backtracking algorithms of ada

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BACKTRACKINGGENERAL METHOD• Problems searching for a set of solutions or which

require an optimal solution can be solved using the backtracking method .

• To apply the backtrack method, the solution must be expressible as an n-tuple(x1,…,xn), where the xi are chosen from some finite set si

• The solution vector must satisfy the criterion function P(x1 , ….. , xn).

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BACKTRACKING (Contd..)

• Suppose there are m n-tuples which are possible candidates for satisfying the function P.

• Then m= m1, m2…..mn where mi is size of set si 1<=i<=n.

• The brute force approach would be to form all of these n-tuples and evaluate each one with P, saving the optimum.




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• The backtracking algorithm has the ability to yield the same answer with far fewer than m-trials.

• In backtracking, the solution is built one component at a time.

• Modified criterion functions Pi (x1...xn) called bounding functions are used to test whether the partial vector (x1,x2,......,xi) can lead to an optimal solution.

• If (x1,...xi) is not leading to a solution, mi+1,....,mnpossible test vectors may be ignored.




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• The constraints may be of two categories.• EXPLICIT CONSTRAINTS are rules which restrict the

values of xi. Examples xi 0 or x1= 0 or 1 or li xi ui.

• All tuples that satisfy the explicit constraints define a possible solution space for I.

• IMPLICIT CONSTRAINTS describe the way in which the xi must relate to each other .

• Implicit constraints allow to find those tuples in the solution space that satisfy the criterion function.




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Example : 8 queens problem

• The problem is to place eight queens on an 8 x 8 chess board so that no two queens attack i.e. no two of them are on the same row, column or diagonal.

• Strategy : The rows and columns are numbered through 1 to 8.

• The queens are also numbered through 1 to 8. • Since each queen is to be on a different row

without loss of generality, we assume queen i is to be placed on row i .




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8 queens problem (Contd..)

• The solution is an 8 tuple (x1,x2,.....,x8) where xi is the column on which queen i is placed.

• The explicit constraints are :Si = {1,2,3,4,5,6,7,8} 1 i n or 1 xi 8 i = 1,.........8

• The solution space consists of 88 8- tuples.




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The implicit constraints are :(i) no two xis can be the same that is, all queens

must be on different columns. (ii) no two queens can be on the same diagonal. (i) reduces the size of solution space from 88 to 8!

8 – tuples.Two solutions are (4,6,8,2,7,1,3,5) and (3,8,4,7,1,6,2,5)




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Q8Q7

Q6Q5

Q4Q3

Q2Q1

87654321




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State space tree representationSolution Space : • Tuples that satisfy the explicit constraints define a solution

space. • The solution space can be organized into a tree. • Each node in the tree defines a problem state. • All paths from the root to other nodes define the state-

space of the problem.• Solution states are those states leading to a tuple in the

solution space. • Answer nodes are those solution states leading to an

answer-tuple( i.e. tuples which satisfy implicit constraints).




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State space tree representation contd..

• The problem may be solved by systematically generating the problem states determining which are solution states, and determining the answer states.

• Let us see the following terminology• LIVE NODE A node which has been generated

and all of whose children are not yet been generated .

• E-NODE (Node being expanded) - The live node whose children are currently being generated .




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• DEAD NODE - A node that is either not to be expanded further, or for which all of its children have been generated.

• DEPTH FIRST NODE GENERATION- In this, as soon as a new child C of the current E-node R is generated, C will become the new E-node. R will become E-node again when C has been fully explored.




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• BOUNDING FUNCTION - will be used to kill live nodes without generating all their children.

• BACTRACKING-is depth – first node generation with bounding functions.

• BRANCH-and-BOUND is a method in which E-node remains E-node until it is dead.




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• BREADTH-FIRST-SEARCH : Branch-and Bound with each new node placed in a queue .The front of the queen becomes the new E-node.

• DEPTH-SEARCH (D-Search) : New nodes are placed in to a stack. The last node added is the first to be explored.




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Example : 4 Queens problem

1. . 2

12

12

3. . . .

1

1 12

3. . 4




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State space tree: 4 Queens problem

1x1 = 1 x1=2

2 18x2=2 3 4 x2=1 x2=3 x2 = 4B 3 8 13 19 24 29

x3=3 x3=4 2 3 B B4 6 14 16 x3 = 1

x4=4 3 B 305 7 15 x4 = 3

B 31




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State space tree: 4 Queens problem contd..

• If (x1….xi) is the path to the current E-node , a bounding function has the criterion that (x1..xi+1) represents a chessboard configuration, in which no queens are attacking.

• A node that gets killed as a result of the bounding function has a B under it.




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State space tree: 4 Queens problem contd..

• We start with root node as the only live node. The path is ( ); we generate a child node 2.

• The path is (1).This corresponds to placing queen 1 on column 1 .

• Node 2 becomes the E node. Node 3 is generated and immediately killed. (because x1=1,x2=2).

• As node 3 is killed, nodes 4,5,6,7 need not be generated.




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State space tree: 4 Queens problem contd..• Node 8 is generated, and the path is (1,3).• Node 8 gets killed as all its children

represent board configurations that cannot lead to answer. We backtrack to node 2 and generate another child node 13.

• But the path (1,4) cannot lead to answer nodes.

• So , we backtrack to 1 and generate the path (2) with node 18. We observe that the path to answer node is (2 4 1 3 )

• Other solution is (3, 1, 4, 2)




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GENERAL BACKTRACKING METHOD

• All answer nodes are to be found• If (x1…..xi) is a path from root to a node then T

(x1,….,xi) be the set of all possible values for Xi+1, such that (x1,x2,…….,xi,xi+1) is also a path from root to a problem state.

• B(x1…xi+1) or Bxi+1is false for the path (x1,..,xi+1) if the path cannot reach an answer node.

• The solution vectors X (1: n) are those values which are generated by T and satisfy Bi+1.




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GENERAL BACKTRACKING METHOD (Contd..)

Procedure backtrack (n) // solution vectors are X(1:n)//// and printed as soon as generated //

// T {X(1),….X(k-1)} gives all possible values of X(k)// given that X(1),…..,X(k-1) have already been chosen // // The predicates Bk (X(1),….X(k) ) determine those

elements //// which satisfy the implicit constraints //// Implicit constraints are “no two X is can be the same” // Integer k, n; local X (1:n)




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GENERAL BACKTRACKING METHOD (Contd..)K 1while K <> 0 doif there remained an untried X(k) such that X(k) T ( X (1) , …..X(k-1) ) and Bk (X (1) ,…,X(k) ) = true then

if (X (1),…, X(k) ) is a path to an answer node then print ( X (1) ,…X (k) ) // and proceed for another//

//solution with untried value of X(k)//K K+1 // consider next set//

endif // end of if there remained ….//

else K K-1 // backtrack to previous component as no value of X(k) //satisfies the constraints//

endifrepeatend Backtrack




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EXAMPLE OF ALGORITHM WITH 4 QUEENS PROBLEMS

k 1 loop 1 X0 =1,1 {1,2,3,4} and B1 (X (1)) = true 1 is not a path to answer node K 1+1=2 repeat K 2loop 2

X(2){2,3,4}and B2 (1,2) is False but there remains untried X(k)=3 and X(K) =4




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EXAMPLE OF ALGORITHM WITH 4 QUEENS PROBLEMS (Contd..)

B2 (1,3)=true , but (1,3) Is not a path toanswer node ; so K K+1=3 There is no X(3) such that B3(1,3,2) or B3 (1,3,4) is true . So backtrack K K-1=2consider X (2) 4




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EXAMPLE OF ALGORITHM WITH 4 QUEENS PROBLEMS (Contd..)

(1,4) is not a path to answer (1,4,2) is not a path to answer B4(1,4,2,3)=false. Thus X(2)=4 is falseWith X(1)=1, we have seen that there is no X(2)satisfying the constraints so KK-1=2-1=1There remained untried X(k)=2,3,4.Repeating with X(1)=2, we observe (2,4,1, 3) is a path to an

answer node . Similarly the other solution is (3,1,4,2)




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BACKTRACKING ALGORITHM WITH RECURSION

Procedure RBACKTRACK (k)// on entering the first K-1 values X(1),…..,X(k-1)// // of the solution vector X(1:n) have been assigned //global n , X(1:n) for each X(k) such that X(k) T ( X (1),..X(k-1) ) Doif Bk (X(1)..,X(k-1), X(k) )= true then {if ( X (1) ,….,X(k) ) is a path to an answer node then print ( X(1),……,X(k) ) endif

if (k<n)CALL RBACKTRACK (k+1) endif }// end of if Bk (X(1)…//repeat // end of for each X(k)…//end RBACKTRACK




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BACKTRACKING ALGORITHM WITH RECURSION (Contd..)

EXAMPLE (RB – RBACKTRACK)Initially RB(1) is called for each X(1) X(1) {1,2,3,4} and B1 (1) is true , but 1 is not an answer node.RB (2) is called X(2) {2,3,4} and B2 (1,3) = true and B2 (1,4) = true(1,3) is not an answer node , RB(3) is called,X(3) = 2,




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BACKTRACKING ALGORITHM WITH RECURSION (Contd..)

but (1,3,2) is not Answer node , so , RB(4) is called B4 (1,3,2,4) = false. (1,3,4,2) is not bounded so , X(2) = 4 is tried (1,4,2,3) is not bounded . With X(1)=1 no solution Now for X(1) = 2,3,4 repeat (2,4,1,3) is an answer node, other paths with X(1) = 2 are not leading to answer node.With X (1) = 3 , (3,1,4,2) is an answer node.X (1) = 4 - no solution.




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EFFICIENCY OF BACKTRACKING (BT) ALGORITHM

• The time required by a backtracking algorithm or the efficiency depends on four factors(i) The time to generate the next X(k); (ii) The number of X(k) satisfying the explicit constraints (iii) The time for bounding functions Bi

(iv) The number of X(k) satisfying the Bi for all i.




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EFFICIENCY OF BACKTRACKING ALGORITHM (Contd..)

• The first three are relatively independent of the problem instance being solved.

• The complexity for the first three is of polynomial complexity .

• If the number of nodes generated is 2n , then the worst case complexity for a backtracking algorithm is O(P(n)2n) where P(n) is a polynomial in n .




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Estimation Of Nodes generated in a BT Algorithm

• Generate a random path in the state space tree.• Let X be a node at level i on this path. • Let mi be the children of X ( at level i+1 ) that do

not get bounded. (i.e. mi are the nodes which can be considered for getting an answer node ).

• Choose randomly one of the mi.• Continue this until this node is either is a leaf or

all its children are bounded.




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Estimation of Nodes generated in a BT Algorithm (Contd..)

• Let m be the no. of unbounded nodes to be generated.

• Let us assume that the bounding functions are static, i.e., the BT algorithm does not change its bounding functions.

• The number of estimated number of unbounded nodes

=1+m1+m1m2+….. +m1m2m3..mi where mi is the estimated no. of nodes at level i+ 1.




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Estimation of Nodes generated in a BT Algorithm (Contd..)

• The number of unbounded nodes on level one is 1.

• The number of unbounded nodes on level 2 is m1

• The total no. of nodes generated till level 2 is 1 + m1

The total number of nodes generated till level i+1 is 1+m1+…+m1…mi .

– The above procedure can be written as an algorithm.




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Estimation of Nodes generated in a BTAlgorithm (Contd..)Procedure Estimate// This procedure follows a random path and estimates//// number of unbounded nodes //m1; r1; k1 loop

Tk {X (k): X(k) T ( X(1) ,….X(k-1)) and Bk (X (1), …., X(k)) is TRUE}If size (Tk) = 0 then exit endif // SIZE returns the // r r * SIZE (Tk) // size of the set Tk // m m+r

X (k) CHOOSE (Tk) // CHOOSE makes a random choice of an element in Tk //

k k+1 repeatreturn (m)

end estimate




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• In implementing the n – queens problem we imagine the chessboard as a two-dimensional array A (1 : n, 1 : n).

• The condition to test whether two queens, at positions (i, j) and (k, l) are on the same row or column is simply to check i = k or j = l

The n-queens problem and solution




backtracking algorithms of ada

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