Proofs That Really Count The Art of Combinatorial

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

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. 2

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 3

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

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’ Step n: Choose the nth person ‘in’ or ‘out’ 2 ways 1 way By multiplication rule, there are 2 n-1 ways to form this committee. 5

: “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} {1, 3, 3, 5, 7, 7} (n = 8, k = 6) or {1, 1, 1, 1} 6

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. 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? 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. 10

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} 8 -sequence: 1. Choose “r” 2. Create k-1 sequence from n+1 numbers converts to 3. Convert 9 -sequence: 11

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

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

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

Fibonacci Nos: Combinatorial Interpretation fn : 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 fn is a Fibonacci number and for n ≥ 2, fn = fn-1 + fn-2 = Fn + 1 15

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

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

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

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

Identity: For n ≥ 1, 3 fn = fn+2 + fn-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 • 20

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

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

Identity: For n ≥ 0, How many tiling pairs have their last fault at cell k? • There are ( fk )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. 23

Identity: For n ≥ 0, 2 n = fn + fn-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. 24

Identity: For n ≥ 0, 2 n = fn + fn-1 + 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) 25

Identity: For n ≥ 0, 2 n = fn + fn-1 + What if “ 00” exists? Let the first occurrence of “ 00” appear in cells k+1, k+2 (k ≤ n-2) fk 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. 26

Identity: For n ≥ 0, 2 n = fn + fn-1 + 01101000000 01101000001 01101000010 01101000011 01101000101 011010001101000111 01101001000 01101001001 01101001010 01101001011 01101001100 01101001101001110 01101001111 16 length-11 binary sequences generate the same 5 -tiling 27

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

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

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

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

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 32

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

Identity: For n ≥ 1, Ln = fn + fn-2 Question: How many tilings of a circular n-board exist? 1. There are Ln circular n-bracelets. 2. Condition on the phase of the tiling: in-phase straightens into an n-tiling, thus fn 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 fn-2 out-of-phase bracelets. 34

Identity: For n ≥ 1, Ln = fn + fn-2 35

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

$Continued Fractions: Comb. Interpretation Define functions p and q such that the continued fraction$

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

Continued Fractions: Comb. Interpretation Let Pn = P(a 0, a 1, a 2, …, an) 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 ai square tiles. • Nothing can be stacked on top of a domino. 38

Continued Fractions: Comb. Interpretation 39

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

Continued Fractions: Comb. Interpretation Pn Qn 41

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 42

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