1. Chapter 13 Backtracking Introduction The 3-coloring problem
The 8-queens problem
The general backtracking method
Branch and bound

2. 13.1 Introduction
In many real world problems, as in most of the NP-hard problems, a solution can be obtained by exhaustively searching through a large but finite number of possibilities.
The need arose for developing systematic techniques of searching, with the hope of cutting down the search space to possibly a much smaller space.
Backtracking is an organized exhaustive search which often avoids searching all possibilities.

3. 13.2 The 3-coloring problem
Problem: Given an undirected graph G=(V,E), it is required to color each vertex in V with one of three colors, say 1,2 and 3, such that no two adjacent vertices have the same color. We call such a coloring legal; otherwise, if two adjacent vertices have the same color, it is illegal.
There are 3n possible colorings to color a graph with n vertices. The set of all possible colorings can be represented by a complete ternary tree called the search tree.
An incomplete coloring of a graph is called partial if no two adjacent colored vertices have the same color.

4. 13.2 The 3-Coloring Problem
A coloring can be represented by an n-tuple ( c1,c2,…,cn) such that
ci∈{1,2,3},1≤i≤n.

5. 13.2 The 3-Coloring Problem
Fig The search three for all possible 3-colorings
for a graph with 3 vertices.

6. 13.2 The 3-Coloring Problem Example 13.1
Consider the graph shown in Fig. 13.2(a), where we are interested in coloring its vertices using the colors {1,2,3}. Figure 13.2(b) shows part of the search tree generated during the process of searching for a legal coloring.

7. 13.2 The 3-Coloring Problem
Fig An Example of using backtracking to
solve the problem 3-COLORING.

8. Input: An undirected graph G=(V,E)
13.2
Algorithm COLORREC
Input: An undirected graph G=(V,E)
Output: A 3-coloring c[1..n] of the vertices of G, where each c[j] is 1,2 or 3.
1. for k←1 to n
c[k] ←0
3. end for
4. flag←false
5. graphcolor(1)
6. If flag then output c
7. else output “no solution”

9. Procedure graphcolor(k)
1. for color=1 to 3
c[k] ←color
If c is a legal coloring then set flag←true and exit
else if c is partial then graphcolor(k+1)
5. end for

10. Input: An undirected graph G=(V,E).
13.2
Algorithm COLORITER
Input: An undirected graph G=(V,E).
Output: A 3-coloring c[1..n] of the vertices of G, where each c[j] is 1,2 or 3.
1. for k←1 to n
c[k] ←0
3. end for
4. flag←false
5. k←1

11. 6. while k≥1
while c[k]≤2
c[k]←c[k]+1
if c is a legal coloring then set flag←true and
exit from the two while loops.
else if c is partial then k←k+1 {advance}
end while
c[k] ←0
k←k-1 {backtrack}
14. end while
15. If flag then output c
16. else output “no solution”

12. 13.2 The 3-coloring problem Time complexity:
We note that O(3n) nodes are generated in the worst case. For each generated node, O(n) work is required to check if the current coloring is legal, partial, or neither. Hence, the overall running time is O(n3n) in the worst case.

13. 13.3 The 8-Queens Problem
Problem: How can we arrange 8 queens on an 88 chessboard so that no two queens can attack each other? Two queens can attack each other if they are in the same row, column or diagonal.
The n-queens problem is defined similarly, where in this case we have n queens and an n n chessboard for an arbitrary value of n>=1.
We will study the 4-queens problem.

14. xi-xj=i-j or xi-xj=j-i
13.3 The 8-Queens Problem
The algorithm:
We used the term legal to mean a placement of n queens that do not attack each other, and the term partial to mean a placement of less than n queens that do not attack each other.
Initially, put n queens in n different rows. Two queens placed at positions xi and xj are in the same column if and only if xi=xj. Two queens are in the same diagonal if and only if
xi-xj=i-j or xi-xj=j-i

15. 13.3 The 8-Queens Problem
Fig Two configurations of the 4-queens problem

16. Algorithm 13.3 4-QUEENS Input: none
Output: A vector x[1..4] corresponding to the solution of the 4-queens problem.
1. for k←1 to 4
x[k] ←0
3. end for
4. flag←false
5. k←1

17. 6. while k≥1
7. while x[k]≤3
x[k] ←x[k]+1
If x is a legal placement then set flag←true and
exit from the two while loops.
else if x is partial then k←k+1 {advance}
end while
x[k] ←0
k←k-1 {backtrack}
14. end while
15. If flag then output x
16. else output “no solution”

18. 13.3 The 8-Queens Problem
Example 13.2 Applying the algorithm produces the solution shown in Fig

19. 13.3 The 8-Queens Problem
Fig An example of using backtracking to
solve the 4-queens problem

20. 13.3 The 8-Queens Problem Time complexity
The brute-force method can be improved to test n! configurations instead of nn.

21. 13.4 The General Backtracking Method
a class of search problems whose solution consists of a vector (x1,x2, ..,xi) satisfying some predefined constraints.
Here:
i is some integer between 0 and n,
n is a constant that dependent on the problem formulation.
each xi in the solution vector belongs to a finite linearly ordered set Xi. Thus, the backtracking algorithm considers the elements of the cartesian product X1X2…Xn in lexicographic order.

22. Initially, the algorithm starts with the empty vector
Initially, the algorithm starts with the empty vector. It then chooses the least element of X1 as x1.
If (x1) is a partial solution, the algorithm proceeds by choosing the least element of X2 as x2.
If(x1,x2) is a partial solution, then he least element of X3 is included;
otherwise x2 is set to the next element in X2.
In general, suppose that the algorithm has detected the partial solution (x1,x2, …xj). It then considers the vector v=(x1,x2,…,xj,xj+1).
We have the following cases:

23. (1) If v represents a final solution to the problem, the algorithm records it as a solution and either terminates in case only one solution is desired or continues to find other solutions.
(2) (The advance step). If v represents a partial solution, the algorithm advances by choosing the least element in the set Xj+2.
(3) if V is neither a final nor a partial solution, we have two subcases:

24. (a) if there are still more elements to choose from in the set Xj+1, the algorithm sets xj+1 to the next member of Xj+1.
(b) (The backtrack step). If there are no more elements to choose from in the set Xj+1, the algorithm backtracks by setting xj to the next member of Xj. If again there are no more elements to choose from in the set Xj, the algorithm backtracks by setting xj-1 to the next member of Xj-1, and so on.

25. 13.4 The general Backtracking Method
Example 13.3
Consider a variant of the PARTITION problem defined as follows. Given a set of n integers X={x1,x2,…,xn} and an integer y, find a subset Y of X whose sum is equal to y.
It is not hard to devise a backtracking algorithm to solve this problem. Note that this problem can be formulated in another way so that the solution is a boolean vector of length n in the obvious way.

26. Algorithm 13.4 BACKTRACKREC
Input: Explicit or implicit description of the sets X1,X2,…,Xn.
Output: A solution vector v=(x1,x2,…,xi),0≤i≤n.
1. v← ()
2. flag←false
3. advance(1)
4. If flag then output v
5. else output “no solution”

27. Procedure advance(k)
1. for each x∈Xk
xk←x; append xk to v
If v is a final solution then set flag←true
and exit
else if v is partial then advance(k+1)
5. end for

28. Algorithm 13.5 BACKTRACKITER
13.4
Algorithm 13.5 BACKTRACKITER
Input: Explicit or implicit description of the sets X1,X2,…,Xn.
Output: A solution vector v=(x1,x2,…,xi), 0<=i<=n
v←()
flag ←false
k ←1

29. while k>=1
while xk is not exhausted
xk ←next element in xk; append xk to v
if v is a final solution then set flag ←true
and exit from the two while loops.
else if v is partial then k ←k+1 {advance}
end while
reset xk so that the next element is the first.
k ←k-1 {backtrack}
if flag then output v
else output “no solution”

30. 13.5 Branch and Bound Similar to backtracking
Concerned with only maximization or minimization of a given function.
A bound is calculated at each node x on the possible value of any solution given by nodes that may later be generated in the subtree rooted at x.
If the bound calculated is worse than a previous bound, the subtree rooted at x is blocked, i.e., none of its children are generated.

31. 13.5 Branch and Bound
E.g. The traveling salesman problem: Given a set of cities and a cost function that is defined on each pair of cities, find a tour of minimum cost. Here a tour is a closed path that visits each city exactly once.
With each partial solution (x1,x2,…,xk), we associate a lower bound y which is interpreted as follows. The cost of any complete tour that visits the cities x1,x2,…,xk in this order must be at least y.

32. 13.5 Branch and Bound
A matrix is the reduction of the original matrix if it reduces the cost matrix so that each row or column contains at least one entry that is equal to 0.
Example see Fig 13.5 and Fig 13.6

33. In general, if the edge included is (ui,v1) and the path from the root contains the two paths u1,u2, … ,ui and v1,v2, … ,vj, then Mvj u1 is set to ∞ , where M is the matrix at the current node.

34. 13.5 Branch and Bound
Fig An instance matrix of the TRAVELING Salesman and its reduction.

40. The solution is:
(1,3) (3,5) (5,4) (4,2) (2,1)