Connect 4 Game Design Java

Problem:

Connect four is a two-player board game in which the players alternately drop colored disks into a seven-column, six-row vertically suspended grid, as shown below.

The objective of the game is to connect four same-colored disks in a row, a column, or a diagonal before your opponent can do likewise. The program prompts two players to drop a RED or YELLOW disk alternately. Whenever a disk is dropped, the program re-displays the board on the console and determines the status of the game (win, draw, or continue).

Output:

| | | | | | | |
| | | | | | | |
| | | | | | | |
| | | | | | | |
| | | | | | | |
| | | | | | | |
---------------
Drop a red disk at column (0–6):
0
| | | | | | | |
| | | | | | | |
| | | | | | | |
| | | | | | | |
| | | | | | | |
|R| | | | | | |
---------------
Drop a yellow disk at column (0–6):
1
| | | | | | | |
| | | | | | | |
| | | | | | | |
| | | | | | | |
| | | | | | | |
|R|Y| | | | | |
---------------
Drop a red disk at column (0–6):
2
| | | | | | | |
| | | | | | | |
| | | | | | | |
| | | | | | | |
| | | | | | | |
|R|Y|R| | | | |
---------------
Drop a yellow disk at column (0–6):
3
| | | | | | | |
| | | | | | | |
| | | | | | | |
| | | | | | | |
| | | | | | | |
|R|Y|R|Y| | | |
---------------
Drop a red disk at column (0–6):
4
| | | | | | | |
| | | | | | | |
| | | | | | | |
| | | | | | | |
| | | | | | | |
|R|Y|R|Y|R| | |
---------------
Drop a yellow disk at column (0–6):
3
| | | | | | | |
| | | | | | | |
| | | | | | | |
| | | | | | | |
| | | |Y| | | |
|R|Y|R|Y|R| | |
---------------
Drop a red disk at column (0–6):
2
| | | | | | | |
| | | | | | | |
| | | | | | | |
| | | | | | | |
| | |R|Y| | | |
|R|Y|R|Y|R| | |
---------------
Drop a yellow disk at column (0–6):
1
| | | | | | | |
| | | | | | | |
| | | | | | | |
| | | | | | | |
| |Y|R|Y| | | |
|R|Y|R|Y|R| | |
---------------
Drop a red disk at column (0–6):
2
| | | | | | | |
| | | | | | | |
| | | | | | | |
| | |R| | | | |
| |Y|R|Y| | | |
|R|Y|R|Y|R| | |
---------------
Drop a yellow disk at column (0–6):
1
| | | | | | | |
| | | | | | | |
| | | | | | | |
| |Y|R| | | | |
| |Y|R|Y| | | |
|R|Y|R|Y|R| | |
---------------
Drop a red disk at column (0–6):

1
| | | | | | | |
| | | | | | | |
| |R| | | | | |
| |Y|R| | | | |
| |Y|R|Y| | | |
|R|Y|R|Y|R| | |
---------------
Drop a yellow disk at column (0–6):
2
| | | | | | | |
| | | | | | | |
| |R|Y| | | | |
| |Y|R| | | | |
| |Y|R|Y| | | |
|R|Y|R|Y|R| | |
---------------
Drop a red disk at column (0–6):
4
| | | | | | | |
| | | | | | | |
| |R|Y| | | | |
| |Y|R| | | | |
| |Y|R|Y|R| | |
|R|Y|R|Y|R| | |
---------------
Drop a yellow disk at column (0–6):
3
| | | | | | | |
| | | | | | | |
| |R|Y| | | | |
| |Y|R|Y| | | |
| |Y|R|Y|R| | |
|R|Y|R|Y|R| | |
---------------
Drop a red disk at column (0–6):
4
| | | | | | | |
| | | | | | | |
| |R|Y| | | | |
| |Y|R|Y|R| | |
| |Y|R|Y|R| | |
|R|Y|R|Y|R| | |
---------------
Drop a yellow disk at column (0–6):
3
| | | | | | | |
| | | | | | | |
| |R|Y|Y| | | |
| |Y|R|Y|R| | |
| |Y|R|Y|R| | |
|R|Y|R|Y|R| | |
---------------
The yello player won.

Solution:

          import java.util.Scanner;    //This will be.. a huge code so we will have to make //lots of methods for each aspect of the game public class ConnectFour {   //We need to first create the basic visual pattern    public static String[][] createPattern()   {     //Although the game is more like a table of 6     //columns and 6 rows, we're going to have to make     //a 2D array of 7 rows and 15 columns because graphically     //there's an extra row to show the ___ at the bottom     //and you have double the columns that show | | |      //between each number       String[][] f = new String[7][15];          //Time to loop over each row from up to down     for (int i =0;i<f.length;i++)     {               //Time to loop over each column from left to right        for (int j =0;j<f[i].length;j++)       {         //Note how it is always the even column         //that has the border and the odd column         //between them that will be either empty or         //have a number          if (j% 2 == 0) f[i][j] ="|";         else f[i][j] = " ";                  //Time to make our lowest row         if (i==6) f[i][j]= "-";       }            }     return f;   }      //Yes, we even need to make a new method for visually   //printing our game, but at least it's not hard to do   public static void printPattern(String[][] f)   {     for (int i =0;i<f.length;i++)     {       for (int j=0;j<f[i].length;j++)       {         System.out.print(f[i][j]);       }       System.out.println();     }   }      //Here's are basic move, making the lowest empty row   //of a specific column have a Red   public static void dropRedPattern(String[][] f)   {     //We need to have the user tell us what column he wants     //to drop a red into     //Note: the user isn't supposed to know that we have 15 columns     //starting at index 0 till 14 but just 6 nice ones     System.out.println("Drop a red disk at column (0–6): ");     Scanner scan = new Scanner (System.in);          //Thankfully, there's a simple formula for converting a 1-2-3-4-5-6      //user column number into a 1-3-5-7-9-11-13     int c = 2*scan.nextInt()+1;         //Now that we know our column, we have to loop     //over each row from the bottom to the top     //till we find the first  empty space, drop, and     //then finish (i.e., break) the move     //Note: although as coders we're used to starting from      //0 to the end, here that wouldn't work so well because     //it would involve multiple if statements, but try it out     //on your own if you want to     for (int i =5;i>=0;i--)     {       if (f[i][c] == " ")       {         f[i][c] = "R";         break;       }            }   }      //Same as the above step, just yellow   public static void dropYellowPattern(String[][] f)   {     System.out.println("Drop a yellow disk at column (0–6): ");     Scanner scan = new Scanner (System.in);     int c = 2*scan.nextInt()+1;     for (int i =5;i>=0;i--)     {       if (f[i][c] == " ")       {         f[i][c] = "Y";         break;       }            }   }       //Here's where it gets hard.   //That's because there are basically four patterns   //of Reds or Yellows that can win the game   //One pattern is a horizontal line of four Rs or Ys,   //another is a vertical line, another is a left-up to right-down   //diagonal line, and the last is left-down to right-up diagonal,   //We thus need to code for each type of line   //and the various places where the line can be   public static String checkWinner(String[][] f)   {          //Time to look for the first type of winning line,     //a horizontal line     //This line can be on any row, so let's loop over      //each row starting from 0 to 5 (since 6 is just ___)     for (int i =0;i<6;i++)     {       //On every row, the four-dotted line can look like       //----_ _,  _----_, or _ _----       //Here, _ can be an empty space or one of the colors        //and - is not empty space  AND every - has the same       // color (R or Y)       //Note: since our R/Y/Empty's can only be in odd places,       //because of how we created the pattern in the first       //method, then our count has to be incremented by 2        //and will start from 0 (which will be 1, ----_ _)        //and stop at 6 (which will be 7, _ _----)       for (int j=0;j<7;j+=2)       {         if ((f[i][j+1] != " ")         && (f[i][j+3] != " ")         && (f[i][j+5] != " ")         && (f[i][j+7] != " ")         && ((f[i][j+1] == f[i][j+3])         && (f[i][j+3] == f[i][j+5])         && (f[i][j+5] == f[i][j+7])))          //If we found a same-colored pattern, we'll return          //the color so that we will know who won           return f[i][j+1];         }     }          //For a vertical line, let's first loop over each     //odd-numbered column by incrementing with 2     //and check for consecutive boxes in the same column     //that are the same color     //Note: make sure you understand the horizontal line's     //codes first or else everything below this point will      //make no sense to you     for (int i=1;i<15;i+=2)     {       //Of course, our lines will look like ----__ but reversed       //and there is need to our rows by 2 but just one       //and we have to start at the vertical version of ----__ and        //and stop at _ _ ---- so it's from 0 to 2       for (int j =0;j<3;j++)       {             if((f[j][i] != " ")             && (f[j+1][i] != " ")             && (f[j+2][i] != " ")             && (f[j+3][i] != " ")             && ((f[j][i] == f[j+1][i])             && (f[j+1][i] == f[j+2][i])             && (f[j+2][i] == f[j+3][i])))               return f[j][i];         }       }          //For the left-up to right-down diagonal line     //We'll have to loop over the 3 uppermost     //rows and then go from left to right column-wise     for (int i=0;i<3;i++)     {              //As expected, our uppermost box will start from 1       //and increase by 2 until it becomes 7 (the 3rd box       //on a row)       //Note how we used 1 instead 0 for the count here       //There's no real reason to use 1 instead of 0 or        //vice versa, since we're still using an odd index       //for the columns and incrementing by 2       for (int j=1;j<9;j+=2)       {             if((f[i][j] != " ")             && (f[i+1][j+2] != " ")             && (f[i+2][j+4] != " ")             && (f[i+3][j+6] != " ")             && ((f[i][j] == f[i+1][j+2])             && (f[i+1][j+2] == f[i+2][j+4])             && (f[i+2][j+4] == f[i+3][j+6])))               return f[i][j];         }       }          //Similar to the method above, but we're just reversing our     //trajectory, i.e. we're starting from the rightmost column     //instead of the leftmost like we did above     for (int i=0;i<3;i++)     {       for (int j=7;j<15;j+=2)       {             if((f[i][j] != " ")             && (f[i+1][j-2] != " ")             && (f[i+2][j-4] != " ")             && (f[i+3][j-6] != " ")             && ((f[i][j] == f[i+1][j-2])             && (f[i+1][j-2] == f[i+2][j-4])             && (f[i+2][j-4] == f[i+3][j-6])))               return f[i][j];         }       }          //If after going over the table and we find no     //same colored lines, then we have to return something     //that says that we didn't find a winning color :P     return null;   }      //The easy part: using these methods   public static void main (String[] args)   {     //Time to make a pattern     String[][] f = createPattern();     //Time to make a condition for our game to keep on     //playing     boolean loop = true;     //We need something to keep track of whose turn it is     int count = 0;     printPattern(f);     while(loop)     {        //Let's say that Red gets the first turn and thus        //every other turn         if (count % 2 == 0) dropRedPattern(f);        else dropYellowPattern(f);        count++;//We need to keep track of the turns        printPattern(f);        //Let's say we want to check for a winner during every        //turn made and say who it is        if (checkWinner(f) != null)        {           if (checkWinner(f) == "R")              System.out.println("The red player won.");           else if (checkWinner(f)== "Y")             System.out.println("The yello player won.");          //Well, if someone one, then the game has to end          loop = false;     }   } } }