Dyna Dynamic mic Analysis Analysis of the of the Co-evo Co - PowerPoint PPT Presentation

Dyna Dynamic mic Analysis Analysis of the of the Co-evo Co evoluti lution on of Ex of Exac act t Cove Covers rs EVOLVE 2011, Luxembourg Jeffrey Horn GECCO June 28, 2005 IEEE SSCI FOCI 2007 Northern Nort hern Mi Michigan Univer chigan Universit sity Depa epartm rtment of ent of Ma Mathematics thematics and C and Computer omputer Science Science Marq arque uette, tte, MI I USA USA jhorn@ jhorn@nmu.edu nmu.edu http://cs http:/ cs.nmu.edu/ .nmu.edu/~jeff ~jeffhorn horn NERL EVOLVE EVOLVE Bour Bourlinst linster er Cas Castle, tle, Luxembourg Luxembourg May May 25 25-27, 2011 27, 2011 1 1 1

OUTL OUTLINE INE 1. The RFS Algorithm EVOLVE 2011, Luxembourg GECCO June 28, 2005 IEEE SSCI FOCI 2007 2. Static Analysis (previous results) 3. Dynamic Analysis (new results) 4. Discussion 2 2 2

EVOLVE 2011, Luxembourg GECCO June 28, 2005 IEEE SSCI FOCI 2007 1. The RFS Algorithm – The Shape Nesting Problem OUTL OUTLINE INE 3 3 3

The Prac The Practica tical l Pro Problem blem of Shap of Shape e Nes Nesting ting Descript iption ion of Description iption of Subs bstrat ate e Piec ece e Shape pe Shape ape EVOLVE 2011, Luxembourg GECCO June 28, 2005 IEEE SSCI FOCI 2007 GA with h Resourc ource-de defin ined ed Fitne ness Shari ring ng (RFS) S) Nestin ing g of Pieces es on Substrat ate 4 4 4

EVOLVE 2011, Luxembourg GECCO June 28, 2005 IEEE SSCI FOCI 2007 Exa Example mple Run Run, , Ar Arbitr bitrar ary y Sha Shape pes 5 5 5

EVOLVE 2011, Luxembourg GECCO June 28, 2005 IEEE SSCI FOCI 2007 Mor More e Exa Example mple Run species_count ≥ 19 Runs 6 6 6

Evo Evolution lution Can Can Be Se Be Select lection ion Only Only generation 1 generation 209 (one generation beyond the initial population ) (8 cooperative species) species_count ≥ 36 species_count ≥ 4 EVOLVE 2011, Luxembourg GECCO June 28, 2005 IEEE SSCI FOCI 2007 species_count ≥ 1 (shows all species) species_count ≥ 2 7 7 7

generation 709 generation 609 (9 cooperative species) (almost 9 cooperative species) EVOLVE 2011, Luxembourg GECCO June 28, 2005 IEEE SSCI FOCI 2007 species_count ≥ 20 species_count ≥ 22 8 8 8

RFS RFS Can Can Out Outpe perfor rform m Some Some Commerc Comme rcial ial Sha Shape pe Nes Nesting ting Soft Softwar ware ProNest, 10 disks RFS, 12 disks OptiNest, 11 disks (default settings) (default settings) EVOLVE 2011, Luxembourg GECCO June 28, 2005 IEEE SSCI FOCI 2007 OptiNest, 12 disks ProNest, 11 disks (optimized settings) (optimized settings) ArtCam, 11 disks 9 9 9 (Horn, IJCCI, 2010)

OUTLINE OUTL INE 1. The RFS Algorithm EVOLVE 2011, Luxembourg – The Shape Nesting Problem GECCO June 28, 2005 IEEE SSCI FOCI 2007 – GA with RFS shared fitness 10 10 10

Fitn Fitnes ess s Ter Terms ms f a is area of shaped piece “a”. It is used as the EVOLVE 2011, Luxembourg f a GECCO June 28, 2005 objective fitness of IEEE SSCI FOCI 2007 “ a” in RFS . Similarly for f b . f b f ab ( = f ba ) f ab is area of overlap (intersection) of “a” and “b”. It is used to calculate the shared fitness of “ a” and “b” in RFS. 11 11 11

Sha Share red d Fitn Fitnes ess s Calcu Calculation lation • Under RFS, shared fitness of an individual x is a function of all individuals in the population: EVOLVE 2011, Luxembourg GECCO June 28, 2005 IEEE SSCI FOCI 2007 • This can be re- written in terms of “species” y : 12 12 12

What is a “species”? • DEFINITION : An individual is a member of the population • DEFINITION : A species is the set of all copies of an individual EVOLVE 2011, Luxembourg GECCO June 28, 2005 IEEE SSCI FOCI 2007 • DEFINITION : the species_count , n X , is the number of copies of a species X in the current population • IMPLICATIONS : – If two individuals differ in their chromosomes, they are members of two distinct species – Any two members of the same species have identical placements on the substrate, and therefore overlap (compete) 100% – We seek the largest possible set of feasible (on substrate), cooperative (non-overlapping) species. 13 13 13

Res Resou ourc rce-de defi fine ned d Fitn Fitnes ess s Sha Sharing ring f  X Shared fitness f Sh,X  n f X XY EVOLVE 2011, Luxembourg GECCO June 28, 2005  IEEE SSCI FOCI 2007 species Y . f A  Example for two f Sh,A  A B n f n f overlapping niches A AB A B f A B A  Example for three f Sh,A   overlapping niches n f n f n f C A AB AC A B C 14 14 14

OUTLINE OUTL INE 1. The RFS Algorithm EVOLVE 2011, Luxembourg – The Shape Nesting Problem GECCO June 28, 2005 IEEE SSCI FOCI 2007 – GA with RFS shared fitness – The Exact Cover Problem 15 15 15

Beg Beginning innings: s: Ni Nich ches es on on a Flat a Flat La Land ndsc scap ape • Horn’s 1997 dissertation • Fitness sharing on a “hat” function (1D flat fitness) EVOLVE 2011, Luxembourg GECCO June 28, 2005 IEEE SSCI FOCI 2007 16 16 16

Con Conve verg rges es to to Exac Exact t Cov Cover er • Pop. Size is 2560. Width of “hat” is 160. Niche radius is 20. So nine “niches” exactly cover the hat. EVOLVE 2011, Luxembourg GECCO June 28, 2005 IEEE SSCI FOCI 2007 17 17 17

Evolved Evo lved Coo Coope pera ration tion • Sets of non-competing (i.e., non-overlapped) species. EVOLVE 2011, Luxembourg GECCO June 28, 2005 IEEE SSCI FOCI 2007 18 18 18

Shape Shape Nest Nesting ing in in One One Dmension Dmension • RFS is equivalent to Fitness Sharing* in one dimension • Convergence to exact cover for 1D shape nesting EVOLVE 2011, Luxembourg GECCO June 28, 2005 IEEE SSCI FOCI 2007 (* Goldberg & Richardson, 1987) 19 19 19

RFS RFS Can Can Find Find Exa Exact ct Cov Cover ers s in 2D in 2D EVOLVE 2011, Luxembourg GECCO June 28, 2005 IEEE SSCI FOCI 2007 Parallel Problem Solving From Nature VII (2002) 20 20 20

RFS wi RFS with th Mut Mutat ation ion (Horn (Horn, , 20 2002 02) Much smaller Pop has pop size (N=500). converged Some globals must on the 16 be discovered by globals, with mutation EVOLVE 2011, Luxembourg mutation still (some are NOT in producing GECCO June 28, 2005 IEEE SSCI FOCI 2007 Initial pop.) some “misfits” Distribution of Entire Population 21 21 21

OUTL OUTLINE INE 1. The RFS Algorithm EVOLVE 2011, Luxembourg GECCO June 28, 2005 IEEE SSCI FOCI 2007 2. Static Analysis (previous results) 3. Dynamic Analysis (new results) 4. Discussion 22 22 22

Niching Niching Equ Equil ilibrium ibrium • Equilibrium under selection: • Under proportionate selection: EVOLVE 2011, Luxembourg GECCO June 28, 2005 IEEE SSCI FOCI 2007 • Solving, • Meaning that all shared fitnesses must be the same: 23 23 23

Fo For Exa r Example, mple, The The Thr Three ee-Niche Niche Cas Case EVOLVE 2011, Luxembourg GECCO June 28, 2005 IEEE SSCI FOCI 2007 Finite population size N 24 24 24

The I The Infi nfinite nite Population Population Model Model EVOLVE 2011, Luxembourg GECCO June 28, 2005 IEEE SSCI FOCI 2007 Divide both sides of equations by N Substitute p x for proportion n x /N 25 25 25

EVOLVE 2011, Luxembourg GECCO June 28, 2005 IEEE SSCI FOCI 2007 A A Sys Syste tem m of Linea of Linear r Equ Equat ations ions 26 26 26

EVOLVE 2011, Luxembourg GECCO June 28, 2005 IEEE SSCI FOCI 2007 Intr Introd oduc ucing ing Pro Prope pert rty y I 27 27 27

EVOLVE 2011, Luxembourg GECCO June 28, 2005 IEEE SSCI FOCI 2007 Int Intro rodu ducin cing g Pr Prop oper erty ty II II 28 28 28

EVOLVE 2011, Luxembourg GECCO June 28, 2005 IEEE SSCI FOCI 2007 App Applying lying Ass Assump umption tions s of of Pro Prop.s p.s I I an and d II II 29 29 29

In 2 In 200 008 8 Gen Gener eraliz alized ed to to Arb Arbitra itrary ry Nu Numb mber er of Sp of Spec ecies ies EVOLVE 2011, Luxembourg GECCO June 28, 2005 IEEE SSCI FOCI 2007 Horn, Parallel Problem Solving From Nature X (2008) 30 30 30

OUTL OUTLINE INE 1. The RFS Algorithm EVOLVE 2011, Luxembourg GECCO June 28, 2005 IEEE SSCI FOCI 2007 2. Static Analysis (previous results) 3. Dynamic Analysis (new results) 4. Discussion 31 31 31

Conve Con verg rgen ence ce to to Equilibrium Equilibrium • For species u, we ask if ? EVOLVE 2011, Luxembourg GECCO June 28, 2005 IEEE SSCI FOCI 2007 • Let substitute for • In our specific case, we ask if 32 32 32

Dista Distanc nce e to to Equ Equil ilibrium ibrium • Let be the current distance of species u from equilibrium EVOLVE 2011, Luxembourg GECCO June 28, 2005 IEEE SSCI FOCI 2007 • Then is the distance of u from equilibrium in the next time step. • We ask if this distance decreases with each time step: ? 33 33 33