exhaustive search and a first approach to avoiding it

Exhaustive Search and a first approach to avoiding it

Brute Force Methods

guarantee best fitness value is found feasible for ‘small’ data sets only

SAT satisfiability problem

set of propositions prop[n] fitness function

logical expression based on propositionsboolean fitness(prop)

try all 2n combinations of T/F for propositions

boolean fitness(boolean b[]) //fitness functionboolean [] prop = new boolean[n];boolean satisfied = sat(n-1);

boolean sat (int index) // find a solution{ if (index==0) //base case

{prop[0] = true;if fitness(prop) return true;prop[0] = false;return fitness(prop);

}prop[index] = true; // recursiveif (sat(index-1)) return true;

prop[index] = false;return sat(index-1);

} ...

..T ..F

.FT.TT .TF .FF

FFTFTT FTF FFFTFTTTT TTF TFF

2

1

0

...

efficiency without risk

pruning the tree: suppose fitness(prop) is

(p0 \/ ~p1) /\ (~p2)

partial evaluation

...

..T ..F

.FT.TT .TF .FF


...

p2

p1

p0

nodes can be ignored: fitness can not be true

TSP travelling salesman directed edges, complete graph:

exhaustive search is the permutation problem

enumerate all n! permutations of n distinct items (cities) {0,1,2,3,…,n-1}

1

2

0

3

ij 0 1 2 3

0 .

1 .

2 .

3 .

rank in path array

3 1 4 2

0 1 2 3D[i][j]

TSP travelling salesman O(nn)input(V) // number of cities (Vertices)int[V] rIP // rankInPath, init to 0==not rankedvisit(1) // generate visiting sequences

void visit(int rank){ for (city = 0 to V-1) if (rIP[city] == 0)

// not yet visited { rIP[city] = rank; if (rank == V) fitness(rIP) // base case else visit(rank+1) //recursive case rIP[city] = 0 }}

fitness(int[] p)

calculate pathlength

from D[i][j]

if bestpath, save p

fitness(int[] p)

calculate pathlength

from D[i][j]

if bestpath, save p

efficiency without risk

fix first city O((n-1)n-1) use sets instead of searching array

O((n-1)!) == O((n-1)n) keep partial fitness values

reduce cost of fitness evaluation apply branch and bound (pruning)

BUT… still O(en)

Branch and bound

...

..T ..F

.FT.TT .TF .FF


...

...

01

012

0123 0132

0

013

02

021

0213 0231

023

03

031

0312 0321

032

pathlength = 845

592 693 519

549126 274

Variations on TSP

undirected edges: D[i][j] == D[j][i] incomplete graph: some D[i][j] = null Euclidean distances (on an airplane)

cities are located on plane at (xi,yi)

D[i][j] is computed from coordinates:D[i][j] = D[j][i]

= sqrt((xi-xj)2 + (yi-yj)2)

other data structures, efficiencies

Continuousproblem spaces

Where isheight of land?1. what scale to sample?

x∈(0,1], y∈(0,1]interval length:0.1: 100 data points0.01: 10,0000.001: 1,000,000

--- x --

- y


Where isheight of land?2. constraints –

ignore waterfewer data points but constraints must

be tested

--- x --

- y


Where isheight of land?3. where to locate

sample --- x --

- y

Continuousproblem spaces - NLP

Non-Linear Programming problems

Typical problems are functions of multiple variables over domains of each

Maximize f(x1,x2,x3,…,xn)

for x1 D∈ 1, x2 D∈ 2, x3 D∈ 3,…, xn D∈ n

NLP problems are NP complete*Linear Programming problems are polynomial

solvable with provable optimum in O(nk)

Linear Programming*-Simplex Method

Fitness function and constraints all linearfitness:

example constraints:

*assuming you are familiar with approximation methods, p.69-76

€

f (X) = a0x0 + a1x

1 + a2x2 + ...+ anx

n

€

x i ≥ 0;0 ≤ i ≤ n

b0x0 + b1x1 + b2x2 + ...+ bnxn ≤ b

c0x0 + c1x1 + c2x2 + ...+ cnxn ≥ c

d0x0 + d1x1 + d2x2 + ...+ dnxn = d

Linear Programming

with these restrictions:feasible solution spaceis a convex polyhedron-> solution is at a vertex2D e.g.:

€

f (X) = a0x0 + a1x

1 + a2x2 + ...+ anx

n

€

x i ≥ 0;0 ≤ i ≤ n

b0x0 + b1x1 + b2x2 + ...+ bnxn ≤ b

c0x0 + c1x1 + c2x2 + ...+ cnxn ≥ c

d0x0 + d1x1 + d2x2 + ...+ dnxn = d

Simplex Method

starts at a vertex (usually xi = 0 ∀i)

and “walks along edges” to adjacentvertex, until optimum is reachedapprox O(n4)

Local Search in Space of Complete Solutions

Complete or Partial Solution Space?

...

..T ..F

.FT.TT .TF .FF


2

1

0

...

space of partial

solutions


space of complete solutions

Partial Solution Space

...

..T ..F

.FT.TT .TF .FF


2

1

0

...

space of partial

solutions

• algorithms are based on tree traversals• must have a partial fitness function• performance is improved by tree ‘pruning’• coming soon to a lecture near you...

Complete Solution Space

...

..T ..F

.FT.TT .TF .FF


...

space of complete solutions

• start with a (random or guided) solution• move to other solutions seeking better fitness• performance is improved by concentrating on

good solutions

Local Search

-ideal solution for convex functions

-partial solution for complex functions

(finds local optimum)

Local search algorithm

localSearch(point P in domain)evaluate fP = fitness(P)

repeat until no improvement in fitness transform P to a neighbouring point P’ evaluate fP’

if fP’ better than fP, P=P’

return P

1 DimensionalConvex space Example

Marking schemes

-25

0

25

50

75

100

0 6 12 18 24 30 36

attendance

grade

Marking schemes

-25

0

25

50

75

100

0 6 12 18 24 30 36

attendance

grade

localSearch(point h=19)evaluate fh = 20(19)-192=19

repeat until no improvement in fitness transform h to a neighbouring point h’ evaluate fh’

if fh’ better than fh, h=h’

return h = 10

m = 20h - h2m = 20h - h2

in this domain, neighbouring points of h are (h+1), (h-1)

h fh

18 3619 1917 5116 6415 7514 8413 9112 9411 9910 1009 99

2 Dimensionalspaces

4 local search neighbours

--- x --

- y

Would 8 neighbours be better?

Would 8 neighbours be better?

Local search with SAT: representation

Any logical expression can be written inConjunctive Normal Form (CNF) a conjunction (/\) of disjunctions (\/) of

propositions and negations:(~P1 \/ P3) /\ (P2 \/ ~P3) /\ (P4 \/ P1)

(A \/ B) -> C= ~(A \/ B)\/ C= (~A /\ ~B) \/ C= (~A \/ C) /\ (~B \/ C)

Local searchwith SAT

What is a “better”point in search space?

What is a “neighbouring” point?

localSearch(point P in domain)evaluate fP = fitness(P)repeat until no better fitness transform P to neighbour pt P’ evaluate fP’ if fP’ better than fP , P=P’return P


P1 P2 P3 P4 (~P1 \/ P3) (P2 \/ ~P3) (P4 \/ ~P1) fitness

T F T F T F F F

FITNESS: number of

TRUE disjuncts

FITNESS: number of

TRUE disjuncts

11

neighbour has one proposition with different

boolean value

neighbour has one proposition with different

boolean value

Local searchwith SAT

P1 P2 P3 P4 (~P1 \/ P3) (P2 \/ ~P3) (P4 \/ ~P1) fitness

T F T F T F F 1



P1 P2 P3 P4 fitness

F F T F 2

P1 P2 P3 P4 fitness

T T T F 2

P1 P2 P3 P4 fitness

T F F F 1

P1 P2 P3 P4 fitness

T F T T 2

Local search with SATbut SAT space is not convex…

procedure GSAT // Michalewicz & Fogelbegin

for i = 1 to MAX_TRIES doT random truth assignmentfor j = 1 to MAX_FLIPS do

if T satisfies fitness return (T)flip a proposition truth value

endendreturn (“no solution found”)

end

Local searchwith TSP

what is a “better”point in search space?

EASY – shorter path

what is a “neighbouring” point?i.e., what change to transform a path to a similar path?

MANY possible definitions of transformation operators.



TSP and local search some neighbourhood definitions depend

on the type of TSP.e.g., 2-Opt:change two non-adjacent edges:ABCDEFG to ABEDCFG

pathlength =pathlength –BC –EF + BE + CFfails when?TSP with directed edges: DE ≠ ED, CD ≠ DC

AB

C

D

E

FG

Neighbourhood/Search tradeoff

TSP neighbourhood definitions

swap adjacent cities O(n)

2-Opt: change two non-adjacent edges O(n2)

3-Opt: change among three non adjacent edges

δ-path (Lin-Kernighan)

AB

C

D

E

FG

AB

C

D

E

3-Opt O(n3)

THREE possible (2-Opt) neighbours:

FOUR possible neighbours:

B C D E F G H J KA

B

E D C H G F

J KAF G H

H G F

C D E

E D C

B

C D E H G F

J KAE D C

H G F

F G H

E D C

δ-path (Lin-Kernighan)

based on 2-Opt with 1 city (A) fixed.

remove KA and consider all δ-paths

(e.g., by adding KE)- record best 2-Opt tour if any (e.g. EF FA)- BUT move to next tour according to best

δ-path length (e.g. including FA)

B C D E F G H J KA A

Lin-Kernighan local search

initial tour T, BestT = Tstart city A, last city L=Kadded_list =null; // added edges cannot be removeddeleted_list=null; // deleted edges can’t be re-addedrepeat

find best 2-Opt neighbour tour (maybe change BestT)move to tour with best δ-path < BestT if anyput added edge in added_list (e.g., KE)put deleted edge in deleted_list(e.g., EF)L = F

until no moves (all edges added or no short δ-path)

B C D E F G H J KA A

Variations on local searchp, point in search space, fitness f(p)neighbourhood Np = {p1,…,pk}

(1) find first pi with f(pi)> f(p)

or(2) find best pj such that f(pj) ≥ f(pi) 1≤ i ≤ k

… pk

p2

p1 f(p)

(2)

(1)

exhaustive search and a first approach to avoiding it

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