ICS 353: Design and Analysis of Algorithms King Fahd University of Petroleum & Minerals Information & Computer Science Department ICS 353: Design and Analysis of Algorithms Backtracking

Reading Assignment M. Alsuwaiyel, Introduction to Algorithms: Design Techniques and Analysis, World Scientific Publishing Co., Inc. 1999. Chapter 13, Sections 1, 2 and 4.

Coping with Hard Problems There are three useful methodologies that could be used to cope with problems having no efficient algorithm to solve: A methodic examination of the implicit state space induced by the problem instance under study. suitable for problems that exhibit good average time complexity. A probabilistic notion of accuracy where a solution is a simple decision maker or test that can accurately perform one task (either passing or failing the alternative) and not say much about the complementary option. An iteration through this test will enable the construction of the solution or the increase in the confidence level in the solution to the desired degree. Obtain an approximate solution Only some classes of hard problems admit such polynomial time approximations. Backtracking belongs to the first category.

A systematic technique of searching Backtracking A systematic technique of searching To reduce the search space Can be considered as an “organized” exhaustive search

3-Coloring Problem Optimization Problem Input: G = (V, E), an undirected graph with n vertices and m edges. Output: A 3-coloring of G, if possible. A coloring can be represented as an n-tuple (c1, c2, ..., cn) The number of different possible colorings for a graph is ........... These possibilities can be represented as a complete ternary tree

3-Coloring: Search Tree Search tree of all possible colorings of a graph with 4 vertices (When could this be for a graph with 5 vertices?)

Example a b c d e

Example In the example Nodes are generated in a depth-first search manner No need to store the whole search tree, just the current active path What is the time complexity of the algorithm in the worst case

Algorithm 3-ColorRec

Algorithm 3-ColorIter 3

The General Backtracking Method Assume that the solution is of the form (x1, x2, ..., xi) where 0  i  n and n is a constant that depends on the problem formulation. Here, the solution is assumed to satisfy certain constraints. i in the case of the 3-coloring problem is fixed. i may vary from one solution to another.

The General Backtracking Method: Example Consider the following version of the Partition Problem Input: X = {x1, x2, ..., xn} a set of n integers, and an integer y. Output: Find a subset Y  X such that the sum of its elements is equal to y. For example, consider X = {10, 20, 30, 40, 50, 60} and y = 60. There is more than one solution to this problem: What are they and what is their length?

The General Backtracking Algorithm (1) Each xi in the solution vector belongs to a finite linearly ordered set Xi. The backtracking algorithm considers the elements of the Cartesian product X1 X2  ...  Xn in lexicographic order. Initially starting with the empty vector. 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: 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. (The advance Step): If v represents a partial solution, the algorithm advances by choosing the least element in the set Xj+2.

The General Backtracking Algorithm (2) If v is neither a final nor a partial solution, we have two sub-cases: 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. (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.

BacktrackRec

BacktrackIter