permutahedra and the saneblidze umble diagonal
Download
Skip this Video
Download Presentation
Permutahedra and the Saneblidze-Umble Diagonal

Loading in 2 Seconds...

play fullscreen
1 / 31

Permutahedra and the Saneblidze-Umble Diagonal - PowerPoint PPT Presentation


  • 129 Views
  • Uploaded on

Permutahedra and the Saneblidze-Umble Diagonal. By Stephen Weaver Directed by Dr. Ron Umble. Computational Geometry the study of algorithms to solve problems in geometry. Permutahedra.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about ' Permutahedra and the Saneblidze-Umble Diagonal' - sai


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
permutahedra and the saneblidze umble diagonal
Permutahedra and the Saneblidze-Umble Diagonal

By Stephen Weaver

Directed by Dr. Ron Umble

slide2

Computational Geometry

the study of algorithms to solve problems in geometry

permutahedra
Permutahedra
  • Geometric shapes based on permutations of the partitions of a set {1,2}  {3}
  • Each permutation corresponds to a face of some permutahedron
  • We use bar notation for convenience {3} {1,2} = 3|12
  • The single partition {1,2,3,…,n} corresponds to the top dimensional face
permutahedra1
Permutahedra
  • The boundary of a face consists of adding a single bar in every possible way

1|2

2|1

12

  • Two faces are adjacent if their boundaries intersect

2|1|3

1|3|2

12|3

1|23

1|2|3

permutahedra examples
Permutahedra - Examples

P2

P1

1|2

2|1

12

1

3|12

P3

3|1|2

3|2|1

P4

13|2

23|1

123

1|3|2

2|3|1

1|23

2|13

1|2|3

12|3

2|1|3

slide6

Permutahedra - Examples

P2

P1

1|2

2|1

12

1

3|12

P3

3|1|2

3|2|1

P4

13|2

23|1

123

1|3|2

2|3|1

1|23

2|13

1|2|3

12|3

2|1|3

slide7

Permutahedra - Examples

P2

P1

1|2

2|1

12

1

3|12

P3

3|1|2

3|2|1

P4

13|2

23|1

123

1|3|2

2|3|1

1|23

2|13

1|2|3

12|3

2|1|3

slide8

Permutahedra - Examples

P2

P1

1|2

2|1

12

1

3|12

P3

3|1|2

3|2|1

P4

13|2

23|1

123

1|3|2

123|4

2|3|1

12|3|4

12|34

1|23

2|13

1|2|3

12|3

2|1|3

diagonal
Diagonal

Given a set S,

Diagonal of S  S { (x,x) | x є S }

S

S

diagonal on p 2 p 2
Diagonal on P2 P2

12  2|1

1|2  2|1

2|1  2|2

1|2  12

2|1  1|2

1|2  1|2

step matrix example
Step Matrix Example

5

6

4

7

8

Reading a Step Matrix

1

9

2

3

1

2

3

4

14|2|3  4|123

slide12

Step Matrix Example

5

6

4

7

8

Reading a Step Matrix

1

9

2

3

1

2

3

4

14|2|3

slide13

Step Matrix Example

5

6

4

7

8

Reading a Step Matrix

1

9

2

3

1

2

3

4

14|2|3  4|123

transforming a step matrix
Transforming a Step Matrix

1

3

1

3

2

5

2

5

4

4

1

3

1

2

2

3

4

5

4

5

s u diagonal on p 3
S-U Diagonal on P3

1

1

2

1

3

1

2

2

2

3

1

2

3

1

3

2

3

3

1|2|3123

1|23 13|2

12|32|13

2|1323|1

13|23|12

1233|2|1

1

1

2

2

3

3

12|323|1

1|233|12

calculating the s u diagonal
Calculating the S-U Diagonal
  • Skip step matrix stage – Use permutations and strings
  • One – to – one correspondence between permutations and step matrices
slide17

Calculating the S-U Diagonal

  • Skip step matrix stage – Use permutations and strings
  • One – to – one correspondence between permutations and step matrices

2

4132

1

3

4

A=4

B=4

slide18

Calculating the S-U Diagonal

  • Skip step matrix stage – Use permutations and strings
  • One – to – one correspondence between permutations and step matrices

2

4132

1

3

4

A=41

B=4|1

slide19

Calculating the S-U Diagonal

  • Skip step matrix stage – Use permutations and strings
  • One – to – one correspondence between permutations and step matrices

2

4132

1

3

4

A=41|3

B=4|13

slide20

Calculating the S-U Diagonal

  • Skip step matrix stage – Use permutations and strings
  • One – to – one correspondence between permutations and step matrices

2

4132

1

3

4

A=41|32

B=4|13|2

14|23  4|13|2

s u diagonal
S-U Diagonal
  • Acts multiplicatively w.r.t. the bar operation

(12|34) = (12) | (34)

= (1|212 + 122|1) | (3|434 + 344|3)

= (1|2|3|412|34) + (1|2|3412|4|3) +

(12|3|42|1|34) + (12|342|1|4|3)

generic diagonal
Generic Diagonal

abc

Three Element

Diagonal

2|134

a|b|c  abc

a|bc  ac|b

ab|c  b|ac

b|ac  bc|a

ac|b  c|ab

abc  c|b|a

ab|c  bc|a

a|bc  c|ab

(2|134) = (2)| (134) =

2|1|3|4  2|134

2|1|34  2|14|3

2|13|4  2|3|14

2|3|14  2|34|1

2|14|3  2|4|13

2|134  2|4|3|1

2|13|4  2|34|1

2|1|34  2|4|13

associativity
Associativity
  • (ab)c = a(bc)
  • m( m(a,b) , c ) = m( a , m(b,c) )
  • m(m x id)(a,b,c) = m(id x m)(a,b,c)
  • S-U Diagonal takes one input and produces two outputs – “comultiplication”
  • ( id) (X) = (id ) (X) ?

Is  “coassociative?”

not coassociative example
Not Coassociative - Example

(x1) (123) + (1x) (123) =

2|1|32|1323|1 + 1|2|32|1323|1

+1|3|213|23|12 + 1|2|313|23|12

+12|32|133|2|1 + 12|32|132|3|1

+12|32|1|323|1 + 12|32|3|123|1

+1|2313|23|2|1 + 1|2313|23|2|1

+1|231|3|23|12 + 1|233|1|23|12

(mod 2)

This measures the error from being coassociative.

homotopy coassociativity
Homotopy Coassociativity
  • Let Vi be the Z2-vector space whose basis is the i dimensional faces of Pn
  • Let  : Vi  Vi-1 be the boundary operation
  • Let H : Vi  (V*  V*  V*) i+1 such thatH+H=(id)+ (id)
  • H acts multiplicatively with respect to bar

H(13|24) = H(13) | H(24)

homotopy function
Homotopy Function
  • H(1) = 0
  • H(12) = 0
  • H(123) = 12|3x2|13x23|1 + 1|23x13|2x3|12
  • H(1234) = ?
calculating h 1234
Calculating H(1234)
  • H = (id) + (id) + H
  • H(1234) = [(id) + (id) + H](1234)
  • X = H(1234) є (V*  V*  V*)4
  • (X) = [(id) + (id) + H](1234)
calculating h 12341
Calculating H(1234)
  • H = (id) + (id) + H
  • H(1234) = [(id) + (id) + H](1234)
  • X = H(1234) є (V*  V*  V*)4
  • (X) = [(id) + (id) + H](1234)

(V* V* V*)4

(V* V* V*)3

calculating h 12342
Calculating H(1234)
  • H = (id) + (id) + H
  • H(1234) = [(id) + (id) + H](1234)
  • X = H(1234) є (V*  V*  V*)4
  • (X) = [(id) + (id) + H](1234)

(V* V* V*)4

120,960 x 73,729

(V* V* V*)3

one solution for h 1234
One Solution for H(1234)

H(1234) =

12|34x24|13x4|2|3|1 + 12|34x24|13x4|3|2|1 + 123|4x3|24|1x34|2|1 + 123|4x3|2|14x34|2|1 + 123|4x3|2|14x3|24|1 + 124|3x4|2|13x4|23|1 + 12|34x24|1|3x4|2|13 + 12|34x24|3|1x4|23|1 + 12|34x2|14|3x24|3|1 + 12|34x2|14|3x2|4|13 + 12|34x2|14|3x4|23|1 + 12|34x2|4|13x4|23|1 + 13|24x34|1|2x4|3|12 + 13|24x3|14|2x34|2|1 + 13|24x3|14|2x3|4|12 + 14|23x4|13|2x4|3|12 + 1|234x14|3|2x4|13|2 + 1|234x14|3|2x4|3|12 + 1|234x4|13|2x4|3|12 + 23|14x3|24|1x34|2|1 + 2|134x24|3|1x4|23|1 + 12|34x2|1|4|3x24|13 + 12|34x2|4|1|3x24|13 + 13|24x3|1|4|2x34|12 + 13|24x3|4|1|2x34|12 + 12|3|4x23|14x34|2|1 + 12|3|4x23|14x3|24|1 + 12|3|4x2|134x24|3|1 + 12|3|4x2|134x4|23|1 + 12|4|3x24|13x4|23|1 + 13|2|4x3|124x34|2|1 + 1|23|4x134|2x4|3|12 + 1|23|4x13|24x34|1|2 + 1|23|4x13|24x3|14|2 + 1|23|4x13|24x4|3|12 + 1|23|4x3|124x34|2|1 + 1|24|3x14|23x4|13|2 + 1|24|3x14|23x4|3|12 + 1|2|34x124|3x4|23|1 + 1|2|34x124|3x4|2|13 + 1|2|34x14|23x4|13|2 + 1|2|34x14|23x4|3|12 + 1|2|34x24|13x4|23|1 + 1|3|24x134|2x4|3|12 + 2|13|4x23|14x34|2|1 + 2|13|4x23|14x3|24|1 + 2|14|3x24|13x4|23|1 + 12|3|4x2|13|4x234|1 + 12|3|4x2|13|4x23|14 + 12|3|4x2|1|34x24|13 + 12|3|4x2|3|14x234|1 + 12|4|3x2|14|3x24|13 + 13|2|4x3|1|24x34|12 + 13|4|2x3|14|2x34|12 + 1|23|4x13|2|4x34|12 + 1|23|4x13|2|4x3|124 + 1|23|4x1|3|24x134|2 + 1|23|4x3|14|2x34|12 + 1|23|4x3|1|24x34|12 + 1|24|3x14|2|3x4|123 + 1|2|34x14|2|3x4|123 + 1|2|34x1|24|3x14|23 + 1|2|34x1|24|3x4|123 + 1|2|34x2|14|3x24|13 + 1|3|24x3|14|2x34|12 + 2|13|4x2|3|14x234|1

future work
Future Work
  • Finding an H with minimal number of terms
  • Modifying / Parallelizing the row reduction algorithm to calculate H for n > 4

Picture on second page found at: http://www.lightstorm3d.de/portfolio/back_to_gaya/stills/programming/collisionDeformer.jpg

Cross 

Partial 

Union 

Delta 

ad