The Art of Counting

The Art of Counting

David M. BressoudMacalester CollegeSt. Paul, MN

BAMA, April 12, 2006

This Power Point presentation can be downloaded from www.macalester.edu/~bressoud/talks

1. Review of binomial coefficients & Pascal’s triangle

2. Slicing cheese

3. A problem inspired by Charles Dodgson (aka Lewis Carroll)

Building the next value from

the previous values

5

2

⎛⎝⎜

⎞⎠⎟

Choose 2 of them

Given 5 objects

How many ways can this be done?

5

2

⎛⎝⎜

⎞⎠⎟

Choose 2 of them

Given 5 objects

How many ways can this be done?ABCDE AB, AC, AD, AE

BC, BD, BE, CD, CE, DE

5

2

⎛⎝⎜

⎞⎠⎟

Choose 2 of them

Given 5 objects

How many ways can this be done?ABCDE AB, AC, AD, AE

BC, BD, BE, CD, CE, DE

4

1

⎛⎝⎜

⎞⎠⎟

4

2

⎛⎝⎜

⎞⎠⎟

0

0

⎛⎝⎜

⎞⎠⎟

=1

1

0

⎛⎝⎜

⎞⎠⎟

=111

⎛⎝⎜

⎞⎠⎟

=1

2

0

⎛⎝⎜

⎞⎠⎟

=121

⎛⎝⎜

⎞⎠⎟

=222

⎛⎝⎜

⎞⎠⎟

=1

3

0

⎛⎝⎜

⎞⎠⎟

=131

⎛⎝⎜

⎞⎠⎟

=332

⎛⎝⎜

⎞⎠⎟

=333

⎛⎝⎜

⎞⎠⎟

=1

4

0

⎛⎝⎜

⎞⎠⎟

=141

⎛⎝⎜

⎞⎠⎟

=442

⎛⎝⎜

⎞⎠⎟

=643

⎛⎝⎜

⎞⎠⎟

=444

⎛⎝⎜

⎞⎠⎟

=1

5

0

⎛⎝⎜

⎞⎠⎟

=151

⎛⎝⎜

⎞⎠⎟

=552

⎛⎝⎜

⎞⎠⎟

=1053

⎛⎝⎜

⎞⎠⎟

=1054

⎛⎝⎜

⎞⎠⎟

=555

⎛⎝⎜

⎞⎠⎟

=1

+

0

0

⎛⎝⎜

⎞⎠⎟

=1

1

0

⎛⎝⎜

⎞⎠⎟

=111

⎛⎝⎜

⎞⎠⎟

=1

2

0

⎛⎝⎜

⎞⎠⎟

=121

⎛⎝⎜

⎞⎠⎟

=222

⎛⎝⎜

⎞⎠⎟

=1

3

0

⎛⎝⎜

⎞⎠⎟

=131

⎛⎝⎜

⎞⎠⎟

=332

⎛⎝⎜

⎞⎠⎟

=333

⎛⎝⎜

⎞⎠⎟

=1

4

0

⎛⎝⎜

⎞⎠⎟

=141

⎛⎝⎜

⎞⎠⎟

=442

⎛⎝⎜

⎞⎠⎟

=643

⎛⎝⎜

⎞⎠⎟

=444

⎛⎝⎜

⎞⎠⎟

=1

5

0

⎛⎝⎜

⎞⎠⎟

=151

⎛⎝⎜

⎞⎠⎟

=552

⎛⎝⎜

⎞⎠⎟

=1053

⎛⎝⎜

⎞⎠⎟

=1054

⎛⎝⎜

⎞⎠⎟

=555

⎛⎝⎜

⎞⎠⎟

=1

+ + ++

+ + +

++

+

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

1 7 21 35 35 21 7 1

1 8 28 56 70 56 28 8 1

1 9 36 84 126 126 84 36 9 1

“Pascal’s”

triangle

Published 1654

“Pascal’s” triangle from Siyuan yujian by Zhu Shihjie, 1303 CE

Dates to Jia Xian circa 1100 CE,

possibly earlier in Baghdad-Cairo or in India.

How many regions do we get if we cut space by 6

planes?

George Pólya(1887–1985)

Let Us Teach Guessing Math Assoc of America, 1965

How many regions do we get if we cut space by 6

planes?

0 planes: 1 region

1 plane: 2 regions

2 planes: 4 regions

3 planes: 8 regions

How many regions do we get if we cut space by 6

planes?

0 planes: 1 region

1 plane: 2 regions

2 planes: 4 regions

3 planes: 8 regions

4 planes: 15 regions

Cut a line by points

0: 1

1: 2

2: 3

3: 4

4: 5

5: 6

6: 7

Cut a line by points

0: 1

1: 2

2: 3

3: 4

4: 5

5: 6

6: 7

Cut a plane by lines

1

2

4

1

2

3

4 56

7

Cut a line by points

0: 1

1: 2

2: 3

3: 4

4: 5

5: 6

6: 7

Cut a plane by lines

1

2

4

7

Cut a line by points

0: 1

1: 2

2: 3

3: 4

4: 5

5: 6

6: 7

Cut a plane by lines

1

2

4

7

Cut a line by points

0: 1

1: 2

2: 3

3: 4

4: 5

5: 6

6: 7

Cut a plane by lines

1

2

4

7

11

Cut a line by points

0: 1

1: 2

2: 3

3: 4

4: 5

5: 6

6: 7

Cut a plane by lines

1

2

4

7

11

16

22

Cut a line by points

0: 1

1: 2

2: 3

3: 4

4: 5

5: 6

6: 7

Cut a plane by lines

1

2

4

7

11

16

22

Cut space by planes

1

2

4

8

4th plane cuts each of the previous 3 planes on a line

Cut a line by points

0: 1

1: 2

2: 3

3: 4

4: 5

5: 6

6: 7

Cut a plane by lines

1

2

4

7

11

16

22

Cut space by planes

1

2

4

8

15

5th plane cuts each of the previous 4 planes on a line

Cut a line by points

0: 1

1: 2

2: 3

3: 4

4: 5

5: 6

6: 7

Cut a plane by lines

1

2

4

7

11

16

22

Cut space by planes

1

2

4

8

15

26

??

1 1 1 1

2 2 2 1 1

3 4 4 1 2 1

4 7 8 1 3 3 1

5 11 15 1 4 6 4 1

6 16 26 1 5 10 10 5 1

7 22 42 1 6 15 20 15 6 1

line by points

plane by lines

space by planes

line by points

plane by lines

space by planes

1 1 1 1

2 2 2 1 1

3 4 4 1 2 1

4 7 8 1 3 3 1

5 11 15 1 4 6 4 1

6 16 26 1 5 10 10 5 1

7 22 42 1 6 15 20 15 6 1

line by points

plane by lines

space by planes

1 1 1 1

2 2 2 1 1

3 4 4 1 2 1

4 7 8 1 3 3 1

5 11 15 1 4 6 4 1

6 16 26 1 5 10 10 5 1

7 22 42 1 6 15 20 15 6 1

line by points

plane by lines

space by planes

1 1 1 1

2 2 2 1 1

3 4 4 1 2 1

4 7 8 1 3 3 1

5 11 15 1 4 6 4 1

6 16 26 1 5 10 10 5 1

7 22 42 1 6 15 20 15 6 1

Number of regions created when space is cut by k planes:

k

0

⎛⎝⎜

⎞⎠⎟+

k1

⎛⎝⎜

⎞⎠⎟+

k2

⎛⎝⎜

⎞⎠⎟+

k3

⎛⎝⎜

⎞⎠⎟

Number of regions created when space is cut by k planes:

k

0

⎛⎝⎜

⎞⎠⎟+

k1

⎛⎝⎜

⎞⎠⎟+

k2

⎛⎝⎜

⎞⎠⎟+

k3

⎛⎝⎜

⎞⎠⎟

Can we make sense of this formula?

What formula gives us the number of finite regions?

What happens in higher dimensional space and what does that mean?

Charles L. Dodgson

aka Lewis Carroll

“Condensation of Determinants,” Proceedings of the Royal Society, London 1866

Bill Mills

Dave Robbins

Howard Rumsey

Institute for Defense Analysis

Alternating Sign Matrix:

•Every row sums to 1

•Every column sums to 1

•Non-zero entries alternate in sign

A5 = 429

Alternating Sign Matrix:

•Every row sums to 1

•Every column sums to 1

•Non-zero entries alternate in sign

Monotone Triangle

Monotone Triangle

12345

1234 1235 1245 1345 2345

123 124 125 134 135 145 234 235 345

12 13 14 15 23 24 25 34 35 45

1 2 3 4 5

12345

1234 1235 1245 1345 2345

123 124 125 134 135 145 234 235 345

12 13 14 15 23 24 25 34 35 45

1 2 3 4 5

3

12345

1234 1235 1245 1345 2345

123 124 125 134 135 145 234 235 345

12 13 14 15 23 24 25 34 35 45

1 2 3 4 5

32 2 4 3 2 5 4 2

12345

1234 1235 1245 1345 2345

123 124 125 134 135 145 234 235 345

12 13 14 15 23 24 25 34 35 45

1 2 3 4 5

3

14

2 2 4 3 2 5 4 2

12345

1234 1235 1245 1345 2345

123 124 125 134 135 145 234 235 345

12 13 14 15 23 24 25 34 35 45

1 2 3 4 5

3

14 723142623714147

2 2 4 3 2 5 4 2

12345

1234 1235 1245 1345 2345

123 124 125 134 135 145 234 235 345

12 13 14 15 23 24 25 34 35 45

1 2 3 4 5

3

14 7

105

23142623714147

2 2 4 3 2 5 4 2

12345

1234 1235 1245 1345 2345

123 124 125 134 135 145 234 235 345

12 13 14 15 23 24 25 34 35 45

1 2 3 4 5

3

14 7

42 105 135 105 42

23142623714147

2 2 4 3 2 5 4 2

12345

1234 1235 1245 1345 2345

123 124 125 134 135 145 234 235 345

12 13 14 15 23 24 25 34 35 45

1 2 3 4 5

3

14 7

42 105 135 105 42

429

23142623714147

2 2 4 3 2 5 4 2

A5 = 429

A10 = 129, 534, 272, 700

A5 = 429

A10 = 129, 534, 272, 700

A20 = 1436038934715538200913155682637051204376827212

= 1.43… 1045

n

1

2

3

4

5

6

7

8

9

An

1

2

7

42

429

7436

218348

10850216

911835460

= 2 3 7

= 3 11 13

= 22 11 132

= 22 132 17 19

= 23 13 172 192

= 22 5 172 193 23

n

1

2

3

4

5

6

7

8

9

An

1

2

7

42

429

7436

218348

10850216

911835460

= 2 3 7

= 3 11 13

= 22 11 132

= 22 132 17 19

= 23 13 172 192

= 22 5 172 193 23

1

1 1

2 3 2

7 14 14 7

42 105 135 105 42

429 1287 2002 2002 1287 429

1

2

7

42

429

7436

1

1 1

2 3 2

7 14 14 7

42 105 135 105 42

429 1287 2002 2002 1287 429

+ + +

1

1 1

2 3 2

7 14 14 7

42 105 135 105 42

429 1287 2002 2002 1287 429

+ + +

1

1 3

1 2 5

1 2 4 5

1 2 3 4 5

1

1 2/2 1

2 2/3 3 3/2 2

7 2/4 14 14 4/2 7

42 2/5 105 135 105 5/2 42

429 2/6 1287 2002 2002 1287 6/2 429

1

1 2/2 1

2 2/3 3 3/2 2

7 2/4 14 5/5 14 4/2 7

42 2/5 105 7/9 135 9/7 105 5/2 42

429 2/6 1287 9/14 2002 16/16 2002 14/9 1287 6/2 429

2/2

2/3 3/2

2/4 5/5 4/2

2/5 7/9 9/7 5/2

2/6 9/14 16/16 14/9 6/2

1+1

1+1 1+2

1+1 2+3 1+3

1+1 3+4 3+6 1+4

1+1 4+5 6+10 4+10 1+5

Numerators:

1+1

1+1 1+2

1+1 2+3 1+3

1+1 3+4 3+6 1+4

1+1 4+5 6+10 4+10 1+5

Conjecture 1:

Numerators:

An,k

An,k+1

=

n−2k−1

⎛⎝⎜

⎞⎠⎟+

n−1k−1

⎛⎝⎜

⎞⎠⎟

n−2n−k−1

⎛⎝⎜

⎞⎠⎟+

n−1n−k−1

⎛⎝⎜

⎞⎠⎟

Conjecture 1:

Conjecture 2 (corollary of Conjecture 1):

An,k

An,k+1

=

n−2k−1

⎛⎝⎜

⎞⎠⎟+

n−1k−1

⎛⎝⎜

⎞⎠⎟

n−2n−k−1

⎛⎝⎜

⎞⎠⎟+

n−1n−k−1

⎛⎝⎜

⎞⎠⎟

An =

3 j +1( )!n+ j( )!j=0

n−1

∏ =1!⋅4!⋅7!L 3n−2( )!n! n+1( )!L 2n−1( )!

For derivation, go to www.macalester.edu/~bressoud/talks

Richard Stanley, M.I.T.

George Andrews, Penn State

length

width

n ≥ L1 > W1 ≥ L2 > W2 ≥ L3 > W3 ≥ …

1979, Andrews’ Theorem: the number of descending plane partitions of size n is

3 j +1( )!n+ j( )!j=0

n−1

∏ =1!⋅4!⋅7!L 3n−2( )!n! n+1( )!L 2n−1( )!

How many ways can we stack 75 boxes into a corner?

Percy A. MacMahon

How many ways can we stack 75 boxes into a corner?

Percy A. MacMahon

# of pp’s of 75 = pp(75) = 37,745,732,428,153

+ q + 3q2

+ 6q3

+ 13q4 + …

Generating function:

1+ pp j( )j=1

∞

∑ qj =1+ q+ 3q2 + 6q3 +13q4 +L

=1

1−qk( )k

k=1

∞

∏

=1

1−q( ) 1−q2( )2

1−q3( )3L

σ 2 k( ) = sum of squares of divisors of k

σ 2 1( ) = 12 = 1; σ 2 2( ) = 12 + 22 = 5;

σ 2 3( ) = 12 + 32 = 10; σ 2 4( ) = 12 + 22 + 42 = 21

σ 2 5( ) = 26; σ 2 6( ) = 12 + 22 + 32 + 62 = 50

For derivation, go to www.macalester.edu/~bressoud/talks

A little algebra turns this generating function into a recursive formula:

j ⋅pp j( ) = σ 2 k( )k=1

j

∑ pp j −k( )

Totally Symmetric Self-Complementary Plane Partitions

Robbins’ Conjecture: The number of TSSCPP’s in a 2n X 2n X 2n box is

3 j +1( )!n+ j( )!j=0

n−1

∏ =1!⋅4!⋅7!L 3n−2( )!

n!⋅n+1( )!L 2n−1( )!

Robbins’ Conjecture: The number of TSSCPP’s in a 2n X 2n X 2n box is

1989: William Doran shows equivalent to counting lattice paths

1990: John Stembridge represents the counting function as a Pfaffian (built on insights of Gordon and Okada)

1992: George Andrews evaluates the Pfaffian, proves Robbins’ Conjecture

3 j +1( )!n+ j( )!j=0

n−1

∏ =1!⋅4!⋅7!L 3n−2( )!

n!⋅n+1( )!L 2n−1( )!

1996

Zeilberger publishes “Proof of the Alternating Sign Matrix Conjecture,” Elect. J. of Combinatorics

Doron Zeilberger, Rutgers

1996 Kuperberg announces a simple proof

“Another proof of the alternating sign matrix conjecture,” International Mathematics Research Notices

Greg Kuperberg

UC Davis

Physicists have been studying ASM’s for decades, only they call them square ice (aka the six-vertex model ).

1996

Zeilberger uses this determinant to prove the original conjecture

“Proof of the refined alternating sign matrix conjecture,” New York Journal of Mathematics

The End

(which is really just the beginning)

This Power Point presentation can be downloaded from www.macalester.edu/~bressoud/talks