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Genetic Algorithms in Problem Solving 1998 7 16

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EVOLVING COMPUTER PROGRAMS (1) t Evolving Lisp Programs Keplers Third Law: P 2 = cA 3 PROGRAM ORBITAL_PERIORD C# Mars # A = 1.52 P = SQRT(A * A * A) PRINT P END ORBITAL_PERIORD (defun orbital_period () ; Mars ; (setf A 1.52) (sqrt (* A (* A A))))

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EVOLVING COMPUTER PROGRAMS (2) Kozas algorithm - Genetic Programming (GP) 1. Choose a set of possible functions and terminals for the program. F = {+, - *, /, }, T = {A} 2. Generate an initial population of random trees (programs) using the set of possible functions and terminals. 3. Calculate the fitness of each program in the population by running it on a set of fitness cases (a set of input for which the correct output is known). 4. Apply selection, crossover, and mutation to the population to form a new population. 5. Steps 3 and 4 are repeated for some number of generations.

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EVOLVING COMPUTER PROGRAMS (3)

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EVOLVING COMPUTER PROGRAMS (4) Block-Stacking Problem u T = {CS, TB, NN} u F = {MS(x), MT(x), DU(exp1, exp2), NOT(exp1), EQ(exp1, exp2) } u (EQ (DU (MT CS) (NOT CS)) (DU (MS NN) (NOT NN)))

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EVOLVING COMPUTER PROGRAMS (5) t Evolving Cellular Automata (CA) Example (N=11, radius = 1) space-time diagram Rule table: neighborhood: 000 001 010 011 100 101 110 111 output bit : 0 0 0 1 0 1 1 1 Lattice: t = 0 1 0 1 0 0 1 1 1 0 1 0 t = 1 0 1 0 0 0 1 1 1 1 0 1

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EVOLVING COMPUTER PROGRAMS (6) Density-classification task (N=149, r =3)

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DATA ANALYSIS AND PREDICTION (1) t Predicting Dynamical Systems individual u C = {($20 Price of Xerox Stock on day 1) ^ ($25 Price of Xerox Stock on day 2 $27) ^ ($22 Price of Xerox Stock on day 3 $25)} crossover, mutation

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DATA ANALYSIS AND PREDICTION (2)

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DATA ANALYSIS AND PREDICTION (3) t Predicting Protein Structure

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EVOLVING NEURAL NETWORKS (1)

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EVOLVING NEURAL NETWORKS (2) t Evolving Weights in a Fixed Network

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EVOLVING NEURAL NETWORKS (3)

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EVOLVING NEURAL NETWORKS (4) t Evolving Network Architectures Direct Encoding

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EVOLVING NEURAL NETWORKS (5) Grammatical Encoding

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Blocks World Problem. The CS Terminal Specifies the block at the top of the stack. Example CS evaluates to E Note: Evaluates to nil if the stack is empty.

Blocks World Problem. The CS Terminal Specifies the block at the top of the stack. Example CS evaluates to E Note: Evaluates to nil if the stack is empty.

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