factorization of permutation - college of william & marycklixx.people.wm.edu › teaching ›...
TRANSCRIPT
![Page 1: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/1.jpg)
Factorization of permutation
Chi-Kwong LiDepartment of Mathematics, The College of William and Mary;
Institute for Quantum Computing, University of Waterloo
Based on the paper:Zejun Huang, Chi-Kwong Li, Sharon H. Li, Nung-Sing Sze,
Chi-Kwong Li Factorization of permutation
![Page 2: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/2.jpg)
Amidakuji/Ghost Leg Drawing
It is a scheme for assigning n people P1, . . . , Pn to n jobsJ1, . . . , Jn “randomly”.
Draw vertical lines from Pi to Ji from i = 1, . . . , n.Draw some horizontal line segments randomly betweenany two vertical lines that are next to each otherso that no horizontal lines meet.To assign a job for Pi, start from the top of thei-th line to the bottom, and make a turn whenevera horizontal segment is encountered.
Questions
Why do we always get an one-one correspondence (bijection)?Can we get all possible job assignments?What is the minimum number of horizontal segments needed for a givenjob assignment?
Chi-Kwong Li Factorization of permutation
![Page 3: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/3.jpg)
Amidakuji/Ghost Leg Drawing
It is a scheme for assigning n people P1, . . . , Pn to n jobsJ1, . . . , Jn “randomly”.
Draw vertical lines from Pi to Ji from i = 1, . . . , n.Draw some horizontal line segments randomly betweenany two vertical lines that are next to each otherso that no horizontal lines meet.To assign a job for Pi, start from the top of thei-th line to the bottom, and make a turn whenevera horizontal segment is encountered.
Questions
Why do we always get an one-one correspondence (bijection)?Can we get all possible job assignments?What is the minimum number of horizontal segments needed for a givenjob assignment?
Chi-Kwong Li Factorization of permutation
![Page 4: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/4.jpg)
Amidakuji/Ghost Leg Drawing
It is a scheme for assigning n people P1, . . . , Pn to n jobsJ1, . . . , Jn “randomly”.
Draw vertical lines from Pi to Ji from i = 1, . . . , n.
Draw some horizontal line segments randomly betweenany two vertical lines that are next to each otherso that no horizontal lines meet.To assign a job for Pi, start from the top of thei-th line to the bottom, and make a turn whenevera horizontal segment is encountered.
Questions
Why do we always get an one-one correspondence (bijection)?Can we get all possible job assignments?What is the minimum number of horizontal segments needed for a givenjob assignment?
Chi-Kwong Li Factorization of permutation
![Page 5: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/5.jpg)
Amidakuji/Ghost Leg Drawing
It is a scheme for assigning n people P1, . . . , Pn to n jobsJ1, . . . , Jn “randomly”.
Draw vertical lines from Pi to Ji from i = 1, . . . , n.Draw some horizontal line segments randomly betweenany two vertical lines that are next to each otherso that no horizontal lines meet.
To assign a job for Pi, start from the top of thei-th line to the bottom, and make a turn whenevera horizontal segment is encountered.
Questions
Why do we always get an one-one correspondence (bijection)?Can we get all possible job assignments?What is the minimum number of horizontal segments needed for a givenjob assignment?
Chi-Kwong Li Factorization of permutation
![Page 6: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/6.jpg)
Amidakuji/Ghost Leg Drawing
It is a scheme for assigning n people P1, . . . , Pn to n jobsJ1, . . . , Jn “randomly”.
Draw vertical lines from Pi to Ji from i = 1, . . . , n.Draw some horizontal line segments randomly betweenany two vertical lines that are next to each otherso that no horizontal lines meet.To assign a job for Pi, start from the top of thei-th line to the bottom, and make a turn whenevera horizontal segment is encountered.
Questions
Why do we always get an one-one correspondence (bijection)?Can we get all possible job assignments?What is the minimum number of horizontal segments needed for a givenjob assignment?
Chi-Kwong Li Factorization of permutation
![Page 7: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/7.jpg)
Amidakuji/Ghost Leg Drawing
It is a scheme for assigning n people P1, . . . , Pn to n jobsJ1, . . . , Jn “randomly”.
Draw vertical lines from Pi to Ji from i = 1, . . . , n.Draw some horizontal line segments randomly betweenany two vertical lines that are next to each otherso that no horizontal lines meet.To assign a job for Pi, start from the top of thei-th line to the bottom, and make a turn whenevera horizontal segment is encountered.
Questions
Why do we always get an one-one correspondence (bijection)?
Can we get all possible job assignments?What is the minimum number of horizontal segments needed for a givenjob assignment?
Chi-Kwong Li Factorization of permutation
![Page 8: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/8.jpg)
Amidakuji/Ghost Leg Drawing
It is a scheme for assigning n people P1, . . . , Pn to n jobsJ1, . . . , Jn “randomly”.
Draw vertical lines from Pi to Ji from i = 1, . . . , n.Draw some horizontal line segments randomly betweenany two vertical lines that are next to each otherso that no horizontal lines meet.To assign a job for Pi, start from the top of thei-th line to the bottom, and make a turn whenevera horizontal segment is encountered.
Questions
Why do we always get an one-one correspondence (bijection)?Can we get all possible job assignments?
What is the minimum number of horizontal segments needed for a givenjob assignment?
Chi-Kwong Li Factorization of permutation
![Page 9: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/9.jpg)
Amidakuji/Ghost Leg Drawing
It is a scheme for assigning n people P1, . . . , Pn to n jobsJ1, . . . , Jn “randomly”.
Draw vertical lines from Pi to Ji from i = 1, . . . , n.Draw some horizontal line segments randomly betweenany two vertical lines that are next to each otherso that no horizontal lines meet.To assign a job for Pi, start from the top of thei-th line to the bottom, and make a turn whenevera horizontal segment is encountered.
Questions
Why do we always get an one-one correspondence (bijection)?Can we get all possible job assignments?What is the minimum number of horizontal segments needed for a givenjob assignment?
Chi-Kwong Li Factorization of permutation
![Page 10: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/10.jpg)
Answer of Question 1
George Polya (1887-1985)If one cannot solve a problem,one can try to solve an easier problem first.
What if there is no horizontal line segment?
What if there is one horizontal line segment?
An easy induction argument!
Chi-Kwong Li Factorization of permutation
![Page 11: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/11.jpg)
Answer of Question 1
George Polya (1887-1985)If one cannot solve a problem,one can try to solve an easier problem first.
What if there is no horizontal line segment?
What if there is one horizontal line segment?
An easy induction argument!
Chi-Kwong Li Factorization of permutation
![Page 12: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/12.jpg)
Answer of Question 1
George Polya (1887-1985)If one cannot solve a problem,one can try to solve an easier problem first.
What if there is no horizontal line segment?
What if there is one horizontal line segment?
An easy induction argument!
Chi-Kwong Li Factorization of permutation
![Page 13: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/13.jpg)
Answer of Question 1
George Polya (1887-1985)If one cannot solve a problem,one can try to solve an easier problem first.
What if there is no horizontal line segment?
What if there is one horizontal line segment?
An easy induction argument!
Chi-Kwong Li Factorization of permutation
![Page 14: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/14.jpg)
Bubble sort
Regard the job assignment as a permutation (a seat assignment)
σ = [i1, . . . , in] =(
1 2 · · · ni1 i2 · · · in
).
Use Coxeter transpositions (i, i+ 1) for i = 1, . . . , n− 1.For any σ, we can determine the total number ι(σ) of inversions of σ.It is the minimum number of Coxeter transpositions needed to generate σ.
Example For σ = [5, 3, 1, 2, 4], total number of inversions is: 4 + 0 + 2 = 6, and
σ → [3, 5, 1, 2, 4]→ [3, 1, 5, 2, 4]→ [3, 1, 2, 5, 4]
→ [3, 1, 2, 4, 5]→ [1, 3, 2, 4, 5]→ [1, 2, 3, 4, 5],So σ = (1, 2)(2, 3)(3, 4)(4, 5)(1, 2)(2, 3).
Answers of Questions 2 and 3
We can always convert a permutation σ to [1, . . . , n] using ι(σ) steps,where ι(σ) is the number of inversions of σ.Worst case occurs at [n, n− 1, . . . , 1]; which requires
(n− 1) + · · ·+ 1 = n(n− 1)/2 steps.
Chi-Kwong Li Factorization of permutation
![Page 15: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/15.jpg)
Bubble sort
Regard the job assignment as a permutation (a seat assignment)
σ = [i1, . . . , in] =(
1 2 · · · ni1 i2 · · · in
).
Use Coxeter transpositions (i, i+ 1) for i = 1, . . . , n− 1.
For any σ, we can determine the total number ι(σ) of inversions of σ.It is the minimum number of Coxeter transpositions needed to generate σ.
Example For σ = [5, 3, 1, 2, 4], total number of inversions is: 4 + 0 + 2 = 6, and
σ → [3, 5, 1, 2, 4]→ [3, 1, 5, 2, 4]→ [3, 1, 2, 5, 4]
→ [3, 1, 2, 4, 5]→ [1, 3, 2, 4, 5]→ [1, 2, 3, 4, 5],So σ = (1, 2)(2, 3)(3, 4)(4, 5)(1, 2)(2, 3).
Answers of Questions 2 and 3
We can always convert a permutation σ to [1, . . . , n] using ι(σ) steps,where ι(σ) is the number of inversions of σ.Worst case occurs at [n, n− 1, . . . , 1]; which requires
(n− 1) + · · ·+ 1 = n(n− 1)/2 steps.
Chi-Kwong Li Factorization of permutation
![Page 16: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/16.jpg)
Bubble sort
Regard the job assignment as a permutation (a seat assignment)
σ = [i1, . . . , in] =(
1 2 · · · ni1 i2 · · · in
).
Use Coxeter transpositions (i, i+ 1) for i = 1, . . . , n− 1.For any σ, we can determine the total number ι(σ) of inversions of σ.
It is the minimum number of Coxeter transpositions needed to generate σ.Example For σ = [5, 3, 1, 2, 4], total number of inversions is: 4 + 0 + 2 = 6, and
σ → [3, 5, 1, 2, 4]→ [3, 1, 5, 2, 4]→ [3, 1, 2, 5, 4]
→ [3, 1, 2, 4, 5]→ [1, 3, 2, 4, 5]→ [1, 2, 3, 4, 5],So σ = (1, 2)(2, 3)(3, 4)(4, 5)(1, 2)(2, 3).
Answers of Questions 2 and 3
We can always convert a permutation σ to [1, . . . , n] using ι(σ) steps,where ι(σ) is the number of inversions of σ.Worst case occurs at [n, n− 1, . . . , 1]; which requires
(n− 1) + · · ·+ 1 = n(n− 1)/2 steps.
Chi-Kwong Li Factorization of permutation
![Page 17: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/17.jpg)
Bubble sort
Regard the job assignment as a permutation (a seat assignment)
σ = [i1, . . . , in] =(
1 2 · · · ni1 i2 · · · in
).
Use Coxeter transpositions (i, i+ 1) for i = 1, . . . , n− 1.For any σ, we can determine the total number ι(σ) of inversions of σ.It is the minimum number of Coxeter transpositions needed to generate σ.
Example For σ = [5, 3, 1, 2, 4], total number of inversions is: 4 + 0 + 2 = 6, and
σ → [3, 5, 1, 2, 4]→ [3, 1, 5, 2, 4]→ [3, 1, 2, 5, 4]
→ [3, 1, 2, 4, 5]→ [1, 3, 2, 4, 5]→ [1, 2, 3, 4, 5],So σ = (1, 2)(2, 3)(3, 4)(4, 5)(1, 2)(2, 3).
Answers of Questions 2 and 3
We can always convert a permutation σ to [1, . . . , n] using ι(σ) steps,where ι(σ) is the number of inversions of σ.Worst case occurs at [n, n− 1, . . . , 1]; which requires
(n− 1) + · · ·+ 1 = n(n− 1)/2 steps.
Chi-Kwong Li Factorization of permutation
![Page 18: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/18.jpg)
Bubble sort
Regard the job assignment as a permutation (a seat assignment)
σ = [i1, . . . , in] =(
1 2 · · · ni1 i2 · · · in
).
Use Coxeter transpositions (i, i+ 1) for i = 1, . . . , n− 1.For any σ, we can determine the total number ι(σ) of inversions of σ.It is the minimum number of Coxeter transpositions needed to generate σ.
Example For σ = [5, 3, 1, 2, 4], total number of inversions is: 4 + 0 + 2 = 6, and
σ → [3, 5, 1, 2, 4]→ [3, 1, 5, 2, 4]→ [3, 1, 2, 5, 4]
→ [3, 1, 2, 4, 5]→ [1, 3, 2, 4, 5]→ [1, 2, 3, 4, 5],
So σ = (1, 2)(2, 3)(3, 4)(4, 5)(1, 2)(2, 3).
Answers of Questions 2 and 3
We can always convert a permutation σ to [1, . . . , n] using ι(σ) steps,where ι(σ) is the number of inversions of σ.Worst case occurs at [n, n− 1, . . . , 1]; which requires
(n− 1) + · · ·+ 1 = n(n− 1)/2 steps.
Chi-Kwong Li Factorization of permutation
![Page 19: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/19.jpg)
Bubble sort
Regard the job assignment as a permutation (a seat assignment)
σ = [i1, . . . , in] =(
1 2 · · · ni1 i2 · · · in
).
Use Coxeter transpositions (i, i+ 1) for i = 1, . . . , n− 1.For any σ, we can determine the total number ι(σ) of inversions of σ.It is the minimum number of Coxeter transpositions needed to generate σ.
Example For σ = [5, 3, 1, 2, 4], total number of inversions is: 4 + 0 + 2 = 6, and
σ → [3, 5, 1, 2, 4]→ [3, 1, 5, 2, 4]→ [3, 1, 2, 5, 4]
→ [3, 1, 2, 4, 5]→ [1, 3, 2, 4, 5]→ [1, 2, 3, 4, 5],So σ = (1, 2)(2, 3)(3, 4)(4, 5)(1, 2)(2, 3).
Answers of Questions 2 and 3
We can always convert a permutation σ to [1, . . . , n] using ι(σ) steps,where ι(σ) is the number of inversions of σ.Worst case occurs at [n, n− 1, . . . , 1]; which requires
(n− 1) + · · ·+ 1 = n(n− 1)/2 steps.
Chi-Kwong Li Factorization of permutation
![Page 20: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/20.jpg)
Bubble sort
Regard the job assignment as a permutation (a seat assignment)
σ = [i1, . . . , in] =(
1 2 · · · ni1 i2 · · · in
).
Use Coxeter transpositions (i, i+ 1) for i = 1, . . . , n− 1.For any σ, we can determine the total number ι(σ) of inversions of σ.It is the minimum number of Coxeter transpositions needed to generate σ.
Example For σ = [5, 3, 1, 2, 4], total number of inversions is: 4 + 0 + 2 = 6, and
σ → [3, 5, 1, 2, 4]→ [3, 1, 5, 2, 4]→ [3, 1, 2, 5, 4]
→ [3, 1, 2, 4, 5]→ [1, 3, 2, 4, 5]→ [1, 2, 3, 4, 5],So σ = (1, 2)(2, 3)(3, 4)(4, 5)(1, 2)(2, 3).
Answers of Questions 2 and 3
We can always convert a permutation σ to [1, . . . , n] using ι(σ) steps,where ι(σ) is the number of inversions of σ.
Worst case occurs at [n, n− 1, . . . , 1]; which requires
(n− 1) + · · ·+ 1 = n(n− 1)/2 steps.
Chi-Kwong Li Factorization of permutation
![Page 21: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/21.jpg)
Bubble sort
Regard the job assignment as a permutation (a seat assignment)
σ = [i1, . . . , in] =(
1 2 · · · ni1 i2 · · · in
).
Use Coxeter transpositions (i, i+ 1) for i = 1, . . . , n− 1.For any σ, we can determine the total number ι(σ) of inversions of σ.It is the minimum number of Coxeter transpositions needed to generate σ.
Example For σ = [5, 3, 1, 2, 4], total number of inversions is: 4 + 0 + 2 = 6, and
σ → [3, 5, 1, 2, 4]→ [3, 1, 5, 2, 4]→ [3, 1, 2, 5, 4]
→ [3, 1, 2, 4, 5]→ [1, 3, 2, 4, 5]→ [1, 2, 3, 4, 5],So σ = (1, 2)(2, 3)(3, 4)(4, 5)(1, 2)(2, 3).
Answers of Questions 2 and 3
We can always convert a permutation σ to [1, . . . , n] using ι(σ) steps,where ι(σ) is the number of inversions of σ.Worst case occurs at [n, n− 1, . . . , 1]; which requires
(n− 1) + · · ·+ 1 = n(n− 1)/2 steps.
Chi-Kwong Li Factorization of permutation
![Page 22: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/22.jpg)
Bubble sort
Regard the job assignment as a permutation (a seat assignment)
σ = [i1, . . . , in] =(
1 2 · · · ni1 i2 · · · in
).
Use Coxeter transpositions (i, i+ 1) for i = 1, . . . , n− 1.For any σ, we can determine the total number ι(σ) of inversions of σ.It is the minimum number of Coxeter transpositions needed to generate σ.
Example For σ = [5, 3, 1, 2, 4], total number of inversions is: 4 + 0 + 2 = 6, and
σ → [3, 5, 1, 2, 4]→ [3, 1, 5, 2, 4]→ [3, 1, 2, 5, 4]
→ [3, 1, 2, 4, 5]→ [1, 3, 2, 4, 5]→ [1, 2, 3, 4, 5],So σ = (1, 2)(2, 3)(3, 4)(4, 5)(1, 2)(2, 3).
Answers of Questions 2 and 3
We can always convert a permutation σ to [1, . . . , n] using ι(σ) steps,where ι(σ) is the number of inversions of σ.Worst case occurs at [n, n− 1, . . . , 1]; which requires
(n− 1) + · · ·+ 1 = n(n− 1)/2 steps.
Chi-Kwong Li Factorization of permutation
![Page 23: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/23.jpg)
A variation of Amidakuji
What if we consider transpositions of the forms (i, i+ 1) and (i, i+ 2)?
How about using transpositions (i, i+ 1), (i, i+ 2), (i, i+ 3), etc.?
An extreme case: Using all (i, j) with 1 ≤ j < n
Decompose σ as product of k disjoint cycles (including fixed points).
Then σ is a product of n− k transpositions.
So, the worst case requires n− 1 steps.
Example. σ =(
1 2 3 4 5 6 7 8 93 4 5 9 6 7 1 8 2
)= (1, 3, 5, 6, 7)(2, 4, 9)(8).
Then σ = (1, 7)(1, 6)(1, 5)(1, 3)(2, 9)(2, 4).
Chi-Kwong Li Factorization of permutation
![Page 24: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/24.jpg)
A variation of Amidakuji
What if we consider transpositions of the forms (i, i+ 1) and (i, i+ 2)?
How about using transpositions (i, i+ 1), (i, i+ 2), (i, i+ 3), etc.?
An extreme case: Using all (i, j) with 1 ≤ j < n
Decompose σ as product of k disjoint cycles (including fixed points).
Then σ is a product of n− k transpositions.
So, the worst case requires n− 1 steps.
Example. σ =(
1 2 3 4 5 6 7 8 93 4 5 9 6 7 1 8 2
)= (1, 3, 5, 6, 7)(2, 4, 9)(8).
Then σ = (1, 7)(1, 6)(1, 5)(1, 3)(2, 9)(2, 4).
Chi-Kwong Li Factorization of permutation
![Page 25: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/25.jpg)
A variation of Amidakuji
What if we consider transpositions of the forms (i, i+ 1) and (i, i+ 2)?
How about using transpositions (i, i+ 1), (i, i+ 2), (i, i+ 3), etc.?
An extreme case: Using all (i, j) with 1 ≤ j < n
Decompose σ as product of k disjoint cycles (including fixed points).
Then σ is a product of n− k transpositions.
So, the worst case requires n− 1 steps.
Example. σ =(
1 2 3 4 5 6 7 8 93 4 5 9 6 7 1 8 2
)= (1, 3, 5, 6, 7)(2, 4, 9)(8).
Then σ = (1, 7)(1, 6)(1, 5)(1, 3)(2, 9)(2, 4).
Chi-Kwong Li Factorization of permutation
![Page 26: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/26.jpg)
A variation of Amidakuji
What if we consider transpositions of the forms (i, i+ 1) and (i, i+ 2)?
How about using transpositions (i, i+ 1), (i, i+ 2), (i, i+ 3), etc.?
An extreme case: Using all (i, j) with 1 ≤ j < n
Decompose σ as product of k disjoint cycles (including fixed points).
Then σ is a product of n− k transpositions.
So, the worst case requires n− 1 steps.
Example. σ =(
1 2 3 4 5 6 7 8 93 4 5 9 6 7 1 8 2
)= (1, 3, 5, 6, 7)(2, 4, 9)(8).
Then σ = (1, 7)(1, 6)(1, 5)(1, 3)(2, 9)(2, 4).
Chi-Kwong Li Factorization of permutation
![Page 27: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/27.jpg)
A variation of Amidakuji
What if we consider transpositions of the forms (i, i+ 1) and (i, i+ 2)?
How about using transpositions (i, i+ 1), (i, i+ 2), (i, i+ 3), etc.?
An extreme case: Using all (i, j) with 1 ≤ j < n
Decompose σ as product of k disjoint cycles (including fixed points).
Then σ is a product of n− k transpositions.
So, the worst case requires n− 1 steps.
Example. σ =(
1 2 3 4 5 6 7 8 93 4 5 9 6 7 1 8 2
)= (1, 3, 5, 6, 7)(2, 4, 9)(8).
Then σ = (1, 7)(1, 6)(1, 5)(1, 3)(2, 9)(2, 4).
Chi-Kwong Li Factorization of permutation
![Page 28: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/28.jpg)
A variation of Amidakuji
What if we consider transpositions of the forms (i, i+ 1) and (i, i+ 2)?
How about using transpositions (i, i+ 1), (i, i+ 2), (i, i+ 3), etc.?
An extreme case: Using all (i, j) with 1 ≤ j < n
Decompose σ as product of k disjoint cycles (including fixed points).
Then σ is a product of n− k transpositions.
So, the worst case requires n− 1 steps.
Example. σ =(
1 2 3 4 5 6 7 8 93 4 5 9 6 7 1 8 2
)= (1, 3, 5, 6, 7)(2, 4, 9)(8).
Then σ = (1, 7)(1, 6)(1, 5)(1, 3)(2, 9)(2, 4).
Chi-Kwong Li Factorization of permutation
![Page 29: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/29.jpg)
A variation of Amidakuji
What if we consider transpositions of the forms (i, i+ 1) and (i, i+ 2)?
How about using transpositions (i, i+ 1), (i, i+ 2), (i, i+ 3), etc.?
An extreme case: Using all (i, j) with 1 ≤ j < n
Decompose σ as product of k disjoint cycles (including fixed points).
Then σ is a product of n− k transpositions.
So, the worst case requires n− 1 steps.
Example. σ =(
1 2 3 4 5 6 7 8 93 4 5 9 6 7 1 8 2
)= (1, 3, 5, 6, 7)(2, 4, 9)(8).
Then σ = (1, 7)(1, 6)(1, 5)(1, 3)(2, 9)(2, 4).
Chi-Kwong Li Factorization of permutation
![Page 30: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/30.jpg)
Some open problems
Let 1 ≤ m < n, and let Gm be the set of transpositions of the form (i, i+ `)with 1 ≤ ` ≤ m.
For a given σ ∈ Sn, find the smallest r such that σ is the product of rtranspositions in Gm.
Determine the optimal (smallest) r∗ = r∗(n,m) so that every σ ∈ Sn is aproduct at most r∗ transpositions in Gm.
To find r∗ and the permutation which is most difficult to get restore, weuse the breadth first search.
Chi-Kwong Li Factorization of permutation
![Page 31: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/31.jpg)
Some open problems
Let 1 ≤ m < n, and let Gm be the set of transpositions of the form (i, i+ `)with 1 ≤ ` ≤ m.
For a given σ ∈ Sn, find the smallest r such that σ is the product of rtranspositions in Gm.
Determine the optimal (smallest) r∗ = r∗(n,m) so that every σ ∈ Sn is aproduct at most r∗ transpositions in Gm.
To find r∗ and the permutation which is most difficult to get restore, weuse the breadth first search.
Chi-Kwong Li Factorization of permutation
![Page 32: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/32.jpg)
Some open problems
Let 1 ≤ m < n, and let Gm be the set of transpositions of the form (i, i+ `)with 1 ≤ ` ≤ m.
For a given σ ∈ Sn, find the smallest r such that σ is the product of rtranspositions in Gm.
Determine the optimal (smallest) r∗ = r∗(n,m) so that every σ ∈ Sn is aproduct at most r∗ transpositions in Gm.
To find r∗ and the permutation which is most difficult to get restore, weuse the breadth first search.
Chi-Kwong Li Factorization of permutation
![Page 33: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/33.jpg)
Some open problems
Let 1 ≤ m < n, and let Gm be the set of transpositions of the form (i, i+ `)with 1 ≤ ` ≤ m.
For a given σ ∈ Sn, find the smallest r such that σ is the product of rtranspositions in Gm.
Determine the optimal (smallest) r∗ = r∗(n,m) so that every σ ∈ Sn is aproduct at most r∗ transpositions in Gm.
To find r∗ and the permutation which is most difficult to get restore, weuse the breadth first search.
Chi-Kwong Li Factorization of permutation
![Page 34: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/34.jpg)
Partial results of the general problem
We have the following list for r∗(n,m) for Sn and (i, i+ `) with ` ≤ m,
n\m 1 2 3 4 5 6 7 8 9 10 112 13 3 24 6 4 35 10 5 5 46 15 [7] 6 6 57 21 [10] 8 7 7 68 28 [14] [10] 9 8 8 79 36 [16] [11] 10 10 9 9 810 45 [19] [14] [12] 11 11 10 10 911 55 [23] [16] [14] 13 12 12 11 11 1012 66 29∗ 20∗ 17∗ 16∗ 14 13 13 12 12 11
where the entries marked by brackets are obtained by computer programming.
Chi-Kwong Li Factorization of permutation
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Another variation (The round table version)
Theorem [Jerrum, 1985], [van Zuylen et. al, 2014]Only use transpositions: (n, 1) and (i, i+ 1) : i = 1, . . . , n− 1.
Given a permutation [p1, . . . , pn].let d = [d1, . . . , dn] = [p1, · · · , pn]− [1, . . . , n] so that
∑di = 0;
modify d to d̃ by replacing (di, dj) by (di − n, n+ dj) if di − dj > n
until pr − ps ≤ n for all r, s.Restore the permutation using this displacement vector d̃ will use theminimum number of steps ι(d̃), which is the generalized inversion numberof p̃ = d̃+ [1, . . . , n].
The number of steps is at most [n2/4] attained at the following permutation:(1) [k + 1, . . . , n, 1, . . . , k] if n = 2k or n = 2k + 1,(2) [k + 2, . . . , n, 1, . . . , k + 1] or [k + 1, . . . , n, 1, . . . , k] if n = 2k + 1.
Example p = [6, 5, 1, 2, 4, 3], d = [5, 3,−2,−2,−1,−3],
d̃ = [−1, 3,−2,−2,−1, 3], p̃ = [0, 5, 1, 2, 4, 9], ι(d̃) = 6,
Note For [4, 5, 6, 1, 2, 3], d = [3, 3, 3,−3,−3,−3] and ι(d) = 9 = [62/4].
Chi-Kwong Li Factorization of permutation
![Page 36: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/36.jpg)
Another variation (The round table version)
Theorem [Jerrum, 1985], [van Zuylen et. al, 2014]Only use transpositions: (n, 1) and (i, i+ 1) : i = 1, . . . , n− 1.Given a permutation [p1, . . . , pn].
let d = [d1, . . . , dn] = [p1, · · · , pn]− [1, . . . , n] so that∑
di = 0;modify d to d̃ by replacing (di, dj) by (di − n, n+ dj) if di − dj > n
until pr − ps ≤ n for all r, s.Restore the permutation using this displacement vector d̃ will use theminimum number of steps ι(d̃), which is the generalized inversion numberof p̃ = d̃+ [1, . . . , n].
The number of steps is at most [n2/4] attained at the following permutation:(1) [k + 1, . . . , n, 1, . . . , k] if n = 2k or n = 2k + 1,(2) [k + 2, . . . , n, 1, . . . , k + 1] or [k + 1, . . . , n, 1, . . . , k] if n = 2k + 1.
Example p = [6, 5, 1, 2, 4, 3], d = [5, 3,−2,−2,−1,−3],
d̃ = [−1, 3,−2,−2,−1, 3], p̃ = [0, 5, 1, 2, 4, 9], ι(d̃) = 6,
Note For [4, 5, 6, 1, 2, 3], d = [3, 3, 3,−3,−3,−3] and ι(d) = 9 = [62/4].
Chi-Kwong Li Factorization of permutation
![Page 37: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/37.jpg)
Another variation (The round table version)
Theorem [Jerrum, 1985], [van Zuylen et. al, 2014]Only use transpositions: (n, 1) and (i, i+ 1) : i = 1, . . . , n− 1.Given a permutation [p1, . . . , pn].
let d = [d1, . . . , dn] = [p1, · · · , pn]− [1, . . . , n] so that∑
di = 0;
modify d to d̃ by replacing (di, dj) by (di − n, n+ dj) if di − dj > n
until pr − ps ≤ n for all r, s.Restore the permutation using this displacement vector d̃ will use theminimum number of steps ι(d̃), which is the generalized inversion numberof p̃ = d̃+ [1, . . . , n].
The number of steps is at most [n2/4] attained at the following permutation:(1) [k + 1, . . . , n, 1, . . . , k] if n = 2k or n = 2k + 1,(2) [k + 2, . . . , n, 1, . . . , k + 1] or [k + 1, . . . , n, 1, . . . , k] if n = 2k + 1.
Example p = [6, 5, 1, 2, 4, 3], d = [5, 3,−2,−2,−1,−3],
d̃ = [−1, 3,−2,−2,−1, 3], p̃ = [0, 5, 1, 2, 4, 9], ι(d̃) = 6,
Note For [4, 5, 6, 1, 2, 3], d = [3, 3, 3,−3,−3,−3] and ι(d) = 9 = [62/4].
Chi-Kwong Li Factorization of permutation
![Page 38: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/38.jpg)
Another variation (The round table version)
Theorem [Jerrum, 1985], [van Zuylen et. al, 2014]Only use transpositions: (n, 1) and (i, i+ 1) : i = 1, . . . , n− 1.Given a permutation [p1, . . . , pn].
let d = [d1, . . . , dn] = [p1, · · · , pn]− [1, . . . , n] so that∑
di = 0;modify d to d̃ by replacing (di, dj) by (di − n, n+ dj) if di − dj > n
until pr − ps ≤ n for all r, s.
Restore the permutation using this displacement vector d̃ will use theminimum number of steps ι(d̃), which is the generalized inversion numberof p̃ = d̃+ [1, . . . , n].
The number of steps is at most [n2/4] attained at the following permutation:(1) [k + 1, . . . , n, 1, . . . , k] if n = 2k or n = 2k + 1,(2) [k + 2, . . . , n, 1, . . . , k + 1] or [k + 1, . . . , n, 1, . . . , k] if n = 2k + 1.
Example p = [6, 5, 1, 2, 4, 3], d = [5, 3,−2,−2,−1,−3],
d̃ = [−1, 3,−2,−2,−1, 3], p̃ = [0, 5, 1, 2, 4, 9], ι(d̃) = 6,
Note For [4, 5, 6, 1, 2, 3], d = [3, 3, 3,−3,−3,−3] and ι(d) = 9 = [62/4].
Chi-Kwong Li Factorization of permutation
![Page 39: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/39.jpg)
Another variation (The round table version)
Theorem [Jerrum, 1985], [van Zuylen et. al, 2014]Only use transpositions: (n, 1) and (i, i+ 1) : i = 1, . . . , n− 1.Given a permutation [p1, . . . , pn].
let d = [d1, . . . , dn] = [p1, · · · , pn]− [1, . . . , n] so that∑
di = 0;modify d to d̃ by replacing (di, dj) by (di − n, n+ dj) if di − dj > n
until pr − ps ≤ n for all r, s.Restore the permutation using this displacement vector d̃ will use theminimum number of steps ι(d̃),
which is the generalized inversion numberof p̃ = d̃+ [1, . . . , n].
The number of steps is at most [n2/4] attained at the following permutation:(1) [k + 1, . . . , n, 1, . . . , k] if n = 2k or n = 2k + 1,(2) [k + 2, . . . , n, 1, . . . , k + 1] or [k + 1, . . . , n, 1, . . . , k] if n = 2k + 1.
Example p = [6, 5, 1, 2, 4, 3], d = [5, 3,−2,−2,−1,−3],
d̃ = [−1, 3,−2,−2,−1, 3], p̃ = [0, 5, 1, 2, 4, 9], ι(d̃) = 6,
Note For [4, 5, 6, 1, 2, 3], d = [3, 3, 3,−3,−3,−3] and ι(d) = 9 = [62/4].
Chi-Kwong Li Factorization of permutation
![Page 40: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/40.jpg)
Another variation (The round table version)
Theorem [Jerrum, 1985], [van Zuylen et. al, 2014]Only use transpositions: (n, 1) and (i, i+ 1) : i = 1, . . . , n− 1.Given a permutation [p1, . . . , pn].
let d = [d1, . . . , dn] = [p1, · · · , pn]− [1, . . . , n] so that∑
di = 0;modify d to d̃ by replacing (di, dj) by (di − n, n+ dj) if di − dj > n
until pr − ps ≤ n for all r, s.Restore the permutation using this displacement vector d̃ will use theminimum number of steps ι(d̃), which is the generalized inversion numberof p̃ = d̃+ [1, . . . , n].
The number of steps is at most [n2/4] attained at the following permutation:(1) [k + 1, . . . , n, 1, . . . , k] if n = 2k or n = 2k + 1,(2) [k + 2, . . . , n, 1, . . . , k + 1] or [k + 1, . . . , n, 1, . . . , k] if n = 2k + 1.
Example p = [6, 5, 1, 2, 4, 3], d = [5, 3,−2,−2,−1,−3],
d̃ = [−1, 3,−2,−2,−1, 3], p̃ = [0, 5, 1, 2, 4, 9], ι(d̃) = 6,
Note For [4, 5, 6, 1, 2, 3], d = [3, 3, 3,−3,−3,−3] and ι(d) = 9 = [62/4].
Chi-Kwong Li Factorization of permutation
![Page 41: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/41.jpg)
Another variation (The round table version)
Theorem [Jerrum, 1985], [van Zuylen et. al, 2014]Only use transpositions: (n, 1) and (i, i+ 1) : i = 1, . . . , n− 1.Given a permutation [p1, . . . , pn].
let d = [d1, . . . , dn] = [p1, · · · , pn]− [1, . . . , n] so that∑
di = 0;modify d to d̃ by replacing (di, dj) by (di − n, n+ dj) if di − dj > n
until pr − ps ≤ n for all r, s.Restore the permutation using this displacement vector d̃ will use theminimum number of steps ι(d̃), which is the generalized inversion numberof p̃ = d̃+ [1, . . . , n].
The number of steps is at most [n2/4] attained at the following permutation:(1) [k + 1, . . . , n, 1, . . . , k] if n = 2k or n = 2k + 1,(2) [k + 2, . . . , n, 1, . . . , k + 1] or [k + 1, . . . , n, 1, . . . , k] if n = 2k + 1.
Example p = [6, 5, 1, 2, 4, 3], d = [5, 3,−2,−2,−1,−3],
d̃ = [−1, 3,−2,−2,−1, 3], p̃ = [0, 5, 1, 2, 4, 9], ι(d̃) = 6,
Note For [4, 5, 6, 1, 2, 3], d = [3, 3, 3,−3,−3,−3] and ι(d) = 9 = [62/4].
Chi-Kwong Li Factorization of permutation
![Page 42: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/42.jpg)
Another variation (The round table version)
Theorem [Jerrum, 1985], [van Zuylen et. al, 2014]Only use transpositions: (n, 1) and (i, i+ 1) : i = 1, . . . , n− 1.Given a permutation [p1, . . . , pn].
let d = [d1, . . . , dn] = [p1, · · · , pn]− [1, . . . , n] so that∑
di = 0;modify d to d̃ by replacing (di, dj) by (di − n, n+ dj) if di − dj > n
until pr − ps ≤ n for all r, s.Restore the permutation using this displacement vector d̃ will use theminimum number of steps ι(d̃), which is the generalized inversion numberof p̃ = d̃+ [1, . . . , n].
The number of steps is at most [n2/4] attained at the following permutation:(1) [k + 1, . . . , n, 1, . . . , k] if n = 2k or n = 2k + 1,(2) [k + 2, . . . , n, 1, . . . , k + 1] or [k + 1, . . . , n, 1, . . . , k] if n = 2k + 1.
Example p = [6, 5, 1, 2, 4, 3], d = [5, 3,−2,−2,−1,−3],
d̃ = [−1, 3,−2,−2,−1, 3], p̃ = [0, 5, 1, 2, 4, 9], ι(d̃) = 6,
Note For [4, 5, 6, 1, 2, 3], d = [3, 3, 3,−3,−3,−3] and ι(d) = 9 = [62/4].
Chi-Kwong Li Factorization of permutation
![Page 43: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/43.jpg)
Some open problems
What if we can use the the circular permutations (i, j) with |j − i| ≤ kwith k = 1, 2, . . . in the round table problem?
One can use L = (1, 2, . . . , n) and S = (1, 2) to generate allpermutations. Then the maximum steps needed are:
S2 S3 S4 S5 S6 S7 S8 S9 S10 S11 S12
1 2 6 11 18 25 35 45 58 71 ???
If we use L, S, L−1, then the maximum steps needed are:
S2 S3 S4 S5 S6 S7 S8 S9 S10 S11
1 2 6 10 15 21 28 36 45 ???
Conjecture We need at most(
n2
)steps, and the worst case is
[2, 1, n, n− 1, . . . , 3].
Chi-Kwong Li Factorization of permutation
![Page 44: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/44.jpg)
Some open problems
What if we can use the the circular permutations (i, j) with |j − i| ≤ kwith k = 1, 2, . . . in the round table problem?One can use L = (1, 2, . . . , n) and S = (1, 2) to generate allpermutations. Then the maximum steps needed are:
S2 S3 S4 S5 S6 S7 S8 S9 S10 S11 S12
1 2 6 11 18 25 35 45 58 71 ???
If we use L, S, L−1, then the maximum steps needed are:
S2 S3 S4 S5 S6 S7 S8 S9 S10 S11
1 2 6 10 15 21 28 36 45 ???
Conjecture We need at most(
n2
)steps, and the worst case is
[2, 1, n, n− 1, . . . , 3].
Chi-Kwong Li Factorization of permutation
![Page 45: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/45.jpg)
Some open problems
What if we can use the the circular permutations (i, j) with |j − i| ≤ kwith k = 1, 2, . . . in the round table problem?One can use L = (1, 2, . . . , n) and S = (1, 2) to generate allpermutations. Then the maximum steps needed are:
S2 S3 S4 S5 S6 S7 S8 S9 S10 S11 S12
1 2 6 11 18 25 35 45 58 71 ???
If we use L, S, L−1, then the maximum steps needed are:
S2 S3 S4 S5 S6 S7 S8 S9 S10 S11
1 2 6 10 15 21 28 36 45 ???
Conjecture We need at most(
n2
)steps, and the worst case is
[2, 1, n, n− 1, . . . , 3].
Chi-Kwong Li Factorization of permutation
![Page 46: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/46.jpg)
Related research
Theoretical computer science?
Determine the optimal sorting algorithmwith the given operations, and determine the worst scenario.
The study of genomics and mutations, i.e.,the change of genetic sequences x1x2x3 · · · , with xi ∈ {A,U,G,C}.
Quantum computing.It is of interest to decompose certain quantum gates into simplerquantum gates (CNOT gates).
Chi-Kwong Li Factorization of permutation
![Page 47: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/47.jpg)
Related research
Theoretical computer science? Determine the optimal sorting algorithmwith the given operations, and determine the worst scenario.
The study of genomics and mutations, i.e.,the change of genetic sequences x1x2x3 · · · , with xi ∈ {A,U,G,C}.
Quantum computing.It is of interest to decompose certain quantum gates into simplerquantum gates (CNOT gates).
Chi-Kwong Li Factorization of permutation
![Page 48: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/48.jpg)
Related research
Theoretical computer science? Determine the optimal sorting algorithmwith the given operations, and determine the worst scenario.
The study of genomics and mutations,
i.e.,the change of genetic sequences x1x2x3 · · · , with xi ∈ {A,U,G,C}.
Quantum computing.It is of interest to decompose certain quantum gates into simplerquantum gates (CNOT gates).
Chi-Kwong Li Factorization of permutation
![Page 49: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/49.jpg)
Related research
Theoretical computer science? Determine the optimal sorting algorithmwith the given operations, and determine the worst scenario.
The study of genomics and mutations, i.e.,the change of genetic sequences x1x2x3 · · · , with xi ∈ {A,U,G,C}.
Quantum computing.It is of interest to decompose certain quantum gates into simplerquantum gates (CNOT gates).
Chi-Kwong Li Factorization of permutation
![Page 50: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/50.jpg)
Related research
Theoretical computer science? Determine the optimal sorting algorithmwith the given operations, and determine the worst scenario.
The study of genomics and mutations, i.e.,the change of genetic sequences x1x2x3 · · · , with xi ∈ {A,U,G,C}.
Quantum computing.It is of interest to decompose certain quantum gates into simplerquantum gates (CNOT gates).
Chi-Kwong Li Factorization of permutation
![Page 51: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/51.jpg)
Let me know if you have any ideas.
Thank you for your attention!
Chi-Kwong Li Factorization of permutation
![Page 52: Factorization of permutation - College of William & Marycklixx.people.wm.edu › teaching › math400 › sorting.pdfDepartment of Mathematics, The College of William and Mary; Institute](https://reader036.vdocument.in/reader036/viewer/2022081406/5f0ebf477e708231d440be02/html5/thumbnails/52.jpg)
Let me know if you have any ideas.
Thank you for your attention!
Chi-Kwong Li Factorization of permutation