# Crossword Solution

The third free response question on the 2016 AP Computer Science exam tasked you with working with a matrix of an arbitrary class. In this case it was a class called Square. When put into a matrix these Square reference formed a crossword puzzle.

You were given a shell of the Square class. When you are given arbitrary classes on an AP exam you will certainly use them in your solutions.

public class Square {
public Square( boolean isBlack, int num ) {
/* Implementation not shown */
}

// There may be instance variables, constructors, and
// methods that are not shown
}

Note that there is no implementation in Square. They are giving you the constructor signature so that you can create new Square instances in your code.

### Part A – toBeLabeled

The first part of this problem has you implementing a method named toBeLabeled. toBeLabeled should return true if the following two criteria are met.

• The square must be white
• The square does not have a white square immediately above to to the left, or both
private boolean toBeLabeled( int r, int c, boolean[][] blackSquares ) {
return (!(blackSquares[r][c]) && ( r == 0 || c == 0 || blackSquares[r-1][c] || blackSquares[r][c-1] );
}

We’re given a boolean matrix called blackSquares that we’ll use to check if a square is black.

To decide if we should label a square the code first looks to see that it’s not black. Since there’s only two options, that means it’s a white square.

Once we know it’s a white square we’ll look and see if it’s in row or column 0. If it is, the square is labeled. If not well look at the row above and column to the left. If either is labeled, then the current square is not labeled.

### Part B – Constructor

The constructor should take a boolean matrix identifying black squares and store a matrix of Square references in the instance variable puzzle.

public Crossword(boolean[][] blackSquares) {
puzzle = new Square[blackSquares.length][blackSquares[0].length];
int num = 1;

for (int r=0; r<blackSquares.length; r++) {
for (int c=0; c<blackSquares[r].length; c++) {
if (blackSquares[r][c]) {
puzzle[r][c] = new Square(true, 0);
}
else {
if (toBeLabeled(r, c, blackSquares) {
puzzle[r][c] = new Square(false, num);
num++;
}
else {
puzzle[r][c] = new Square(false, 0);
}
}
}
}

First step is to instantiate puzzle so that it is a 2d array of Square references. It should be the same size as the blackSquares parameter. Since this is an AP-A free response we can assume that the matrix is rectangular. All rows will have the same number of columns as row 0.

We also need a variable to keep track of the label number. In the example I call it num.

Now we iterate through all of the cells using a nest for loop.

If the square is black, we know it won’t be labeled. So we’ll create a new instance of a black square with a label of 0.

If the square is white we need to check if it should be labeled. Do that by calling the toBeLabeled method created in Part A. If it should be labeled then create a new Square instance with a label equal to num and then increment num for next time. If it should not be labeled use 0 as the label.

### Complete Crossword Class

This is the same code as the examples above, but all together in one file.

public class Crossword {

private Square[][] puzzle;

private boolean toBeLabeled( int r, int c, boolean[][] blackSquares ) {
return (!(blackSquares[r][c]) &&
( r == 0 || c == 0 || blackSquares[r-1][c] || blackSquares[r][c-1] );
}

public Crossword(boolean[][] blackSquares) {
puzzle = new Square[blackSquares.length][blackSquares[0].length];
int num = 1;

for (int r=0; r<blackSquares.length; r++) {
for (int c=0; c<blackSquares[r].length; c++) {
if (blackSquares[r][c]) {
puzzle[r][c] = new Square(true, 0);
}
else {
if (toBeLabeled(r, c, blackSquares) {
puzzle[r][c] = new Square(false, num);
num++;
}
else {
puzzle[r][c] = new Square(false, 0);
}
}
}
}
}