Polyhedral homotopy continuation method

Transcript

1. Polyhedral Homotopy Continuation Method For Solving Systems of Polynomial Equations Tianran Chen Department of Mathematics Michigan State University April 14, 2011 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 1 / 1
2. Homotopy Continuation Method P(x) = 0 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 2 / 1
3. Homotopy Continuation Method P(x) = 0 P(x) = 0 Q(x) = 0 “Solvable” T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 2 / 1
4. Homotopy Continuation Method P(x) = 0 P(x) = 0 Q(x) = 0 “Solvable” H(x, t) H(x, 0) = 0 H(x, 1) = 0 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 2 / 1
5. Homotopy Continuation Method P(x) = 0 P(x) = 0 Q(x) = 0 “Solvable” H(x, t) H(x, 0) = 0 H(x, 1) = 0 Solutions of H(x, t) = 0 t = 0 t = 1 Nonsingular T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 2 / 1
6. Assumption Potential problems in homotopy continuation methods: T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 3 / 1
7. Assumption Potential problems in homotopy continuation methods: T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 3 / 1
8. Assumption Potential problems in homotopy continuation methods: T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 3 / 1
9. Assumption Potential problems in homotopy continuation methods: T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 3 / 1
10. Assumption Potential problems in homotopy continuation methods: T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 3 / 1
11. Assumption Potential problems in homotopy continuation methods: T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 3 / 1
12. Background To solve the system of polynomial equations P(x1, x2) = 2 x4 1 x5 2 + 3 x3 1 x2 2 + 4 x1 1 x1 2 + 5 x2 2 + 6 = 0 5 x3 1 x4 2 + 6 x3 1 x2 2 + 7 x2 1 x2 2 + 8 x2 2 + 9 = 0 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 4 / 1
13. Background To solve the system of polynomial equations P(x1, x2) = 2 x4 1 x5 2 + 3 x3 1 x2 2 + 4 x1 1 x1 2 + 5 x2 2 + 6 = 0 5 x3 1 x4 2 + 6 x3 1 x2 2 + 7 x2 1 x2 2 + 8 x2 2 + 9 = 0 One can use the linear homotopy x9 1 − a x7 2 − b 2 x4 1 x5 2 + 3 x3 1 x2 2 + 4 x1 1 x1 2 + 5 x2 2 + 6 5 x3 1 x4 2 + 6 x3 1 x2 2 + 7 x2 1 x2 2 + 8 x2 2 + 9 with randomly chosen complex numbers a and b. T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 4 / 1
14. Background To solve the system of polynomial equations P(x1, x2) = 2 x4 1 x5 2 + 3 x3 1 x2 2 + 4 x1 1 x1 2 + 5 x2 2 + 6 = 0 5 x3 1 x4 2 + 6 x3 1 x2 2 + 7 x2 1 x2 2 + 8 x2 2 + 9 = 0 One can use the linear homotopy H(x1 , x2 , t) = (1 − t) x9 1 − a x7 2 − b + t 2 x4 1 x5 2 + 3 x3 1 x2 2 + 4 x1 1 x1 2 + 5 x2 2 + 6 5 x3 1 x4 2 + 6 x3 1 x2 2 + 7 x2 1 x2 2 + 8 x2 2 + 9 = 0 0 with randomly chosen complex numbers a and b. T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 4 / 1
15. Background To solve the system of polynomial equations P(x1, x2) = 2 x4 1 x5 2 + 3 x3 1 x2 2 + 4 x1 1 x1 2 + 5 x2 2 + 6 = 0 5 x3 1 x4 2 + 6 x3 1 x2 2 + 7 x2 1 x2 2 + 8 x2 2 + 9 = 0 One can use the linear homotopy H(x1 , x2 , t) = (1 − t) x9 1 − a x7 2 − b + t 2 x4 1 x5 2 + 3 x3 1 x2 2 + 4 x1 1 x1 2 + 5 x2 2 + 6 5 x3 1 x4 2 + 6 x3 1 x2 2 + 7 x2 1 x2 2 + 8 x2 2 + 9 = 0 0 with randomly chosen complex numbers a and b. H(x1, x2, 1) = P(x1, x2) H(x1, x2, 0) = x9 1 − a x7 2 − b T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 4 / 1
16. Background To solve the system of polynomial equations P(x1, x2) = 2 x4 1 x5 2 + 3 x3 1 x2 2 + 4 x1 1 x1 2 + 5 x2 2 + 6 = 0 5 x3 1 x4 2 + 6 x3 1 x2 2 + 7 x2 1 x2 2 + 8 x2 2 + 9 = 0 One can use the linear homotopy H(x1 , x2 , t) = (1 − t) x9 1 − a x7 2 − b + t 2 x4 1 x5 2 + 3 x3 1 x2 2 + 4 x1 1 x1 2 + 5 x2 2 + 6 5 x3 1 x4 2 + 6 x3 1 x2 2 + 7 x2 1 x2 2 + 8 x2 2 + 9 = 0 0 with randomly chosen complex numbers a and b. H(x1, x2, 1) = P(x1, x2) H(x1, x2, 0) = x9 1 − a x7 2 − b We need to follow 9 × 7 = 63 paths. T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 4 / 1
17. Problem While the equation P(x1, x2) = 2 x4 1 x5 2 + 3 x3 1 x2 2 + 4 x1 1 x1 2 + 5 x2 2 + 6 = 0 5 x3 1 x4 2 + 6 x3 1 x2 2 + 7 x2 1 x2 2 + 8 x2 2 + 9 = 0 only has 15 solutions, 63 paths have to be traced. T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 5 / 1
18. Problem While the equation P(x1, x2) = 2 x4 1 x5 2 + 3 x3 1 x2 2 + 4 x1 1 x1 2 + 5 x2 2 + 6 = 0 5 x3 1 x4 2 + 6 x3 1 x2 2 + 7 x2 1 x2 2 + 8 x2 2 + 9 = 0 only has 15 solutions, 63 paths have to be traced. t = 0 t = 1 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 5 / 1
19. Polyhedral Homotopy (Huber & Sturmfels 1995) The parameter t can also be introduced in a nonlinear fashion. T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 6 / 1
20. Polyhedral Homotopy (Huber & Sturmfels 1995) The parameter t can also be introduced in a nonlinear fashion. Multiply each term by a random power of the new parameter t H(x1, x2, t) = h1 = 2 x4 1 x5 2 t3.6 + 3 x3 1 x2 2 t2.1 + 4 x1 1 x1 2 t4.8 + 5 x2 2 t2.4 + 6 t3.1 h2 = 5 x3 1 x4 2 t2.3 + 6 x3 1 x2 2 t3.1 + 7 x2 1 x2 2 t4 + 8 x2 2 t2.6 + 9 t1.6 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 6 / 1
21. Polyhedral Homotopy (Huber & Sturmfels 1995) The parameter t can also be introduced in a nonlinear fashion. Multiply each term by a random power of the new parameter t H(x1, x2, t) = h1 = 2 x4 1 x5 2 t3.6 + 3 x3 1 x2 2 t2.1 + 4 x1 1 x1 2 t4.8 + 5 x2 2 t2.4 + 6 t3.1 h2 = 5 x3 1 x4 2 t2.3 + 6 x3 1 x2 2 t3.1 + 7 x2 1 x2 2 t4 + 8 x2 2 t2.6 + 9 t1.6 Number of paths: 15 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 6 / 1
22. Polyhedral Homotopy (Huber & Sturmfels 1995) The parameter t can also be introduced in a nonlinear fashion. Multiply each term by a random power of the new parameter t H(x1, x2, t) = h1 = 2 x4 1 x5 2 t3.6 + 3 x3 1 x2 2 t2.1 + 4 x1 1 x1 2 t4.8 + 5 x2 2 t2.4 + 6 t3.1 h2 = 5 x3 1 x4 2 t2.3 + 6 x3 1 x2 2 t3.1 + 7 x2 1 x2 2 t4 + 8 x2 2 t2.6 + 9 t1.6 Number of paths: 15 (Much better!) . . . was 63 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 6 / 1
23. Polyhedral Homotopy (Huber & Sturmfels 1995) The parameter t can also be introduced in a nonlinear fashion. Multiply each term by a random power of the new parameter t H(x1, x2, t) = h1 = 2 x4 1 x5 2 t3.6 + 3 x3 1 x2 2 t2.1 + 4 x1 1 x1 2 t4.8 + 5 x2 2 t2.4 + 6 t3.1 h2 = 5 x3 1 x4 2 t2.3 + 6 x3 1 x2 2 t3.1 + 7 x2 1 x2 2 t4 + 8 x2 2 t2.6 + 9 t1.6 Number of paths: 15 (Much better!) . . . was 63 H(x1, x2, 1) = P(x1, x2) T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 6 / 1
24. Polyhedral Homotopy (Huber & Sturmfels 1995) The parameter t can also be introduced in a nonlinear fashion. Multiply each term by a random power of the new parameter t H(x1, x2, t) = h1 = 2 x4 1 x5 2 t3.6 + 3 x3 1 x2 2 t2.1 + 4 x1 1 x1 2 t4.8 + 5 x2 2 t2.4 + 6 t3.1 h2 = 5 x3 1 x4 2 t2.3 + 6 x3 1 x2 2 t3.1 + 7 x2 1 x2 2 t4 + 8 x2 2 t2.6 + 9 t1.6 Number of paths: 15 (Much better!) . . . was 63 H(x1, x2, 1) = P(x1, x2) H(x1, x2, 0) ≡ 0 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 6 / 1
25. Polyhedral Homotopy (Huber & Sturmfels 1995) The parameter t can also be introduced in a nonlinear fashion. Multiply each term by a random power of the new parameter t H(x1, x2, t) = h1 = 2 x4 1 x5 2 t3.6 + 3 x3 1 x2 2 t2.1 + 4 x1 1 x1 2 t4.8 + 5 x2 2 t2.4 + 6 t3.1 h2 = 5 x3 1 x4 2 t2.3 + 6 x3 1 x2 2 t3.1 + 7 x2 1 x2 2 t4 + 8 x2 2 t2.6 + 9 t1.6 Number of paths: 15 (Much better!) . . . was 63 H(x1, x2, 1) = P(x1, x2) H(x1, x2, 0) ≡ 0 Problem: Cannot identify the starting points. T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 6 / 1
26. Example: One Dimensional Case Start with a simple example: Want to solve P(x) = 5x5 + 4x4 + 3x3 + 2x + 7 = 0 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 7 / 1
27. Example: One Dimensional Case Start with a simple example: Want to solve P(x) = 5x5 + 4x4 + 3x3 + 2x + 7 = 0 Pick random powers of t, 5x5 + 4x4 + 3x3 + 2x + 7 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 7 / 1
28. Example: One Dimensional Case Start with a simple example: Want to solve P(x) = 5x5 + 4x4 + 3x3 + 2x + 7 = 0 Pick random powers of t, 5x5t1.3 + 4x4t0.8 + 3x3t1.9 + 2xt1.2 + 7t1.1 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 7 / 1
29. Example: One Dimensional Case Start with a simple example: Want to solve P(x) = 5x5 + 4x4 + 3x3 + 2x + 7 = 0 Pick random powers of t, H(x, t) := 5x5t1.3 + 4x4t0.8 + 3x3t1.9 + 2xt1.2 + 7t1.1 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 7 / 1
30. Example: One Dimensional Case Start with a simple example: Want to solve P(x) = 5x5 + 4x4 + 3x3 + 2x + 7 = 0 Pick random powers of t, H(x, t) := 5x5t1.3 + 4x4t0.8 + 3x3t1.9 + 2xt1.2 + 7t1.1 Then H(x, 1) ≡ P(x) T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 7 / 1
31. Example: One Dimensional Case Start with a simple example: Want to solve P(x) = 5x5 + 4x4 + 3x3 + 2x + 7 = 0 Pick random powers of t, H(x, t) := 5x5t1.3 + 4x4t0.8 + 3x3t1.9 + 2xt1.2 + 7t1.1 Then H(x, 1) ≡ P(x) H(x, 0) ≡ 0 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 7 / 1
32. Example: One Dimensional Case Start with a simple example: Want to solve P(x) = 5x5 + 4x4 + 3x3 + 2x + 7 = 0 Pick random powers of t, H(x, t) := 5x5t1.3 + 4x4t0.8 + 3x3t1.9 + 2xt1.2 + 7t1.1 Then H(x, 1) ≡ P(x) H(x, 0) ≡ 0 Problem: cannot identify the starting point T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 7 / 1
33. Example: One Dimensional Case Start with a simple example: Want to solve P(x) = 5x5 + 4x4 + 3x3 + 2x + 7 = 0 Pick random powers of t, H(x, t) := 5x5t1.3 + 4x4t0.8 + 3x3t1.9 + 2xt1.2 + 7t1.1 Then H(x, 1) ≡ P(x) H(x, 0) ≡ 0 Problem: cannot identify the starting point Cause: all terms have t T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 7 / 1
34. An “Incorrect” Fix First instinct: let’s factor out the smallest power of t H(x, t) = 5x5t1.3 + 4x4t0.8 + 3x3t1.9 + 2xt1.2 + 7t1.1 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 8 / 1
35. An “Incorrect” Fix First instinct: let’s factor out the smallest power of t H(x, t) = 5x5t1.3 + 4x4t0.8 + 3x3t1.9 + 2xt1.2 + 7t1.1 = t0.8(5x5t0.5 + 4x4 + 3x3t1.1 + 2xt0.4 + 7t0.3) T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 8 / 1
36. An “Incorrect” Fix First instinct: let’s factor out the smallest power of t H(x, t) = 5x5t1.3 + 4x4t0.8 + 3x3t1.9 + 2xt1.2 + 7t1.1 = t0.8(5x5t0.5 + 4x4 + 3x3t1.1 + 2xt0.4 + 7t0.3) At t = 0: 5x5t0.5 + 4x4 + 3x3t1.1 + 2xt0.4 + 7t0.3 → 4x4 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 8 / 1
37. An “Incorrect” Fix First instinct: let’s factor out the smallest power of t H(x, t) = 5x5t1.3 + 4x4t0.8 + 3x3t1.9 + 2xt1.2 + 7t1.1 = t0.8(5x5t0.5 + 4x4 + 3x3t1.1 + 2xt0.4 + 7t0.3) At t = 0: 5x5t0.5 + 4x4 + 3x3t1.1 + 2xt0.4 + 7t0.3 → 4x4 The equation 4x4 = 0 has only trivial solutions. T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 8 / 1
38. An “Incorrect” Fix First instinct: let’s factor out the smallest power of t H(x, t) = 5x5t1.3 + 4x4t0.8 + 3x3t1.9 + 2xt1.2 + 7t1.1 = t0.8(5x5t0.5 + 4x4 + 3x3t1.1 + 2xt0.4 + 7t0.3) At t = 0: 5x5t0.5 + 4x4 + 3x3t1.1 + 2xt0.4 + 7t0.3 → 4x4 The equation 4x4 = 0 has only trivial solutions. (All singular) T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 8 / 1
39. An “Incorrect” Fix First instinct: let’s factor out the smallest power of t H(x, t) = 5x5t1.3 + 4x4t0.8 + 3x3t1.9 + 2xt1.2 + 7t1.1 = t0.8(5x5t0.5 + 4x4 + 3x3t1.1 + 2xt0.4 + 7t0.3) At t = 0: 5x5t0.5 + 4x4 + 3x3t1.1 + 2xt0.4 + 7t0.3 → 4x4 The equation 4x4 = 0 has only trivial solutions. (All singular) Root of the problem: The equation has only one term. T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 8 / 1
40. Binomial Equation Equation of 2 terms: T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 9 / 1
41. Binomial Equation Equation of 2 terms: can be solved easily, T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 9 / 1
42. Binomial Equation Equation of 2 terms: can be solved easily, no matter the degree. T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 9 / 1
43. Binomial Equation Equation of 2 terms: can be solved easily, no matter the degree. 3x100 + 2x93 = 0 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 9 / 1
44. Binomial Equation Equation of 2 terms: can be solved easily, no matter the degree. 3x100 + 2x93 = 0 3x100 = −2x93 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 9 / 1
45. Binomial Equation Equation of 2 terms: can be solved easily, no matter the degree. 3x100 + 2x93 = 0 3x100 = −2x93 x100−93 = −2/3 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 9 / 1
46. Binomial Equation Equation of 2 terms: can be solved easily, no matter the degree. 3x100 + 2x93 = 0 3x100 = −2x93 x100−93 = −2/3 x7 = −2/3 solutions are nonzero, nonsingular T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 9 / 1
47. Binomial Equation Equation of 2 terms: can be solved easily, no matter the degree. 3x100 + 2x93 = 0 3x100 = −2x93 x100−93 = −2/3 x7 = −2/3 solutions are nonzero, nonsingular P = 5x5 + 4x4 + 3x3 + 2x + 7 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 9 / 1
48. Binomial Equation Equation of 2 terms: can be solved easily, no matter the degree. 3x100 + 2x93 = 0 3x100 = −2x93 x100−93 = −2/3 x7 = −2/3 solutions are nonzero, nonsingular P = 5x5 + 4x4 + 3x3 + 2x + 7 ↓ ↓ ↓ ↓ ↓ (5 ) (4 ) (3 ) (1 ) (0 ) T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 9 / 1
49. Binomial Equation Equation of 2 terms: can be solved easily, no matter the degree. 3x100 + 2x93 = 0 3x100 = −2x93 x100−93 = −2/3 x7 = −2/3 solutions are nonzero, nonsingular H = 5x5t1.3 + 4x4t0.8 + 3x3t1.9 + 2xt1.2 + 7t1.1 ↓ ↓ ↓ ↓ ↓ (5 ) (4 ) (3 ) (1 ) (0 ) T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 9 / 1
50. Binomial Equation Equation of 2 terms: can be solved easily, no matter the degree. 3x100 + 2x93 = 0 3x100 = −2x93 x100−93 = −2/3 x7 = −2/3 solutions are nonzero, nonsingular H = 5x5t1.3 + 4x4t0.8 + 3x3t1.9 + 2xt1.2 + 7t1.1 ↓ ↓ ↓ ↓ ↓ (5, 1.3) (4, 0.8) (3, 1.9) (1, 1.2) (0, 1.1) T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 9 / 1
51. Binomial Equation Equation of 2 terms: can be solved easily, no matter the degree. 3x100 + 2x93 = 0 3x100 = −2x93 x100−93 = −2/3 x7 = −2/3 solutions are nonzero, nonsingular H = 5x5t1.3 + 4x4t0.8 + 3x3t1.9 + 2xt1.2 + 7t1.1 ↓ ↓ ↓ ↓ ↓ (5, 1.3) (4, 0.8) (3, 1.9) (1, 1.2) (0, 1.1) ˆ 5 ˆ 4 ˆ 3 ˆ 1 ˆ 0 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 9 / 1
52. Newton Polygon x t 0 1 2 3 4 5 1 2 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 10 / 1
53. Newton Polygon x t 0 1 2 3 4 5 1 2 ˆ 5 = (5, 1.3) ˆ 4 = (4, 0.8) ˆ 3 = (3, 1.9) ˆ 1 = (1, 1.2) ˆ 0 = (0, 1.1) T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 10 / 1
54. Newton Polygon x t 0 1 2 3 4 5 1 2 ˆ 5 = (5, 1.3) ˆ 4 = (4, 0.8) ˆ 3 = (3, 1.9) ˆ 1 = (1, 1.2) ˆ 0 = (0, 1.1) T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 10 / 1
55. Newton Polygon x t 0 1 2 3 4 5 1 2 ˆ 5 = (5, 1.3) ˆ 4 = (4, 0.8) ˆ 3 = (3, 1.9) ˆ 1 = (1, 1.2) ˆ 0 = (0, 1.1) T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 10 / 1
56. Newton Polygon x t 0 1 2 3 4 5 1 2 ˆ 5 = (5, 1.3) ˆ 4 = (4, 0.8) ˆ 3 = (3, 1.9) ˆ 1 = (1, 1.2) ˆ 0 = (0, 1.1) ˆ α = (0.075, 1) T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 10 / 1
57. Newton Polygon x t 0 1 2 3 4 5 1 2 ˆ 5 = (5, 1.3) ˆ 4 = (4, 0.8) ˆ 3 = (3, 1.9) ˆ 1 = (1, 1.2) ˆ 0 = (0, 1.1) ˆ α = (0.075, 1) ˆ 5, ˆ α = 1.675 ˆ 4, ˆ α = 1.1 ˆ 3, ˆ α = 2.125 ˆ 1, ˆ α = 1.275 ˆ 0, ˆ α = 1.1 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 10 / 1
58. Change of Variables Change of variables with α = 0.075 ˆ α = (α, 1) = (0.075, 1) x = ytα T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 11 / 1
59. Change of Variables Change of variables with α = 0.075 ˆ α = (α, 1) = (0.075, 1) x = ytα at t = 1 x = y (H(x, 1) ≡ P(x)) T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 11 / 1
60. Change of Variables Change of variables with α = 0.075 ˆ α = (α, 1) = (0.075, 1) x = ytα at t = 1 x = y (H(x, 1) ≡ P(x)) H(x, t) = 5x5t1.3 + 4x4t0.8 + 3x3t1.9 + 2xt1.2 + 7t1.1 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 11 / 1
61. Change of Variables Change of variables with α = 0.075 ˆ α = (α, 1) = (0.075, 1) x = ytα at t = 1 x = y (H(x, 1) ≡ P(x)) H(x, t) = 5(ytα)5t1.3 + 4(ytα)4t0.8 + 3(ytα)3t1.9 + 2(ytα)t1.2 + 7t1.1 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 11 / 1
62. Change of Variables Change of variables with α = 0.075 ˆ α = (α, 1) = (0.075, 1) x = ytα at t = 1 x = y (H(x, 1) ≡ P(x)) H(x, t) = 5(ytα)5t1.3 + 4(ytα)4t0.8 + 3(ytα)3t1.9 + 2(ytα)t1.2 + 7t1.1 = 5y5t5α+1.3 + · · · T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 11 / 1
63. Change of Variables Change of variables with α = 0.075 ˆ α = (α, 1) = (0.075, 1) x = ytα at t = 1 x = y (H(x, 1) ≡ P(x)) H(x, t) = 5(ytα)5t1.3 + 4(ytα)4t0.8 + 3(ytα)3t1.9 + 2(ytα)t1.2 + 7t1.1 = 5y5t (5,1.3),(α,1) + · · · T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 11 / 1
64. Change of Variables Change of variables with α = 0.075 ˆ α = (α, 1) = (0.075, 1) x = ytα at t = 1 x = y (H(x, 1) ≡ P(x)) H(x, t) = 5(ytα)5t1.3 + 4(ytα)4t0.8 + 3(ytα)3t1.9 + 2(ytα)t1.2 + 7t1.1 = 5y5t ˆ 5, ˆ α + · · · T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 11 / 1
65. Change of Variables Change of variables with α = 0.075 ˆ α = (α, 1) = (0.075, 1) x = ytα at t = 1 x = y (H(x, 1) ≡ P(x)) H(x, t) = 5(ytα)5t1.3 + 4(ytα)4t0.8 + 3(ytα)3t1.9 + 2(ytα)t1.2 + 7t1.1 = 5y5t ˆ 5, ˆ α + 4y4t ˆ 4, ˆ α + 3y3t ˆ 3, ˆ α + 2y1t ˆ 1, ˆ α + 7t ˆ 0, ˆ α T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 11 / 1
66. Change of Variables ˆ 5, ˆ α = 1.675 ˆ 4, ˆ α = 1.1 ˆ 3, ˆ α = 2.125 ˆ 1, ˆ α = 1.275 ˆ 0, ˆ α = 1.1 H(x, t) = 5(ytα)5t1.3 + 4(ytα)4t0.8 + 3(ytα)3t1.9 + 2(ytα)t1.2 + 7t1.1 = 5y5t ˆ 5, ˆ α + 4y4t ˆ 4, ˆ α + 3y3t ˆ 3, ˆ α + 2y1t ˆ 1, ˆ α + 7t ˆ 0, ˆ α T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 11 / 1
67. Change of Variables ˆ 5, ˆ α = 1.675 ˆ 4, ˆ α = 1.1 ˆ 3, ˆ α = 2.125 ˆ 1, ˆ α = 1.275 ˆ 0, ˆ α = 1.1 H(x, t) = 5(ytα)5t1.3 + 4(ytα)4t0.8 + 3(ytα)3t1.9 + 2(ytα)t1.2 + 7t1.1 = 5y5t ˆ 5, ˆ α + 4y4t ˆ 4, ˆ α + 3y3t ˆ 3, ˆ α + 2y1t ˆ 1, ˆ α + 7t ˆ 0, ˆ α = 5y5t1.675 + 4y4t1.1 + 3y3t2.125 + 2y1t1.275 + 7t1.1 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 11 / 1
68. Change of Variables ˆ 5, ˆ α = 1.675 ˆ 4, ˆ α = 1.1 ˆ 3, ˆ α = 2.125 ˆ 1, ˆ α = 1.275 ˆ 0, ˆ α = 1.1 H(x, t) = 5(ytα)5t1.3 + 4(ytα)4t0.8 + 3(ytα)3t1.9 + 2(ytα)t1.2 + 7t1.1 = 5y5t ˆ 5, ˆ α + 4y4t ˆ 4, ˆ α + 3y3t ˆ 3, ˆ α + 2y1t ˆ 1, ˆ α + 7t ˆ 0, ˆ α = 5y5t1.675 + 4y4t1.1 + 3y3t2.125 + 2y1t1.275 + 7t1.1 = t1.1(5y5t1.675−1.1 + 4y4 + 3y3t2.125−1.1 + 2y1t1.275−1.1 + 7) T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 11 / 1
69. Change of Variables ˆ 5, ˆ α = 1.675 ˆ 4, ˆ α = 1.1 ˆ 3, ˆ α = 2.125 ˆ 1, ˆ α = 1.275 ˆ 0, ˆ α = 1.1 H(x, t) = 5(ytα)5t1.3 + 4(ytα)4t0.8 + 3(ytα)3t1.9 + 2(ytα)t1.2 + 7t1.1 = 5y5t ˆ 5, ˆ α + 4y4t ˆ 4, ˆ α + 3y3t ˆ 3, ˆ α + 2y1t ˆ 1, ˆ α + 7t ˆ 0, ˆ α = 5y5t1.675 + 4y4t1.1 + 3y3t2.125 + 2y1t1.275 + 7t1.1 = t1.1(5y5t0.575 + 4y4 + 3y3t1.025 + 2y1t0.175 + 7) T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 11 / 1
70. Change of Variables ˆ 5, ˆ α = 1.675 ˆ 4, ˆ α = 1.1 ˆ 3, ˆ α = 2.125 ˆ 1, ˆ α = 1.275 ˆ 0, ˆ α = 1.1 H(x, t) = 5(ytα)5t1.3 + 4(ytα)4t0.8 + 3(ytα)3t1.9 + 2(ytα)t1.2 + 7t1.1 = 5y5t ˆ 5, ˆ α + 4y4t ˆ 4, ˆ α + 3y3t ˆ 3, ˆ α + 2y1t ˆ 1, ˆ α + 7t ˆ 0, ˆ α = 5y5t1.675 + 4y4t1.1 + 3y3t2.125 + 2y1t1.275 + 7t1.1 = t1.1(5y5t0.575 + 4y4 + 3y3t1.025 + 2y1t0.175 + 7) = t1.1[4y4 + 7 + (5y5t0.575 + 3y3t1.025 + 2y1t0.175)] T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 11 / 1
71. Change of Variables ˆ 5, ˆ α = 1.675 ˆ 4, ˆ α = 1.1 ˆ 3, ˆ α = 2.125 ˆ 1, ˆ α = 1.275 ˆ 0, ˆ α = 1.1 H(x, t) = 5(ytα)5t1.3 + 4(ytα)4t0.8 + 3(ytα)3t1.9 + 2(ytα)t1.2 + 7t1.1 = 5y5t ˆ 5, ˆ α + 4y4t ˆ 4, ˆ α + 3y3t ˆ 3, ˆ α + 2y1t ˆ 1, ˆ α + 7t ˆ 0, ˆ α = 5y5t1.675 + 4y4t1.1 + 3y3t2.125 + 2y1t1.275 + 7t1.1 = t1.1(5y5t0.575 + 4y4 + 3y3t1.025 + 2y1t0.175 + 7) = t1.1[4y4 + 7 + (5y5t0.575 + 3y3t1.025 + 2y1t0.175)] = t1.1[4y4 + 7 + (terms with positive powers of t)] T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 11 / 1
72. Change of Variables ˆ 5, ˆ α = 1.675 ˆ 4, ˆ α = 1.1 ˆ 3, ˆ α = 2.125 ˆ 1, ˆ α = 1.275 ˆ 0, ˆ α = 1.1 H(x, t) = 5(ytα)5t1.3 + 4(ytα)4t0.8 + 3(ytα)3t1.9 + 2(ytα)t1.2 + 7t1.1 = 5y5t ˆ 5, ˆ α + 4y4t ˆ 4, ˆ α + 3y3t ˆ 3, ˆ α + 2y1t ˆ 1, ˆ α + 7t ˆ 0, ˆ α = 5y5t1.675 + 4y4t1.1 + 3y3t2.125 + 2y1t1.275 + 7t1.1 = t1.1(5y5t0.575 + 4y4 + 3y3t1.025 + 2y1t0.175 + 7) = t1.1[4y4 + 7 + (5y5t0.575 + 3y3t1.025 + 2y1t0.175)] = t1.1[4y4 + 7 + (terms with positive powers of t)] T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 11 / 1
73. Change of Variables ˆ 5, ˆ α = 1.675 ˆ 4, ˆ α = 1.1 ˆ 3, ˆ α = 2.125 ˆ 1, ˆ α = 1.275 ˆ 0, ˆ α = 1.1 H(x, t) = 5(ytα)5t1.3 + 4(ytα)4t0.8 + 3(ytα)3t1.9 + 2(ytα)t1.2 + 7t1.1 = 5y5t ˆ 5, ˆ α + 4y4t ˆ 4, ˆ α + 3y3t ˆ 3, ˆ α + 2y1t ˆ 1, ˆ α + 7t ˆ 0, ˆ α = 5y5t1.675 + 4y4t1.1 + 3y3t2.125 + 2y1t1.275 + 7t1.1 = t1.1(5y5t0.575 + 4y4 + 3y3t1.025 + 2y1t0.175 + 7) = t1.1[4y4 + 7 + (5y5t0.575 + 3y3t1.025 + 2y1t0.175)] = t1.1[4y4 + 7 + (terms with positive powers of t)] Hα(y, t) = t−1.1H(ytα, t) T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 11 / 1
74. Change of Variables ˆ 5, ˆ α = 1.675 ˆ 4, ˆ α = 1.1 ˆ 3, ˆ α = 2.125 ˆ 1, ˆ α = 1.275 ˆ 0, ˆ α = 1.1 H(x, t) = 5(ytα)5t1.3 + 4(ytα)4t0.8 + 3(ytα)3t1.9 + 2(ytα)t1.2 + 7t1.1 = 5y5t ˆ 5, ˆ α + 4y4t ˆ 4, ˆ α + 3y3t ˆ 3, ˆ α + 2y1t ˆ 1, ˆ α + 7t ˆ 0, ˆ α = 5y5t1.675 + 4y4t1.1 + 3y3t2.125 + 2y1t1.275 + 7t1.1 = t1.1(5y5t0.575 + 4y4 + 3y3t1.025 + 2y1t0.175 + 7) = t1.1[4y4 + 7 + (5y5t0.575 + 3y3t1.025 + 2y1t0.175)] = t1.1[4y4 + 7 + (terms with positive powers of t)] Hα(y, t) = t−1.1H(ytα, t) = 4y4 + 7 + (terms with positive powers of t) T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 11 / 1
75. Starting System Hα(y, t) = 4y4 + 7 + (5y5t0.575 + 3y3t1.025 + 2y1t0.175) T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 12 / 1
76. Starting System Hα(y, t) = 4y4 + 7 + (5y5t0.575 + 3y3t1.025 + 2y1t0.175) = 4y4 + 7 + (terms with positive powers of t) T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 12 / 1
77. Starting System Hα(y, t) = 4y4 + 7 + (5y5t0.575 + 3y3t1.025 + 2y1t0.175) = 4y4 + 7 + (terms with positive powers of t) Hα(y, 0) = 0 4y4 + 7 = 0 y4 = −7/4 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 12 / 1
78. Starting System Hα(y, t) = 4y4 + 7 + (5y5t0.575 + 3y3t1.025 + 2y1t0.175) = 4y4 + 7 + (terms with positive powers of t) Hα(y, 0) = 0 4y4 + 7 = 0 y4 = −7/4 can be solved and it has 4 solutions. T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 12 / 1
79. Starting System Hα(y, t) = 4y4 + 7 + (5y5t0.575 + 3y3t1.025 + 2y1t0.175) = 4y4 + 7 + (terms with positive powers of t) Hα(y, 0) = 0 4y4 + 7 = 0 y4 = −7/4 can be solved and it has 4 solutions. Nonsingular T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 12 / 1
80. Starting System Hα(y, t) = 4y4 + 7 + (5y5t0.575 + 3y3t1.025 + 2y1t0.175) = 4y4 + 7 + (terms with positive powers of t) Hα(y, 0) = 0 4y4 + 7 = 0 y4 = −7/4 can be solved and it has 4 solutions. Nonsingular T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 12 / 1
81. Another Possible α x t 0 1 2 3 4 5 1 2 ˆ 5 ˆ 4 ˆ 3 ˆ 1 ˆ 0 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 13 / 1
82. Another Possible α x t 0 1 2 3 4 5 1 2 ˆ 5 ˆ 4 ˆ 3 ˆ 1 ˆ 0 ˆ α = (−0.5, 1) T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 13 / 1
83. Another Possible α x t 0 1 2 3 4 5 1 2 ˆ 5 ˆ 4 ˆ 3 ˆ 1 ˆ 0 ˆ α = (−0.5, 1) ˆ 5, ˆ α = −1.2 ˆ 4, ˆ α = −1.2 ˆ 3, ˆ α = 0.4 ˆ 1, ˆ α = 0.7 ˆ 0, ˆ α = 1.1 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 13 / 1
84. Another Possible α With α = −0.5 ˆ α = (−0.5, 1) H(ytα, t) = 5y5t−1.2 + 4y4t−1.2 + 3y3t0.4 + 2y1t0.7 + 7t1.1 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 14 / 1
85. Another Possible α With α = −0.5 ˆ α = (−0.5, 1) H(ytα, t) = 5y5t−1.2 + 4y4t−1.2 + 3y3t0.4 + 2y1t0.7 + 7t1.1 = t−1.2[5y5 + 4y4 + (3y3t1.6 + 2y1t1.9 + 7t2.3)] T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 14 / 1
86. Another Possible α With α = −0.5 ˆ α = (−0.5, 1) H(ytα, t) = 5y5t−1.2 + 4y4t−1.2 + 3y3t0.4 + 2y1t0.7 + 7t1.1 = t−1.2[5y5 + 4y4 + (3y3t1.6 + 2y1t1.9 + 7t2.3)] = t−1.2[5y5 + 4y4 + (terms with positive powers of t)] T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 14 / 1
87. Another Possible α With α = −0.5 ˆ α = (−0.5, 1) H(ytα, t) = 5y5t−1.2 + 4y4t−1.2 + 3y3t0.4 + 2y1t0.7 + 7t1.1 = t−1.2[5y5 + 4y4 + (3y3t1.6 + 2y1t1.9 + 7t2.3)] = t−1.2[5y5 + 4y4 + (terms with positive powers of t)] Hα(y, t) := t1.2H(ytα, t) = 5y5 + 4y4 + (terms with positive powers of t) T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 14 / 1
88. Another Possible α With α = −0.5 ˆ α = (−0.5, 1) H(ytα, t) = 5y5t−1.2 + 4y4t−1.2 + 3y3t0.4 + 2y1t0.7 + 7t1.1 = t−1.2[5y5 + 4y4 + (3y3t1.6 + 2y1t1.9 + 7t2.3)] = t−1.2[5y5 + 4y4 + (terms with positive powers of t)] Hα(y, t) := t1.2H(ytα, t) = 5y5 + 4y4 + (terms with positive powers of t) Hα(y, 0) = 5y5 + 4y4 = 0 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 14 / 1
89. Another Possible α With α = −0.5 ˆ α = (−0.5, 1) H(ytα, t) = 5y5t−1.2 + 4y4t−1.2 + 3y3t0.4 + 2y1t0.7 + 7t1.1 = t−1.2[5y5 + 4y4 + (3y3t1.6 + 2y1t1.9 + 7t2.3)] = t−1.2[5y5 + 4y4 + (terms with positive powers of t)] Hα(y, t) := t1.2H(ytα, t) = 5y5 + 4y4 + (terms with positive powers of t) Hα(y, 0) = 5y5 + 4y4 = 0 y = −4/5 t = 0 t = 1 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 14 / 1
90. Final Solution Together, we get all 5 solutions of P(x) = 0. t = 0 t = 1 α = 0.075 t = 0 t = 1 α = −0.5 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 15 / 1
91. Summary: Key Ingredients Binomial equations: 4y4 + 7 = 0 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 16 / 1
92. Summary: Key Ingredients Binomial equations: 4y4 + 7 = 0 Change of variables: x = ytα T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 16 / 1
93. Summary: Key Ingredients Binomial equations: 4y4 + 7 = 0 Change of variables: x = ytα Newton polygon: T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 16 / 1
94. Summary: Key Ingredients Binomial equations: 4y4 + 7 = 0 Change of variables: x = ytα Newton polygon: Normal vector of lower edges: (α, 1) T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 16 / 1
95. The Two Dimensional Case Go back to our original problem P(x1, x2) = 2 x4 1 x5 2 + 3 x3 1 x2 2 + 4 x1 1 x1 2 + 5 x2 2 + 6 = 0 5 x3 1 x4 2 + 6 x3 1 x2 2 + 7 x2 1 x2 2 + 8 x2 2 + 9 = 0 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 17 / 1
96. The Two Dimensional Case Go back to our original problem P(x1, x2) = 2 x4 1 x5 2 + 3 x3 1 x2 2 + 4 x1 1 x1 2 + 5 x2 2 + 6 = 0 5 x3 1 x4 2 + 6 x3 1 x2 2 + 7 x2 1 x2 2 + 8 x2 2 + 9 = 0 H(x1, x2, t) = h1 = 2 x4 1 x5 2 t3.6 + 3 x3 1 x2 2 t2.1 + 4 x1 1 x1 2 t4.8 + 5 x2 2 t2.4 + 6 t3.1 h2 = 5 x3 1 x4 2 t2.3 + 6 x3 1 x2 2 t3.1 + 7 x2 1 x2 2 t4 + 8 x2 2 t2.6 + 9 t1.6 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 17 / 1
97. The Two Dimensional Case Go back to our original problem P(x1, x2) = 2 x4 1 x5 2 + 3 x3 1 x2 2 + 4 x1 1 x1 2 + 5 x2 2 + 6 = 0 5 x3 1 x4 2 + 6 x3 1 x2 2 + 7 x2 1 x2 2 + 8 x2 2 + 9 = 0 H(x1, x2, t) = h1 = 2 x4 1 x5 2 t3.6 + 3 x3 1 x2 2 t2.1 + 4 x1 1 x1 2 t4.8 + 5 x2 2 t2.4 + 6 t3.1 h2 = 5 x3 1 x4 2 t2.3 + 6 x3 1 x2 2 t3.1 + 7 x2 1 x2 2 t4 + 8 x2 2 t2.6 + 9 t1.6 H(x1, x2, 1) = P(x1, x2) T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 17 / 1
98. The Two Dimensional Case Go back to our original problem P(x1, x2) = 2 x4 1 x5 2 + 3 x3 1 x2 2 + 4 x1 1 x1 2 + 5 x2 2 + 6 = 0 5 x3 1 x4 2 + 6 x3 1 x2 2 + 7 x2 1 x2 2 + 8 x2 2 + 9 = 0 H(x1, x2, t) = h1 = 2 x4 1 x5 2 t3.6 + 3 x3 1 x2 2 t2.1 + 4 x1 1 x1 2 t4.8 + 5 x2 2 t2.4 + 6 t3.1 h2 = 5 x3 1 x4 2 t2.3 + 6 x3 1 x2 2 t3.1 + 7 x2 1 x2 2 t4 + 8 x2 2 t2.6 + 9 t1.6 H(x1, x2, 1) = P(x1, x2) H(x1, x2, 0) ≡ 0 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 17 / 1
99. Binomial System System of equations each of two terms    c1xa1 1 xa2 2 + c2xa 1 1 xa2 2 = 0 d1xb1 1 xb2 2 + d2xb 1 1 xb2 2 = 0 can be solved efficiently. T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 18 / 1
100. Binomial System System of equations each of two terms    c1xa1 1 xa2 2 + c2xa 1 1 xa2 2 = 0 d1xb1 1 xb2 2 + d2xb 1 1 xb2 2 = 0 can be solved efficiently. Question: How to make H(x1, x2, 0) a binomial system? T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 18 / 1
101. Binomial System System of equations each of two terms    c1xa1 1 xa2 2 + c2xa 1 1 xa2 2 = 0 d1xb1 1 xb2 2 + d2xb 1 1 xb2 2 = 0 can be solved efficiently. Question: How to make H(x1, x2, 0) a binomial system? Idea: Change of variables. T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 18 / 1
102. Newton Polytopes Example (One dimensional case) 5x5t1.3 + 4x4t0.8 + 3x3t1.9 + 2xt1.2 + 7t1.1 ↓ ↓ ↓ ↓ ↓ (5, 1.3) (4, 0.8) (3, 1.9) (1, 1.2) (0, 1.1) T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 19 / 1
103. Newton Polytopes Example (One dimensional case) 5x5t1.3 + 4x4t0.8 + 3x3t1.9 + 2xt1.2 + 7t1.1 ↓ ↓ ↓ ↓ ↓ (5, 1.3) (4, 0.8) (3, 1.9) (1, 1.2) (0, 1.1) 2x4 1 x5 2 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 19 / 1
104. Newton Polytopes Example (One dimensional case) 5x5t1.3 + 4x4t0.8 + 3x3t1.9 + 2xt1.2 + 7t1.1 ↓ ↓ ↓ ↓ ↓ (5, 1.3) (4, 0.8) (3, 1.9) (1, 1.2) (0, 1.1) 2x4 1 x5 2 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 19 / 1
105. Newton Polytopes Example (One dimensional case) 5x5t1.3 + 4x4t0.8 + 3x3t1.9 + 2xt1.2 + 7t1.1 ↓ ↓ ↓ ↓ ↓ (5, 1.3) (4, 0.8) (3, 1.9) (1, 1.2) (0, 1.1) 2x4 1 x5 2 −→ (4, 5, ) T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 19 / 1
106. Newton Polytopes Example (One dimensional case) 5x5t1.3 + 4x4t0.8 + 3x3t1.9 + 2xt1.2 + 7t1.1 ↓ ↓ ↓ ↓ ↓ (5, 1.3) (4, 0.8) (3, 1.9) (1, 1.2) (0, 1.1) 2x4 1 x5 2 t3.6 −→ (4, 5, 3.6) T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 19 / 1
107. Newton Polytopes Example (One dimensional case) 5x5t1.3 + 4x4t0.8 + 3x3t1.9 + 2xt1.2 + 7t1.1 ↓ ↓ ↓ ↓ ↓ (5, 1.3) (4, 0.8) (3, 1.9) (1, 1.2) (0, 1.1) 2x4 1 x5 2 t3.6 −→ (4, 5, 3.6) 2x4 1 x5 2 + 3x3 1 x2 2 + 4x1 1 x1 2 + 5x2 2 + 6 ↓ ↓ ↓ ↓ ↓ (4, 5 ) (3, 2 ) (1, 1 ) (0, 2 ) (0, 0 ) T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 19 / 1
108. Newton Polytopes Example (One dimensional case) 5x5t1.3 + 4x4t0.8 + 3x3t1.9 + 2xt1.2 + 7t1.1 ↓ ↓ ↓ ↓ ↓ (5, 1.3) (4, 0.8) (3, 1.9) (1, 1.2) (0, 1.1) 2x4 1 x5 2 t3.6 −→ (4, 5, 3.6) 2x4 1 x5 2 t3.6 + 3x3 1 x2 2 t2.1 + 4x1 1 x1 2 t4.8 + 5x2 2 t2.4 + 6t3.1 ↓ ↓ ↓ ↓ ↓ (4, 5, 3.6) (3, 2, 2.1) (1, 1, 4.8) (0, 2, 2.4) (0, 0, 3.1) T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 19 / 1
109. Newton Polytopes (4, 5, 3.6) ∈ R3 (4, 5) ∈ R2 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 20 / 1
110. Newton Polytopes (4, 5, 3.6) ∈ R3 (4, 5) ∈ R2 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 20 / 1
111. Newton Polytopes (4, 5, 3.6) ∈ R3 (4, 5) ∈ R2 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 20 / 1
112. Newton Polytopes T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 21 / 1
113. Supporting Planes Example In the 1-dimensional case, we used normal vectors of “lower edges”: x t 0 1 2 3 4 5 1 2 ˆ 5 ˆ 4 ˆ 3 ˆ 1 ˆ 0 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 22 / 1
114. Supporting Planes Example In the 1-dimensional case, we used normal vectors of “lower edges”: x t 0 1 2 3 4 5 1 2 ˆ 5 ˆ 4 ˆ 3 ˆ 1 ˆ 0 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 22 / 1
115. Supporting Planes Example In the 1-dimensional case, we used normal vectors of “lower edges”: x t 0 1 2 3 4 5 1 2 ˆ 5 ˆ 4 ˆ 3 ˆ 1 ˆ 0 Generalize to higher dimention T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 22 / 1
116. Supporting Planes •, ˆ α minimized at three vertices T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 23 / 1
117. Supporting Planes •, ˆ α minimized at three vertices T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 23 / 1
118. Supporting Planes •, ˆ α minimized at three vertices T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 23 / 1
119. Supporting Planes •, ˆ α minimized at three vertices T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 23 / 1
120. Supporting Planes •, ˆ α minimized at one vertex T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 24 / 1
121. Supporting Planes •, ˆ α minimized at one vertex T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 24 / 1
122. Supporting Planes •, ˆ α minimized at one vertex T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 24 / 1
123. Supporting Planes •, ˆ α minimized at one vertex T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 24 / 1
124. Supporting Planes •, ˆ α minimized at one vertex T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 24 / 1
125. Supporting Planes •, ˆ α minimized at two vertices T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 25 / 1
126. Supporting Planes •, ˆ α minimized at two vertices T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 25 / 1
127. Supporting Planes •, ˆ α minimized at two vertices T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 25 / 1
128. Supporting Planes •, ˆ α minimized at two vertices T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 25 / 1
129. Finding α Want to find: ˆ α = (α1, α2, 1) such that •, ˆ α minimizes at exactly two vertices on each Newton polytope T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 26 / 1
130. Finding α Want to find: ˆ α = (α1, α2, 1) such that •, ˆ α minimizes at exactly two vertices on each Newton polytope T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 26 / 1
131. Finding α Want to find: ˆ α = (α1, α2, 1) such that •, ˆ α minimizes at exactly two vertices on each Newton polytope T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 26 / 1
132. Finding α Want to find: ˆ α = (α1, α2, 1) such that •, ˆ α minimizes at exactly two vertices on each Newton polytope T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 26 / 1
133. Finding α Want to find: ˆ α = (α1, α2, 1) such that •, ˆ α minimizes at exactly two vertices on each Newton polytope T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 26 / 1
134. Finding α Want to find: ˆ α = (α1, α2, 1) such that •, ˆ α minimizes at exactly two vertices on each Newton polytope T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 26 / 1
135. Finding α Want to find: ˆ α = (α1, α2, 1) such that •, ˆ α minimizes at exactly two vertices on each Newton polytope T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 26 / 1
136. Finding α Want to find: ˆ α = (α1, α2, 1) such that •, ˆ α minimizes at exactly two vertices on each Newton polytope T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 26 / 1
137. Finding α Want to find: ˆ α = (α1, α2, 1) such that •, ˆ α minimizes at exactly two vertices on each Newton polytope T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 26 / 1
138. Change of Variables With ˆ α = (α1, α2, 1) x1 = y1tα1 x2 = y2tα2 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 27 / 1
139. Change of Variables With ˆ α = (α1, α2, 1) x1 = y1tα1 x2 = y2tα2 ↓ T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 27 / 1
140. Change of Variables With ˆ α = (α1, α2, 1) x1 = y1tα1 x2 = y2tα2 ↓ X = Y tα T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 27 / 1
141. Change of Variables With ˆ α = (α1, α2, 1) x1 = y1tα1 x2 = y2tα2 ↓ X = Y tα Xa tω T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 27 / 1
142. Change of Variables With ˆ α = (α1, α2, 1) x1 = y1tα1 x2 = y2tα2 ↓ X = Y tα Xa tω = xa1 1 xa2 2 tω T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 27 / 1
143. Change of Variables With ˆ α = (α1, α2, 1) x1 = y1tα1 x2 = y2tα2 ↓ X = Y tα Xa tω = xa1 1 xa2 2 tω = (y1 tα1 )a1 · (y2 tα2 )a2 · tω T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 27 / 1
144. Change of Variables With ˆ α = (α1, α2, 1) x1 = y1tα1 x2 = y2tα2 ↓ X = Y tα Xa tω = xa1 1 xa2 2 tω = (y1 tα1 )a1 · (y2 tα2 )a2 · tω = ya1 1 ta1α1 · ya2 2 ta2α2 · tω T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 27 / 1
145. Change of Variables With ˆ α = (α1, α2, 1) x1 = y1tα1 x2 = y2tα2 ↓ X = Y tα Xa tω = xa1 1 xa2 2 tω = (y1 tα1 )a1 · (y2 tα2 )a2 · tω = ya1 1 ta1α1 · ya2 2 ta2α2 · tω T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 27 / 1
146. Change of Variables With ˆ α = (α1, α2, 1) x1 = y1tα1 x2 = y2tα2 ↓ X = Y tα Xa tω = xa1 1 xa2 2 tω = (y1 tα1 )a1 · (y2 tα2 )a2 · tω = ya1 1 ta1α1 · ya2 2 ta2α2 · tω T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 27 / 1
147. Change of Variables With ˆ α = (α1, α2, 1) x1 = y1tα1 x2 = y2tα2 ↓ X = Y tα Xa tω = xa1 1 xa2 2 tω = (y1 tα1 )a1 · (y2 tα2 )a2 · tω = ya1 1 ta1α1 · ya2 2 ta2α2 · tω T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 27 / 1
148. Change of Variables With ˆ α = (α1, α2, 1) x1 = y1tα1 x2 = y2tα2 ↓ X = Y tα Xa tω = xa1 1 xa2 2 tω = (y1 tα1 )a1 · (y2 tα2 )a2 · tω = ya1 1 ta1α1 · ya2 2 ta2α2 · tω = ya1 1 ya2 2 ta1α1+a2α2+ω T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 27 / 1
149. Change of Variables With ˆ α = (α1, α2, 1) x1 = y1tα1 x2 = y2tα2 ↓ X = Y tα Xa tω = xa1 1 xa2 2 tω = (y1 tα1 )a1 · (y2 tα2 )a2 · tω = ya1 1 ta1α1 · ya2 2 ta2α2 · tω = ya1 1 ya2 2 ta1α1+a2α2+ω a1α1 + a2α2 + ω T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 27 / 1
150. Change of Variables With ˆ α = (α1, α2, 1) x1 = y1tα1 x2 = y2tα2 ↓ X = Y tα Xa tω = xa1 1 xa2 2 tω = (y1 tα1 )a1 · (y2 tα2 )a2 · tω = ya1 1 ta1α1 · ya2 2 ta2α2 · tω = ya1 1 ya2 2 ta1α1+a2α2+ω a1α1 + a2α2 + ω = a1 · α1 + a2 · α2 + ω · 1 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 27 / 1
151. Change of Variables With ˆ α = (α1, α2, 1) x1 = y1tα1 x2 = y2tα2 ↓ X = Y tα Xa tω = xa1 1 xa2 2 tω = (y1 tα1 )a1 · (y2 tα2 )a2 · tω = ya1 1 ta1α1 · ya2 2 ta2α2 · tω = ya1 1 ya2 2 ta1α1+a2α2+ω a1α1 + a2α2 + ω = a1 · α1 + a2 · α2 + ω · 1 =   a1 a2 ω   ,   α1 α2 1   T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 27 / 1
152. Change of Variables With ˆ α = (α1, α2, 1) x1 = y1tα1 x2 = y2tα2 ↓ X = Y tα Xa tω = xa1 1 xa2 2 tω = (y1 tα1 )a1 · (y2 tα2 )a2 · tω = ya1 1 ta1α1 · ya2 2 ta2α2 · tω = ya1 1 ya2 2 ta1α1+a2α2+ω a1α1 + a2α2 + ω = a1 · α1 + a2 · α2 + ω · 1 =   a1 a2 ω   ,   α1 α2 1   = ˆ a, ˆ α T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 27 / 1
153. Change of Variables With ˆ α = (α1, α2, 1) x1 = y1tα1 x2 = y2tα2 ↓ X = Y tα Xa tω = xa1 1 xa2 2 tω = (y1 tα1 )a1 · (y2 tα2 )a2 · tω = ya1 1 ta1α1 · ya2 2 ta2α2 · tω = ya1 1 ya2 2 ta1α1+a2α2+ω a1α1 + a2α2 + ω = a1 · α1 + a2 · α2 + ω · 1 =   a1 a2 ω   ,   α1 α2 1   = ˆ a, ˆ α ˆ a =   a1 a2 ω   T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 27 / 1
154. Change of Variables With ˆ α = (α1, α2, 1) x1 = y1tα1 x2 = y2tα2 ↓ X = Y tα Xa tω = xa1 1 xa2 2 tω = (y1 tα1 )a1 · (y2 tα2 )a2 · tω = ya1 1 ta1α1 · ya2 2 ta2α2 · tω = ya1 1 ya2 2 ta1α1+a2α2+ω a1α1 + a2α2 + ω = a1 · α1 + a2 · α2 + ω · 1 =   a1 a2 ω   ,   α1 α2 1   = ˆ a, ˆ α ˆ a =   a1 a2 ω   ˆ α =   α1 α2 ω   T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 27 / 1
155. Change of Variables With ˆ α = (α1, α2, 1) x1 = y1tα1 x2 = y2tα2 ↓ X = Y tα Xa tω = xa1 1 xa2 2 tω = (y1 tα1 )a1 · (y2 tα2 )a2 · tω = ya1 1 ta1α1 · ya2 2 ta2α2 · tω = ya1 1 ya2 2 ta1α1+a2α2+ω a1α1 + a2α2 + ω = a1 · α1 + a2 · α2 + ω · 1 =   a1 a2 ω   ,   α1 α2 1   = ˆ a, ˆ α ˆ a =   a1 a2 ω   ˆ α =   α1 α2 ω   Xa tω = Ya t ˆ a, ˆ α T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 27 / 1
156. Change of Variables: h1 Xa tω → ↓ Ya t ˆ a, ˆ α T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 28 / 1
157. Change of Variables: h1 Xa tω → a ω = ˆ a ↓ Ya t ˆ a, ˆ α T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 28 / 1
158. Change of Variables: h1 Xa tω → a ω = ˆ a ↓ Ya t ˆ a, ˆ α ˆ α =   α1 α2 1   T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 28 / 1
159. Change of Variables: h1 Xa tω → a ω = ˆ a ↓ Ya t ˆ a, ˆ α ˆ α =   α1 α2 1   h1 = c1Yat ˆ a, ˆ α + c2Ya t ˆ a , ˆ α + terms with higher power of t T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 28 / 1
160. Change of Variables: h2 Xa tω → ↓ Ya t ˆ a, ˆ α T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 29 / 1
161. Change of Variables: h2 Xa tω → a ω = ˆ a ↓ Ya t ˆ a, ˆ α T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 29 / 1
162. Change of Variables: h2 Xa tω → a ω = ˆ a ↓ Ya t ˆ a, ˆ α ˆ α =   α1 α2 1   T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 29 / 1
163. Change of Variables: h2 Xa tω → a ω = ˆ a ↓ Ya t ˆ a, ˆ α ˆ α =   α1 α2 1   h2 = d1Ybt ˆ b, ˆ α + d2Yb t ˆ b , ˆ α + terms with higher power of t T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 29 / 1
164. Change of Variables X = Y tα T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 30 / 1
165. Change of Variables X = Y tα H(X, t) = h1 = c1Yat ˆ a, ˆ α + c2Ya t ˆ a , ˆ α + terms with higher power of t h2 = d1Ybt ˆ b, ˆ α + d2Yb t ˆ b , ˆ α + terms with higher power of t T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 30 / 1
166. Change of Variables X = Y tα H(X, t) = h1 = c1Yatβ1 + c2Ya tβ1 + terms with higher power of t h2 = d1Ybt ˆ b, ˆ α + d2Yb t ˆ b , ˆ α + terms with higher power of t T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 30 / 1
167. Change of Variables X = Y tα H(X, t) = h1 = c1Yatβ1 + c2Ya tβ1 + terms with higher power of t h2 = d1Ybtβ2 + d2Yb tβ2 + terms with higher power of t T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 30 / 1
168. Change of Variables X = Y tα H(X, t) = h1 = c1Yatβ1 + c2Ya tβ1 + terms with higher power of t h2 = d1Ybtβ2 + d2Yb tβ2 + terms with higher power of t Hα(Y, t) = t−β1 h1 = t−β2 h2 = T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 30 / 1
169. Change of Variables X = Y tα H(X, t) = h1 = c1Yatβ1 + c2Ya tβ1 + terms with higher power of t h2 = d1Ybtβ2 + d2Yb tβ2 + terms with higher power of t Hα(Y, t) = t−β1 h1 = t−β2 h2 = T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 30 / 1
170. Change of Variables X = Y tα H(X, t) = h1 = c1Yatβ1 + c2Ya tβ1 + terms with higher power of t h2 = d1Ybtβ2 + d2Yb tβ2 + terms with higher power of t Hα(Y, t) = t−β1 h1 = c1Ya + c2Ya + terms with positive power of t t−β2 h2 = T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 30 / 1
171. Change of Variables X = Y tα H(X, t) = h1 = c1Yatβ1 + c2Ya tβ1 + terms with higher power of t h2 = d1Ybtβ2 + d2Yb tβ2 + terms with higher power of t Hα(Y, t) = t−β1 h1 = c1Ya + c2Ya + terms with positive power of t t−β2 h2 = T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 30 / 1
172. Change of Variables X = Y tα H(X, t) = h1 = c1Yatβ1 + c2Ya tβ1 + terms with higher power of t h2 = d1Ybtβ2 + d2Yb tβ2 + terms with higher power of t Hα(Y, t) = t−β1 h1 = c1Ya + c2Ya + terms with positive power of t t−β2 h2 = d1Yb + d2Yb + terms with positive power of t T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 30 / 1
173. Starting System Hα(Y, t) = c1 Ya + c2 Ya + terms with positive power of t d1 Yb + d2 Yb + terms with positive power of t T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 31 / 1
174. Starting System Hα(Y, t) = c1 Ya + c2 Ya + terms with positive power of t d1 Yb + d2 Yb + terms with positive power of t Hα(Y, 0) = T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 31 / 1
175. Starting System Hα(Y, t) = c1 Ya + c2 Ya + terms with positive power of t d1 Yb + d2 Yb + terms with positive power of t Hα(Y, 0) = c1Ya + c2Ya + terms with positive power of 0 d1Yb + d2Yb + terms with positive power of 0 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 31 / 1
176. Starting System Hα(Y, t) = c1 Ya + c2 Ya + terms with positive power of t d1 Yb + d2 Yb + terms with positive power of t Hα(Y, 0) = c1Ya + c2Ya d1Yb + d2Yb T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 31 / 1
177. Starting System Hα(Y, t) = c1 Ya + c2 Ya + terms with positive power of t d1 Yb + d2 Yb + terms with positive power of t Hα(Y, 0) = c1Ya + c2Ya d1Yb + d2Yb Starting system: c1Ya + c2Ya = 0 d1Yb + d2Yb = 0 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 31 / 1
178. Starting System Hα(Y, t) = c1 Ya + c2 Ya + terms with positive power of t d1 Yb + d2 Yb + terms with positive power of t Hα(Y, 0) = c1Ya + c2Ya d1Yb + d2Yb Starting system: c1Ya + c2Ya = 0 d1Yb + d2Yb = 0 is a binomial system, which can be solved. T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 31 / 1
179. Starting System Hα(Y, t) = c1 Ya + c2 Ya + terms with positive power of t d1 Yb + d2 Yb + terms with positive power of t Hα(Y, 0) = c1Ya + c2Ya d1Yb + d2Yb Starting system: c1Ya + c2Ya = 0 d1Yb + d2Yb = 0 is a binomial system, which can be solved. We get 12 solutions → starting points. T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 31 / 1
180. Other Choices of α There are also other choices of α. ˆ α =   2 3 18 1   ⇒ Hα(y1, y2, 0) = 2y4 1 y5 2 + 5y2 2 5y3 1 y4 2 + 8y2 2 ⇒ 1 solution ˆ α =   1 3 0 1   ⇒ Hα(y1, y2, 0) = 2y4 1 y5 2 + 3y3 1 y2 2 5y3 1 y4 2 + 6y3 1 y2 2 ⇒ 2 solutions Together we have 12 + 1 + 2 = 15 paths lead us to the 15 solutions of P(X) T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 32 / 1
181. Mixed Cell Computation Recall: we want to find ˆ α = (α, 1) s.t. •, ˆ α minimizes at exactly two vertices over each Newton polytope. T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 33 / 1
182. Lower Edges Trying to find ˆ α = (α, 1). T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 34 / 1
183. Lower Edges Trying to find ˆ α = (α, 1). ˆ α is an “upward-pointing” vector. T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 34 / 1
184. Lower Edges Trying to find ˆ α = (α, 1). ˆ α is an “upward-pointing” vector. The linear function •, ˆ α can only minimize on the “lower hull” T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 34 / 1
185. Lower Edges Trying to find ˆ α = (α, 1). ˆ α is an “upward-pointing” vector. The linear function •, ˆ α can only minimize on the “lower hull” So we only need to consider “lower edges” T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 34 / 1
186. Lower Edges Trying to find ˆ α = (α, 1). ˆ α is an “upward-pointing” vector. The linear function •, ˆ α can only minimize on the “lower hull” So we only need to consider “lower edges” System of linear inequalities “Linear Programming” T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 34 / 1
187. Test For Edges Can ˆ a1 and ˆ a2 define a lower edge? Is there an ˆ α = (α, 1) such that ˆ a1 , ˆ α = ˆ a2 , ˆ α T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 35 / 1
188. Test For Edges Can ˆ a1 and ˆ a2 define a lower edge? Is there an ˆ α = (α, 1) such that ˆ a1 , ˆ α = ˆ a2 , ˆ α < ˆ ai , ˆ α T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 35 / 1
189. Test For Edges Can ˆ a1 and ˆ a2 define a lower edge? Is there an ˆ α = (α, 1) such that ˆ a1 , ˆ α = ˆ a2 , ˆ α < ˆ ai , ˆ α for i 1, 2 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 35 / 1
190. Test For Edges Can ˆ a1 and ˆ a2 define a lower edge? Is there an ˆ α = (α, 1) such that ˆ a1 , ˆ α = ˆ a2 , ˆ α < ˆ ai , ˆ α for i 1, 2 Linear programming T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 35 / 1
191. Enumerate Lower Edges Lower edges Lower edges (ˆ a1, ˆ a2), (ˆ a1, ˆ a5), (ˆ a2, ˆ a3), · · · (ˆ b1, ˆ b2), (ˆ b1, ˆ b4), (ˆ b3, ˆ b5), · · · T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 36 / 1
192. The Enumeration Algorithm lower lower lower edges edges edges (ˆ a1, ˆ a2) (ˆ b1, ˆ b2) (ˆ c1, ˆ c2) (ˆ a1, ˆ a5) (ˆ b1, ˆ b4) (ˆ c3, ˆ c5) . . . . . . . . . T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 37 / 1
193. The Enumeration Algorithm lower lower lower edges edges edges (ˆ a1, ˆ a2) → (ˆ b1, ˆ b2) (ˆ c1, ˆ c2) (ˆ a1, ˆ a5) (ˆ b1, ˆ b4) (ˆ c3, ˆ c5) . . . . . . . . . Can the edges (ˆ a1, ˆ a2), (ˆ b1, ˆ b2) share the same upward pointing normal vector ˆ α = (α, 1)? T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 37 / 1
194. The Enumeration Algorithm lower lower lower edges edges edges (ˆ a1, ˆ a2) → (ˆ b1, ˆ b2) → (ˆ c1, ˆ c2) (ˆ a1, ˆ a5) (ˆ b1, ˆ b4) (ˆ c3, ˆ c5) . . . . . . . . . Can the edges (ˆ a1, ˆ a2), (ˆ b1, ˆ b2), (ˆ c1, ˆ c2) share the same upward pointing normal vector ˆ α = (α, 1)? T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 37 / 1
195. The Enumeration Algorithm lower lower lower edges edges edges (ˆ a1, ˆ a2) (ˆ b1, ˆ b2) (ˆ c1, ˆ c2) (ˆ a1, ˆ a5) (ˆ b1, ˆ b4) (ˆ c3, ˆ c5) . . . . . . . . . Can the edges (ˆ a1, ˆ a2), (ˆ b1, ˆ b4) share the same upward pointing normal vector ˆ α = (α, 1)? T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 37 / 1
196. The Enumeration Algorithm lower lower lower edges edges edges (ˆ a1, ˆ a2) (ˆ b1, ˆ b2) (ˆ c1, ˆ c2) (ˆ a1, ˆ a5) (ˆ b1, ˆ b4) (ˆ c3, ˆ c5) . . . . . . . . . Linear inequalities Linear Programming T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 37 / 1
197. The Enumeration Algorithm lower lower lower edges edges edges (ˆ a1, ˆ a2) (ˆ b1, ˆ b2) (ˆ c1, ˆ c2) (ˆ a1, ˆ a5) (ˆ b1, ˆ b4) (ˆ c3, ˆ c5) . . . . . . . . . Highly serial Linear inequalities Linear Programming T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 37 / 1
198. Parallel Mixed Cell Computation (Why?) The curve following algorithm is naturally parallel T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 38 / 1
199. Parallel Mixed Cell Computation (Why?) The curve following algorithm is naturally parallel Mixed cells T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 38 / 1
200. Parallel Mixed Cell Computation (Why?) The curve following algorithm is naturally parallel Mixed cells S = Fraction of the time spent on serial algorithm T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 38 / 1
201. Parallel Mixed Cell Computation (Why?) The curve following algorithm is naturally parallel Mixed cells S = Fraction of the time spent on serial algorithm Maximum speed up = 1 S (no matter how many processors are used) T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 38 / 1
202. Parallel Mixed Cell Computation We adopted a task-based paradigm. T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 39 / 1
203. Parallel Mixed Cell Computation We adopted a task-based paradigm. Inspired by graph traverse algorithms T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 39 / 1
204. Parallel Mixed Cell Computation We adopted a task-based paradigm. Inspired by graph traverse algorithms Group computation as a graph of interdependent tasks T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 39 / 1
205. Parallel Mixed Cell Computation We adopted a task-based paradigm. Inspired by graph traverse algorithms Group computation as a graph of interdependent tasks Reformuate the problem as a generalized graph traverse problem T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 39 / 1
206. Benchmarks: Results on Multi-core Architecture We define the speedup using p processor cores (threads) to be Speedup := Tserial Tp T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 40 / 1
207. Benchmarks: Results on Multi-core Architecture We define the speedup using p processor cores (threads) to be Speedup := Tserial Tp (maximum = p) T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 40 / 1
208. Benchmarks: Results on Multi-core Architecture We define the speedup using p processor cores (threads) to be Speedup := Tserial Tp (maximum = p) Testcase Mixed Vol. Tserial p = 2 p = 4 p = 6 p = 8 cyclic15 35,243,520 8.9 hr 1.92 3.76 5.59 7.33 eco20 262,144 3.5 hr 1.95 3.81 5.79 7.65 gaukwa8 410,338,673 2.8 hr 1.95 3.78 5.62 7.36 katsura15 32,730 1.0 hr 1.95 3.84 5.75 7.68 noon21 10,460,353,161 56 min 1.94 3.81 5.63 7.35 sonic8 6,533,120 56 min 1.99 3.94 5.89 7.80 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 40 / 1
209. NUMA Architecture (Sonic8) 4 8 12 16 20 24 4 8 12 16 20 24 Ideal speedup Actual speedup Tserial p = 4 p = 8 p = 12 p = 16 p = 20 p = 24 53.3min 3.908 7.712 11.364 14.848 17.860 19.44 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 41 / 1
210. Cluster Computing based on MPI Speedup for cyclic15 on clusters using MPI with up to 64 nodes 8 16 24 32 40 48 56 64 8 16 24 32 40 48 56 64 Ideal speedup Actual speedup T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 42 / 1
211. Distributed Computing: VortexAC6 Distributed computing: large number of computers cooperate on solving a single problem via the Internet. T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 43 / 1
212. Distributed Computing: VortexAC6 Distributed computing: large number of computers cooperate on solving a single problem via the Internet. Using this approach, we were able to compute the mixed volume of a very large system: VortexAC6 T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 43 / 1
213. Distributed Computing: VortexAC6 Distributed computing: large number of computers cooperate on solving a single problem via the Internet. Using this approach, we were able to compute the mixed volume of a very large system: VortexAC6 Number of mixed cells 12, 103, 484 Mixed volume 27, 550, 213 Number of processors used 145 Total CPU time 6157.70 hours (8.5 months) Wall clock time 50.52 hours (2 days) T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 43 / 1
214. Thank You! T.R. Chen (MSU) Polyhedral Homotopy April 14, 2011 44 / 1