Conway's Game of Life

Hi there,

I am a first year Programming student, and this is my first time asking for assistance. Due to Programming not being what I expected, this will probably be the last time I ever do any form of Programming. I have been put off for life.

Basically we have been given a task where we have to write a C version of John Conway's Game of Life, using a 20x20 board and the pattern shown in my code (below).

Code:

#include <stdio.h> #include <string.h> void display (char b1[][21], char b2[][21]); void new_board (char b1[][21], char b2[][21]); void copy_back (char b1[][21], char b2[][21]); int next_cells (int p, int q); char b1[20][21] = {"**..................", "**..................", "....................", "....................", "....................", "....................", "....................", "........***.........", "........*...........", ".........*..........", "....................", "....................", "....................", "....................", "....................", "....................", "....................", "....................", "....................", "...................."}; char b2[20][21]; int main() { char c; display(b1, b2); printf("Hi! Would you like to play the Game of Life? (y/n)"); c = getchar(); while (c == 'y') { new_board(b1, b2); copy_back(b1, b2); display(b1, b2); c = getchar(); } } void display(char b1[][21], char b2[][21]) { int i; for (i=0; i<20; i++) { printf("%s \n", b1[i]); } } void new_board(char b1[][21], char b2[][21]) { int i, j, p; for (i=0; i<20; i++) { for (j=0; j<20; j++) { p = next_cells(i,j); if(b1[i][j]=='*') { if (p ==2 || p == 3) { b2[i][j] = '*'; } else { b2[i][j] = '.'; } } else { if (p == 3) { b2[i][j] = '*'; } else { b2[i][j] = '.'; } } } } b2[i][j] = '\0'; } void copy_back(char b1[][21], char b2[][21]) { int i; for (i=0; i<20; i++) { strcpy(b1[i], b2[i]); } } int next_cells(int p, int q) { int count; if (p!=0 && q!=19 && b1[p-1][q+1] == '*') { count = p + q; } return count; }

Unable to get the program working myself, I spoke to the lecturer who gave the following tips, however I have been unable to implement them successfully. Some extra guidance would be really amazing.

* I should use getch rather than getchar (since it doesn't clear the buffer).

* Basically my next_calls function is awful, and I am only using 1 IF when I should be using another 7, because I am only counting one of the neighboring cells.

* I need to increment my count variable.

* In every one of the 8 IFs, I should be protecting against the possibility that [p][q] is an edge entry on b1.

What did you expect programming to be? The other month I made a Conway's GOL program and loved it. It's such a shame to hear that you didn't enjoy it.

Quote:

* I should use getch rather than getchar (since it doesn't clear the buffer).
"getch()" doesn't wait for a new line character. When you use getchar(), the function waits for a new line character and then leaves the new line character in the stdin. You will need to include conio.h.

* Basically my next_calls function is awful, and I am only using 1 IF when I should be using another 7, because I am only counting one of the neighboring cells.
You need to look at
row -1, column -1
row 0, column -1
row +1, column -1
row -1, column 0
row +1, column 0
row -1, column +1
row 0, column +1
row +1, column +1

That is every neighbour.

* I need to increment my count variable.
Are you sure that they didn't say initialise? i.e. count=0;

* In every one of the 8 IFs, I should be protecting against the possibility that [p][q] is an edge entry on b1.
I can see that you are trying to do that with the p!=0 and q!=19, but you should test if p!=19 and q!=0 as well

I would have enjoyed my Programming class more, if I didn't have the added stress of three other papers throwing assignments at me. I can understand why there is such a high drop out/failure rate, especially for this class. I can't learn under pressure.

Thanks for the information you provided Click_Here.

I also thought that the fact that your teacher said that your function was "awful" is a bit rough - Words like that can curb anyone's enthusiasm.
Sure, it need a bit of work, but a good teacher should make you want to work on it.

OK, so now I got that out of the way!

I think that the first thing you should do is pick a random spot in the middle, give it (say) 3 neighbours and see if you can get your next_cell function to find them - Move them around and see if you are not missing anywhere around the element.

Only after you have that working, pick a spot near the edge - add if statements that can tell if the element is at the edge - "if current element is not on column 0, check column -1"

If you get into any troubles with your code, post it and we'll help you.

Quote:

Originally Posted by Click_here

"getch()" doesn't wait for a new line character. When you use getchar(), the function waits for a new line character and then leaves the new line character in the stdin. You will need to include conio.h.

Not quite. The terminal (or some equivalent, depending on OS) waits for a complete line, then getchar() reads ones character from that line. The rest of the characters are left in the buffer.

See FAQ > Flush the input buffer - Cprogramming.com for a method to clear the input buffer without using non-standard hacks like fflush(stdin) and getch().

Quote:

Originally Posted by christop

fflush(stdin)

You should never use fflush(stdin).
FAQ > Why fflush(stdin) is wrong - Cprogramming.com

I need to protect against the possibility that [p][q] is an edge entry on b1. I not sure how I go about writing this.

You have already started this in your code:

Code:

if (p!=0 && q!=19 && b1[p-1][q+1] == '*') {

p == 0 is on the edge -> q == 19 is on the edge

Don't forget to check for p == 19 and p == 0 (other edges)

I'm just going to have to hand this in as is. This particular assignment will fail, but I won't end up with an incomplete. I know what this program has to do, and understand how the game works - I just can't write it.

It might help writing in comments saying what you want the program to do at particular points

/* Check to see if not on the edge */
/* Check to see if row-1 column -1 is set */

That way the marker can give you extra points even though the program isn't working.

It might also help if after you hand the program in, that you ask your teacher for a time to go through your program with you and get it working - I'm sure that they'll appreciate the effort.

When I did my own version of this program a few months ago, I used a common method for dealing with edge conditions. Simply add an extra border to the array, and define those cell as always being "dead." This way, when you're testing an edge, you can use the same tests you use everywhere else since you have an extra layer (i.e. you're protected against going out of bounds of your array).

Code:

5x5 board (5x5 array) that requires special cases on edge conditions: ..... ..... ..... ..... ..... 5x5 board (7x7 array) with extra border so you will never encounter an edge condition: xxxxxxx x.....x x.....x x.....x x.....x x.....x xxxxxxx x = not part of the board, always dead

I understand what you are saying, however I really would not have a clue about how to add an extra border to the array let alone finish this assignment. I have another one I have to start. Thanks for your input anyway.

Quote:

Originally Posted by Anhur

* I should use getch rather than getchar (since it doesn't clear the buffer).

That's just wrong ( both normally and ethically )!
"getch()" doesn't care about the buffer, it grabs the character the user just pressed.
"getchar()" doesn't clear the buffer, it just grabs one character from it( and removes it of course ).
( Your teacher told you these?? OMG!!! :mad: )

You can use "getchar()" to clear the input buffer, but inside a loop, not by itself.

Would someone be able to help me write the IF Statements?

Quote:

Originally Posted by Anhur

Would someone be able to help me write the IF Statements?

I haven't been following your thread Anhur, but I'm familiar with Conway's Game of Life. Can you be more specific about your problem?

Are you having trouble with your edges (going out of bounds), or ?

There need to be 8 IF statements within my next_calls function. One IF for each of the neighboring cells. At the moment I have only looked at [p-1][q+1], which I think is the north east cell.

What are the variables p and q - are those row and col, then?

Been awhile since you posted your code. Please post your most recent code.

The if() statements are not difficult. All of them are simply slight modifications of the one you have, as listed earlier in this thread.

Code:

if(neighbor cell at 12 o'cock) is the same as you, sameVariable++ else differentVariable++ if(neighbor cell at 2 o'cock) is the same as you, sameVariable++ else differentVariable++ if(neighbor cell at 3 o'cock) is the same as you, sameVariable++ else differentVariable++ if(neighbor cell at 4 o'cock) is the same as you, sameVariable++ else differentVariable++ if(neighbor cell at 6 o'cock) is the same as you, sameVariable++ else differentVariable++ if(neighbor cell at 8 o'cock) is the same as you, sameVariable++ else differentVariable++ if(neighbor cell at 9 o'cock) is the same as you, sameVariable++ else differentVariable++ if(neighbor cell at 10 o'cock) is the same as you, sameVariable++ else differentVariable++

The only thing left is to add the test for the edges. Post up your code and I'll be back in one hour.

My code hasn't changed. I just can't get my head around all the errors.
I can't even write the additional IF statements.

Hey there again

I suggest that you spend some time thinking about how to find if the element north is on not worrying about whether the element is at an edge or not (that can come in later) -> Hint: Use array notation.

Can you make a if statement that says, "if Element North is on, printf("On"); Else printf("Off");" - Show us your code that does this when you have done that

After that you need to think: Can you then do it for East, South and West? And then North-East, South-East, South-West and North-West.

Code:

int next_cells(int p, int q) { int count; if (p != 0 && q != 19 && b1[p-1][q+1] == '*') { count = p + q; } else { if (p != 0 && q != 19 && [p-1][q-1] == '*') { count = p + q; } } } return count; }

Is this even remotely correct?

I thought that I'd clarify my North/South notation

[1][2][3]
[4][5][6]
[7][8][9]

If '5' is the element under question: '2' is north, '8' South, '6' East, '3' North-East, ect...

Think -> If this was a 2D array, how would you test to see if the element North-West was on?

You are starting to be on the right track -

You are using p as the row and q as the column -> You should rename them "row" and "col" so it is easier for you to visualise the problem

b1[p-1][q+1] is North-East

Some pointers on your code:

Code:

int next_cells(int p, int q) { int count; //Needs to start at 0 /* The checks here are to make sure 0 <= p <= 19 * and 0 <= q <= 19. That is why you are testing * for p != 0 and q != 19. I think that you have * missed that point in your second test */ if (p != 0 && q != 19 && b1[p-1][q+1] == '*') { /* Why isn't this "count = count + 1; ?*/ count = p + q; } else { /* This should not be in an 'else' for the first * 'if' statement, but another 'if' statement all * together. As it is at the moment, this code * will only excecute if North-East is not on */ if (p != 0 && q != 19 && [p-1][q-1] == '*') { /* Why isn't this "count = count + 1; ?*/ count = p + q; } } } return count; }

Quote:

Originally Posted by Anhur

Code:

int next_cells(int p, int q) { int count; if (p != 0 && q != 19 && b1[p-1][q+1] == '*') { count = p + q; } else { if (p != 0 && q != 19 && [p-1][q-1] == '*') { count = p + q; } } } return count; }

Is this even remotely correct?

Well... :cool:

I'll use r for row and c for column. P and q don't work as clearly, imo.

Code:

int CountNeighbors(int r, int c) { //array is global? if not, add it as a parameter here int count = 0; if(r>0) { //12 o'clock if(array[r-1][c] == '*') count++; } if(r>0 && c<SIZE-1) { //2 o'clock SIZE-1 checks that c will not go out of bounds, in the next line of code if(array[r-1][c+1] == '*') count++; } //and add the rest of the if statements here, working clockwise (or in some ordered fashion) //and lastly return count; }

Thanks guys!
I'll implement all of these changes now, and I'll post the new code as soon as I am done. :)

Code:

#include <stdio.h> #include <string.h> #include <conio.h> void display(char b1[][21], char b2[][21]); void new_board(char b1[][21], char b2[][21]); void copy_back(char b1[][21], char b2[][21]); int count_neighbours(int r, int c); char b1[20][21] = {"**..................", "**..................", "....................", "....................", "....................", "....................", "....................", "........***.........", "........*...........", ".........*..........", "....................", "....................", "....................", "....................", "....................", "....................", "....................", "....................", "....................", "...................."}; char b2[20][21] = {"**..................", "**..................", "....................", "....................", "....................", "....................", "....................", "........***.........", "........*...........", ".........*..........", "....................", "....................", "....................", "....................", "....................", "....................", "....................", "....................", "....................", "...................."}; int main() { char c; display(b1, b2); printf("Hi! Would you like to play the Game of Life? (y/n)"); c = getchar(); while (c == 'y') { new_board(b1, b2); copy_back(b1, b2); display(b1, b2); c = getchar(); } } void display(char b1[][21], char b2[][21]) { int i; for (i=0; i<20; i++) { printf("%s \n", b1[i]); } } void new_board(char b1[][21], char b2[][21]) { int i, j, p; for (i=0; i<20; i++) { for (j=0; j<20; j++) { p = count_neighbours(i,j); if(b1[i][j]=='*') { if (p ==2 || p == 3) { b2[i][j] = '*'; } else { b2[i][j] = '.'; } } else { if (p == 3) { b2[i][j] = '*'; } else { b2[i][j] = '.'; } } } } b2[i][j] = '\0'; } void copy_back(char b1[][21], char b2[][21]) { int i; for (i=0; i<20; i++) { strcpy(b1[i], b2[i]); } } int count_neighbours(int r, int c) { int count = 0; if(r > 0) { if(array[r-1][c] == '*'); count++; } if(r > 0) { if(array[r-1][c+1] == '*'); count++; } if(r > 0) { if(array[r][c+1] == '*'); count++; } if(r > 0) { if(array[r+1][c+1] == '*'); count++; } if(r > 0) { if(array[r+1][c] == '*'); count++; } if(r > 0) { if(array[r+1][c-1] == '*'); count++; } if(r > 0) { if(array[r][c-1] == '*'); count++; } if(r > 0) { if(array[r-1][c-1] == '*'); count++; } }

You have (hopefully) counted up the "neighborhood" around each cell. But now you need to return that count, so the game can be adjusted, and reflect the changes.

maybe something like:

Code:

pseudo code idea: for(each row) { for(each column) { count = count_Neighbors(row, col) if(count > Some Amount) { code to handle that } } }

This should all go inside your main while game loop (or in a function called from that while loop).

Code:

int main() { char c; display(b1, b2); printf("Hi! Would you like to play the Game of Life? (y/n)"); c = getchar(); while (c == 'y') { new_board(b1, b2); copy_back(b1, b2); display(b1, b2); c = getchar(); for (r) { for (c) { count = count_neighbours(r, c); if (count = ?) ? } }

Not
if( count = ?)

but
if( count == ?) //you need TWO equal signs for a comparison in C

Look at the rules of your version of the Game of Life, and see what should be used in place of the ? Some number should trigger a change in the board cell value.

You need to run through the specifics of a for loop:

Code:

for(r=0;r<numberOfRowsOnYourBoard;r++) { for(c=0;c<numberOfColumnsOnYourBoard;c++) {

I don't understand what you mean by trigger, there is no such word in my instruction sheet. We have never covered anything like that in class.

Code:

int main() { char c; display(b1, b2); printf("Hi! Would you like to play the Game of Life? (y/n)"); c = getchar(); while (c == 'y') { new_board(b1, b2); copy_back(b1, b2); display(b1, b2); c = getchar(); for (r = 0; 20; r++) { for (c = 0; 20; c++) { count = count_neighbours(r, c); if (count == ) ? } }

Quote:

Originally Posted by Anhur

I don't understand what you mean by trigger, there is no such word in my instruction sheet. We have never covered anything like that in class.

By "trigger", I mean there will be a number given. When the count of neighbors reaches that given number, or more, then the cells in the game of life, which are mostly surrounded by different neighbors, will change.

For instance, if most of my neighboring cells become a '.' then I will become a '.' also (maybe).

So, read up on Conway's Game of Life, or look in your class notes or assignment sheet. Because the Game of Life doesn't have just one set of rules - there are several.

You need to learn what count is needed to cause a cell to change, AND what that change should be.

Google is your friend, if all else fails.

Think of it like sport - there is auto racing, but there is not just ONE kind of auto racing - there is Formula 1, there is Indy, there is stock car, there is funny car. Each has different rules, etc.

You can't finish the program until you know the rules for your version of the game.

"A live square stays alive if it has exactly two or exactly three live neighbours, otherwise it dies"
"A dead square becomes alive if it has exactly three live neighbours, otherwise it stays dead."

What char are you using to show that a square is alive? And which one do you want to show that a square is dead? '.' and '*', are all that I see, with * being alive, is that right?

So put those numbers into the if statement in the main game while loop, and see what you can work up.

I honestly would not have a clue on how to write that. I was hoping to have this program written and working correctly by now, but seeing as it's due in a few hours, I'll just have to hand in what I have done.

I recommend you drop this class, before the deadline for dropping, has passed. Programming takes a LOT of time/study/practice. Perhaps you could take it up later, after you set it up so you have more time for it, and spent some time with some of the excellent tutorials that are on-line (including whole video classes).

I don't make this recommendation lightly. When I saw the if statements you made up, and they were all the same, and your comment above, I knew it was not possible to succeed here, at this time.

Perhaps another time.

Quote:

I recommend you drop this class, before the deadline for dropping, has passed.

I disagree (and I am a little shocked to hear that from the most patient advice giver on here) - This is clearly your first attempt at programming and there is only one way to learn (by doing it). You need to work harder.

If you can't get the program working before the deadline, still hand it in - You shouldn't quit because something is hard, you should learn from it.

After the deadline, your teaching will sit there and give you all the answers. Make sure that you can write the if statements for a problem that you visualise. Ring them up and tell them that you want to go through your program with them and get it working - Even though you won't get any marks for it any more.

When the student says he can't learn because he's got other important stuff to do (papers), and he shows that with the lack of code in his replies - I have to accept that as the truth.

Doesn't mean I like it, and doesn't mean he can't learn to program if he had more time to study and practice.