dise ño de algoritmos (heurísticas modernas) gabriela ochoa ldcb.ve/~gabro

Clase 2, Parte 1: Conceptos Básicos de Búsqueda. Algoritmo de Ascenso de Colina. Enunciado Tarea 1Diseño de Algoritmos (Heurísticas Modernas)

Gabriela Ochoa

http://www.ldc.usb.ve/~gabro/

Basic Concepts

Representation: Encodes alternative candidate solutions for manipulation

Objective: describes the purpose to be fulfilled

Evaluation Function: returns a specific value that indicates the quality of any particular solution given the representation

Search Problem

Definition: Given a search space S and its feasible part F in S, find x Є F such that eval(x) ≤ eval(y), for all y Є F (minimization)

The point x that satisfies the above condition is called global solution

The terms “search problem” and “optimization problem” are considered synonymous. The search for the best solution is the optimization problem

Boolean satisfiability problem (SAT) An instance of the problem: is defined by a Boolean

expression written using only AND, OR, NOT, variables, and parentheses.

The question is: given the expression, is there some assignment of TRUE and FALSE values to the variables that will make the entire expression true?

SAT is of central importance in various areas of computer science, including theoretical computer science, algorithmics, artificial intelligence, hardware design and verification.

Computational Complexity of SAT SAT is NP-complete. In fact, it was the first known

NP-complete problem, as proved by Stephen Cook in 1971

The problem remains NP-complete even if all expressions are written in conjunctive normal form with 3 variables per clause (3-CNF) (x11 OR x12 OR x13) AND (x21 OR x22 OR x23) AND (x31 OR x32 OR x33) AND ...

where each x is a variable, with or without a NOT in front of it, and each variable can appear multiple times in the expression.

Problem Formulation (SAT)

Let us consider a problem of size 30 (i.e. 30 variables) Representation

1 True, 0 False, Binary String of length 30 Search Space

2 choices for each variable, taken over 30 variables, generates 230 possibilities

Objective To find the vector of bits such that the compound Boolean

statement is satisfied (made true) Evaluation Function?

Not enough information to take the objective

Travelling salesman problem (TSP) Given a number of cities and the

costs of travelling from one to the other, what is the cheapest roundtrip route that visits each city and then returns to the starting city?

An equivalent formulation in terms of graph theory is: Find the Hamiltonian cycle with the least weight in a weighted graph.

Problem Formulation (TSP)

Let us consider a problem of size 30 (i.e. 30 cities) Representation: Permutation of natural numbers 1,… ,30

where each number corresponds to a city to be visited in sequence

Search Space: Permutations of all cities. Symmetric TSP, circuit the same regardless the starting city: (n-1)!/2

Objective: Minimize the total distance traversed, visiting each city once, and returning to the starting city. Min Sum(dist(x,y))

Evaluation Function: Map each tour to its corresponding total distance

Neighbourhoods and local Optima Region of the search space that is “near” to

some particular point in that space

S. xN(x)

A search space S, a potential solution x, and its neighbourhood N(x)

Defining Neighbourhoods 1

Define a distance function dist on the search space S Dist: S x S → R N(x) = {y Є S: dist(x,y) ≤ ε }

Examples: Euclidean distance, for search spaces defined

over continuous variables Hamming distance, for search spaces definced

over binary strings (e.g. SAT)

Use a mapping m, that defines a neighbourhood for any point x Є S

2-swap mapping: generates a new set of potential solutions from a given solution x

Solutions are generated by swapping two cities from a given tour

Every solution has n(n-1)/2 neighbours Example: 2 4 5 3 1 → 2 3 5 4 1,

Defining Neighbourhoods, TSP

1-flip mapping: generates a new set of potential solutions from a given solution x

Solutions are generated by flipping a single bit in the given bit string

Every solution has n neighbours Example: 1 1 0 0 1 → 0 1 0 0 1 ( 1st bit)

Defining Neighbourhoods, SAT

Gaussian Distribution: for each variable defines a neighbourhood

Mean: the current point, Std. dev.: 1/6 of the range of the variable

x = (x1, …, xn), where li ≤ xi ≤ ui

x’ = xi + N(0,σi), where σi = (ui - li)/6

N(0,σi) is an independent random Gaussian number with mean zero and std. dev. σi

Defining Neighbourhoods, Real Numbers

Local Optimum

A potential solution x Є S is a local optimum with respect to the neighbourhood N, if and only if eval(x) ≤ eval(y), for all y Є N(x)

Métodos de Ascenso de Colina - 1 Usan una técnica de mejoramiento iterativo Comienzan a partir de un punto (punto actual) en el

espacio de búsqueda En cada iteración, un nuevo punto es seleccionado de

la vecindad del punto actual Si el nuevo punto es mejor, se transforma en em punto

actual, sino otro punto vecino es seleccionado y evaluado

El método termina cuando no hay mejorías, o cuando se alcanza un numero predefinido de iteraciones

Hillclimbing Methods - 2

May converge to local optima usually have to start search from various

starting points Initial starting points may be chosen,

randomly according to some regular pattern based on other information (e.g. results of a

prior search)

Hillclimbing Methods - 3

Variations of hillclimbing algorithms differ in the way a new string is selected for comparisons with the current string

One version of simple (iterated) hillclimbing method is the steepest ascent hillclimbing

Hillclimbing Methods - 4

Example problem:

The search space is a set of binary strings v of length 30

The objective function f (to be maximized):f(v)=|11*one(v)-150|

where one(v) returns the number of ones in v.e.g. v1=(110111101111011101101111010101)

f(v1) = |11*22 - 150| = 92

Hillclimbing Methods - 5procedure iterated hillclimber

begin

t 0 repeat

local FALSE select a curent string vc at random

evaluate vc

repeat

form 30 new strings in the neigborhood of vc by

flipping single bits of vc

select vn from the set of new strings with the

largest value of the objective function f

if f(vc) < f(vn) then vc vn

else local TRUE until local

t t+1 until t=MAX

end

Hillclimbing Methods - 6

success/failure of each iteration depends on starting point success defined as returning a local or a global

optimum in problems with many local optima a global

optimum may not be found

Hillclimbing Methods - 7

Weaknesses: Usually terminate at solutions that are local optima No information as to how much the discovered local

optimum deviates from the global (or even other local optima)

Obtained optimum depends on starting point Usually no upper bound on computation time

Hillclimbing Methods - 8

Advantages: Very easy to apply (only a representation, the

evaluation function and a measure that defines the neigborhood around a point is needed)