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Permutahedra and the Saneblidze-Umble Diagonal

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.

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Permutahedra and the Saneblidze-Umble Diagonal

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  1. Permutahedra and the Saneblidze-Umble Diagonal By Stephen Weaver Directed by Dr. Ron Umble

  2. Computational Geometry the study of algorithms to solve problems in geometry

  3. 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

  4. 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

  5. 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

  6. 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

  7. 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

  8. 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

  9. Diagonal Given a set S, Diagonal of S  S { (x,x) | x є S } S S

  10. 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

  11. 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

  12. Step Matrix Example 5 6 4 7 8 Reading a Step Matrix 1 9 2 3 1 2 3 4 14|2|3

  13. 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

  14. Transforming a Step Matrix 1 3 1 3 2 5 2 5 4 4 1 3 1 2 2 3 4 5 4 5

  15. 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

  16. Calculating the S-U Diagonal • Skip step matrix stage – Use permutations and strings • One – to – one correspondence between permutations and step matrices

  17. 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

  18. 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

  19. 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

  20. 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

  21. 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)

  22. 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

  23. 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?”

  24. 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.

  25. 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)

  26. Homotopy Function • H(1) = 0 • H(12) = 0 • H(123) = 12|3x2|13x23|1 + 1|23x13|2x3|12 • H(1234) = ?

  27. 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)

  28. 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

  29. 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

  30. 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

  31. 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 

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