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

1

1

2

n

k

nk

1

2

n

1

n

k

k

4

Identity:

Q: Count ways to create a committee of even size from n people?

1. For 2k ≤ n,

1

0

22

n

k

n

k

0

...0 2 4 2 2k

n n n n n

k k

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 1st person ‘in’ or ‘out’ 2 ways

Step 2: Choose the 2nd person ‘in’ or ‘out’ 2 ways

Step n-1: Choose the (n-1)th person ‘in’ or ‘out’ 2 ways

Step n: Choose the nth person ‘in’ or ‘out’ 1 way

By multiplication rule, there are 2n-1 ways to form this committee.

1

0

22

n

k

n

k

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}

n

k

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.

1

1

n nk n

k k

n

k

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.

1

1

n nk n

k k

Hence, there are such sequences.1

1

nn

k

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Identity:

Example: n = 5, k = 9, and our underlined value is r = , then weare choosing a length 8 sequence from {1, 2, 3, 4, 5, 6}

1

1

n nk n

k k

1 1 2 3 3 5 6 6

1 1 2 3 3 52 2 2

2

1. Choose “r”

2. Create k-1 sequence from n+1 numbers

3. Convert

8-sequence:

converts to

9-

sequence:

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Fibonacci Numbers – a number sequence defined as

• F0 = 0, F1 = 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

Fibonacci Numbers

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Fibonacci Nos: Combinatorial Interpretation

fn : Counts the ways to tile an n-board with squares and dominoes.

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Example: n = 4, f4 = 5

Fibonacci Nos: Combinatorial Interpretation

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Fibonacci Nos: Combinatorial Interpretation

fn : Counts the ways to tile an n-board with squares and dominoes.

Define f-1 = 0 and let f0 = 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

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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

Fibonacci Nos: Combinatorial Interpretation

Q: How many ways to tile an n-board with squares and dominoes?

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Identity: For n ≥ 0, f0 + f1 + f2 + … + fn = fn+2 -1

1. By definition there are fn + 2 tilings of an (n+2)-board; excluding the “all-squares” tiling leaves fn + 2 – 1.

Q: How many tilings of an (n+2)-board have at least 1 domino?

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Identity: For n ≥ 0, f0 + f1 + f2 + … + fn = fn+2 -1

2. Consider the last domino (in spots k+1 & k+2).

• fk 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?

1 2 3 n n+1 n+2

1 2 3 n n+1 n+2

1 2 3 n n+1 n+2

1 2 3 n n+1 n+2

1 2 3 n n+1 n+2

...

...

...

...

...

... ......

f0

f1

f2

fn-1

fn

Cells 1, 2, …, k k+1 k+2Summing over all possible locations of k gives LHS.

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Identity: For n ≥ 1, 3fn = 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.

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Identity: For n ≥ 1, 3fn = 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

n-tiling

n-tiling

(n-1)-tiling

(n-2)-tiling

n-tiling

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Identity: For n ≥ 0,

We say there is a fault at cell i, if both tilings are breakable at cell i.

2

10

n

k n nk

f f f

1 2 3 4 5 6 7 8 9 10

1 2 3 4 5 6 7 8 9 10

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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.

2

10

n

k n nk

f f f

1 2 3 ... n n+1

1 2 3 ... n

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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:

2

10

n

k n nk

f f f

… k

... k-1, k

Summing over all possible locations of k gives LHS.

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Identity: For n ≥ 0, 2n = fn + fn-1 +

Q: How many binary sequences of length n exist?

1. There are 2n 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.

22

0

2n

n kk

k

f

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Example:

The binary sequence 011101011 maps to the 9-tiling shown below.

22

0

2n

n kk

k

f

01 1 1 01 01 1

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, 2n = fn + fn-1 +

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Identity: For n ≥ 0, 2n = fn + fn-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.

22

0

2n

n kk

k

f

0 0... ...2 3 4 k1 k+3 n-1 nk+1 k+2

fk

2n-2-k

2n-2-k

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

2

0

2n

n kk

k

f

16 length-11 binary sequences generate the same 5-tiling

01101000000 0110100100001101000001 0110100100101101000010 0110100101001101000011 0110100101101101000100 0110100110001101000101 0110100110101101000110 0110100111001101000111 01101001111

01 1 01

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Lucas Numbers

Lucas Numbers – a number sequence defined as

• L0 = 2, L1 = 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

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Lucas Nos: Combinatorial Interpretation

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

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Lucas Nos: Combinatorial Interpretation

“out-of-phase” – a tiling where a domino covers cells n and 1

“in-phase” – all other tilings

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Lucas Nos: Combinatorial Interpretation

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

Let l0 = 2, and l1 = 1. Then for n ≥ 2,

ln = ln-1 + ln-2 = Ln

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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

1

23

4

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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

ln-1 tilings ending in a square and ln-2 tilings ending in a dominoHence, ln = ln-1 + ln-2 = Ln

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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.

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Identity: For n ≥ 1, Ln = fn + fn-2

n

n

36

Continued Fractions

Given a0 ≥ 0, a1 ≥ 1, a2 ≥ 1, …, an ≥ 1, define [a0, a1, a2, …, an] to be the fraction in lowest terms for

For example, [2, 3, 4] =

0

1

2

3

11

11

...

1

n

aa

aa

a

1 30

21 1334

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Continued Fractions: Comb. Interpretation

Define functions p and q such that the continued fraction

[a0, a1, a2, …, an] =

when reduced to lowest terms.

0 1 2

0 1 2

( , , ,..., )

( , , ,..., )n n

n n

p a a a a p

q a a a a q

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Continued Fractions: Comb. Interpretation

Let Pn = P(a0, a1, a2, …, 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.

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Continued Fractions: Comb. Interpretation

0

a0

a1

a2

a3

an-1

an

1 2 3 n-1 n...

...

...

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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(a0, a1, a2, …, an) count the number of ways to tile an n-board with dominoes and stackable square tiles.

Define Qn = P(a1, a2, …, an).

Then 0 1 2[ , , , ... , ]n nn

n n

P pa a a a

Q q

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Continued Fractions: Comb. Interpretation

0

a0

a1

a2

a3

an-1

an

1 2 3 n-1 n...

...

...

a1

a2

a3

an-1

an

1 2 3 n-1 n...

...

...

Qn

Pn

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Continued Fractions: Comb. Interpretation

0

3

7

15

1 2

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

7

15

1 2

333

106

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What else?

• Linear Recurrences

• Binomial Identities

• Stirling Numbers

• Continued Fractions

• Harmonic Numbers

• Number Theory

Includes many open identities…

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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