© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Data Structures for Java William H. Ford William R. Topp Chapter 29 Assorted.

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Building Combinations The function C(n, k) evaluates the number of different combinations of n items taken k at a time. Mathematics provides a definition of the function in terms of factorials. The function C(n, k) evaluates the number of different combinations of n items taken k at a time. Mathematics provides a definition of the function in terms of factorials. C(n,k) can be computed using a recursive function C(n,k) can be computed using a recursive function

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Building Combinations (continued) In general, we can evaluate C(n,k) by adding the total of ways to select k items from n-1 (C(n-1,k)) and the total number in of ways to select k-1 items from n-1 items (C(n-1, k-1), assuming item n is added to each combination. This defines the recursive step for C(n,k). In general, we can evaluate C(n,k) by adding the total of ways to select k items from n-1 (C(n-1,k)) and the total number in of ways to select k-1 items from n-1 items (C(n-1, k-1), assuming item n is added to each combination. This defines the recursive step for C(n,k). C(n,k) = C(n-1, k) + C(n-1, k-1);// recursive step

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Building Combinations (concluded) If k = n, there is only 1 possible combination, namely the one with all n items. Having k = 1 is the other extreme, and each of the n singleton collections consisting of one of the n items is a valid combination. We allow k = 0 and define C(n,0) to be 1. If k = n, there is only 1 possible combination, namely the one with all n items. Having k = 1 is the other extreme, and each of the n singleton collections consisting of one of the n items is a valid combination. We allow k = 0 and define C(n,0) to be 1.

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.comm() public static int comm(int n, int k) { // stopping condition if (n == k || k == 0) return 1; else if (k == 1) // stopping condition return n; else // recursive step return comm(n-1,k) + comm(n-1,k-1); }

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Finding All Subsets The power set of an n ‑ element set S is the set of all subsets of S. The size of the power set is 2 n. The power set of an n ‑ element set S is the set of all subsets of S. The size of the power set is 2 n.

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Finding All Subsets (continued) Start with set S of size n. Remove one of the elements, x, from the set to create a new set S' with n ‑ 1 elements. Start with set S of size n. Remove one of the elements, x, from the set to create a new set S' with n ‑ 1 elements. The new set, S', has the form S' = S ‑ {x}. The power set of S' contains 2 n-1 subsets. The new set, S', has the form S' = S ‑ {x}. The power set of S' contains 2 n-1 subsets. Adding the element x back into each of these subsets creates a second collection of 2 n-1 subsets. Adding the element x back into each of these subsets creates a second collection of 2 n-1 subsets. The power set of S is the union of the power set for S' and the elements in the second collection. The power set of S is the union of the power set for S' and the elements in the second collection.

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Finding All Subsets (concluded) The "recursive" part of the algorithm involves creating the power set of S'. Just repeat the process by removing an element from S' to form (S')' which is a set with n-2 elements. The stopping condition occurs when removing an element leaves the empty set {}. The power set of empty set is the one-element set whose only member is {} The "recursive" part of the algorithm involves creating the power set of S'. Just repeat the process by removing an element from S' to form (S')' which is a set with n-2 elements. The stopping condition occurs when removing an element leaves the empty set {}. The power set of empty set is the one-element set whose only member is {}

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Listing Permutations A permutation on n items (1, 2,..., n) is an ordered arrangement of the items. The number of permutations of n items is n! = n*(n-1)*(n-2)…(2)(1). A permutation on n items (1, 2,..., n) is an ordered arrangement of the items. The number of permutations of n items is n! = n*(n-1)*(n-2)…(2)(1).

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Listing Permutations (continued) The recursive algorithm for n! traverses a recursion tree. Permutations are leaf nodes at the bottom level. The recursive algorithm for n! traverses a recursion tree. Permutations are leaf nodes at the bottom level.

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Listing Permutations (continued) Index 0: Index 0: Use a loop to create n-1 copies of the array by exchanging each element in the index range [1, n) with the element at index 0. The initial array and the copies define all of the arrays that have distinct values at index 0 and correspond to the elements at level 1 in the tree. Use a loop to create n-1 copies of the array by exchanging each element in the index range [1, n) with the element at index 0. The initial array and the copies define all of the arrays that have distinct values at index 0 and correspond to the elements at level 1 in the tree.

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Listing Permutations (continued) Index 1: Index 1: For each of the n arrays with distinct elements at index 0, a recursive call is made with the array and index 1 (next index). A loop takes the array argument and creates n- 2 new arrays by exchanging each element in the range [2,n) with the element at index 1. For each array at level 1, the array and the n- 2 copies represent n-1 distinct ordering of elements at indices 0 to 1. For each of the n arrays with distinct elements at index 0, a recursive call is made with the array and index 1 (next index). A loop takes the array argument and creates n- 2 new arrays by exchanging each element in the range [2,n) with the element at index 1. For each array at level 1, the array and the n- 2 copies represent n-1 distinct ordering of elements at indices 0 to 1.

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Listing Permutations (continued) A recursive call is made for the each array and for all of the copies that are created during the loop (iterative process). As a result, permutations are created and then displayed by following the different paths in the tree from left to right. The array at the end of each path is a permutation. The following shows the order in which each permutation is created. A recursive call is made for the each array and for all of the copies that are created during the loop (iterative process). As a result, permutations are created and then displayed by following the different paths in the tree from left to right. The array at the end of each path is a permutation. The following shows the order in which each permutation is created.

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.permute() private static void permute(Object[] permList, int index) { int i, j, arrSize = permList.length; Object temp; if (index == arrSize-1) // display the permutation System.out.println(Arrays.toString(permList)); else { Object[] newPermList = new Object[arrSize]; for (i = 0; i < arrSize; i++) newPermList[i] = permList[i]; // find all permutations over the // range (index, arrSize) permute(newPermList, index+1);

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. permute() (concluded) // exchange permList[index] with permList[i] // for i=index+1 to the end of the array and // find all permutations for (i=index+1; i < arrSize; i++) { temp = permList[i]; permList[i] = permList[index]; permList[index] = temp; newPermList = new Object[arrSize]; for (j = 0; j < arrSize; j++) newPermList[j] = permList[j]; permute(newPermList, index+1); }

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Traveling Salesman Problem The Traveling Salesman Problem (TSP) describes an interesting and challenging combinatorial problem which can be stated very simply. A salesman is responsible for n cities in a territory that has paths that connect any two cities. The problem is to identify a tour (a simple cycle) that has the salesman visit each city just once and finish up where he started. The order of visits should minimize the total distance traveled. The Traveling Salesman Problem (TSP) describes an interesting and challenging combinatorial problem which can be stated very simply. A salesman is responsible for n cities in a territory that has paths that connect any two cities. The problem is to identify a tour (a simple cycle) that has the salesman visit each city just once and finish up where he started. The order of visits should minimize the total distance traveled.

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Traveling Salesman Problem (continued) The traveling salesman problem has a variety of applications. For instance, a school district must arrange bus routes to pick up students at a variety of stops and deliver them to the school. A robot arm must be programmed to move efficiently over a circuit board to drill holes and solder connections. The traveling salesman problem has a variety of applications. For instance, a school district must arrange bus routes to pick up students at a variety of stops and deliver them to the school. A robot arm must be programmed to move efficiently over a circuit board to drill holes and solder connections.

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Traveling Salesman Problem (continued) Possible Tours: (n-1) * (n-2) *... * 2 * 1 = (n-1)! Possible Tours: (n-1) * (n-2) *... * 2 * 1 = (n-1)! For large n, the number of possible tours is so large that an exhaustive search of all possible paths is unrealistic. No one has found an algorithm to solve the general problem with polynomial runtime efficiency. For large n, the number of possible tours is so large that an exhaustive search of all possible paths is unrealistic. No one has found an algorithm to solve the general problem with polynomial runtime efficiency.

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Traveling Salesman Problem (concluded) Traveling Salesman problem is NP- complete. There is likely no solution with polynomial running time. Traveling Salesman problem is NP- complete. There is likely no solution with polynomial running time. We can use permutations to solve the traveling salesman problem for relatively small n. Create an array with the starting city in position 0 and the other cities occupying the other positions. Check each permutation and find a tour of minimum distance. We can use permutations to solve the traveling salesman problem for relatively small n. Create an array with the starting city in position 0 and the other cities occupying the other positions. Check each permutation and find a tour of minimum distance.

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Dynamic Programming Fibonacci Sequence: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, … Fibonacci Sequence: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, … Recursive definition: Recursive definition:

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Recursive fib() // recursive computation of Fibonacci number n public static int fib(int n) { // stopping conditions if (n <= 1) return n; else // recursive step return fib(n-1) + fib(n-2); }

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Dynamic Programming (continued) Running time for the recursive method fib(n) is exponential due to the redundant function calls. Running time for the recursive method fib(n) is exponential due to the redundant function calls.

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Dynamic Programming (continued) Recursive solution for fib(n) fails to partition problem into non-overlapping subproblems. Recursive solution for fib(n) fails to partition problem into non-overlapping subproblems.

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Top-Down Dynamic Programming Top-down dynamic programming gains efficiency by storing previously computed values during a recursive process. Top-down dynamic programming gains efficiency by storing previously computed values during a recursive process. Use an array that stores the return values from intermediate method calls. The recursive step first checks the array to see whether the result has already been computed. If so, it directly accesses the value. If not, the recursive step executes, then adds the result to the array for later access. Use an array that stores the return values from intermediate method calls. The recursive step first checks the array to see whether the result has already been computed. If so, it directly accesses the value. If not, the recursive step executes, then adds the result to the array for later access.

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.fibDyn() // computation of the nth Fibonacci number using // top down dynamic programming to avoid redundant // recursive method calls public static int fibDyn(int n, int[] fibList) { int fibValue; // check for a previously computed result and return if (fibList[n] >= 0) return fibList[n]; // otherwise execute the recursive algorithm // to obtain the result

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. fibDyn() (concluded) // stopping conditions if (n <= 1) fibValue = n; else // recursive step fibValue = fibDyn(n-1, fibList) + fibDyn(n-2, fibList); // store the result and return its value fibList[n] = fibValue; return fibValue; }

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Running Time of fibDyn() The figure shows that n+1 method calls reach the bottom of the tree followed by n-2 on the way back up. The number of method calls to compute fibDyn(n) is n + 1+ (n-2) = 2n-1. The figure shows that n+1 method calls reach the bottom of the tree followed by n-2 on the way back up. The number of method calls to compute fibDyn(n) is n + 1+ (n-2) = 2n-1.

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Combinations with Dynamic Programming Top-down dynamic programming can be used to improve the recursive algorithm for C(n,k). Top-down dynamic programming can be used to improve the recursive algorithm for C(n,k). Store intermediate results in the matrix commMat, which has n+1 rows and k+1 columns, where commMat[i][j] is the solution for C(i,j). Initially all of the values in the matrix are set to -1. Each recursive call to commDyn(n, k) first checks whether the value commMat[n][k] is nonzero. If so, it uses the matrix element as its return value Store intermediate results in the matrix commMat, which has n+1 rows and k+1 columns, where commMat[i][j] is the solution for C(i,j). Initially all of the values in the matrix are set to -1. Each recursive call to commDyn(n, k) first checks whether the value commMat[n][k] is nonzero. If so, it uses the matrix element as its return value

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.commDyn() // computation of C(n,k) using top down dynamic // programming to avoid redundant recursive // method calls public static int commDyn(int n, int k, int[][] commMat) { int returnValue; // check if value is already computed if (commMat[n][k] >= 0) return commMat[n][k];

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. commDyn() (concluded) if (n == k || k == 0) returnValue = 1; else if (k == 1) returnValue = n; else // carry out the recursive step returnValue = commDyn(n-1,k,commMat) + commDyn(n-1,k-1,commMat); // before returning, assign value to the matrix commMat[n][k] = returnValue; return returnValue; }

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Bottom-Up Dynamic Programming Bottom-up dynamic programming builds the solution in order, starting at the lowest level and moving upward, using values from the previous level. Bottom-up dynamic programming builds the solution in order, starting at the lowest level and moving upward, using values from the previous level.

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Bottom-Up Dynamic Programming (continued) The combinations formula C(n,k) = C(n-1, k) + C(n-1, k-1) indicates that we can compute intermediate values for entries in row i by using results from row i-1. The following table displays the matrix for n = 6, k = 6. When n = k, the table of entries is often called Pascal's Triangle. The combinations formula C(n,k) = C(n-1, k) + C(n-1, k-1) indicates that we can compute intermediate values for entries in row i by using results from row i-1. The following table displays the matrix for n = 6, k = 6. When n = k, the table of entries is often called Pascal's Triangle.

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Bottom-Up Dynamic Programming (continued) Bottom-up dynamic programming very efficiently computes C(n,k) by building the Pascal triangle. Bottom-up dynamic programming very efficiently computes C(n,k) by building the Pascal triangle.

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.commDynB() // computation of C(n,k) using bottom-up // dynamic programming public static int commDynB(int n, int k) { // create an n+1 by k+1 matrix int[][] commMat = new int[n+1][k+1]; int i, j;

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. commDynB() (continued) // rows range from 0 through n for (i=0; i <= n; i++) // only generate columns 0 through k for (j = 0; j <= Math.min(i,k); j++) // commMat[i][j] = 1 when j == 0 or j == i if (j == 0 || i == j) commMat[i][j] = 1; else commMat[i][j] = commMat[i-1][j-1] + commMat[i-1][j]; // return the entry commMat(n,k) return commMat[n][k]; }

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Knapsack Problem Given a knapsack with limited capacity, measured in volume, is to hold a set of items that have specified sizes and values. Find a subset of objects that will fit into the knapsack and provide the maximum value. The problem is a prototype for many important applications; for instance, transport companies want to load a truck, freight car, or ship with cargo that returns a maximum profit. Given a knapsack with limited capacity, measured in volume, is to hold a set of items that have specified sizes and values. Find a subset of objects that will fit into the knapsack and provide the maximum value. The problem is a prototype for many important applications; for instance, transport companies want to load a truck, freight car, or ship with cargo that returns a maximum profit.

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Knapsack Problem (continued) A brute force solution of the knapsack problem checks all subsets of items, which is an exponential algorithm. A brute force solution of the knapsack problem checks all subsets of items, which is an exponential algorithm. A dynamic programming solution applies the principle of optimality to each item in the collection. The principle states that, no matter what the first decision, the remaining decisions must be optimal with respect to any state in the algorithm that results from the first decision. A dynamic programming solution applies the principle of optimality to each item in the collection. The principle states that, no matter what the first decision, the remaining decisions must be optimal with respect to any state in the algorithm that results from the first decision.

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Knapsack Problem (continued) Solve the knapsack problem by using the principle of optimality in the generation of the knapsack matrix. Solve the knapsack problem by using the principle of optimality in the generation of the knapsack matrix. The mathematical definition of maxValueMat[i][cap] is where a j = 1 if item j is in the subset and a j = 0 if itemj is not in the subset. After we build the matrix, matValueMat[n][capacity] is the solution to the problem. The mathematical definition of maxValueMat[i][cap] is where a j = 1 if item j is in the subset and a j = 0 if itemj is not in the subset. After we build the matrix, matValueMat[n][capacity] is the solution to the problem.

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Knapsack Problem (continued // keep the same max value by default maxValueMat[i][cap] = maxValueMat[i-1][cap]; // test if itemList[i] fits into the knapsack if (cap-itemList[i].size >= 0) { // test if maximum value increases testMax = maxValueMat[i-1][cap-itemList[i].size] + itemList[i].value; // if yes, assign new max if (testMax > maxValueMat[i-1][cap]) maxValueMat[i][cap] = testMax; } Use the following algorithm to determine maxValueMat[i][cap]

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. The Item Class The class Item describes the items in the knapsack. Two integer variables define the size and value of an object. To simplify access, we declare the members public. A constructor creates an instance of the class using two arguments for the size and value. The class Item describes the items in the knapsack. Two integer variables define the size and value of an object. To simplify access, we declare the members public. A constructor creates an instance of the class using two arguments for the size and value.

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Item Class // specifies the size and value of an item public class Item { public int size, value; public Item(int size, int value) { this.size = size; this.value = value; }

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Knapsack Class public class Knapsack { // capacity of the knapsack private int capacity; // the list of items private Item[] itemList; // the number of items (itemList.length) private int numItems; // the knapsack matrix private int[][] maxValueMat; // initialize variables and build the knapsack matrix public Knapsack(Item[] list, int cap) {... }

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Knapsack Class (concluded) // builds maxValueMat for specified capacity private void buildMaxValueMat() {... } // displays capacity, items in knapsack, max value, // and unused capacity public void displayKnapsack() {... } }

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.buildMaxValueMat() // builds maxValueMat for specified capacity private void buildMaxValueMat() { int i, cap, testMax; // compute entries in the matrix for (i = 1; i <= numItems; i++) for (cap = 1; cap <= capacity; cap++) { // initially assume the max value for capacity // cap is produced by using items from 1 to i-1 maxValueMat[i][cap] = maxValueMat[i-1][cap];

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. buildMaxValueMat() (concluded) // test if itemList[i] fits into the knapsack if (cap-itemList[i].size >= 0) { // test if maximum value increases testMax = maxValueMat[i-1][cap-itemList[i].size] + itemList[i].value; // if yes, assign new max if (testMax > maxValueMat[i-1][cap]) maxValueMat[i][cap] = testMax; }

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. displayKnapsack() Identify items in the knapsack solution by going backwards from the item in the knapsack matrix, row n, column cap. Identify items in the knapsack solution by going backwards from the item in the knapsack matrix, row n, column cap.

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. displayKnapsack() (continued) // displays capacity, items in knapsack, max value, // and unused capacity public void displayKnapsack() { int i = numItems, cap = capacity; // create label with capacity and maximum value System.out.println("Capacity: " + capacity + " Value: " + maxValueMat[numItems][capacity]); // list items in the knapsack by reading // from maxValueMat System.out.println("Contents: ");

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. displayKnapsack() (concluded) while (i > 0) { // if values in successive rows are not equal, // itemList[i] is part of the solution if (maxValueMat[i][cap] != maxValueMat[i-1][cap]) { System.out.println(" item" + i + '(' + itemList[i].size + ',' + itemList[i].value + ')'); // look for maximum value remaining space cap -= itemList[i].size; } i--; } System.out.println(" Unused capacity: " + cap); }

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Program 29.1 import java.util.Scanner; public class Program29_1 { public static void main(String[] args) { // array of Item objects for the knapsack Item[] itemList = {new Item(2,1), new Item(3,4), new Item(4,3), new Item(5,6), new Item(6,8)}; // use for keyboard input Scanner keyIn = new Scanner(System.in); int capacity;

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Program 29.1 (concluded) System.out.print("Enter the capacity: "); capacity = keyIn.nextInt(); // create a knapsack object Knapsack ks = new Knapsack(itemList,capacity); // display the solution ks.displayKnapsack(); System.out.println(); }

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Program 29.1 (Runs) Run 1: Enter the capacity: 12 Capacity: 12 Value: 14 Contents: item5(6,8) item4(5,6) Unused capacity: 1 Run 2: Enter the capacity: 19 Capacity: 19 Value: 21 Contents: item5(6,8) item4(5,6) item3(4,3) item2(3,4) Unused capacity: 1

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Evaluating the Knapsack Problem Solution Running time for the knapsack problem is O(nC), where n is the number of items, and C the capacity. Running time for the knapsack problem is O(nC), where n is the number of items, and C the capacity. The amount of computation depends critically on C. For instance, if C = n the algorithm is O(n 2 ), but if C = 2 n, the algorithm has running time O(n2 n ). The amount of computation depends critically on C. For instance, if C = n the algorithm is O(n 2 ), but if C = 2 n, the algorithm has running time O(n2 n ).

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Backtracking With recursive backtracking, we move forward and possibly several steps backward until reaching a solution. With recursive backtracking, we move forward and possibly several steps backward until reaching a solution. The 8-Queens problem attempts to position eight queens on a chessboard in such a way that no two queens can attack each other. The 8-Queens problem attempts to position eight queens on a chessboard in such a way that no two queens can attack each other. 8-Queens has an elegant solution implemented by using backtracking. 8-Queens has an elegant solution implemented by using backtracking.

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. The 8-Queens Problem A queen can be attacked by a queen in the same row or column and by a queen on one of the two diagonals. A queen can be attacked by a queen in the same row or column and by a queen on one of the two diagonals.

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. The 8-Queens Problem (continued) We build a solution column by column, starting with column 0. In each succeeding column, col, we move from row 0 to row 1, then to row 2, and so forth, until we find a "safe" cell where the queen can be positioned without begin vulnerable to attack by any of the other queens already positioned on the board. We build a solution column by column, starting with column 0. In each succeeding column, col, we move from row 0 to row 1, then to row 2, and so forth, until we find a "safe" cell where the queen can be positioned without begin vulnerable to attack by any of the other queens already positioned on the board.

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. The 8-Queens Problem (continued) No safe cell may exist in column col. No safe cell may exist in column col. The approach fails in column 5, because placing a queen in any of the eight rows leaves it vulnerable to attack by one of the five queens on the board. Backtrack back to column 4. The approach fails in column 5, because placing a queen in any of the eight rows leaves it vulnerable to attack by one of the five queens on the board. Backtrack back to column 4. Move forward to another row in column col if a queen at (row,col) does not lead to a solution. Move forward to another row in column col if a queen at (row,col) does not lead to a solution.

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. The 8-Queens Problem (continued) We move forward to column 5 and continue the algorithm. Placing the queen in cell (7,4) is a short-lived success, because it does not permit any queen to be safely positioned in column 5. We move forward to column 5 and continue the algorithm. Placing the queen in cell (7,4) is a short-lived success, because it does not permit any queen to be safely positioned in column 5. Backtrack may involve going back more than one column to find a solution. Backtrack may involve going back more than one column to find a solution.

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Implementing the 8-Queens Solution The method placeQueens() implements the backtracking strategy. The method placeQueens() implements the backtracking strategy. safeLocation() checks if a queen at (row,col) is safe from attack by queens in columns 0 through col-1. safeLocation() checks if a queen at (row,col) is safe from attack by queens in columns 0 through col-1.

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.safeLocation() // determine if cell (row,col) is safe from // attack by queens in cells (queenList[0],0),..., // (queenList[col-1],col-1) public static boolean safeLocation(int row, int col, int[] queenList) { int qRow, qCol;

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. safeLocation() (concluded) // check previous columns only for (qCol = 0; qCol < col; qCol++) { qRow = queenList[qCol]; if (qRow == row) // same row return false; else if (qCol == col) // same col return false; // can they attack on a diagonal? else if(qCol-qRow == col-row || qCol+qRow == col+row) return false; } return true; }

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. placeQueens() The static method placeQueens() returns a boolean value that indicates when a current partial solution will lead to a complete solution of the problem. That is, when it is possible to take the current arrangement of queens in columns 0 to col-1 (defined by queenList) and locate safe cells for the queens in the remaining columns. The static method placeQueens() returns a boolean value that indicates when a current partial solution will lead to a complete solution of the problem. That is, when it is possible to take the current arrangement of queens in columns 0 to col-1 (defined by queenList) and locate safe cells for the queens in the remaining columns.

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. placeQueens() (continued) // cells (queenList[0],0),..., (queenList[col-1],col-1) // are safe positions; determine if the solution can // be extended to columns col, col+1,..., 7 public static boolean placeQueens( int[] queenList, int col) { int row; boolean foundLocation; if (col == 8) // stopping condition foundLocation = true; else { foundLocation = false; // start with row 0 row = 0;

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. placeQueens() (continued) while (row < 8 && !foundLocation) { // check whether cell (row, col) is safe; // if so, assign row to queenList and call // placeQueens() for next column; otherwise, // go to the next row if (safeLocation(row,col,queenList) == true) { // found good location queenList[col] = row; // recursive step; try to place queens in // columns col+1 through 7 foundLocation = placeQueens(queenList,col+1); if (!foundLocation) // use next row since current one does // not lead to a solution row++; }

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. placeQueens() (concluded) else // current row fails; go to the next row row++; } // end while } // pass success or failure back to previous col return foundLocation; }

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. queens() The method queens() facilitates calling the recursive method placeQueens(). The method queens() facilitates calling the recursive method placeQueens().

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. queens() (concluded) // place a queen in (row, 0) and try to find a solution // to the 8-Queens problem for columns 1, 2,..., 7 public static boolean queens(int[] queenList, int row) { // place first queen at (row,0) queenList[0] = row; // locate remaining queens in columns 1 through 7 if (placeQueens(queenList, 1)) return true; else return false; }

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Displaying a ChessBoard Class ChessBoard simulates a chess board for the 8-Queens problem. Class ChessBoard simulates a chess board for the 8-Queens problem. The ChessBoard class has a private instance variable board which is an 8 by 8 two ‑ dimension array of boolean values. The ChessBoard class has a private instance variable board which is an 8 by 8 two ‑ dimension array of boolean values. The method setQueens() has the array argument queenList which identifies the location of queens on the board as (row,col) pairs and creates a representation of a chess board by assigning the [row][col] entry a value true if a queen is present and the value false to indicate a free cell. The method setQueens() has the array argument queenList which identifies the location of queens on the board as (row,col) pairs and creates a representation of a chess board by assigning the [row][col] entry a value true if a queen is present and the value false to indicate a free cell.

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Program 29.2 import java.util.Scanner; public class Program29_2 { public static void main (String[] args) { int row; // the array needed by the 8-Queens algorithm int[] queenList = new int[8]; // board will display the solution ChessBoard board = new ChessBoard(); Scanner keyIn = new Scanner(System.in); // enter a starting row for queen in column 0 System.out.print("Enter row for queen " + "in column 0: "); row = keyIn.nextInt(); System.out.println();

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Program 29.2 (concluded) // see if there is a solution if (Queens.queens(queenList, row)) { // insert the solution into the chessboard board.setQueens(queenList); // display the solution board.drawBoard(); } else System.out.println("No solution"); }

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Program 29.2 (Run #1) Enter row for queen in column 0: 2 0 1 2 3 4 5 6 7 0 - Q - - - - - - 1 - - - - - Q - - 2 Q - - - - - - - 3 - - - - - - Q - 4 - - - Q - - - - 5 - - - - - - - Q 6 - - Q - - - - - 7 - - - - Q - - -

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Program 29.2 (Run #2) Enter row for queen in column 0: 5 0 1 2 3 4 5 6 7 0 - Q - - - - - - 1 - - - Q - - - - 2 - - - - - Q - - 3 - - - - - - - Q 4 - - Q - - - - - 5 Q - - - - - - - 6 - - - - - - Q - 7 - - - - Q - - -

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. 8-Queens GUI Application Chapter 29 of the software supplement contains classes ChessBoardGUI and QueensGUI that implement a graphical solution to the 8-Queens problem. Chapter 29 of the software supplement contains classes ChessBoardGUI and QueensGUI that implement a graphical solution to the 8-Queens problem.

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Data Structures for Java William H. Ford William R. Topp Chapter 29 Assorted.

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© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Data Structures for Java William H. Ford William R. Topp Chapter 29 Assorted.

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Presentation on theme: "© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Data Structures for Java William H. Ford William R. Topp Chapter 29 Assorted."— Presentation transcript:

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