1. Proofs That Really Count The Art of Combinatorial Proof Bradford Greening, Jr. Rutgers University - Camden

2. 2 Theme Show elegant counting proofs for several mathematical identities. Proof Techniques Pose a counting question Answer it in two different ways. Both answers solve the same counting question, so they must be equal.

3. 3 Identity: For n ≥ 0, Q:Number of ways to choose 2 numbers from {0, 1, 2, …, n}? 1.By definition, 2.Condition on the larger of the two chosen numbers. If larger number = k, smaller number is from {0, 1, …, k – 1} Summing over all k, the total number of selections is

4. 4 Identity: Q:Count ways to create a committee of even size from n people? 1.For 2k ≤ n,

5. 5 Identity: Q:Count ways to create a committee of even size from n people? 2.A committee of even size can be formed as follows: Step 1: Choose the 1 st person ‘in’ or ‘out’2 ways Step 2: Choose the 2 nd person ‘in’ or ‘out’2 ways Step n-1: Choose the (n-1) th person ‘in’ or ‘out’2 ways Step n: Choose the n th person ‘in’ or ‘out’1 way By multiplication rule, there are 2 n-1 ways to form this committee.

6. 6 : “n multi-choose k” Counts the ways to choose k elements from a set of n elements with repetition allowed {1, 2, 3, 4, 5, 6, 7, 8} (n = 8, k = 6) {1, 3, 3, 5, 7, 7}or{1, 1, 1, 1, 1, 1}

7. 7 : “n multi-choose k” # of ways to choose k elements from a set of n elements with repetition allowed equivalent to number of ways of distributing k identical balls into n distinct bins.

8. 8 Identity: Q: How many ways to create a non-decreasing sequence of length k with numbers from {1, 2, 3, …, n} and underline 1 term? 1.By definition, there are ways to create the sequence, then k ways to choose the underlined term. {1, 1, 1, 1, 1, 2, 2, 2, 3, 3, …, n, n, n, n}

9. 9 Identity: Q: How many ways to create a non-decreasing sequence of length k with numbers from {1, 2, 3, …, n} and underline 1 term? 1.There are ways to create the sequence, then k ways to choose the underlined term.

10. 10 Identity: Q: How many ways to create a non-decreasing sequence of length k with numbers from {1, 2, 3, …, n} and underline 1 term? 2.Determine the value that will be underlined, let it be r. Make a non-decreasing sequence of length k-1 from {1, 2, 3, …, n+1}. Convert this sequence: Any r’s chosen get placed to the left of our underlined r. Any n+1’s chosen get converted to r’s and placed to the right of our r. Hence, there are such sequences.

11. 11 Identity: Example: n = 5, k = 9, and our underlined value is r =, then we are choosing a length 8 sequence from {1, 2, 3, 4, 5, 6} 1.Choose “r” 2.Create k-1 sequence from n+1 numbers 3.Convert 8-sequence: converts to 9-sequence:

12. 12 Fibonacci Numbers – a number sequence defined as F 0 = 0, F 1 = 1, and for n ≥ 2, F n = F n-1 + F n-2 i.e. 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144… 5 + 8 Fibonacci Numbers

13. 13 Fibonacci Nos: Combinatorial Interpretation f n : Counts the ways to tile an n-board with squares and dominoes.

14. 14 Example: n = 4, f 4 = 5 Fibonacci Nos: Combinatorial Interpretation

15. 15 Fibonacci Nos: Combinatorial Interpretation f n : Counts the ways to tile an n-board with squares and dominoes. Define f -1 = 0 and let f 0 = 1 count the empty tiling of 0-board. Then f n is a Fibonacci number and for n ≥ 2, f n = f n-1 + f n-2 = F n + 1

16. 16 If the first tile is a square, there are f n – 1 ways to complete sequence. If the first tile is a domino, there are f n – 2 ways to complete sequence. Hence, f n = f n – 1 + f n – 2 = F n + 1 Fibonacci Nos: Combinatorial Interpretation Q: How many ways to tile an n-board with squares and dominoes?

17. 17 Identity: For n ≥ 0, f 0 + f 1 + f 2 + … + f n = f n+2 -1 1.By definition there are f n + 2 tilings of an (n+2)-board; excluding the “all-squares” tiling leaves f n + 2 – 1. Q: How many tilings of an (n+2)-board have at least 1 domino?

18. 18 Identity: For n ≥ 0, f 0 + f 1 + f 2 + … + f n = f n+2 -1 2.Consider the last domino (in spots k+1 & k+2). f k ways to tile first k spots 1 way to tile remaining spots Q: How many tilings of an (n+2)-board have at least 1 domino? Summing over all possible locations of k gives LHS.

19. 19 Identity: For n ≥ 1, 3f n = f n+2 + f n-2 Set 1: Tilings of an n-board; by definition, |Set 1| = f n Set 2: Tilings of an (n+2)-board or an (n-2)-board; by definition, |Set 2| = f n+2 + f n-2 Create a 1-to-3 correspondence between the set of n-tilings and the set of (n+2)-tilings and (n-2)-tilings.

20. 20 Identity: For n ≥ 1, 3f n = f n+2 + f n-2 For each n-tiling, make 3 new tilings by adding a domino by adding two squares a. if n-tiling ends in a square, put a domino before the last square. b. if n-tiling ends in a domino, remove the domino

21. 21 Identity: For n ≥ 0, We say there is a fault at cell i, if both tilings are breakable at cell i.

22. 22 Identity: For n ≥ 0, Q: How many tilings of an n-board and (n+1)-board exist? 1.By definition, f n f n+1 tilings exist. 2.Place the (n+1)-board directly above the n-board. Consider the location of the last fault.

23. 23 Identity: For n ≥ 0, How many tiling pairs have their last fault at cell k? There are ( f k ) 2 ways to tile the first k cells. 1 fault free way to tile the remaining cells: Summing over all possible locations of k gives LHS.

24. 24 Identity: For n ≥ 0, 2 n = f n + f n-1 + Q: How many binary sequences of length n exist? 1.There are 2 n binary sequences of length n. 2.For each binary sequence define a tiling as follows: “1” is equivalent to a square in the tiling. “01” is equivalent to a domino.

25. 25 Example: The binary sequence 011101011 maps to the 9-tiling shown below. If no “00” exists, this gives a unique tiling of length n (if the sequence ended in “1”) n-1 (if the sequence ended in 0) Identity: For n ≥ 0, 2 n = f n + f n-1 +

26. 26 Identity: For n ≥ 0, 2 n = f n + f n-1 + What if “00” exists? Let the first occurrence of “00” appear in cells k+1, k+2 (k ≤ n-2) Match this sequence to the k-tiling defined by the first k terms of the sequence. (Note: k > 0, then the kth digit must be “1”) Each k-tiling will be counted times. fkfk

27. 27 Identity: For n ≥ 0, 2 n = f n + f n-1 + 16 length-11 binary sequences generate the same 5-tiling 0110100000001101001000 0110100000101101001001 0110100001001101001010 0110100001101101001011 0110100010001101001100 0110100010101101001101 0110100011001101001110 0110100011101101001111

28. 28 Lucas Numbers Lucas Numbers – a number sequence defined as L 0 = 2, L 1 = 1, and for n ≥ 2, L n = L n-1 + L n-2 i.e. 2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, … 11+18

29. 29 Lucas Nos: Combinatorial Interpretation l n : Counts the ways to tile a circular n-board (called bracelets) with curved squares and dominoes.

30. 30 Lucas Nos: Combinatorial Interpretation “out-of-phase” – a tiling where a domino covers cells n and 1 “in-phase” – all other tilings

31. 31 Lucas Nos: Combinatorial Interpretation l n : Counts the ways to tile a circular n-board (called bracelets) with curved squares and dominoes. Let l 0 = 2, and l 1 = 1. Then for n ≥ 2, l n = l n-1 + l n-2 = L n

32. 32 Lucas Nos: Combinatorial Interpretation Q: How many ways to tile a circular n-board? Note that the first tile can be a square covering cell 1 a domino covering cells 1 and 2 a domino covering cells n and 1

33. 33 Lucas Nos: Combinatorial Interpretation Consider the last tile (the tile counterclockwise before the first tile) Since the first tile determines the phase, fixing the last tile shows us l n-1 tilings ending in a square and l n-2 tilings ending in a domino Hence, l n = l n-1 + l n-2 = L n

34. 34 Identity: For n ≥ 1, L n = f n + f n-2 Question: How many tilings of a circular n-board exist? 1.There are L n circular n-bracelets. 2.Condition on the phase of the tiling: in-phase straightens into an n-tiling, thus f n in-phase bracelets out-of-phase: must have a domino covering cells n and 1 cells 2 to n-1 can be covered as a straight (n-2)-board, thus f n-2 out-of-phase bracelets.

35. 35 Identity: For n ≥ 1, L n = f n + f n-2

36. 36 Continued Fractions Given a 0 ≥ 0, a 1 ≥ 1, a 2 ≥ 1, …, a n ≥ 1, define [a 0, a 1, a 2, …, a n ] to be the fraction in lowest terms for For example, [2, 3, 4] =

37. 37 Continued Fractions: Comb. Interpretation Define functions p and q such that the continued fraction [a 0, a 1, a 2, …, a n ] = when reduced to lowest terms.

38. 38 Continued Fractions: Comb. Interpretation Let P n = P(a 0, a 1, a 2, …, a n ) count the number of ways to tile an (n+1)-board with dominoes and stackable square tiles. Height Restrictions: The ith cell may be covered by a stack of up to a i square tiles. Nothing can be stacked on top of a domino.

39. 39 Continued Fractions: Comb. Interpretation

40. 40 Continued Fractions: Comb. Interpretation Recall P n counts the number of ways to tile an n+1 board with dominoes and stackable square tiles. Let Q n = Q(a 0, a 1, a 2, …, a n ) count the number of ways to tile an n-board with dominoes and stackable square tiles. Define Q n = P(a 1, a 2, …, a n ). Then

41. 41 Continued Fractions: Comb. Interpretation QnQn PnPn

42. 42 Continued Fractions: Comb. Interpretation For example, the beginning of the “π-board” given by [3, 7, 15] can be tiled in 333 ways: all squares = 315 ways stack of squares, domino = 3 ways domino, stack of squares = 15 ways Removing the initial cell, the [7, 15]-board can be tiled in 106 ways: all squares = 105 ways domino = 1 way Thus [3, 7, 15] = ≈ 3.1415

43. 43 What else? Linear Recurrences Binomial Identities Stirling Numbers Continued Fractions Harmonic Numbers Number Theory Includes many open identities…

44. 44 References All material from “Proofs That Really Count: The Art of Combinatorial Proof” By Arthur T. Benjamin, Harvey Mudd College and Jennifer J. Quinn, Occidental College ©2003