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.
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.
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;
}
}
}
}
