Proofs That Count

Proofs That Really CountThe Art of Combinatorial ProofBradford Greening, Jr.Rutgers University - Camden

ThemeShow 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.

Identity: For n 0, Q:Number of ways to choose 2 numbers from {0, 1, 2, , n}?

By definition,

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

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

For 2k n,

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

A committee of even size can be formed as follows: Step 1: Choose the 1st person in or out2 waysStep 2: Choose the 2nd person in or out2 ways

Step n-1: Choose the (n-1)th person in or out2 waysStep n: Choose the nth person in or out1 way

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

: 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 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.

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?

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}


There are ways to create the sequence, then k ways to choose the underlined term.


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 rs chosen get placed to the left of our underlined r. Any n+1s chosen get converted to rs and placed to the right of our r.

Hence, there are such sequences.

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}Choose rCreate k-1 sequence from n+1 numbersConvert8-sequence: converts to 9-sequence:

1

1

2

3

3

5

6

6

1

1

2

3

3

5

2

2

2

2

Fibonacci Numbers

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 Nos: Combinatorial Interpretation

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

Fibonacci Nos: Combinatorial Interpretation Example: n = 4, f4 = 5

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

Fibonacci Nos: Combinatorial Interpretation 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 + 1Q: How many ways to tile an n-board with squares and dominoes?

Identity: For n 0, f0 + f1 + f2 + + fn = fn+2 -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?

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

Consider the last domino (in spots k+1 & k+2).fk ways to tile first k spots1 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.

1

2

3

n

n+1

n+2

1

2

3

n

n+1

n+2

1

2

f0

f1

f2

fn-1

3

n

n+1

n+2

1

fn

2 3

n

n+1

n+2

1 2

3

n

n+1

n+2

...

...

...

...

...

...

...

...

Cells 1, 2, , k

k+1 k+2

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.

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

For each n-tiling, make 3 new tilingsby 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

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

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2 3

4

5

6 7

8 9

10

1

2 3

4 5

6

7

8

9 10

Identity: For n 0, Q: How many tilings of an n-board and (n+1)-board exist?

By definition, fn fn+1 tilings exist.

Place the (n+1)-board directly above the n-board.Consider the location of the last fault.

1

2

3

1

2

3

...

n

n+1

...

n

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.

...

k-1, k

k

Identity: For n 0, 2n = fn + fn-1 + Q: How many binary sequences of length n exist?

There are 2n binary sequences of length n.

For each binary sequence define a tiling as follows:1 is equivalent to a square in the tiling.01 is equivalent to a domino.

Identity: For n 0, 2n = 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)

01

1

1

01

01

1

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.fk

...

...

1

2

3

4

k

k+3

n-1

n

0

0

k+1

k+2

2n-2-k

2n-2-k

Identity: For n 0, 2n = fn + fn-1 + 16 length-11 binary sequences generate the same 5-tiling01101000000011010010000110100000101101001001011010000100110100101001101000011011010010110110100010001101001100011010001010110100110101101000110011010011100110100011101101001111

01

1

01

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

Lucas Nos: Combinatorial Interpretation

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

Lucas Nos: Combinatorial Interpretationout-of-phase a tiling where a domino covers cells n and 1

in-phase all other tilings

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

Lucas Nos: Combinatorial InterpretationQ: How many ways to tile a circular n-board?

Note that the first tile can bea square covering cell 1a domino covering cells 1 and 2a domino covering cells n and 1

1

2

3

4

Lucas Nos: Combinatorial InterpretationConsider 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

Identity: For n 1, Ln = fn + fn-2Question: How many tilings of a circular n-board exist?

There are Ln circular n-bracelets.

Condition on the phase of the tiling:in-phase straightens into an n-tiling, thus fn in-phase braceletsout-of-phase: must have a domino covering cells n and 1cells 2 to n-1 can be covered as a straight (n-2)-board,thus fn-2 out-of-phase bracelets.

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

n

n

Continued FractionsGiven 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] =

Continued Fractions: Comb. Interpretation

Define functions p and q such that the continued fraction

[a0, a1, a2, , an] =

when reduced to lowest terms.


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.


V.P.

Continued Fractions: Comb. InterpretationRecall 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

Continued Fractions: Comb. InterpretationQnPn

V.P.

V.P.

Continued Fractions: Comb. InterpretationFor example, the beginning of the-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

V.P.

V.P.

What else?Linear Recurrences

Binomial Identities

Stirling NumbersContinued Fractions

Harmonic Numbers

Number Theory

Includes many open identities

ReferencesAll material from

Proofs That Really Count: The Art of Combinatorial Proof ByArthur T. Benjamin, Harvey Mudd CollegeandJennifer J. Quinn, Occidental College 2003

We have built a proof by induction to show the relationship between the Fibonacci sequence and the tilings of the n-board. This visual representation of tiling an n-board will allow us to clearly give proofs of many identities as we shall see in the coming slides. Here we touch on a new method of proof. We interpret the left side of the identity as the size of a set, the right side of the identity as the size of a different set, and show by creating a correspondence that they are equal. Here we touch on a new method of proof. We interpret the left side of the identity as the size of a set, the right side of the identity as the size of a different set, and show by creating a correspondence that they are equal. For the next proof, we define the notion of a fault.One fault free way to tile the remaining cells: all dominoes save for a single square in the k+1th cell of the row with odd tail length. The boards will still be of length n+1 and n, respectively, however, we must place a square in the odd-length tiling (whether it is the n-tiling or the (n+1)-tiling we do not care) to preserve its original odd length.For example, the length 11 binary sequence 01101001001 maps to the 5-tiling domino-square-domino, however so would any binary sequence of form 0110100abcd (a, b, c, d = 0 or 1)The red numbers are what gives the 5-tiling, the blue numbers are the first instance of 00Sequence is different b/c of different initial conditionsNote: Since we can now cover cells n and 1 by a domino, we must define this as a different Why should we condition on the last tile instead of the first? Out-of-phase: must have domino covering cells n and 1, by definition.Pictorial representation of proof on previous slide.Given without proof to save time.Open in the sense that the identities have not been proven combinatorially.

Proofs That Count

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