You will write a Java program that implements Conway’s Game of Life, a simple cellular automaton...
You will write a Java program that implements Conway’s Game of Life, a simple cellular
automaton discussed in class. See for example:
http://www.bitstorm.org/gameoflife/
Our version has a 10 x 10 grid, numbered like this:
0 1 2 3 4 5 6 7 8 9
0
1
2
3
4
5
6
7
8
9
The grid is represented by a 10 x 10 2dimensional integer array. If the grid point (i, j) is
"populated", the array element [i][j] contains 1; otherwise it contains 0. Elements along
the edges, i == 0 or 9, or j == 0 or 9, are always unpopulated.
When we display the grid, a populated cell is indicated by a ‘#’; an unpopulated cell is
indicated by a space.
What your program should do:
• Prompt the user to enter a list of (i,j) pairs, both nonnegative integers
(stop when a negative integer is read for either i or j)
• Prompt the user to enter the number of time steps
• Initialize the grid based on the (i,j) pairs entered by the user (set the
corresponding array elements to 1)
• Display the initial state of the grid (call the displayGrid() method)
• For each time step,
* update the grid according to Conway’s rules (call the updateGrid()
method)
* display the grid (call the displayGrid() method)
We follow Conway’s standard rules for updating the cells, as follows. A "neighbor" is a
vertically, horizontally, or diagonally adjacent cell that is populated. A cell has anywhere
from 0 to 8 neighbors.
• For a cell that is "populated", if the cell has <= 1 neighbors, or >= 4 neighbors, it
dies (becomes 0). Otherwise, it survives (remains 1).
• For a cell that is not populated, if the cell has exactly 3 neighbors, it becomes
populated (becomes 1).
• Cells on the edge always remain unpopulated (0).
The displayGrid() method has prototype:
void displayGrid(int mat[][]);
It displays the borders of the grid (see sample runs below), and prints the 10 x 10 grid of
cells. Populated cells are indicated with a ‘#’ sign, unpopulated cells with a space.
The updateGrid() method has prototype:
void updateGrid(int mat[][]);
mat is the 2dimensional array that contains the current state of the grid. The method
counts the neighbors in each cell of the grid, and updates the cells in the grid according to
Conway’s rules.
You can obviously enter arbitrary initial conditions to test your program. Some fun
examples adapted from the bitstorm.org site:
Glider: initial populated cells at (3, 4) (4, 2) (4, 4) (5, 3) (5, 4)
5 cell row: initial populated cells at (6,4) (6,5) (6,6) (6,7) (6,8)
Small explosion: initial populated cells at (4, 5) (4,6) (5,4) (5,5) (5, 7) (6, 5) (6,6)
Note that these will not behave exactly the same as the bitstorm.org display when the
populated cells run into the boundary conditions.
Sample runs:
unixlab% java Life
Please enter list of (i,j) pairs for populated cells
(negative i or j to quit): 6 4 6 5 6 6 6 7 6 8 -1 -1
Enter number of time steps: 5
Initial grid:
0123456789
0
1
2
3
4
5
##### 6
7
8
9
Time step 1
0123456789
0
1
2
3
4
### 5
### 6
### 7
8
9
Time step 2
0123456789
0
1
2
3
# 4
# # 5
# # 6
# # 7
# 8
9
Time step 3
0123456789
0
1
2
3
# 4
### 5
## ## 6
### 7
# 8
9
Time step 4
0123456789
0
1
2
3
### 4
# # 5
# # 6
# # 7
### 8
9
Time step 5
0123456789
0
1
2
# 3
### 4
# # # 5
### ## 6
# # # 7
### 8
9
unixlab% java Life
Please enter list of (i,j) pairs for populated cells
(negative i or j to quit): 4 5
4 6
5 4
5 5
5 7
6 5
6 6
-1 -1
Enter number of time steps: 5
Initial grid:
0123456789
0
1
2
3
## 4
## # 5
## 6
7
8
9
Time step 1
0123456789
0
1
2
3
### 4
# # 5
### 6
7
8
9
Time step 2
0123456789
0
1
2
# 3
### 4
# # 5
### 6
# 7
8
9
Time step 3
0123456789
0
1
2
### 3
### 4
# # 5
### 6
### 7
8
9
Time step 4
0123456789
0
1
# 2
# # 3
# # 4
# # 5
# # 6
# # 7
# 8
9
Time step 5
0123456789
0
1
# 2
### 3
## ## 4
### ### 5
## ## 6
### 7
# 8
9
Homework Answers
Program:
![for (int c= 1; c < cols-1; c++) //detects for 1s, when find 1, calc number of neighbors if (mat [r] [c] -1) num1 for = 0 ; (i](http://img.homeworklib.com/questions/02f191b0-3839-11eb-b140-03927fa3d0c9.png?x-oss-process=image/resize,w_560)
Sample output:
Code to copy:
//Import statements.
import java.util.Scanner;
//ConwaysGameofLife class.
public class ConwaysGameofLife
{
//Main function.
public static void main(String[] args)
{
//Variable declaration.
final int rows = 10, cols =10;
int[][] tab = new int[rows][cols];
//Create an object to the Scanner.
Scanner sc = new Scanner(System.in);
int tempI, tempJ, steps;
//Prompt the user to enter the list of pairs.
System.out.println("Please enter list of (i,j) pairs for populated cells (negative i or j to quit): ");
while (true)
{
tempI = sc.nextInt();
tempJ = sc.nextInt();
if (tempI >= 0 || tempJ >= 0){
tab[tempI][tempJ] = 1;
} else {
break;
}
}
//Prompt the user to enter the time steps.
System.out.println("Enter number of time steps: ");
steps = sc.nextInt();
//display, update with pause and display step #
for (int i = 0; i <= steps; i++)
{
if (i == 0)
{
System.out.println("Initial grid:");
}
else
{
System.out.println("Time step " + i);
}
displayGrid(tab, rows, cols);
updateGrid(tab, rows, cols);
try
{
Thread.sleep(1000);
}
catch (Exception e)
{
System.out.println("Error dumb ass lmao");
}
}
}
//display # with cells with 1s, otherwise blank space
public static void displayGrid(int mat[][], int rows, int cols)
{
for (int i = 0; i < rows; i++)
{
if (i < 10)
{
System.out.print(i);
} else {
System.out.print(i%10);
}
}
System.out.print("\n");
for (int j = 0; j < rows; j++)
{
for (int k = 0; k < cols; k++)
{
if (mat[j][k] == 1){
System.out.print("#");
} else {
System.out.print(" ");
}
}
System.out.print(" " + j + "\n");
}
}
//update cells method
public static void updateGrid(int mat[][], int rows, int cols)
{
int numN = 0;
int temp[][] = new int[rows][cols];
//sync temp to mat
for (int a = 0; a < rows; a ++)
{
for (int b = 0; b < cols; b++)
{
temp[a][b] = mat[a][b];
}
}
for (int r = 1; r < rows-1; r++)
{
for (int c = 1; c < cols-1; c++)
{
//detects for 1s, when find 1, calc number of neighbors
if (mat[r][c] == 1)
{
numN = 0;
for (int j = r-1; j <= r+1; j++)
{
for (int k = c - 1; k <= c + 1; k++)
{
if (mat[j][k] == 1)
{
numN += 1;
}
}
}
if (numN-1 <= 1 || numN-1 >= 4)
{
temp[r][c] = 0;
}
} else {
//otherwise find if there are 3 neighbors nearby
numN = 0;
for (int j = r-1; j <= r+1; j++)
{
for (int k = c - 1; k <= c + 1; k++)
{
if (mat[j][k] == 1)
{
numN += 1;
}
}
}
if (numN == 3)
{
temp[r][c] = 1;
}
}
}
}
//sync mat to temp
for (int a = 0; a < rows; a ++)
{
for (int b = 0; b < cols; b++)
{
mat[a][b] = temp[a][b];
}
}
}
}
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