Checking if Element Exists in Boolean Array

Question

I took a programming class, and I'm revisiting old programs that I did not quite get right. This one is a Game Of Life program, and I have a question about code cleanup.

I need to make sure that an array element is in bounds before checking whether its neighbor's boolean value is true or false. I have a statement to check if firstGen[0][0]'s top-left (up one row, left one column) is in bounds. Is there an easier or more elegant way to check if an element is in bounds or to restrict the element checks to the boundaries of a given array without using four && conditionals per if statement?

Note that I have only changed the first if statement thus far, so there may be errors elsewhere. I also excluded the boundary checks for the other neighbors.

    public static boolean[][] generation(boolean[][] firstGen)
    {
    int length = firstGen.length;
    boolean[][] newGen = new boolean[length][length];

    for (int j = 0; j < firstGen[0].length; j++)
        { for (int i = 1; i < firstGen.length; i++)
            {
                int count = 0;
                if  ((i-1 >= 0) && (i-1 < length) && (j-1 >= 0) && (j-1 < length)) //top-left element exists
                    { if  (newGen[i-1][j-1] == true) count++; } //increment `count` if top-left element is true

                if ((newGen[i][j] == false) && (count == 3)) newGen[i][j] = true;
                else if ((newGen[i][j] == true) && (count == 1)) newGen[i][j] = false;
                else if ((newGen[i][j] == true) && (count > 3)) newGen[i][j] = false;
                else break;
             }
        }
        return newGen;
      }

assylias · Accepted Answer · 2013-05-30 14:45:05Z

3

If i and j are in bounds, then you know for sure that i - 1 < length and j - 1 < length are both true.

Also:

i - 1 >= 0 can be written i > 0
if (condition == true) can be rewritten if (cond)

So you could replace:

if  ((i-1 >= 0) && (i-1 < length) && (j-1 >= 0) && (j-1 < length)) //top-left element exists
    { if  (newGen[i-1][j-1] == true) count++; } //increment `count` if top-left element is true

by:

//increment `count` if top-left element is true
if  (i > 0 && j > 0 && newGen[i-1][j-1]) count++;

answered May 30, 2013 at 14:45

assylias

330k84 gold badges680 silver badges806 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Marko Topolnik Over a year ago

+1 And suddenly it's so simple :-) I would additionally recommend what he'll need in the next step: an elegant way to go through all the neighbors without code repetition.

rice2007 Over a year ago

@assylias Sometimes Java solutions are so simple. They're the kind tat make you just wanna bang your head into a wall for make things unnecessarily complicated. I would also like to thank you for the reminder about true and false boolean conditionals. Again, such a derp thing.

rice2007 Over a year ago

@Marko I'm not sure I have it quite solved, but I have a mental solution to handle that next step as well. I'm thinking nested for loops.

Nealon · Accepted Answer · 2013-05-30 15:12:33Z

That's the best way I can think of to check if its out of bounds, but an alternative method in general, and one that I think gives programs like the Game of Life more exciting outcomes, is adding periodic boundaries. Basically this means that if you walk off one edge, you end up on the other side (like in pac-man). It sounds complicated, but really all it takes is the % function, which returns the remainder of division between the two numbers given.

So:

27 % 5 = 2;

So for adding periodic boundries you would update x and y positions like this:

x = (x + xStep + horizontalSize) % horizontalSize;
y = (y + yStep + verticalSize) % verticalSize;

Where xStep and yStep are +1 or -1 depending on what direction you want to go. (this works nicely with a for loop) The addition of the size is to make sure you go below zero when you get close to borders.

Then you never have to worry about messy border conditions, everything simply overlaps. No need to check each and every border. I hope this makes sense. Please ask for clarification if not. I've used this more for random walker programs but the idea is the same.

Collectives™ on Stack Overflow

Checking if Element Exists in Boolean Array

2 Answers 2

3 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

3 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related