Could someone explain this code to me

vincentstrebelle · March 18, 2020, 12:49pm

Hi everyone,
I wonder if someone can explaine to me how this code work :

int locX, locY;
int offsetX, offsetY;
 
final int unit = 32;
byte[][] terrain = new byte[10][10];

void setup() {
  size(321, 321);
 
  for (int i = 0; i < 10; i++) {
    for (int j = 0; j < 10; j++) {
      terrain[i][j] = byte(ceil(noise(i*0.05,j*0.05)*6));
    }
  }
}

void draw() {
  for (int i = 0; i < 10; i++) {
    for (int j = 0; j < 10; j++) {
      offsetX = i;
      offsetY = j;
 
      // Out of bounds
      if (offsetX < 0 || offsetX > 9 || offsetY < 0 || offsetY > 9) fill(0xff808080); //grey
 
      // Draw based on terrain
      else {
        if (terrain[offsetX][offsetY]>5){
          fill(0xff005000); //dark green
        }
        else if (terrain[offsetX][offsetY]>4){
          fill(0xff008000); //light green
        }
        else if (terrain[offsetX][offsetY]>3){
          fill(0xffc0c000); //yellow
        }
        else {
          fill(0xff0000c0); //blue
        }
      }
      rect(i*unit, j*unit, unit, unit);
    }
  }
}

I understand (i think) that in the setup(), each “i” and “j” get a random number between “i” * 0.05 and “j” *0.05 and then multiply by 6.

And then in the draw(), it check if the values of terrain[offsetX][offsetY] is more than a number i choose and color the rectangles.

So i get that everything is based on : byte[][] terrain = new byte[10][10];
but i dont understand precisely how it’s work and how to controle the result.

Thanks for your times.

Chrisir · March 18, 2020, 1:00pm

first of all terrain is a 2D array

similar to a chess board you need to indexes to get one of its cell.

The code below shows the indexes x,y

see https://www.processing.org/tutorials/2darray/


int locX, locY;
int offsetX, offsetY;

final int unit = 32;
String[][] terrain = new String[10][10];

void setup() {
  size(321, 321);

  for (int i = 0; i < 10; i++) {
    for (int j = 0; j < 10; j++) {
      terrain[i][j] = i+"," +j;
    }
  }
}

void draw() {
  for (int i = 0; i < 10; i++) {
    for (int j = 0; j < 10; j++) {
      offsetX = i;
      offsetY = j;


      fill(0xff808080); //grey


      rect(i*unit, j*unit, unit, unit);
      fill(0); 
      text(    terrain[i][j], 
        i*unit+4, j*unit + 17);
    }
  }
}

vincentstrebelle · March 18, 2020, 1:11pm

ok so each rect has a specific x and y indexes.
So this part of the code :

for (int i = 0; i < 10; i++) {
    for (int j = 0; j < 10; j++) {
      terrain[i][j] = byte(ceil(noise(i*0.05,j*0.05)*10));
    }
  }

create a random number based of the index of each rect.
Am i right ?

Chrisir · March 18, 2020, 1:13pm

Yes.

see https://www.processing.org/reference/noise_.html

ceil then rounds this float value up (we get an int) and
byte() converts it to byte, which is not really necessary

see


int locX, locY;
int offsetX, offsetY;

final int unit = 32;
int[][] terrain = new int[10][10];

void setup() {
  size(321, 321);

  for (int i = 0; i < 10; i++) {
    for (int j = 0; j < 10; j++) {
      terrain[i][j] = ceil(noise(i*0.05, j*0.05)*6);
    }
  }
}

void draw() {
  for (int i = 0; i < 10; i++) {
    for (int j = 0; j < 10; j++) {
      offsetX = i;
      offsetY = j;

      // Out of bounds
      if (offsetX < 0 || offsetX > 9 || offsetY < 0 || offsetY > 9) fill(0xff808080); //grey

      // Draw based on terrain
      else {
        if (terrain[offsetX][offsetY]>5) {
          fill(0xff005000); //dark green
        } else if (terrain[offsetX][offsetY]>4) {
          fill(0xff008000); //light green
        } else if (terrain[offsetX][offsetY]>3) {
          fill(0xffc0c000); //yellow
        } else {
          fill(0xff0000c0); //blue
        }
      }
      rect(i*unit, j*unit, unit, unit);
    }
  }
}

vincentstrebelle · March 18, 2020, 1:23pm

ok so each rect get one random number based on his indexes. (ex: the top left rect get the number 4)
Then this number (4 in my example) is checked by the conditions and get a color.

Am i still right ?

vincentstrebelle · March 18, 2020, 1:24pm

why is it not necessary ?

Chrisir · March 18, 2020, 1:26pm

Yes you are right with the colors.

It might help to read the reference of noise()

Chrisir · March 18, 2020, 1:26pm

You could also work with int instead of byte. I changed my post above to demonstrate this

Chrisir · March 18, 2020, 1:27pm

the sketch has the purpose to make a landscape / terrain with water, desert, grass and forest or whatever I guess

vincentstrebelle · March 18, 2020, 1:31pm

yes, so if i get it right. To controle the result of my code i need to calculate and know the range of the number given by the randomization

Chrisir · March 18, 2020, 1:34pm

You could store the result (the terrain) on the hard drive and load it from there.

Also noiseSeed seems to control noise to get the same result every time.

see https://www.processing.org/reference/noiseSeed_.html

Chrisir · March 18, 2020, 1:40pm

here is a text file that my version just produced

or

or

Chrisir

vincentstrebelle · March 18, 2020, 1:40pm

and if i take the bottom right rect as an example, his index is 9,9. How the randomization work ?
if the rect indexes are 1,9 does it take a random number between 1 * 0.05 and 9 * 0.05 or does it work differently ? And so for the bottom right rect does it take a random number between 9 * 0.05 and 9 * 0.05 ?

it’s the randomization part of the code that i don’t understand completly

Chrisir · March 18, 2020, 1:52pm

it would help to read the reference about noise.

I posted the link above

see noise() / Reference / Processing.org

It’s not a random.

Think of it more as a landscape of random values.

with this formula : noise(i*0.05, j*0.05)

we retrieve the value at point i*0.05, j*0.05 (x,y) in this landscape.
The result is a constant between 0 and 1. We multiply it by 6 so it’s between 0 and 6 and then we use ceil() to round it up. So it’s an integer between 1 and 6.

But 1,2,3 are all blue:

       else {
          fill(0xff0000c0); //blue
        }

No, it’s the random value at 9 * 0.05 and 9 * 0.05 in the 2D random landscape.

the resulting values are 1,2,3,4,5,6 or so.

here is a new code.

It saves the terrain (file gets overwritten every time !!!) and you can hit Space Bar to renew


int locX, locY;
int offsetX, offsetY;

final int unit = 32;
int[][] terrain = new int[10][10];

void setup() {
  size(321, 321);

  randomizeNoise();
}

void draw() {
  for (int i = 0; i < 10; i++) {
    for (int j = 0; j < 10; j++) {
      offsetX = i;
      offsetY = j;

      // Draw based on terrain
      if (terrain[offsetX][offsetY]>5) {
        fill(0xff005000); //dark green
      } else if (terrain[offsetX][offsetY]>4) {
        fill(0xff008000); //light green
      } else if (terrain[offsetX][offsetY]>3) {
        fill(0xffc0c000); //yellow
      } else {
        fill(0xff0000c0); //blue
      }

      rect(i*unit, j*unit, 
        unit, unit);

      fill(0); 
      text(terrain[i][j], 
        i*unit+4, j*unit + 17);
    }
  }

  String[] s1=new String[10]; 
  for (int i = 0; i < 10; i++) {
    s1[i] = ""; 
    for (int j = 0; j < 10; j++) {
      s1[i] += terrain[i][j];
    }
  }

  saveStrings( "TT.txt", s1);
  //noLoop();
}

// ==============================================================

void keyPressed() {
  // SPACE BAR or any key 
  randomizeNoise();
}

void randomizeNoise() {
  noiseSeed( int(random(11111) ) ); 
  for (int i = 0; i < 10; i++) {
    for (int j = 0; j < 10; j++) {
      terrain[i][j] = ceil(noise(i*0.05, j*0.05)*6);
    }
  }
}
//

vincentstrebelle · March 18, 2020, 2:01pm

Thanks, now i have a better understanding of the code. i readed the noise reference but I did not understand it well. Until now thank to you.

Chrisir · March 18, 2020, 2:03pm

I am not an expert on noise() but it seems like a more slow change of values than with random.

when you replace the noise line with
terrain[i][j] = ceil(random(0, 6));

it works technically, BUT you receive a very wacky and turbulent landscape:

(no long smooth connected patches of one kind of terrain!)

Chrisir · March 18, 2020, 2:07pm

when you want to use the landscape maybe you can save with one program (my code above) and load with another one.

Or you produce 10 different maps in setup and find a way to identify the best. But how to measure this?

You could make sure that 3,4,5,6 are all present in the map and are all over 3 occurrences for example.

vincentstrebelle · March 18, 2020, 2:10pm

that’s a good idea, now than i have a better understanding of this code i will dive into it and find a good way to use it.

Thank you.

Chrisir · March 18, 2020, 3:50pm

just for the sake of completeness:

this version loads a txt file.

You have to copy it from the folder of the previous sketch.

(the letters in the text file are upside down (or grid is rotated by 180°), but the terrains are the same in both Sketches. You can edit the text file in a text editor when you know what number signifies which type or terrain, but it’s rotated)

Chrisir

// THIS VERSION LOADS 

int locX, locY;
int offsetX, offsetY;

final int unit = 32;
int[][] terrain = new int[10][10];

void setup() {
  size(321, 321);

  String[] stringsFromFile = loadStrings("TT.txt");

  int x=0; 
  for (String s1 : stringsFromFile) {
    for (int i=0; i < s1.length(); i++) {
      terrain[x][i]= (int) (s1.charAt(i)-48);
    }
    x++;
  }
  //
}// func 

void draw() {
  for (int i = 0; i < 10; i++) {
    for (int j = 0; j < 10; j++) {
      offsetX = i;
      offsetY = j;

      // Draw based on terrain
      if (terrain[offsetX][offsetY]>5) {
        fill(0xff005000); //dark green
      } else if (terrain[offsetX][offsetY]>4) {
        fill(0xff008000); //light green
      } else if (terrain[offsetX][offsetY]>3) {
        fill(0xffc0c000); //yellow
      } else {
        fill(0xff0000c0); //blue
      }

      rect(i*unit, j*unit, 
        unit, unit);

      fill(0); 
      text(terrain[i][j], 
        i*unit+4, j*unit + 17);
    }
  }

  String[] s1=new String[10]; 
  for (int i = 0; i < 10; i++) {
    s1[i] = ""; 
    for (int j = 0; j < 10; j++) {
      s1[i] += terrain[i][j];
    }
  }
}
//

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Could someone explain this code to me

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