then the dimension off(Ui) remains the same, and the dim�(Ui) \ �(Uj) = dim�(Ui \ Uj). i.e. these spaces avoidthe kernel of the map, whose dimension is dimW�2k. Any random k-dimensional subspace, islikely to avoid it. It is also likely to avoid every Ui+Uj , which guarantees that the dimensionsof the pairwise intersections of the Ui and the �(Ui) are the same.
Lemma 18.6. If F is large enough, then givenW;Ui as above, there exists K �W of dimension
dimW � 2k such that K \ (Ui + Uj) = f0gExercise 18.7. Find out how large jFj has to be.
Project using this \generic" subspace as a kernel, onto a 2k-dimensional space. This givesthe desired representation, proving Lemma 18.4.
Suppose now that we look for a projective representation as in the Lemma. Every subspaceof it can be represented by a k�2k matrix, where the rows form a basis of this subspace. Thenthe intersection condition is that if we put the two matrices on top of one another, then theresulting matrix is singular if and only if the corresponding vertices are adjacent. Now ifwe are looking for such a projective representation, then all entries are variables ti;r;s. Thedeterminants of the above matrices, (that we want to be 0 or not zero according to adjacencyof the corresponding vertices), are polynomials in the ti;r;s, where 1 � i � n, 1 � r � k,1 � s � 2k. So we have 2k2n variables. The conditions are that certain polynomials are zeroin some cases and nonzero in some other cases. So this corresponds to a zero-pattern of these
18.2. THEOREM OF MILNOR-THOM 83
�n2
�polynomials. A zero entry corresponds to an edge, and � corresponds to a non-edge. So
the number of distinct graphs that can be represented in such a space is exactly equal to thenumber of zero-patterns.
Suppose at least an " fraction of the 2(n
2) graphs can be so represented. Then
e�n2
�2k
2k2n
!2k2n
� Z�f � "2(
n
2)�; so
�en2k
2k2n
�4k2n
> "2n(n�1);
�2n
k
�4k2n> "22n(n�1));
(2n)4k2
> "2=n2n�1:
Taking logs, we get that 4k2 log2(2n) > (n� 1)� 2n log(1=") so that
4k2 >n� 1
1 + log2 n� 2
nlog(1="):
Assuming " > c�n, we get that asymptotically k & 12
qn
logn .
This �nishes the proof of the Theorem 18.2.
Over the real numbers, there is a much stronger result. There we have the concept of asign-pattern. Again given polynomials as above, whenever we plug in something and get allnonzero numbers, then we get a sign pattern by looking at whether the number is positive ornegative. Then we have the theorem:
Theorem 18.8 (Warren(1968)). The number of sign patterns is <�4emdn
�n.
(As before, m is the number of the polynomials, d is the degree, and n is the number ofvariables.) An application of this: If you cut up Rn with hypersurfaces, each region of thecomplement corresponds to a sign-pattern.
18.2 Theorem of Milnor-Thom
A real algebraic set is the set of common roots of a set of polynomials in real a�ne space.
Theorem 18.9 (Milnor (1964), Thom(1965)). Let V � Rn be the set of common roots of
the polynomials f1; : : : ; fm 2 R[x1; : : : ; xn], where deg fi � d. Then the number of connected
components of V is � d(2d� 1)n�1 � (2d)n.
84 CHAPTER 18.
In fact, the same bound holds for the sum of the Betti numbers, which gives much more (i .e,it counts higher-dimensional holes). There are computer science applications of these higherBetti numbers too. Notice that the bound does not depend on the number of polynomials, butit depends exponentially on the number of variables.
Exercise 18.10. Show Warren's theorem using the Milnor-Thom Theorem.
Exercise 18.11. From Warren's theorem deduce the theorem on number of zero-patterns forthe reals and complex numbers, or any �eld of characteristic 0 (with a di�erent constant thane).
Chapter 1919th day, Friday 8/6/04 (Scribe: Charilaos Skiadas)
19.1 Matrix rigidity (Valiant 1978)
Let A be a matrix over a �eld F. Our aim is to explore how we can reduce its rank withchanging as few entries as possible. For that, denote by Rr(A) the minimum number of entriesthat need to be changed in A in order to obtain a matrix of rank less than or equal to r. Inother words, this is equal to minfweight(C) : rk(A� C) � rg where the weight of a matrix isthe number of nonzero entries. We are particularly interested in the case when r = �(n).
Exercise 19.1. For the identity matrix we have Rr(I) = n� r (Prove that fewer changes donot su�ce!)
Proposition 19.2. For every n� n matrix A we have Rr(A) � (n� r)2.
For instance, this would tell us that Rn=2 � n2=4. For the proof of the proposition, we canwithout loss of generality assume that rk(A) � r, so we can assume that A has a nonsingularr� r minor B, and let us for simplicity assume it appears in the upper left corner. If we focuson the �rst r rows of A for the moment, we see that the columns of B are linearly independent.Let A0 be the r�n matrix consisting of the �rst r rows of A. Hence, all of the other columns ofA0 can be written as linear combinations of the columns of B. So there are numbers ci;j suchthat ai;j =
Prk=1 ai;kck;j for i = 1; : : : ; r and j = r + 1; : : : ; n. Now, if we simply change the
entries in the (n� r)� (n� r) bottom right corner, and rede�ne them according to the aboveequation (we de�ne a0i;j =
Prk=1 ai;kck;j for i; j = r+1; : : : ; n), then this equation now hold for
all j, in other words all the columns of A are linear combinations of the �rst r columns.These�rst r columns are linearly independent, since the columns of B are. This gives us a matrixwith rank r, and we had to make (n� r)2 changes, hence the proposition is proved.
Theorem 19.3. Over in�nite �elds, almost all matrices satisfy Rr(A) = (n� r)2.
The meaning of \almost all" is that while dimMn(F) = n2, the \dimension" of the set(\variety") of matrices with Rr(A) < (n� r)2 is strictly less than n2.
85
86 CHAPTER 19.
Think of all n� n matrices, Mn(F), and let Dr be the subset consisting of all matrices ofrank r or less. Then this is de�ned by a set of algebraic equations, namely all (r+1)� (r+1)
minors are equal to 0. There are� nr+1
�2such equations. Such a set, an \a�ne variety", has
an intuitive notion of dimension. We want to �nd what that is for Dr. Notice that all ofthe equations above are independent since, for any (r + 1)� (r + 1) minor, the matrix whichis the identity on that minor and 0 everywhere else satis�es all but one of the equations.A better way to measure dimension is how many independent directions there are for smallchanges: Suppose the rank of a matrix is r, and that the top left r � r minor is nonsingular.We would like to know how many parameters we can change slightly, while keeping the rankr. We can freely slightly change all entries in the �rst r rows and in the �rst r columnsand still keep the top left r � r minor non-singular. But all the other (n � r)2 entries arecompletely determined, if we want to keep the rank no more than r. So we end up withrn+ r(n� r) = r(2n� r) = 2rn� r2 = n2 � (n� r)2 free choices. So this is the dimension ofDr. (Dr is a �nite union of spaces of the above form, for the various choices of minors. But a�nite union of sets of a given dimension still has the same dimension as them. This is only trueover an in�nite �eld!) Suppose now that we permit m entries to be changed before getting amatrix of rank r. Then each of these entries increases the dimension by 1, giving us one morefree variable, hence the whole process will increase the dimension overall by m. Denote Er;m
to be the set of matrices B such that there exists a matrix A 2 Dr with weight(B � A) � m.Then by the above discussion we have dimEr;m � dimDr + m. So for m < n2 � dimDr,almost all matrices satisfy Rr(A) > mn since the set Er;m = fA jRr(A) � mg has dimension� n2 � dimDr +m < n2. So for almost all matrices, Rr(A) � n2 � dimDr = (n � r)2, andsince it was also less than or equal to it, it is actually equal to it. Let us emphasize again, thatthis only works over in�nite �elds.
Exercise 19.4. Prove that if F is a �xed �nite �eld, then almost all matrices satisfy
Rn2(A) > c
n2
log n:
\Almost all" here means that the proportion of matrices satisfying the inequality goes to 0 asn!1.
What we need is explicit families of matrices with high rigidity (Rcn(A) > n1+"). Thisis not known for any �xed " > 0. We have some examples of non-explicit families: Forinstance, a matrix with independent transcendental entries. In other words, a \generic matrix"is not explicit. Independent transcendental entries means that if f(a1;1; : : : ; an;n) = 0 for somef 2 Z[ai;j ], then f = 0.
The big question here is: Can one construct matrices with high rigidity with integer entries,where the integer entries have at most polynomially increasing number of digits (i. e., thenumber of digits is at most nc for some absolute constant c)?
All nonzero Vandermonde matrices are candidates. A special case of interest is Vander-monde matrix Vn(1; !; !
2; : : : ; !n�1), where ! is a primitive n-th root of unity (this is known
19.1. MATRIX RIGIDITY (VALIANT 1978) 87
as the discrete Fourier Transform (DFT) matrix.) One way to generalize this is as follows:Recall that a character of a �nite abelian group is a homomorphism � : G! fz 2 C; jzj = 1g.Exercise 19.5. Show that a �nite abelian group of order n has exactly n characters.
The character table of an abelian group is an n�nmatrix with rows indexed by charactersand columns indexed by elements of G. The entries are evaluations of the characters on theelements.
Exercise 19.6. Show that the characters of a �nite abelian group are orthogonal (i. e., therows of the character table are orthogonal).
For example, if Zn = Z=nZ, all the characters are of the form �i(a) = !ai, where ! is asabove. Then the above matrix is simply the character table of of Zn. Another example wouldbe G = Z2 � � � �Z2 = (Fk2;+). Then jGj = 2k. If a; b 2 G, then �b(a) = (�1)ab is a character,so every element of G de�nes a character, and the character table in this case is the Sylvestermatrix. More generally, all Hadamard matrices are expected to be good candidates for rigidity.
The best known result at this time is:
Theorem 19.7 (S. Lokam). R n17(P ) > cn2, for P the matrix made out of the square roots
of the �rst n2 prime numbers.
Unfortunately, this is not exactly explicit. The dimension of the extension of Q by thesenumbers is 2n
2, so this �eld has exponentially large dimension.
Exercise 19.8. The square roots of all square-free numbers are linearly independent over Q.
This is the best we can say at the moment.
Open question: Give a nonlinear lower bound on Rcn for the generic Vandermonde matrix.Lokam's proof gives Rp
n=2(Vn(x1; : : : ; xn)) > cn2.
88 CHAPTER 19.
Chapter 2020th day, Monday 8/9/04 (Scribe: Charilaos Skiadas)
20.1 Matrix Rigidity continued
Matrix rigidity Let A be a matrix over a �eld F. Recall from last time that Rr(A) =minfweight(C) : rk(A � C) � rg, the rigidity function. We \proved" that if F is in�nite,then for almost all n� n matrices A we have that Rr(A) = (n� r)2. So, if r < (1� �)n thenRr(A) = (n2) for almost all matrices. We don't know any \explicit" examples of families ofsuch matrices. But we know
Theorem 20.1 (S. Lokam). Rpn(Vn(x1; : : : ; xn)) > cn2, where the xi are independent tran-
scendentals.
For r > 2pn, even a n1+� lower bound is not known. We will see the idea for the proof by
looking instead at the matrix in the following theorem:
Theorem 20.2. Let P be the matrix with entries the square roots of the �rst n2 primes. then
R n17> cn2
We will use that the square roots of all square-free integers are linearly independentover Q. The main tool in the proof is the Shoup-Smolensky invariant of a set of num-bers. This is de�ned as follows: St(a1; : : : ; an) = rkQfai1 � � � aip : 1 � i1 < � � � < it �mg. For example, S2(
p2;p3;p5;p7) is the rank of fp6;p10;p14;p15;p21;p35g = 6,
while S3(p2;p3;p5;p7) = 4. If A is an n � m matrix with entries ai;j , then St(A) =
Stfa1;1; : : : ; an;mg. Also let S�t (a1; : : : ; an) is same as above, only with repetitions permitted.Obviously St � S�t .
Lemma 20.3. S�t (AB) � St(A)St(B).
Proof: In AB, if we take the product of t entries, then each such entry is a sum of products ofpairs. So a t-wise product is going to be a sum of t-wise products from A times t-wise productsfrom B. So the space on the left hand side will be generated by those products. This provesthe above bound.
89
90 CHAPTER 20.
Corollary 20.4. If A 2Mn(F) with rk(A) = r, then S�t (A) ��nr+t�1
t�1�2.
Proof: The rank of A is less than or equal to r if and only if A can be represented as A = BC,where B has r columns and C has r rows. Hence S�t (A) � S�t (B)S�t (C). Now each of thoseis no greater than the number of all t-wise products with repetition, which gives us the abovebinomial estimate. (Recall the cookies problem, where not everyone necessarily gets a cookie).In general, if B is an n� r matrix, then S�t (B) �
�nr+t�1t�1
�.
Lemma 20.5. With P as in the theorem above (the one containing square roots of primes),
suppose weight(C) = w. Then St(P � C) � �n2�wt �.
The proof is immediate (using the fact that square roots of the square-free positive integers
are linearly independent). So if w = Rr(P ), we have that�n2�w
t
� � St(P �C) � S�t (P �C) ��nr+tt
�2. Setting t = nr, we get that
�n2�wnr
�nr<�n2�w
nr
� � �2nrnr
�2< 16nr. This gives us that
n2�wnr < 16. We �nd that w � n2(1� 16r
n ). So we have shown that Rr(P ) � n2(1� 16rn ), which
proves the theorem, if we set r = n17 .
Exercise 20.6. Prove with a similar argument Lokam's theorem, thatRpn=2(Vn(x1; : : : ; xn)) >
cn2.
Exercise 20.7. Prove the same when xi = p1=ni .
Note, that dimQQ(p1=n1 ; : : : ; p
1=nn ) = nn, so this is not very \explicit" either.
A linear code of length n and dimension k over Fq is an k-dimensional subspace C in Fnq .
The information rate is kn . The idea is that our original message is in F
kq , and is encoded using
the encoding Fkq�= C. The minimum weight of C is the minimum of the weights of all x 2 C,
x 6= 0. The Hamming distance of x; y 2 Fnq is the weight of x�y. For any x 6= y with x; y 2 C,dist(x; y) � minweight(C). Let d = minweight(C), the coding distance. In noisy transmission,up to d � 1 errors can be recognized, and up to d�1
2 errors can be uniquely corrected. Thegoal of algebraic coding theory is to �nd codes with large information rate and large codingdistance, and of course that are explicit.
Suppose C � Fnq , dimC = k. Then dimC? = n � k. A basis of C? will be given by an� k � n matrix B.
Exercise 20.8. minweight(C) � d i� the rows of B are d� 1-wise linearly independent.
OPEN QUESTION: Do there exist good cyclic codes (cyclic codes are those where a cyclicpermutation of the entries preserves the space.) Good means k
n = (1), d = �(n).
In Fnq , we have q vectors that were n-wise independent, namely the elements of the form
(1; a; a2; : : : ; ad�1). This will give us a code in Fqq. Then, dimC? = d, rate is 1 � dq . If we do
this over a large �eld, we get excellent rate and error-correction. These are good codes overlarge �eld. What we want though is a family of codes which is good over a �xed �nite �eld.Most important are the binary codes, over F2.
Chapter 2121th day, Wednesday 8/11/04 (Scribe: Ivona Bez�akov�a)
Exercise 21.1. Let u1; : : : ; u5, ui =��!OAi be �ve pairwise perpendicular unit vectors in R3
where O is the origin and the vertices Ai form a regular pentagon. Let us assume that all theAi have the same x-coordinate. Let c = (1; 0; 0). Prove: cos(c; ui) = 1= 4
p5.
Hint. Spherical cosine formula: Let (a; b; c) be a spherical triangle with arc lengths a, b, c andlet A be the angle of b and c. Then cos a = cos b cos c+ sin b sin c cosA.
21.1 Linear Programming
Notation 21.2. Let a; b 2 Rn. We denote a � b if ai � bi for every i 2 [n].
Given is a real k � n matrix A, and vectors b 2 Rk, c 2 Rn. Let x 2 Rn be an unknownvector. A linear program speci�es the constraints on x (a system of k + n linear inequalitiesin n unknowns): Ax � b and x � 0. The goal, or the objective function is to maximize cTx.
In a summarized form, here is the linear program and its dual:
Linear Program (LP): Dual Linear Program (DLP)
Constraints:Ax � bx � 0
AT y � c
y � 0
Objective: max cTx min bT y
Where A 2 Rk�n; b 2 Rk; c; x 2 Rn AT 2 Rn�k; b; y 2 Rk; c 2 Rn
De�nition 21.3. The LP is feasible if there exists a solution. For a feasible LP we denote theoptimum solution by optLP := max cTx.
Theorem 21.4 (LP duality). If LP and dual LP are feasible, then optLP = optdual LP .
Exercise 21.5. Prove the � direction of the LP duality theorem.
De�nition 21.6. Integer linear programming (ILP) looks for integral optima under linearconstraints. In other words, all coordinates of x must be integers.
91
92 CHAPTER 21.
Clearly,optILP � optLP = optdual LP � optI dual LP
As an example we present the following ILP for �(G), the size of the maximum independentset in a graph G. There will be an unknown variable xi for every vertex i. The LP:P
i2C xi � 1 for every clique Cxi � 0
maxP
i xi
Integer optimum is exactly �(G) (why?). We denote the fractional optimum by ��(G).
Dual variables are: yC for every clique C. The dual LP:PC3i � 1 for every vertex iyC � 0
minP
C yC
Integral optimum of the dual LP is the minimum clique cover (why?), i.e. the minimumnumber of cliques covering V (G). Notice that this is exactly the chromatic number of thecomplement of G, �(G). Fractional optimum is denoted ��(G).
Therefore,�(G) � ��(G) = ��(G) � �(G)
Notice that the inequality �(G) � �(G) is trivial. However, using the linear programmingmethod we get an interesting quantity in between.
Exercise 21.7. Find the values of �(C5); ��(C5); �(C5); �
�(C5).
Exercise 21.8. Work out ��(C7) from scratch (not using the above inequality).
21.2 Shannon Capacity of a graph
Notation 21.9. We denote the adjacency of two vertices i; j by i � j. If i is adjacent or equalto j, we write i �= j.
De�nition 21.10. Let G;H be two graphs. The strong product of G and H, denoted G �H isde�ned as follows. The vertex set V (G�H) := V (G)�V (H), while the adjacency (i; j) � (i0; j0)holds if (i; j) 6= (i0; j0) and i �= i0 (in G) and j �= j0 (in H).
De�nition 21.11. Let G be a graph and let Gn be its n-th strong power. Shannon capacity
of G, denoted �(G), is de�ned as
�(G) = limn!1
np�(Gn)
21.2. SHANNON CAPACITY OF A GRAPH 93
Exercise 21.12. Prove: �(G �H) � �(G)�(H)
Lemma 21.13. Let f : N! R+ be a function satisfying f(nm) � f(n)f(m). Then limn!1 npf(n)
exists and is equal to supn!1 npf(n).
Exercise 21.14. Prove the lemma.
Corollary 21.15. For every k, �(G) � kp�(Gk)
Corollary 21.16. �(G) � �(G)
Exercise 21.17. Prove: �(G) � �(G)
The Shannon's capacity of C5 is bounded by 2 � �(C5) � 3. We will �nd its exact valueshortly.
Open problem. The exact value of �(C7).
Lemma 21.18. If g(G) is a graph function s.t.
1. for every G, �(G) � g(G)
2. g(G �H) � g(G)g(H)
then �(G) � g(G).
Exercise 21.19. Prove: ��(G �H) = ��(G)��(H)Hint. � directly, � dually.
Corollary 21.20. �(G) � ��(G)
Corollary 21.21. �(C5) � 5=2.
We will use a bootstrapping method to tighten the lower bound. By Corollary 21.15 weknow that �(C5) �
p�(C2
5 ).
Exercise 21.22. Prove: �(C25 ) = 5
Theorem 21.23 (Lov�asz). Let G be a vertex transitive and self-complementary graph, i.e.
the automorphism group of G is transitive and G �= G. Then �(G) =pn.
In particular, the theorem implies that �(C5) =p5.
Exercise 21.24. Prove: If G is self-complementary, i.e. G �= G, then �(G) � pn.
94 CHAPTER 21.
21.2.1 Lov�asz's theta function (Lov�asz's capacity)
De�nition 21.25. Orthonormal representation of a graph G is de�ned as follows. Everyvertex i is assigned a d-dimensional unit vector ui 2 Rd, kuik = 1. If i 6� j, then ui and uj areperpendicular.
De�nition 21.26. Lov�asz capacity, denoted �(G) is de�ned as the minimum over all orthonor-mal representations ui of G of the quantity minkck=1maxi=1;:::;n 1=hc; uii.Theorem 21.27. For every graph G,
1. �(G) � �(G) and
2. �(G �H) = �(G)�(H).
Moreover, �(G) can be approximated in polynomial time.
Corollary 21.28. �(G) � �(G)
Exercise 21.29. Prove: �(G) � �(G) � �(G).
By the �rst exercise, �(C5) �p5. Recall that
p5 � �(C5) � �(C5) �
p5 and therefore
�(C5) =p5.
Lemma 21.30. If u1; : : : ; uk are orthonormal, then kck2 �Pihc; uii.Exercise 21.31. Prove the lemma.
To prove part 1 of Theorem 21.27, take a suitable subset of vertices, use the above lemma(a variant of the Pytagorean theorem) and the de�nition of Lov�asz's capacity. To completethe proof of �(C5) =
p5 it su�ces to show the � inequality in part 2 of Theorem 21.27. This
will be done using tensor products.
De�nition 21.32. Tensor product of two vectors a 2 Rk and b 2 R`, denoted a � b, is ak`-dimensional vector, where (ik + j)-th coordinate is aibj .
Exercise 21.33. Let a; b 2 Rk, and u; v 2 R`. Prove: ha � u; b � vi = ha; bihu; vi.Corollary 21.34. If a1; : : : ; ak is an orthonormal representation of G and u1; : : : ; ul is an
orthonormal representation of H then ai � uj is an orthonormal representation of G �H.
Corollary 21.35. �(G �H) � �(G)�(H)
Chapter 2222th day, Friday 8/13/04 (Scribe: Charilaos Skiadas)
22.1 0; 1-measures
Exercise 22.1. Prove the following theorem. (Hint: Use Zorn's Lemma.)
Theorem 22.2. If is an in�nite set, then there exists a �nite additive nontrivial 0; 1 measure
on 2, � : 2 ! f0; 1g, such that �() = 1. Nontrivial means that �(fag) = 0 for all a.
This is a hint to the in�nite switch problem.
22.2 Sign-Rigidity (Paturi-Simon)
Let M be a sign matrix, i.e., a matrix consisting of �1 entries A matrix A of the samedimensions as M realizes M if the sign of ai;j is equal to mi;j . The sign-rank of M is theminimum rank of a matrix realizing M .
Exercise 22.3 (Alon-Frankl-R�odl, 1984). Use Warren's theorem to prove that almost alln� n sign-matrices have sign-rank greater than or equal to n
32 .
Finding explicit matrices that satify the preceding example is a hard problem. No particularexamples are known.
Conjecture 22.4. Hadamard matrices have sign-rank � cn.
For applications, all we need would be an explicit matrix with rank greater than n� forsome �xed � > 0. The best known explicit bound was (log n) until 2002.
Theorem 22.5 (Forster 2002). Let X � Rk such that jXj � k and the elements of X are
in general position (i.e., k-wise linearly independent). Then there exists an invertible k � k
matrix A such thatP
x2X(Ax)(Ax)T
(Ax)T (Ax)= aIk, a constant multiple of the identity matrix.
95
96 CHAPTER 22.
If we assume that the theorem is true, the value of a can easily be found by taking thetrace on both sides (recall that the trace of the product of two matrices is independent of theorder). This gives ak =
Px2X 1 = jXj
Exercise 22.6 (Forster). IfA realizesM , then rk(A) � njjM jj . Recall that jjM jj = maxjjxjj=1 jjMxjj =p
�max(MTM) by the spectral theorem.
Corollary 22.7. If H is a Hadamard matrix, then sign-rank(H) � pn:
Proof: If A realizes H, then rk(A) � njjHjj . Since H
TH = nI, we get that jjHjj2 = �max(nI) =n
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