Backtracking Introduction to Backtracking Backtracking is a n

Backtracking Introduction to Backtracking

Backtracking is a n o i technique used to ct n solve problems with a Ju large search space, by systematically trying and eliminating possibilities. A standard example of backtracking would be going through a maze. ◦ At some point in a maze, you might have two options of which direction to go: Portion B Backtracking Portion A

Backtracking One strategy would on i t nc be to try going through Ju Portion A of the maze. At this point in time you know that Portion A will NOT lead you out of the maze, so you then start searching in Portion B Portion A you get stuck before you find your way out, then you "backtrack" to the junction. Portion B If

Backtracking Clearly, at a single junction you could have even more than 2 choices. The backtracking strategy says to try each choice, one after the other, ◦ if you ever get stuck, "backtrack" to the junction and try the next choice. If you try all choices and Ju n o i nct C B A

Backtracking – Eight Queens Problem Find an arrangement of 8 queens on a single chess board such that no two queens are attacking one another. In chess, queens can move all the way down any row, column or diagonal (so long as no pieces are in the way). ◦ Due to the first two restrictions, it's clear that each row and column of the board will have exactly one queen.

Backtracking – Eight Queens Problem The backtracking strategy is as follows: Q Q 1) Place a queen on the first available square in row 1. 2) Move onto the next row, placing a queen on the first available square there (that doesn't conflict with the previously placed queens). Q Q Continue… 3) Continue in this fashion until either: a) b) you have solved the problem, or you get stuck. When you get stuck, remove the queens that got you there, until you get to a row where there is another valid square to try. Animated Example: http: //www. hbmeyer. de/backtrac k/achtdamen/eight. htm#up

Backtracking – Eight Queens Problem When we carry out backtracking, an easy way to visualize what is going on is a tree that shows all the different possibilities that have been tried.

Backtracking – Eight Queens Problem The neat thing about coding up backtracking, is that it can be done recursively, without having to do all the bookkeeping at once. ◦ Instead, the stack or recursive calls does most of the bookkeeping ◦ (ie, keeping track of which queens we've placed, and which combinations we've tried so far, etc. )

perm[] - stores a valid permutation of queens from index 0 to location-1. location – the column we are placing the next queen used. List[] – keeps track of the rows in which the queens have already been placed. void solve. It. Rec(int perm[], int location, struct onesquare used. List[]) { if (location == SIZE) { print. Sol(perm); } for (int i=0; i<SIZE; i++) { if (used. List[i] == false) { if (!conflict(perm, location, i)) { Found a solution to the problem, so print it! Loop through possible rows to place this queen. Only try this row if it hasn’t been used Check if this position conflicts with any previous queens on the diagonal perm[location] = i; used. List[i] = true; solve. It. Rec(perm, location+1, used. List); used. List[i] = false; } } 1) mark the queen in this row 2) mark the row as used 3) solve the next column location recursively 4) un-mark the row as used, so we can get ALL possible valid solutions.

Backtracking – 8 queens problem Analysis Another possible brute-force algorithm is generate the permutations of the numbers 1 through 8 (of which there are 8! = 40, 320), ◦ and uses the elements of each permutation as indices to place a queen on each row. ◦ Then it rejects those boards with diagonal attacking positions. The backtracking algorithm, is a slight improvement on the permutation method, ◦ constructs the search tree by considering one row of the board at a time, eliminating most non-solution board positions at a very early stage in their construction. ◦ Because it rejects row and diagonal attacks even on incomplete boards, it examines only 15, 720 possible queen placements. A further improvement which examines only 5, 508 possible queen placements is to combine the permutation based method with the early pruning method: ◦ The permutations are generated depth-first, and the search space is pruned if the partial permutation produces a diagonal attack

Magic Number Squares A 3 x 3 magic number square has 9 numbers {1, 2, … 9} that must be arranged in such a way that every row, column, and diagonal has the same sum. ◦ For a 3 x 3 magic number square the sum is 3*(3*3+1)/2 = 15. 15 15

Magic Number Square It is possible to have different sized magic number squares of size nxn, ◦ where numbers {1, 2, … n*n} must be arranged in such a way that every row and column has the same sum. ◦ For a nxn magic number square the sum is n*(n*n+1)/2

Magic Number Square Say we want to create a class Magic. Square that takes in an int n and returns all valid Magic Number Squares of size nxn. ◦ Example, n =43 9 returns 8 magic squares: 2 7 6 2 8 1 6 9 5 1 4 3 8 3 5 7 8 1 6 2 9 4 7 5 3 6 1 8 7 5 3 2 9 4 4 3 8 9 5 1 2 7 6 6 7 2 1 5 9 8 3 4 3 5 7 4 9 2 8 3 4 1 5 9 6 7 2

Magic Number Square How would we start? ◦ The plan like we did with Eight Queens is just to try a number in the first position. Then we can recursively solve the board at the next position given that the number has been placed. Then we systematically “throw out” boards that do not meet our constraints. ◦ What do we need to do that? First we will want an empty board that we will fill in one square at a time. Then we will want to mark which numbers in the set (1, … , n*n) have already been used. We need to calculate the sum that each row, column, and diagonal must add up to.

public class Magic. Square { private int[][] square; private boolean[] possible; private int total. Sqs; private int sum; private int numsquares; • square[][] will hold the numbers we place. • possible[] will hold the numbers that have still not bee used yet. • sum will store what value the rows/columns/diagonals must sum to. // Creates an empty Magic. Square object. public Magic. Square(int n) { // Fill in an empty square; 0 represents unfilled. square = new int[n][n]; for (int i=0; i<n; i++) for (int j=0; j<n; j++) square[i][j] = 0; // Start with all numbers being possible. total. Sqs = n*n; possible = new boolean[total. Sqs]; for (int i=0; i<total. Sqs; i++) possible[i] = true; sum = n*(n*n+1)/2; numsquares = 0; } square[][] is size nxn initialize all values to 0, it’s empty to start! There are n*n total numbers, so possible[] is size n*n. At first all of the numbers can be used so the entire array is set to true.

// Recursive method which fills in the square at (row, col). // Try each possible number for this square. public void fill(int row, int col) { for (int i=0; i<total. Sqs; i++) { // Screen away incorrect squares. if (possible[i]) { all possible values for this spot if (!check. Rows() || Try !check. Cols() || !check. Diags()) square[row][col] = i+1; possible[i] = false; return; // Go to the next square. // Filled everything, so print !! int newcol = col+1; Find {the new row and col, if (row == square. length) so we can recursively fill the next spot. System. out. println(this); numsquares++; int newrow = row; if (newcol == square. length) { newrow++; return; newcol = 0; } } Base Cases: Either we broke the constraints, Recursively call fill() on our new spot. so we throw that board out. OR We’re DONE!! Undo this square for future recursive calls, so that we can try ALL valid boards } } // Recursively fill(newrow, newcol); // Undo this square[row][col] = 0; possible[i] = true;

// Recursive method which fills in the square at (row, col). // Try each possible number for this square. public void fill(int row, int col) { for (int i=0; i<total. Sqs; i++) { // Screen away incorrect squares. if (possible[i]) { square[row][col] = i+1; if (!check. Rows() || !check. Cols() || !check. Diags()) possible[i] = false; return; // Go to the next square. // Filled everything, so print !! int newcol = col+1; if (row == square. length) { int newrow = row; System. out. println(this); if (newcol == square. length) { numsquares++; newrow++; return; newcol = 0; } } // Recursively fill(newrow, newcol); // Undo this square[row][col] = 0; possible[i] = true; } }

// Returns true iff all the rows are okay. public boolean check. Rows() { // Try each row. for (int i=0; i<square. length; i++) { int test = 0; boolean un. Filled = false; Determine whether each row is filled and the row’s sum. // Add up the values in this row. for (int j=0; j<square[i]. length; j++) { test += square[i][j]; if (square[i][j] == 0) un. Filled = true; } } // If the row is filled and adds up to the wrong number, // this is an invalid square. If the row is filled and adds if (!un. Filled && test != sum) up to the wrong number return false; then our row constraint is violated, // Never found proof of an invalid row. return true; } and we return false. At this point we have checked all rows, and none of them violated our constraint.

Mazes and Backtracking A final example of something that can be solved using backtracking is a maze. ◦ From your start point, you will iterate through each possible starting move. ◦ From there, you recursively move forward. ◦ If you ever get stuck, the recursion takes you back to where you were, and you try the next possible move. In dealing with a maze, to make sure you don't try too many possibilities, ◦ one should mark which locations in the maze have been visited already so that no location in the maze gets visited twice. ◦ (If a place has already been visited, there is no point in trying to reach the end of the maze from there

References Slides adapted from Arup Guha’s Computer Science II Lecture notes: http: //www. cs. ucf. edu/~dmarino/ucf/cop 350 3/lectures/ Additional material from the textbook: Data Structures and Algorithm Analysis in Java (Second Edition) by Mark Allen Weiss Additional images: www. wikipedia. com xkcd. com