The Principle of Overcountingfaculty.cord.edu/ahendric/2010Fall325/notBurnside.pdf · The...
Transcript of The Principle of Overcountingfaculty.cord.edu/ahendric/2010Fall325/notBurnside.pdf · The...
The Principle of Overcounting
In combinatorics, we often have to count in a two-step procedure:
1 Use a “wrong” procedure to count, which yields a numberthat’s too high.
2 Then figure out how wrong our answer was, and reduce it tothe correct number.
Example
How many subsets of {A,B,C,D,E} have exactly 3 elements?We start by counting the ways to choose three elements:
5
·
4
·
3 = 60
The Principle of Overcounting
In combinatorics, we often have to count in a two-step procedure:
1 Use a “wrong” procedure to count, which yields a numberthat’s too high.
2 Then figure out how wrong our answer was, and reduce it tothe correct number.
Example
How many subsets of {A,B,C,D,E} have exactly 3 elements?We start by counting the ways to choose three elements:
5
·
4
·
3 = 60
The Principle of Overcounting
In combinatorics, we often have to count in a two-step procedure:
1 Use a “wrong” procedure to count, which yields a numberthat’s too high.
2 Then figure out how wrong our answer was, and reduce it tothe correct number.
Example
How many subsets of {A,B,C,D,E} have exactly 3 elements?We start by counting the ways to choose three elements:
5
·
4
·
3 = 60
The Principle of Overcounting
In combinatorics, we often have to count in a two-step procedure:
1 Use a “wrong” procedure to count, which yields a numberthat’s too high.
2 Then figure out how wrong our answer was, and reduce it tothe correct number.
Example
How many subsets of {A,B,C,D,E} have exactly 3 elements?We start by counting the ways to choose three elements:
5 ·
4
·
3 = 60
The Principle of Overcounting
In combinatorics, we often have to count in a two-step procedure:
1 Use a “wrong” procedure to count, which yields a numberthat’s too high.
2 Then figure out how wrong our answer was, and reduce it tothe correct number.
Example
How many subsets of {A,B,C,D,E} have exactly 3 elements?We start by counting the ways to choose three elements:
5 · 4 ·
3 = 60
The Principle of Overcounting
In combinatorics, we often have to count in a two-step procedure:
1 Use a “wrong” procedure to count, which yields a numberthat’s too high.
2 Then figure out how wrong our answer was, and reduce it tothe correct number.
Example
How many subsets of {A,B,C,D,E} have exactly 3 elements?We start by counting the ways to choose three elements:
5 · 4 · 3
= 60
The Principle of Overcounting
In combinatorics, we often have to count in a two-step procedure:
1 Use a “wrong” procedure to count, which yields a numberthat’s too high.
2 Then figure out how wrong our answer was, and reduce it tothe correct number.
Example
How many subsets of {A,B,C,D,E} have exactly 3 elements?We start by counting the ways to choose three elements:
5 · 4 · 3 = 60
The Principle of Overcounting
ABC
ABD
ABE
ACB
ACD
ACE
ADB
ADC
ADE
AEB
AEC
AED
BAC
BAD
BAE
BCA
BCD
BCE
BDA
BDC
BDE
BEA
BEC
BED
CAB
CAD
CAE
CBA
CBD
CBE
CDA
CDB
CDE
CEA
CEB
CED
DAB
DAC
DAE
DBA
DBC
DBE
DCA
DCB
DCE
DEA
DEB
DEC
EAB
EAC
EAD
EBA
EBC
EBD
ECA
ECB
ECD
EDA
EDB
EDC
The Principle of Overcounting
The Principle of Overcounting
ABC
ABD
ABE
ACB
ACD
ACE
ADB
ADC
ADE
AEB
AEC
AED
BAC
BAD
BAE
BCA
BCD
BCE
BDA
BDC
BDE
BEA
BEC
BED
CAB
CAD
CAE
CBA
CBD
CBE
CDA
CDB
CDE
CEA
CEB
CED
DAB
DAC
DAE
DBA
DBC
DBE
DCA
DCB
DCE
DEA
DEB
DEC
EAB
EAC
EAD
EBA
EBC
EBD
ECA
ECB
ECD
EDA
EDB
EDC
{A,B,C}{A,B,D}{A,B,E}{A,C ,D}{A,C ,E}{A,D,E}{B,C ,D}{B,C ,E}{B,D,E}{C ,D,E}
The Principle of Overcounting
ABC
ABD
ABE
ACB
ACD
ACE
ADB
ADC
ADE
AEB
AEC
AED
BAC
BAD
BAE
BCA
BCD
BCE
BDA
BDC
BDE
BEA
BEC
BED
CAB
CAD
CAE
CBA
CBD
CBE
CDA
CDB
CDE
CEA
CEB
CED
DAB
DAC
DAE
DBA
DBC
DBE
DCA
DCB
DCE
DEA
DEB
DEC
EAB
EAC
EAD
EBA
EBC
EBD
ECA
ECB
ECD
EDA
EDB
EDC
{A,B,C}{A,B,D}{A,B,E}{A,C ,D}{A,C ,E}{A,D,E}{B,C ,D}{B,C ,E}{B,D,E}{C ,D,E}
60
6= 10 =
(53
)
Another Example
Another Example
How many ways to seat three people around a round table?
Initial (wrong) answer: 3! = 6.
Another Example
Another Example
How many ways to seat three people around a round table?
3
Initial (wrong) answer: 3! = 6.
Another Example
Another Example
How many ways to seat three people around a round table?
3
2
Initial (wrong) answer: 3! = 6.
Another Example
Another Example
How many ways to seat three people around a round table?
3
21
Initial (wrong) answer: 3! = 6.
Another Example
Another Example
How many ways to seat three people around a round table?
3
21
Initial (wrong) answer: 3! = 6.
Another Example
Another Example
Another Example
Another Example
Correct answer:3!
3= 2.
A More Complicated Example
Example
You have two colors, red and blue. How many ways are there topaint the corners of a scalene triangle?
Initial answer: 2 · 2 · 2 = 8.
A More Complicated Example
Example
You have two colors, red and blue. How many ways are there topaint the corners of a scalene triangle?
2
Initial answer: 2 · 2 · 2 = 8.
A More Complicated Example
Example
You have two colors, red and blue. How many ways are there topaint the corners of a scalene triangle?
2
2
Initial answer: 2 · 2 · 2 = 8.
A More Complicated Example
Example
You have two colors, red and blue. How many ways are there topaint the corners of a scalene triangle?
2
2
2
Initial answer: 2 · 2 · 2 = 8.
A More Complicated Example
Example
You have two colors, red and blue. How many ways are there topaint the corners of a scalene triangle?
2
2
2
Initial answer: 2 · 2 · 2 = 8.
A More Complicated Example
Example
You have two colors, red and blue. How many ways are there topaint the corners of an equilateral triangle?
Initial (wrong) answer: 2 · 2 · 2 = 8.
A More Complicated Example
A More Complicated Example
A More Complicated Example
A More Complicated Example
Correct answer: 4.
An Even More Complicated Example
An Even More Complicated Example
You have two colors, red and blue. How many ways are there tocolor the corners of a square?
An Even More Complicated Example
An Even More Complicated Example
You have two colors, red and blue. How many ways are there tocolor the corners of a square?
22
2 2
Wrong answer: 24 = 16
An Even More Complicated Example
An Even More Complicated Example
An Even More Complicated Example
An Even More Complicated Example
Correct answer: 6.
Absurdly Complicated Example
Absurdly Complicated Example
How many ways are there to paint the corners of a square withfour colors? Count colorings as the same if one can be rotated orflipped to look like the other.
The initial (wrong) answer: 44 = 256.
I don’t want to draw all those out,let alone figure out where the repeats are.
We need a faster way to predict the answer, withoutexamining all the possibilities by hand.
The answer: group actions!
R90
R90
Observations
In our example:
Every coloring we wanted was in the whole list of 24
possibilities.
Some of those were equivalent and shouldn’t be countedseparately.
We considered two colorings equivalent when one could betransformed into the other by a rotation.
That is, we were looking at the orbits of the colorings underthe action of the group of rotations {R0,R90,R180,R270}.Our problem: Counting the orbits.
The Set-Up
G acts on S where
S = all coloringsG = symmetry group of the figure
“distinguishable” = in different orbits
Combinatorics Strategy
1 Think of all possible ways (including duplicates).
2 Decide what makes two ways count as equivalent.
3 Decide what group action has those equivalence classes as itsorbits.
4 Use the not-Burnside Theorem to count the orbits.
Combinatorics Strategy
1 Think of all possible ways (including duplicates).
2 Decide what makes two ways count as equivalent.
3 Decide what group action has those equivalence classes as itsorbits.
4 Use the not-Burnside Theorem to count the orbits.
Combinatorics Strategy
1 Think of all possible ways (including duplicates).
2 Decide what makes two ways count as equivalent.
3 Decide what group action has those equivalence classes as itsorbits.
4 Use the not-Burnside Theorem to count the orbits.
Combinatorics Strategy
1 Think of all possible ways (including duplicates).
2 Decide what makes two ways count as equivalent.
3 Decide what group action has those equivalence classes as itsorbits.
4 Use the not-Burnside Theorem to count the orbits.
The not-Burnside Theorem
Theorem (not-Burnside)
Let G act on S. For each g ∈ G, let ψ(g) = |Fix(g)|. Then thenumber of orbits of S under the action of G is
1
|G |∑g∈G
ψ(g).
Proof Sketch:
Count {(g , s) ∈ G × S : s · g = s} in two different ways. �
The not-Burnside Theorem
Theorem (not-Burnside)
Let G act on S. For each g ∈ G, let ψ(g) = |Fix(g)|. Then thenumber of orbits of S under the action of G is
1
|G |∑g∈G
ψ(g).
Proof Sketch:
Count {(g , s) ∈ G × S : s · g = s} in two different ways. �
Applying not-Burnside
Even More Complicated Example, revisited
You have two colors, red and blue. How many ways are there tocolor the corners of a square? Count two colorings as the same ifthey can be rotated to resemble one another.
Our symmetry group is {R0,R90,R180,R270} ≤ D8.
g ψ(g)
R0
16
R90
2
R180
4
R270
224
Thus the number of distinguishable colorings is
1
|G |∑g∈G
ψ(g) =
1
4· 24 = 6.
Applying not-Burnside
Even More Complicated Example, revisited
You have two colors, red and blue. How many ways are there tocolor the corners of a square? Count two colorings as the same ifthey can be rotated to resemble one another.
Our symmetry group is {R0,R90,R180,R270} ≤ D8.
g ψ(g)
R0 16R90
2
R180
4
R270
224
Thus the number of distinguishable colorings is
1
|G |∑g∈G
ψ(g) =
1
4· 24 = 6.
Applying not-Burnside
Even More Complicated Example, revisited
You have two colors, red and blue. How many ways are there tocolor the corners of a square? Count two colorings as the same ifthey can be rotated to resemble one another.
Our symmetry group is {R0,R90,R180,R270} ≤ D8.
g ψ(g)
R0 16R90 2R180
4
R270
224
Thus the number of distinguishable colorings is
1
|G |∑g∈G
ψ(g) =
1
4· 24 = 6.
Applying not-Burnside
Even More Complicated Example, revisited
You have two colors, red and blue. How many ways are there tocolor the corners of a square? Count two colorings as the same ifthey can be rotated to resemble one another.
Our symmetry group is {R0,R90,R180,R270} ≤ D8.
g ψ(g)
R0 16R90 2R180 4R270
224
Thus the number of distinguishable colorings is
1
|G |∑g∈G
ψ(g) =
1
4· 24 = 6.
Applying not-Burnside
Even More Complicated Example, revisited
You have two colors, red and blue. How many ways are there tocolor the corners of a square? Count two colorings as the same ifthey can be rotated to resemble one another.
Our symmetry group is {R0,R90,R180,R270} ≤ D8.
g ψ(g)
R0 16R90 2R180 4R270 2
24
Thus the number of distinguishable colorings is
1
|G |∑g∈G
ψ(g) =
1
4· 24 = 6.
Applying not-Burnside
Even More Complicated Example, revisited
You have two colors, red and blue. How many ways are there tocolor the corners of a square? Count two colorings as the same ifthey can be rotated to resemble one another.
Our symmetry group is {R0,R90,R180,R270} ≤ D8.
g ψ(g)
R0 16R90 2R180 4R270 2
24
Thus the number of distinguishable colorings is
1
|G |∑g∈G
ψ(g) =
1
4· 24 = 6.
Applying not-Burnside
Even More Complicated Example, revisited
You have two colors, red and blue. How many ways are there tocolor the corners of a square? Count two colorings as the same ifthey can be rotated to resemble one another.
Our symmetry group is {R0,R90,R180,R270} ≤ D8.
g ψ(g)
R0 16R90 2R180 4R270 2
24
Thus the number of distinguishable colorings is
1
|G |∑g∈G
ψ(g) =1
4· 24 = 6.
Applying not-Burnside
Absurdly Complicated Example, revisited
How many ways are there to paint the corners of a square withfour colors? Count colorings as the same if one can be rotated orflipped to look like the other.
Note that our symmetry group is D8.
Applying not-Burnside
g ψ(g)
R0
256
R90
4
R180
16
R270
4
H
16
V
16
D1
64
D2
64440
Thus the number of distinguishable colorings is
1
|D8|∑g∈D8
ψ(g) =
1
8· 440 = 55.
Applying not-Burnside
g ψ(g)
R0 256R90
4
R180
16
R270
4
H
16
V
16
D1
64
D2
64440
Thus the number of distinguishable colorings is
1
|D8|∑g∈D8
ψ(g) =
1
8· 440 = 55.
Applying not-Burnside
g ψ(g)
R0 256R90 4R180
16
R270
4
H
16
V
16
D1
64
D2
64440
Thus the number of distinguishable colorings is
1
|D8|∑g∈D8
ψ(g) =
1
8· 440 = 55.
Applying not-Burnside
g ψ(g)
R0 256R90 4R180 16R270
4
H
16
V
16
D1
64
D2
64440
Thus the number of distinguishable colorings is
1
|D8|∑g∈D8
ψ(g) =
1
8· 440 = 55.
Applying not-Burnside
g ψ(g)
R0 256R90 4R180 16R270 4H
16
V
16
D1
64
D2
64440
Thus the number of distinguishable colorings is
1
|D8|∑g∈D8
ψ(g) =
1
8· 440 = 55.
Applying not-Burnside
g ψ(g)
R0 256R90 4R180 16R270 4H 16V
16
D1
64
D2
64440
Thus the number of distinguishable colorings is
1
|D8|∑g∈D8
ψ(g) =
1
8· 440 = 55.
Applying not-Burnside
g ψ(g)
R0 256R90 4R180 16R270 4H 16V 16D1
64
D2
64440
Thus the number of distinguishable colorings is
1
|D8|∑g∈D8
ψ(g) =
1
8· 440 = 55.
Applying not-Burnside
g ψ(g)
R0 256R90 4R180 16R270 4H 16V 16D1 64D2
64440
Thus the number of distinguishable colorings is
1
|D8|∑g∈D8
ψ(g) =
1
8· 440 = 55.
Applying not-Burnside
g ψ(g)
R0 256R90 4R180 16R270 4H 16V 16D1 64D2 64
440
Thus the number of distinguishable colorings is
1
|D8|∑g∈D8
ψ(g) =
1
8· 440 = 55.
Applying not-Burnside
g ψ(g)
R0 256R90 4R180 16R270 4H 16V 16D1 64D2 64
440
Thus the number of distinguishable colorings is
1
|D8|∑g∈D8
ψ(g) =
1
8· 440 = 55.
Applying not-Burnside
g ψ(g)
R0 256R90 4R180 16R270 4H 16V 16D1 64D2 64
440
Thus the number of distinguishable colorings is
1
|D8|∑g∈D8
ψ(g) =1
8· 440 = 55.