HEPGAME: Physics, Artificial Intelligence, and the Simplification of Expressions

Transcription

1 HEPGAME: Physics, Artificial Intelligence, and the Simplification of Expressions Aske Plaat LIACS/LCDS Leiden University December 17, 2014 based on [Ruijl, Vermaseren, Plaat, Van den Herik] presented at ICAART 2014 This work is supported in part by ERC Advanced Grant no , HEPGAME awarded on for the period 1 July 2013 to 1 July 2018

2 Overview Overview ERC Team CERN HEPGAME Expressions HEPGAME Horner s rule (refresher) (Go) SA-

3 ERC Team Overview ERC Team CERN HEPGAME Expressions Jos Vermaseren Mathias Ritzman Takahiro Ueda Franz Herzog Ben Ruijl Ali Mirsoleimani Aske Plaat Jaap van den Herik

4 CERN Overview ERC Team CERN HEPGAME Expressions

5 4July2012 Overview ERC Team CERN HEPGAME Expressions

6 Champagne at Nikhef Overview ERC Team CERN HEPGAME Expressions

7 HEPGAME Overview ERC Team CERN HEPGAME Expressions Goal HEPGAME is about solving very large expressions in High Energy Physics using combinatorial methods from Artificial Intelligence.

8 Example of a short expression Overview ERC Team CERN HEPGAME Expressions +32o 3 n 2 m + 32o 3 n 2 l 48o 3 n 2 km 48o 3 n 2 kl + 32o 4 j 2 m + 32o 4 j 2 l + 64o 4 ijm 128o 4 ijh + 64o 4 ijl + 64o 4 i 2 m 128o 4 i 2 h + 64o 4 i 2 l 128o 4 gjh + 64o 4 gim 128o 4 gih + 64o 4 gil + 32o 4 g 2 m + 32o 4 g 2 l 64o 4 km 64o 4 kl +32o 4 kjm + 32o 4 kjl 32o 4 kj 2 m 32o 4 kj 2 l + 64o 4 kim 192o 4 kih + 64o 4 kil 64o 4 kijm + 128o 4 kijh 64o 4 kijl 64o 4 ki 2 m + 128o 4 ki 2 h 64o 4 ki 2 l + 96o 4 kgm 192o 4 kgh + 96o 4 kgl + 128o 4 kgjh 64o 4 kgim + 128o 4 kgih 64o 4 kgil 32o 4 kg 2 m 32o 4 kg 2 l + 64o 4 k 2 m + 64o 4 k 2 l 64o 4 k 2 jm 64o 4 k 2 jl 64o 4 k 2 im 64o 4 k 2 il 64o 4 k 2 gm 64o 4 k 2 gl 32o 4 k 3 m 32o 4 k 3 l + 48fo 2 n 2 m + 32fo 2 n 2 h + 48fo 2 n 2 l 48fo 2 n 2 jm + 64fo 2 n 2 jh 48n 3 h 2 48fo 2 n 2 jl 96fo 2 n 2 im 96fo 2 n 2 il 64fo 2 n 2 gh 48fo 2 n 2 km 48fo 2 n 2 kl + 256fo 3 jh + 32fo 3 j 2 m 128fo 3 j 2 h + 32fo 3 j 2 l 32fo 3 j 3 m 32fo 3 j 3 l 64fo 3 im + 256fo 3 ih 64fo 3 il + 128fo 3 ijm 448fo 3 ijh +128fo 3 ijl 128fo 3 ij 2 m + 64fo 3 ij 2 h 128fo 3 ij 2 l + 192fo 3 i 2 m 384fo 3 i 2 h + 192fo 3 i 2 l 192fo 3 i 2 jm +256fo 3 i 2 jh 192fo 3 i 2 jl 128fo 3 i 3 m + 128fo 3 i 3 h 128fo 3 i 3 l + 64fo 3 gm + 64fo 3 gl 448fo 3 gjh +64fo 3 gj 2 h + 192fo 3 gim 576fo 3 gih + 192fo 3 gil 64fo 3 gijm + 384fo 3 gijh 64fo 3 gijl 128fo 3 gi 2 m +128fo 3 gi 2 h 128fo 3 gi 2 l + 32fo 3 g 2 m 64fo 3 g 2 h + 32fo 3 g 2 l 32fo 3 g 2 jm + 128fo 3 g 2 jh 32fo 3 g 2 jl 64fo 3 g 2 im 64fo 3 g 2 ih 64fo 3 g 2 il 64fo 3 g 3 h 64fo 3 km + 128fo 3 kh 64fo 3 kl + 32fo 3 kjm 448fo 3 kjh +32fo 3 kjl 96fo 3 kj 2 m + 256fo 3 kj 2 h 96fo 3 kj 2 l + 128f 2 o 2 i 2 m 384f 2 o 2 i 2 h + 128f 2 o 2 i 2 l 384f 2 o 2 i 2 jm

9 Why optimize? Why optimize? Horner schemes Common subexpression elimination Optimization is important: Expressions will be evaluated many times in numerical integration This may take weeks! Reducing the number of operations may save many days

10 Horner schemes Why optimize? Horner schemes Common subexpression elimination Say we have x 3 y 2 + x 2 y + x 3 z x x x y y + x x y + x x x z 9 Horner scheme: x 2 (y + x(y 2 + z)) x x (y + x (y y + z)) 4 Other possibility: x 3 z + y(x 2 (1 + xy)) x x x z + y (x x (1 + x y)) 7 Optimal order problem is NP-hard

11 Horner schemes Why optimize? Horner schemes Common subexpression elimination Say we have x 3 y 2 + x 2 y + x 3 z x x x y y + x x y + x x x z 9 Horner scheme: x 2 (y + x(y 2 + z)) x x (y + x (y y + z)) 4 Other possibility: x 3 z + y(x 2 (1 + xy)) x x x z + y (x x (1 + x y)) 7 Optimal order problem is NP-hard

12 Horner schemes Why optimize? Horner schemes Common subexpression elimination Say we have x 3 y 2 + x 2 y + x 3 z x x x y y + x x y + x x x z 9 Horner scheme: x 2 (y + x(y 2 + z)) x x (y + x (y y + z)) 4 Other possibility: x 3 z + y(x 2 (1 + xy)) x x x z + y (x x (1 + x y)) 7 Optimal order problem is NP-hard

13 Horner schemes Why optimize? Horner schemes Common subexpression elimination Say we have x 3 y 2 + x 2 y + x 3 z x x x y y + x x y + x x x z 9 Horner scheme: x 2 (y + x(y 2 + z)) x x (y + x (y y + z)) 4 Other possibility: x 3 z + y(x 2 (1 + xy)) x x x z + y (x x (1 + x y)) 7 Optimal order problem is NP-hard

14 Why optimize? Horner schemes Common subexpression elimination Common SubExpression Elimination (CSEE) Find most common subexpressions and build a replacement tree b (a + e) is found CSEE reduces both and + + b + b c + a e a e

15 Games, Minimax, Chess Selection Expansion Simulation Backpropagation UCT The power resides in - minimax plus enhancements, i.e., evaluation in each node of the tree is possible

16 Chess Minimax Tree Selection Expansion Simulation Backpropagation UCT Size: O(10 46 ) 1. Search 2. Since the tree is too large we need heuristic evaluation

17 Selection Expansion Simulation Backpropagation UCT Monte Carlo Tree Search () Successes with Go inspired us to apply to Horner schemes

18 Selection Expansion Simulation Backpropagation UCT

19 Selection Expansion Simulation Backpropagation UCT Monte Carlo Tree Search () Successes with Go inspired us to apply to Horner schemes Build a state tree selectively Each node is a variable Idea Use to find near optimal Horner scheme, which reduces the number of operations the most

20 Selection Selection Expansion Simulation Backpropagation UCT x z w

21 Expansion Selection Expansion Simulation Backpropagation UCT x z w y

22 Simulation/Playout Selection Expansion Simulation Backpropagation UCT x z w y Random scheme Number of operations after Horner + CSEE

23 Backpropagation Selection Expansion Simulation Backpropagation UCT

24 Selection Expansion Simulation Backpropagation UCT Upper confidence bounds applied to trees (UCT) UCT best child: argmax children c of s Û x(c) 2lnn(s) n(c) + 2C p n(c) x(c) is score of c n(s) visits at s C p is exploration-exploitation constant

25 The problems with UCT Selection Expansion Simulation Backpropagation UCT Tuning C p is expensive Exploration is mostly done at the top of the tree The final iterations are wasted on exploration Idea Spend first iterations exploring, but start exploiting later on. Let C p decrease with iteration number i.

26 SA-UCT Selection Expansion Simulation Backpropagation UCT Inspiration from Physics: Simulated Annealing Introduce parameter: T = C p N i N (1) N: total number of updates i: currentupdatenumber C p : exploration-exploitation constant at i = 0

27 SA-UCT Selection Expansion Simulation Backpropagation UCT SA-UCT: argmax children c of s Û x(c) 2lnn(s) n(c) + 2 T n(c) Role of T similar to temperature in Simulated Annealing

28 Sensitivity analysis Sensitivity analysis Two parameters: N: number of tree updates C p :foruct C p = T (0): forsa-uct Question How does behave when we vary N and C p?

29 Sensitivity analysis Left: too much exploitation (local minima) Right: too much exploration (di usion) Middle: just right Number of operations Number of operations C p C p (a) UCT (b) SA-UCT Figure : Expression HEP( ) at N = 300

30 Sensitivity analysis Number of operations Number of operations C p C p (a) UCT (b) SA-UCT Figure : Expression HEP( ) at N = 1000

31 Sensitivity analysis Number of operations Number of operations C p C p (a) UCT (b) SA-UCT Figure : Expression HEP( ) at N = 3000

32 Future work The end SA-UCT SA-UCT widens the region for C p for which good results are obtained (in the above example: 10 times larger) With larger N results are more pronounced

33 Our results compared Future work The end For res(6,7) polynomial with 587,880 operations: Early experiments using frequency occurrence + CSE: 8 times fewer operations Mathematica and Maple: optimization is di Haggies: 6 times fewer operations : 16 times fewer operations Practical use Numerical integration will be 16 times faster cult to measure

34 Future work Future work The end Gauging the C p parameter automatically Exploit domain specific knowledge Apply to more general expressions with tensors Future plans We are looking for large expressions from various fields!

35 Future work The end Thank you for your attention