Tic Tac Toe

[metachief]

What can be done more efficiently?

Turing:

% Meta Chief % May 20, 2009 % TicTacToe /*  A two player game of tic tac toe  */ var mouseX, mouseY, mouseB : int % Stores mouse properties type BoardType : array 1 .. 3, 1 .. 3 of int % Grid points type var board : BoardType %Stores grid points const squareSize := 40 % Size of each square on the board proc InitBoard % Assigns all grid points to a value of zero     for a : lower (board) .. upper (board)         for b : lower (board) .. upper (board)             board (a, b) := 0         end for     end for end InitBoard proc DrawBoard % Depending on the value of a grid point, the board is drawn     for a : lower (board) .. upper (board)         for b : lower (board) .. upper (board)             drawbox (a * squareSize, b * squareSize, a * squareSize + squareSize, b * squareSize + squareSize, black)             if board (a, b) = 1 then  % If the grid point value is 1 an x is drawn                 drawline (a * squareSize, b * squareSize, a * squareSize + squareSize, b * squareSize + squareSize, black)                 drawline (a * squareSize + squareSize, b * squareSize, a * squareSize, b * squareSize + squareSize, black)             elsif board (a, b) = 2 then % If the grid point value is 2 a circle is drawn                 drawoval (a * squareSize + squareSize div 2, b * squareSize + squareSize div 2, squareSize div 2, squareSize div 2, black)             end if         end for     end for end DrawBoard var turn : int := 1 % Stores the players' turn var turnCount : int := 0 % Counts the number of turns in total proc UseMouse % Allow the user to use the mouse to change the value on the grid     for a : lower (board) .. upper (board)         for b : lower (board) .. upper (board)             var x, y : int             if Mouse.ButtonMoved ("down") then                 var buttonnumber, buttonupdown : int                 Mouse.ButtonWait ("down", x, y, buttonnumber,                     buttonupdown)                 if mouseX div squareSize <= upper (board) and mouseY div squareSize <= upper (board) and mouseX div squareSize >= lower (board) and mouseY div squareSize >= lower (board) then                     % If the mouse in on the board (to prevent out of range error)                     if board (mouseX div squareSize, mouseY div squareSize) = 0 then                         board (mouseX div squareSize, mouseY div squareSize) := turn % Turn represents an x or a cricle depending on the turn                         if turn = 1 then                             turn := 2                         elsif turn = 2 then                             turn := 1                         end if                         turnCount += 1                     end if                 end if             end if         end for     end for end UseMouse fcn CheckWin : int % Checks possivle win conditions     for i : 1 .. 2         if board (1, 1) = i and board (2, 2) = i and board (3, 3) = i then             result i         elsif board (1, 1) = i and board (1, 2) = i and board (1, 3) = i then             result i         elsif board (1, 1) = i and board (2, 1) = i and board (3, 1) = i then             result i         elsif board (2, 1) = i and board (2, 2) = i and board (2, 3) = i then             result i         elsif board (3, 1) = i and board (3, 2) = i and board (3, 3) = i then             result i         elsif board (3, 1) = i and board (2, 2) = i and board (1, 3) = i then             result i         elsif board (1, 3) = i and board (2, 3) = i and board (3, 3) = i then             result i         elsif board (1, 2) = i and board (2, 2) = i and board (3, 2) = i then             result i         end if     end for     result 0 end CheckWin proc DrawWord (x, y, c : int, word : string) % Draws words with specified position and color     var fontID : int     fontID := Font.New ("mono:12:bold")     Font.Draw (word, x - (Font.Width (word, fontID) div 2), y, fontID, c)     Font.Free (fontID) end DrawWord proc ShowTurn % Displays the turn of the players     DrawWord (maxx div 2, maxy - 20, black, "Player " + intstr (turn)) end ShowTurn proc Main     var winID : int     winID := Window.Open ("offscreenonly,graphics:200,200,position:centre,centre")     InitBoard % Initialises the game grid     loop         mousewhere (mouseX, mouseY, mouseB)         UseMouse         DrawBoard         ShowTurn         if CheckWin = 1 then             cls             DrawWord (maxx div 2, maxy div 2, black, "Player " + intstr (CheckWin) + " won!")             exit         elsif CheckWin = 2 then             cls             DrawWord (maxx div 2, maxy div 2, black, "Player " + intstr (CheckWin) + " won!")             exit         elsif turnCount = 9 and CheckWin = 0 then             cls             DrawWord (maxx div 2, maxy div 2, black, "Tie!")             exit         end if         View.Update         cls     end loop end Main % Main Main

[Zren]

Other than converting the CheckWin if statements into one long unreadable line. Though that would make it faster to type, and less readability.
It's nicely done for a simple game of tic tac toe.

Though I'm all about code shortening so we could make this:

Turing:

if CheckWin = 1 then             cls             DrawWord (maxx div 2, maxy div 2, black, "Player " + intstr (CheckWin) + " won!")             exit         elsif CheckWin = 2 then             cls             DrawWord (maxx div 2, maxy div 2, black, "Player " + intstr (CheckWin) + " won!")             exit end if

Into this:

Turing:

if CheckWin = 1 | CheckWin = 2 then             cls             DrawWord (maxx div 2, maxy div 2, black, "Player " + intstr (CheckWin) + " won!")             exit end if

| can be replaced with or by the way.

[DemonWasp]

In InitBoard:
- Be aware that lower ( board ) and upper ( board ) will return only the length of the first dimension. While this code will work for square arrays (which is all you need), it won't work for anything non-square.


In CheckForWin:
- You don't need to check for both players, you just need to check that the parts of the array have the same value. If you compare (1,1) to (2,2) and find them to be equal and that (2,2) = (3,3), then you know that someone has won. You can then just use the value of any one of those squares to see who won.
- You could also use for loops over the array to shorten the amount of code you'd have to write.


In DrawWord:
- With frequently-used resources like fonts, it's probably best to create them once, cache them, then free them only when you're really done with them. Right now, every time you DrawWord, Turing has to load the requested font before it can draw.


Not bad though.

[A.J]

Good job, metachief

Well, I would suggest that you store the tic tac toe board in a one dimensional array instead of a two dimensional array.

So, for example, the position (1, 1). would become 1, and the position (1, 2) would become 2, etc...
(something like this:
____________
| 1 | 2 | 3 |
| 4 | 5 | 6 |
| 7 | 8 | 9 |
____________

Using a simple hashing function to hash coordinates on the board to a single integer would be pretty easy (for example, in this case, given x and y, (x, y) on the board is y + 3*(x-1)). And, this can make it easier to work with.