Recursion: Backtracking Dr. Jicheng Fu Department of Computer Science University of Central Oklahoma

Objectives (Chapter 5) Understand backtracking algorithms and use them to solve problems Use recursive functions to implement backtracking algorithms See how the choice of data structures can affect the efficiency of a program

The Eight-Queens Puzzle How to place eight queens on a chess board so that no queen can take another Remember that in chess, a queen can take another piece that lies on the same row, the same column or the same diagonal (either direction) There seems to be no analytical solutions

Solutions do exists Requires luck coupled with trial and error Or much exhaustive computation

Solve the Puzzle How will you solve the problem? Put the queens on the board one by one in some systematic/logical order Make sure that the queen placed will not be taken by another already on the board If you are lucky to put eight queens on the board, one solution is found Otherwise, some queens need to be removed and placed elsewhere to continue the search for a solution

Why put the queens in a systematic order? Make sure that you will not consider the same situation again The whole process is suitable for recursive programming A large problem can be divided into small subproblems with a similar nature

An Outline of the Recursive Function Recursive function: solve_from Given a configuration of queens on the chessboard, search for all solutions Class Queens Represent a particular configuration of queens on the chessboard Code sketch solve_from (Queens configuration) { if configuration already contains eight queens Print configuration else for every square p that is unguarded by configuration { add a queen on square p to configuration; solve_from(configuration); remove the queen from square p of configuration; } Stop Condition Emphasize the all back tracking algorithms use this exact format. If a student uses something else, it won’t be correct. Emphasize the importance of the for loop. The essence of back-tracking is to first figure out the steps needed to build the partial solution towards the final solution. Secondly, at each step, figure out all the options to try one at a time. It is a brute force approach. Sub-problems. Need to try all possibilities

How to find the next square to try? There must be a queen, exactly one queen in each row There can never be more than one queen in each row Therefore, we can place the queens one row at a time in order What we have decided is actually a systematic way to solve the problem

Example: Four Queens X: Guarded squares ?: Other squares that have not been tried

Backtracking Backtracking algorithms Search for a solution by constructing partial solutions that remain consistent with the requirements of the problem, and then extending a partial solution toward completion When inconsistency occurs, the algorithms backs up (backtracks) by removing the most recently constructed part of the solution, and then trying another possibility

Suitable for implementations using Recursion, or Stacks Only the most recent part is used Useful for situations where many possibilities may first appear, but few survive further tests Scheduling problems Arranging sports tournaments Designing a compiler Parsing

Main program int main( ) /* Pre: The user enters a valid board size. Post: All solutions to the n-queens puzzle for the selected board size are printed. Uses: The class Queens and the recursive function solve_from. */ { int board_size; print_information( ); cout << "What is the size of the board? " << flush; cin >> board_size; if (board size < 0 || board size > max_board) cout << "The number must be between 0 and “ << max_board << endl; else { Queens configuration(board_size); // Initialize empty configuration. solve_from(configuration); // Find all solutions extending configuration. }

The configuration: Queens class Constructor set the user-selected board size and initialize the empty Queens object print print the solutions unguarded bool Queens :: unguarded(int col) const; Post: Returns true or false according as the square in the first unoccupied row (row count) and column col is not guarded by any queen. Why no row number?

insert remove is_solved void Queens :: insert(int col); Pre: The square in the first unoccupied row (row count) and column col is not guarded by any queen. Post: A queen has been inserted into the square at row count and column col; row count has been incremented by 1. remove void Queens :: remove(int col); Pre: There is a queen in the square in row count - 1 and column col. Post: The above queen has been removed; count has been decremented by 1. is_solved bool Queens :: is_solved( ) const; Post: The function returns true if the number of queens already placed equals board size; otherwise, it returns false.

The backtracking function: solve_from void solve_from(Queens &configuration) /* Pre: The Queens configuration represents a partially completed arrangement of non-attacking queens on a chessboard. Post: All n-queens solutions that extend the given configuration are printed. The configuration is restored to its initial state. Uses: The class Queens and the function solve_from , recursively. */ { if (configuration.is_solved( )) configuration.print( ); else for (int col = 0; col < configuration.board_size; col++) if (configuration.unguarded(col)) { configuration.insert(col); solve_from(configuration); // Recursively continue to add queens. configuration.remove(col); } See the advantage of using OOP rather than procedure-based.

Implementation of the Queens Class Store the chessboard as a 2-dimensional array with entries indicating the locations of the queens Queens class const int max_board = 30; class Queens { public: Queens(int size); bool is_solved( ) const; void print( ) const; bool unguarded(int col) const; void col); void reminsert(int ove(int col); int board_size; // dimension of board = maximum number of queens private: int count; // current number of queens = first unoccupied row bool queen_square[max_board][max_board]; };

is_solved, remove and print are also trivial Constructor Queens :: Queens(int size) /* Post: The Queens object is set up as an empty configuration on a chessboard with size squares in each row and column. */ { board_size = size; count = 0; for (int row = 0; row < board_size; row++) for (int col = 0; col < board_size; col++) queen_square[row][col] = false; } insert void Queens :: insert(int col) /* Pre: The square in the first unoccupied row (row count ) and column col is not guarded by any queen. Post: A queen has been inserted into the square at row count and column col ; count has been incremented by 1. */ { queen_square[count++][col] = true; } is_solved, remove and print are also trivial A part of the big 2-d Boolean array is used.

unguarded bool Queens :: unguarded(int col) const /* Post: Returns true or false according as the square in the first unoccupied row (row count ) and column col is not guarded by any queen. */ { int i; bool ok = true; // turns false if there is a queen in column or diagonal for (i = 0; ok && i < count; i++) // Check upper part of column ok = !queen_square[i][col]; for (i = 1; ok && count - i >= 0 && col - i >= 0; i++) // Check upper-left diagonal ok = !queen_square[count - i][col - i]; for (i = 1; ok && count - i >= 0 && col + i < board size; i++) // Check upper-right diagonal ok = !queen_square[count - i][col + i]; return ok; }

Exercise Describe a rectangular maze by indicating its paths and walls within an array. Write a backtracking program to find a way through the maze You are a tournament director and need to arrange a round robin tournament among N = 2k players. In this tournament, everyone plays exactly one game each day; after N - 1 days, a match occurred between every pair of players. Write a backtracking program to do this.

Review and Refinement The time increases rapidly with the board size First refinement Use the 2-dimensional array to keep track of all the squares that are guarded by queens For each square, keep a count of the number of queens guarding the square Faster, but we still need the loops to update the guard counts for each square

Second refinement Objective Key idea Eliminate all loops Key idea Each row, column and diagonal can contain at most one queen Keep track of unguarded squares by using three bool arrays: col_free, upward_free and downward_free Diagonals from the lower left to the upper right are called upward diagonals Diagonals from the upper left to the lower right are called downward diagonals An integer array is used to record the column number for the queens in each row

Eliminate loops It is trivial to identify each column How to identify each diagonal For upward diagonals, the row and column indices have a constant sum The sum ranges from 0 to 2  board_size – 2 The sum can be used to identify each upward diagonals The square in row i and column j is in upward diagonal number i + j For downward diagonals, the difference of the row and column indices is constant The difference ranges from -board_size+1 to board_size–1 The downward diagonal can be numbered using the difference The square in row i and column j is in downward diagonal number i - j + board_size - 1

Refined Implementation Revised Queens class class Queens { public: Queens(int size); bool is_solved( ) const; void print( ) const; bool unguarded(int col) const; void insert(int col); void remove(int col); int board_size; private: int count; bool col_free[max_board]; bool upward_free[2 * max_board - 1]; bool downward_free[2 * max_board - 1]; // column number of queen in each row int queen_in_row[max_board]; };

Constructor Queens :: Queens(int size) /* Post: The Queens object is set up as an empty configuration on a chessboard with size squares in each row and column. */ { board_size = size; count = 0; for (int i = 0; i < board_size; i++) col_free[i] = true; for (int j = 0; j < (2 * board size - 1); j++) upward_free[j] = true; for (int k = 0; k < (2 * board size - 1); k++) downward_free[k] = true; }

Insertion void Queens :: insert(int col) /* Pre: The square in the first unoccupied row (row count ) and column col is not guarded by any queen. Post: A queen has been inserted into the square at row count and column col; count has been incremented by 1. */ { queen_in_row[count] = col; col_free[col] = false; upward_free[count + col] = false; downward_free[count - col + board_size - 1] = false; count++; }

Unguarded bool Queens :: unguarded(int col) const /* Post: Returns true or false according as the square in the first unoccupied row (row count ) and column col is not guarded by any queen. */ { return col_free[col] && upward_free[count + col] && downward_free[count - col + board_size - 1]; }

Evaluation New Implementation 2-D array Implementation

Analysis of Backtracking Effectiveness of Backtracking Naïve approach Consider all configurations: put all 8 queens on the board and reject illegal configurations There can be only one queen in each row 88 = 16,777,216 There can be only one queen in each column 8! = 40,320 Our program is even better, since it rejects squares in guarded diagonals

Part of the recursion tree for the eight-queens problem

Lower Bounds Number of solutions For n-queens problem, the amount of work done by backtracking problem still grows very fast To place a queen in each of the first n/4 rows, backtracking algorithm investigates a minimum of n(n - 3)(n - 6) … (n – 3 * n/4) positions n (n - 3) (n - 6) … (n – 3 * n / 4) > ( n / 4) (n/4) Exponential growth Number of solutions Not bounded by any polynomial in n Even not bounded by any exponential form kn, where k is a constant It is proved to be an unsolved problem