Turing Connect Four Stacking Problem

[Michael3176]

What is it you are trying to achieve?
I do not know how to make counters stack on top of the other, for example if the user chooses i column twice, i do not know how to recognize that on Turing.
I am trying make my connect four error traps and make it so the user cannot choose a full coloumn.

What is the problem you are having?
My error traps dont work properly and i dont know how to make counters stack on top of the other

Post any relevant code (You may choose to attach the file instead of posting the code if it is too long)
<Answer Here>

Turing:

import GUI var mainWin := Window.Open ("position:300;300, graphics:660;500") var coloumn : string var player : int := 1 var done : boolean forward proc mainMenu proc title     cls     locate (1, 36)     put "Connect Four"     put "" end title proc switch     if player = 1 then         player := 2     else         player := 1     end if end switch proc pauseProgram     var reply : string (1)     put ""     put "Press any key to continue...." ..     getch (reply) end pauseProgram proc intro     title     var menuButton := GUI.CreateButton (295, 50, 0, "Main Menu", mainMenu) end intro proc instructions     title     put         "Connect Four is a two player game on which playes take turns dropping tokens into a 7x6 grid falling vertically. Players win by getting four tokens in a row in any straight direction. Click on one of the top cirlces to drop a token."     var menuButton := GUI.CreateButton (295, 50, 0, "Main Menu", mainMenu) end instructions proc checkWin     done := false     if whatdotcolour (65, 35) = red and whatdotcolour (150, 35) = red and whatdotcolour (240, 35) = red and whatdotcolour (330, 35) = red then         locate (2, 1)         put "Player one has won."         done := true     elsif             whatdotcolour (65, 35) = blue and whatdotcolour (150, 35) = blue and whatdotcolour (240, 35) = blue and whatdotcolour (330, 35) = blue then         locate (2, 1)         put "Player two has won."         done := true     else         done := false     end if end checkWin proc processing     if coloumn = "1" then         if player = 1 then             drawfilloval (65, 35, 20, 20, red)             %draw the rest of ovals ontop         else             drawfilloval (65, 35, 20, 20, blue)         end if     elsif coloumn = "2" then         if player = 1 then             drawfilloval (150, 35, 20, 20, red)             %draw the rest of ovals ontop         else             drawfilloval (150, 35, 20, 20, blue)         end if     elsif coloumn = "3" then         if player = 1 then             drawfilloval (240, 35, 20, 20, red)         else             drawfilloval (240, 35, 20, 20, blue)         end if     elsif coloumn = "4" then         if player = 1 then             drawfilloval (330, 35, 20, 20, red)         else             drawfilloval (330, 35, 20, 20, blue)         end if     elsif coloumn = "5" then         if player = 1 then             drawfilloval (420, 35, 20, 20, red)         else             drawfilloval (420, 35, 20, 20, blue)         end if     elsif coloumn = "6" then         if player = 1 then             drawfilloval (505, 35, 20, 20, red)         else             drawfilloval (505, 35, 20, 20, blue)         end if     else         if coloumn = "7" then             if player = 1 then                 drawfilloval (595, 35, 20, 20, red)             else                 drawfilloval (595, 35, 20, 20, blue)             end if         end if     end if     switch end processing proc display     for x : 10 .. 310 by 50         drawline (20, x, 635, x, black)     end for     for x : 20 .. 640 by 88         drawline (x, 10, x, 310, black)     end for end display proc userInput     title     display     loop         locate (2, 1)         put "Please enter a coloumn number: " ..         get coloumn         if coloumn not= "1" and coloumn not= "2" and coloumn not= "3" and coloumn not= "4" and coloumn not= "5" and coloumn not= "6" and coloumn not= "7" then             put "Please enter coloums 1-9 only: " ..             get coloumn         else             processing             display             checkWin         end if         exit when done = true     end loop end userInput proc goodBye     title     put "Made by the."     delay (1300)     Window.Close (mainWin) end goodBye body proc mainMenu     title     var intrusctionsButton := GUI.CreateButton (295, 300, 100, "Instructions", instructions)     var newGameButton := GUI.CreateButton (295, 200, 100, "New Game", userInput)     var exitB := GUI.CreateButtonFull (295, 100, 100, "Exit", GUI.Quit, 0, KEY_ESC, false) end mainMenu intro loop     exit when GUI.ProcessEvent end loop goodBye

[Panphobia]

Make an array of integers, and add one to each column until it is full, and then if counter[a] > whatever then give him a message to choose an empty column, or just dont process anything if it is over, and just return to the current persons turn

[Insectoid]

I suggest you create a 2D array of int to represent the game board. When coordinate x, y is selected, you set that position in the array to 1 or 2 depending on the player that selected it. If a player tries to select a position that is anything but 0, force him to pick a new spot.

[Raknarg]

http://compsci.ca/v3/viewtopic.php?t=14333

[Michael3176]

Could you give me a code example, Im new to turing and dont quite understand, can arrays use (x,y)coordinates?

[Panphobia]

Arrays don't specifically store just xy coords, but if you make a 2D array of integers in each slot you can store an x or a y coord, like I would make a 2D array of how many slots there are in connect four in one column times 2 so like counter(6)(0)=x counter(6)(1)=y

[Raknarg]

Ok. Lets say I wanted to make a map of 25 tiles. I could do something like this:

Turing:

var map : array 1 .. 5, 1 .. 5 of int

Lets assume that all the values are at zero (they're not, we're assuming we already set their values). Now if we draw the map like a coordinated plane, we would end up with something like this:

code:

(y)  5 | 0 0 0 0 0  4 | 0 0 0 0 0  3 | 0 0 0 0 0  2 | 0 0 0 0 0  1 | 0 0 0 0 0      1 2 3 4 5 (x)

cool. Now lets make this a little more relatable. Lets say this is connect four, and I dropped a black token in the centre. Lets say that on this map, 0 is empty, 1 is black and 2 is red. We now have this:

code:

(y)  5 | 0 0 0 0 0  4 | 0 0 0 0 0  3 | 0 0 0 0 0  2 | 0 0 0 0 0  1 | 0 0 1 0 0      1 2 3 4 5 (x)

The next turn, red wants to drop it directly to the left. Now we have this:

code:

(y)  5 | 0 0 0 0 0  4 | 0 0 0 0 0  3 | 0 0 0 0 0  2 | 0 0 0 0 0  1 | 0 2 1 0 0      1 2 3 4 5 (x)

Looks good. Now here comes the part you're having trouble with. Lets say black wants to put another token in the middle. It wont simply drop to the bottom. It's got to land on top of the old black piece. But how will the problem know?

It's easy!

5 | 0 0 0 0 0

Looks like there's nothing on this line. Maybe there's another space down.

4 | 0 0 0 0 0

Nope. Lets keep going.

3 | 0 0 0 0 0

Nope.

2 | 0 0 0 0 0

Nope again.

1 | 0 2 1 0 0

Look now. The center space is taken up. Therefore, the next piece would land on top of it.

Now taking that information into account, we have this:

code:

(y)  5 | 0 0 0 0 0  4 | 0 0 0 0 0  3 | 0 0 0 0 0  2 | 0 0 1 0 0  1 | 0 2 1 0 0      1 2 3 4 5 (x)

Hope that makes more sense now.

[Michael3176]

so how would i set up a 2d array of the board using code? Then i could indicate everytime the user would choose a coloumn, indicate that the coodinate was chosen using an array right, and if that coordinate was chosen, if they choose that coloumn again, place it ontop, but how would i now recognize when they have chosen a coloumn twice, would i make a "tries" variable and keep adding one, and if it was two, then locate above it?

[Raknarg]

You set up the array basically how I just did, however many slots you want as x, and however many slots as y. Basically the player is just choosing which x coordinate they want to put their token in.

So all you have to do is use a for loop to search through the array. At anytime if the slot at the x of the players choice is not zero, then you know you need to place a token above it. If it's the top of the row, then they need to make another choice.

[Michael3176]

ok so i did a setup proc like this

procedure setup
for x : 1 .. 7
for y : 1 .. 6
board (x, y) := 0
end for
end for
end setup

and in my processing i did

if coloumn = "1" then
if player = 1 then
drawfilloval (65, 35, 20, 20, red)
board (1, 1) := 1
else
drawfilloval (65, 35, 20, 20, blue)
board (1, 1) := 2
end if
now im stuck i dont no what to write to make them stack

[Raknarg]

First o all, do all your drawing outside the loop. You usually want to have them all together, so that your main program flows like this:

code:

loop      input ()      processing ()      output/drawing () end loop

It would be easier if you rewrote it, but we dont necessarily have to do that...

so now when you get to this part, you have an x coordinate, but you need to find your y coordinate. We can do exactly what we talked about.

so in each statement, you have a for loop to search the column.

Turing:

if coloumn = "1" then      y := search_column()           if player = 1 then           drawfilloval (65, 35, 20, 20, red)           board (1, y) := 1      else           drawfilloval (65, 35, 20, 20, blue)           board (1, y) := 2           end if

Now you might be thinking there should be an easier way to do this, and you would be correct. This is set up poorly and could be done much easier.

[Raknarg]

Or another neat way is using another array to keep track of the next slot open in each column. So like an array 1 .. 7, and all the elements start out at one. Each selection, you set the y coordinate of the token the open slot number of that array.

so

map (x, slotOpen (x)) := player
slotOpen (x) += 1

Something like that

[Michael3176]

So in the first idea you had, what would the for loop look like in code that your talking about, and I don't quite understand how the second one would work

[jr5000pwp]

So I wanted to make an example to show you a similar implementation of a 2D array, but I might have went slightly overboard.
This is a column based basin filler. You can set the rows and columns, as well as the maximum amount of water per cell. It fills by randomly picking columns and filling a random amount from 0 to MaxLevel*2, that's why it slows down near the end.
Hopefully you can learn something from this.

Turing:

%Only updates the screen when View.Update is called to provide smooth animation setscreen ("offscreenonly, nocursor") %Number of columns, beware of odd output from too many const numColumns : int := 20 %Number of rows, beware of odd output from too many const numRows : int := 24 %The maximum level a cell can reach const maxLevel : real := 99 %A 16x16 array of real numbers representing the water levels of different cells var basin : array 1 .. numColumns, 1 .. numRows of real %The array representing if a column is full var columnFull : array 1 .. numColumns of boolean %The amount of digits in the maxLevel, used for formatting var numDigits : int := 1 %This procedure initializes the basin array procedure initArray     %Loop through all the columns     for x : 1 .. numColumns         %Set the column to not-full         columnFull (x) := false         %Loop through all the rows         for y : 1 .. numRows             %Initialize the cell to 0             basin (x, y) := 0.0         end for     end for end initArray %Determines the number of digits in the maxLevel for formatting procedure setNumDigits     %Starts off with numbers with 1 digit (Less than 10)     var digitsCounter : int := 1     var digits : int := 10     %Loop until you find the number of digits     loop         %If the maxlevel is less than the current digits counter then we've found the number of digits         if maxLevel < digits then             numDigits := digitsCounter             exit         end if         %Multiplying by 10 will add a 0 on to the end of the number         digits *= 10         %Increment the number of digits         digitsCounter += 1     end loop end setNumDigits %This function checks if all cells have been filled. function isFull : boolean     %Loop through all columns     for x : 1 .. numColumns         %If the column isn't full, then the basin isn't full         if not columnFull (x) then             result false         end if     end for     %If no columns are not full, then they are all full and the basin is full     result true end isFull %This function takes an amount and a column and attempts to add that amount to the column, it returns the amoun that couldn't be added. function addWater (amount : real, x : int) : real     %If the column is already full then just result the amount input     if columnFull (x) then         result amount     else         %The variable representing the remaining fluid, used to partially fill cells         var amountToAdd : real := amount         %Loop through each row of the column         for i : 1 .. numRows             %If there is still room left in the cell             if basin (x, i) not= maxLevel then                 %Determine how much we can add to the cell, if there is enough room for all of the water, do so, otherwise fill the cell to the max                 var amountIncrement : real := min (amountToAdd, maxLevel - basin (x, i))                 %Add the amount to the cell                 basin (x, i) += amountIncrement                 %Remove the amount from the remaining fluid                 amountToAdd -= amountIncrement                 %Once you have used all the remaining fluid, there is no point to continue on the column, so exit                 exit when amountToAdd = 0.0             end if         end for         %If there is still water remaining this column is now identified as full         if amountToAdd not= 0.0 then             columnFull (x) := true         end if         %Result the remaining amount of water         result amountToAdd     end if end addWater %This procedure draws the basin in text procedure drawBasin     %Loop through all columns     for x : 1 .. numColumns         %Loop through all rows         for y : 1 .. numRows             %Set the cursor position to be of this tile             locate (y, (x - 1) * (2 + numDigits) + 1)             %Print the amount in the basin rounded to the nearest 10th. There's probably a better way to do this             put round (basin (x, numRows - y + 1) * 10) / 10         end for     end for end drawBasin %This is the main procedure, it will gradually fill the basin procedure fillBasin     %Loop until the basin is full     loop         %Choose a random column         var x : int := Rand.Int (1, numColumns)         %Call the function to add the water and store the remaining water in a variable in case we want to use it later         var amountResulted : real := addWater (Rand.Real * maxLevel * 2, x)         %Draw the basin         drawBasin         %Update the screen         View.Update         %When the basin is full, stop running the program         exit when isFull         %Clear the screen         cls     end loop end fillBasin %Initialize the arrays first to avoid errors initArray %Determine the number of digits used for formatting setNumDigits %Fill the basin fillBasin

[Raknarg]

Turing:

for decreasing i : 6 . . 0    if i = 0 then       slots (choicex, 1) = player       valid_move = true       exit    if slots (choicex, i) not= 0 then       if i =  6 then          valid_move = false          exit       else          slots (choicex, i + 1) = player          valid_move = true          exit       end if    end if end for

This is the first idea. You go through the column. If the slot is empty, move on. If it reaches a filled slot, and its the top slot, no moves are possible. Otherwise, set the token to the slot above. If you reach 0 and it's empty, put it in the first slot.