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

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

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 : “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 : “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 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 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 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 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 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… Fibonacci Numbers

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

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

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 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 Identity: For n ≥ 0, f 0 + f 1 + f 2 + … + f n = f n 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 Identity: For n ≥ 0, f 0 + f 1 + f 2 + … + f n = f n 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 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 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 Identity: For n ≥ 0, We say there is a fault at cell i, if both tilings are breakable at cell i.

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 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 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 Example: The binary sequence 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 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 Identity: For n ≥ 0, 2 n = f n + f n length-11 binary sequences generate the same 5-tiling

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

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

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

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 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 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 Continued Fractions: Comb. Interpretation

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 Continued Fractions: Comb. Interpretation QnQn PnPn

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] = ≈

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

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