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The Art of Counting. David M. Bressoud Macalester College St. Paul, MN BAMA, April 12, 2006. This Power Point presentation can be downloaded from www.macalester.edu/~bressoud/talks. Review of binomial coefficients & Pascal’s triangle Slicing cheese - PowerPoint PPT Presentation
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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
