### CS51 - Spring 2010 - Lecture 18

• http://www.xkcd.com/816/

• Exercise 15.4.4

You are on a team that is developing a game, in which the player must escape from a monster. Both the monster and the player move through a space that looks like a grid. In a single move, the monster can move one space vertically, horizontally, or diagonally. The player can only move vertically or horizontally. However, the play can do something that the monster cannot do. The monster can never move off the grid. If it is at the right edge of the grid, for example, it cannot move to the right. The player, on the other hand, can move to the right, re-entering the grid on the opposite side. The player can make analogous moves off the left edge of the grid as well.

a. Write a method moveRight that takes an int parameter col that is the player's current column. The method should retrun an int that is the column into which the player should move if going right. You may assume that a constant COLS has been defined that gives the total number of columns in the grid.

b. Now write a method moveLeft.

• preregistration pizza

• Back to TicTacToe...
- Using a multidimensional array, how can we keep track of the board?
- private int NUM_ROWS = 3;
- private int NUM_COLS = 3;
- 3 by 3 array of integers: int[][] marks = new int[NUM_ROWS][NUM_COLS];
- private int EMPTY_MARK = 0;
- private int X_MARK = 1;
- private int O_MARK = 2;
- how does the game play progress?
- user clicks: onMouseClick
- check if we've clicked within the board
- figure out which square the user clicked
- figure out if it's a valid square to add the symbol to
- place the symbol (if appropriate)
- check for a win
- figuring out which square the user clicked
- given...
LEFT // x value of the left of the board
TOP // y value of the top of the board
CELL_SIZE // the size of the an individual cell in the board
- draw a picture of the window with the nested board
- if we know that the user clicked at x, y, which we know is in the board, how do we calculate the row/col indices
x = x - LEFT;
y = y - TOP;
int col = (int)(x/CELL_SIZE);
int row = (int)(y/CELL_SIZE);
- How do we check for a win? (group work?)
- need to check four cases:
- rows all the same
- columns all the same
- top-left to bottom-right diagonal all the same
- top-right to bottom-left diagonal all the same
- Note: when we're checking for a win, we're checking if a particular player has won for the last move
- we know the mark the player made (i.e. X or O)
- we know the row and col position where the mark was made
- look at TicTacToe code
- in addition to checking for a win, the win methods also draw the winning line
- what would be another representation for the baord?
- option 1: two boolean 2d arrays, one if there is a mark or not and one whether it was an X or an O
private boolean[][] isMarked;
private boolean[][] isX;
- option 2: create another class that represents an entry, say TicTacToeEntry
private TicTacToeEntry[][] board;

• magic square
- a magic square is an n by n matrix
- containing the numbers 1 through n^2
- such that all the rows, columns and diagonals sum to the same number

• take a few minutes and try to do a 3 by 3 magic square

• show MagicSquares demo
- How can we represent a magic square?
- an n by n integer array
- int[][] magicArray = new magicArray[n][n];
- if n is odd, you can build a magic square with the following rules
- put a 1 in the center of the bottom row
- place the next number (i.e. 2 follows 1, etc) in
- the cell one slot below and one slot to the right
- if you fall off the bottom or right, wrap around
- if that cell already has a number, put it in the slot above the original position (again, with wraparound)

• magic square
- a magic square is an n by n matrix
- containing the numbers 1 through n^2
- such that all the rows, columns and diagonals sum to the same number

• show MagicSquares demo
- How can we represent a magic square?
- an n by n integer array
- int[][] magicArray = new magicArray[n][n];
- if n is odd, you can build a magic square with the following rules
- put a 1 in the center of the bottom row
- place the next number (i.e. 2 follows 1, etc) in
- the cell one slot below and one slot to the right
- if you fall off the bottom or right, wrap around
- if that cell already has a number, put it in the slot above the original position (again, with wraparound)
- for an n by n integer array, how would we put a 1 in the center of the bottom row?
- magicArray[n-1, n/2] = 1;
- given row and col, what would be the square one slot below and right?
- row+1
- col+1
- how do we deal with wrap around for an n by n array?
- row + 1 % n
- col + 1 % n
- with wrap around, what is the column above col for an n by n array?
- col - 1 % n
- (col - 1 + n) % n

• show MagicSquares code (specifically MagicSquare.fillSquare)

• Nibbles lab
- Nibbles in real life :)
- Snake is a 1D array and is an ActiveObject
- NibblesField is a 2D array
- keeps track of the state of the board
- also keeps track of what's going on on the screen
- 2 other support classes
- Position
- Direction
- look at moving the snake
- stretch and shrink
- two different ways of implementing
- how can we use stretch and shrink to move?
- lab has two parts
- code the Snake class (given a working implementation of the NibblesField class)
- once you've done this, code up the NibblesField class
- design
- only have to do a design for the Snake class, but make sure to think about the second part of the lab too
- may work in groups on the design (and only need to provide one design)

• CS course options next semester
- Descriptions: http://www.cs.pomona.edu/courses.html
- CS62: Data Structures and Advances Programming
- CS55: Discrete Mathematics