proofs that really count the art of combinatorial proof bradford greening, jr. rutgers university -...
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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
11
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:
12
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
13
Fibonacci Nos: Combinatorial Interpretation
fn : Counts the ways to tile an n-board with squares and dominoes.
14
Example: n = 4, f4 = 5
Fibonacci Nos: Combinatorial Interpretation
15
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?
17
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.
19
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.
20
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
22
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
23
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.
24
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
29
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
31
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
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
1
23
4
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
ln-1 tilings ending in a square and ln-2 tilings ending in a dominoHence, ln = ln-1 + ln-2 = Ln
34
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.
35
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
37
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
38
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.
39
Continued Fractions: Comb. Interpretation
0
a0
a1
a2
a3
an-1
an
1 2 3 n-1 n...
...
...
40
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
41
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
42
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
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