Ch 8, Backtracking

8/2/2019 Ch 8, Backtracking

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17/05/1433Prepared by Dr. Zakir H.Ahmed

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Analysis of Algorithms

Code: CS-436

Chapter - 08Backtracking

Dr. Zakir H. Ahmed

Phone No. 2582172


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This Chapter Contains the followingTopics:


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Introduction

Suppose you have to make a series of

decisions, among various choices, where You dont have enough information to know

what to choose

Each decision leads to a new set ofchoices

Some sequence of choices (possibly morethan one) may be a solution to yourproblem

Backtracking is a methodical way of trying outvarious sequences of decisions, until you findone that works


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Solving a maze

Given a maze, find a path from start to finish

At each intersection, you have to decide betweenthree or fewer choices:

Go straight

Go left

Go right

You dont have enough information to choosecorrectly

Each choice leads to another set of choices

One or more sequences of choices may (or maynot) lead to a solution

Many types of maze problem can be solved with

backtracking


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Coloring a map

You wish to color a map with not more than four

colors red, yellow, green, blue

Adjacent countries must be indifferent colors

You dont have enough information to choosecolors



Many coloring problems can be solved withbacktracking


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Solving a puzzle

In this puzzle, all holes but one are filled withwhite pegs

You can jump over one peg with another

Jumped pegs are removed The object is to remove all but the last peg

You dont have enough information to jumpcorrectly



Many kinds of puzzle can be solved withbacktracking


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Backtracking (animation)

start ?

?

dead end

dead end

??

dead end

dead en

?

success!

dead end


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N-Queens Problem

Place n-queens on an nnboard so that no pairof queens attacks eachother, that is, so that notwo of them are on thesame row, column, or

diagonal.

Let us first consider n=4.

Then Variables:

x1, x2 , x3 , x4

Domains:

{1, 2, 3, 4}

Constraints:

xi x

j and

| xi - xj | | i-j|.4321

x1

x2

x3

x4


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Permutation tree for 4-queens problem

x1

x

2

x3

x4

1 2 3 4

(1,2,3,4) (4,3,2,1)(1,2,4,3) (4,3,1,2)

2 3 4

3 4

4 3

21 3

1

21

2


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Backtrack Solution

1 1. . . 2

x1

x2

x3

x4

1

2B

3 4

B B

1. . 2

. . . .


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Backtrack Solution

x1

x2

x3

x4

1 2

(2,4,1,3)

2B

3 4

B

B

4

1

3

B

2

B

B

B

12

. 3

. . . .

1. . . 2

3

. . 4

1. . . 2


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Backtracking Process

Let (x1, x2, , xi) be a path from the root to a

node in a state space tree. Let T(x1, x2, , xi) be the set of all possible

values for xi+1 such that (x1, x2, , xi+1) also apath to a problem state.

T(x1, x2, , xi) = 0.

We assume the existence of bounding functionBi+1 (expressed as predicates) such that ifBi+1(x1, x2, , xi+1) is false for a path (x1, x2, ,xi+1) from the root not to a problem state, thenthe path cant be extended to reach an answernode.

Thus candidates for position i+1 of the solutionvector (x1, x2, , xn) are those values whichare generated by T and satisfy Bi+1.

The algorithm given in next slide presents arecursive formulation of the backtracking

technique.


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Backtracking Algorithm

Following describes the backtracking process

using recursion. On entering, the first k-1 values x[1], x[2], ,

x[k-1] of solution vector x[1:n] have beenassigned. X[ ] and n are global.

Algorithm Backtrack (k){ for (each x[k] T(x[1], x[2], , x[k-1])) do

{ if (Bk(x[1], x[2], , x[k]) 0) then

{ if (x[1], x[2], , x[k] is a path to an

answer node) then

write (x[1:k]);

if (k


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N-Queens Problem

We observed from the 4-queens problem that we

can let (x1, , xn) represent a solution in which xi isthe column of i-th row where the i-th queen isplaced.

Suppose two queens are placed at positions (i,j)and (k,l).

They are in the same diagonal only if

i-j = k-l or i+j = k+l

=>j-l = i-k =>j-l = k-i

Therefore two queens lie on the same diagonal ifand only if |j-l| = |i-k|.

Algorithm Place(k,i) returns a Boolean value that isTRUE if the k-th queen is placed in column i.

It tests both whether i is distinct from all previousvalues x[1], ., x[k-1] and whether there is noother queen on the same diagonal.

Its computing time is O(k-1). Using Algorithm Place(), we can refine the general

backtrack method given in previous slide, and givea precise solution to the n-queens problem.


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An Algorithm

Algorithm Place(k, i)

{ for j:=1 to k-1 doif ((x[j] = i) or (Abs(x[j] - i) = Abs(j k))) then

return FALSE;

return TRUE;

}

Algorithm NQueens(k, n)

{ for i:=1 to n do

if (Place(k,i)) then{ x[k] := i;

if (k = n) then

write (x[1: n]);

else

NQueens(k+1, n);}

}


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Chapter-08

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Ch 8, Backtracking

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