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Magic Number Square Tentaizu Approach Backtracking Magic Number Square Tentaizu Approach

Magic Number Squares A 3x3 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 3x3 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 MagicSquare that takes in an int n and returns all valid Magic Number Squares of size nxn. Example, n = 3 returns 8 magic squares: 2 7 6 9 5 1 4 3 8 4 9 2 3 5 7 8 1 6 8 1 6 3 5 7 4 9 2 2 9 4 7 5 3 6 1 8 6 1 8 7 5 3 2 9 4 8 3 4 1 5 9 6 7 2 4 3 8 9 5 1 2 7 6 6 7 2 1 5 9 8 3 4

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.

initialize all values to 0, it’s empty to start! public class MagicSquare { private int[][] square; private boolean[] possible; private int totalSqs; private int sum; private int numsquares; // Creates an empty MagicSquare object. public MagicSquare(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. totalSqs = n*n; possible = new boolean[totalSqs]; for (int i=0; i<totalSqs; i++) possible[i] = true; sum = n*(n*n+1)/2; numsquares = 0; } 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. 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.

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

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

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

Magic Number Square Decision Tree Shown on the board…

Tentaizu Approach The game is played on a 7x7 board. Exactly 10 of the 49 squares are each hiding a star. Your task is to determine which squares are hiding the stars. Other squares in the board provide clues: A number in a square indicates how many stars lie next to the square No square with a number in it contains a star, but a star may appear in a square with no adjacent numbers.

Tentaizu Approach Break down the problem into smaller steps Read in the board Determine the squares with numbers Determine the squares that are blank Place the stars in a backtracking method based on the possible locations Check that the current or finished board meets the requirements of the game Ten stars are present Each numbered square has the correct number of adjacent stars

Optional Homework Problem Get together with someone in the class and brainstorm how you will solve the Tentaizu assignment. Turn in a paper describing your strategy. What matrices and/or arrays will you need? What constraints will you check against? How will you know when these constraints are violated? How will you know when the board is complete? Show a decision tree diagramming how your Tentaizu algorithm will work. (Like how we did with the Magic Number Square)

References Slides adapted from Arup Guha’s Computer Science II Lecture notes: http://www.cs.ucf.edu/~dmarino/ucf/cop3503/le ctures/ 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