How do you calculate walls between tiles in a 3d environment?

Question

I have an int[9,8]tiles that stores numbers from 1 to 9 depending on a previous selection by the player. So for example:

 // 1 = blue tile, 2 = yellow tile, 3 = red tile
 tiles[0,0] = 1;
 tiles[0,1] = 1;
 tiles[0,2] = 1;
 .
 .
 tiles[2,0] = 2;
 tiles[2,1] = 2;
 tiles[2,2] = 2;
 .
 .
 tiles[5,0] = 3;
 tiles[5,1] = 3;
 tiles[5,1] = 3;

 // Which looks something like this in the diagram below:
 // 1 1 1 1 1 1 1 1 1
 // 1 1 1 1 1 1 1 1 1
 // 2 2 2 2 2 1 1 1 1
 // 2 2 2 2 2 1 1 1 1
 // 2 2 2 2 2 3 3 3 3
 // 3 3 3 3 3 3 3 3 3
 // 3 3 3 3 3 3 3 3 3

The result is something like this:

Top Image without Walls, Bottom Image with Walls

The red lines marked on the image are the walls that will go inbetween each of the colour areas. I know how to loop through each tile and work out if the tile next to it in any of the four directions (North, South, East, West) is a different colour and hence a wall is required. I don't know an efficient way to store the information and cancel out any walls that are found from looping through every tile.

Obviously the first time I scan a tile with the one below registering a different colour will indicate a wall. But then when I look at that 2nd tile in my loop, it will also register the need for a wall. So 2 walls will be created.

I would like to know if there is any easy way to calculate this wall information another way, but more importantly how can I store this information in say another int[,] array so I can generate the walls from there. Every tile in my game is a 1x1 quad rotated to point upwards. So the int array relates to the X and Z coordinates of the tile itself.

Any help would be greatly appreciated. It's 6:12am and I have been up all night trying to figure this out for a game I am making.

Answer 1

Best Answer

Answer by Cherno · May 19, 2018 at 08:22 PM

One solution would be to store the wall information in a four-dimensional bool-array. Since a wall sits between two tiles, you need the coordinates of two neighboring tiles. The dimensions or the array would be:

0: tile A x-coordinate 1: tile A y-coordinate 2: tile B x-coordinate 3: tile B y-coordinate

The array would have lengths equal to the x and y lengths of your grid + 1 (for the non-existent tile at max + 1 coordinates, which doesn't have to correspond to tiles for the process of checking for walls). For border cases, you would always set the bool to true instead of attempting to access the tiles array which would be out of bounds).

 if(tiles[0,3] == blue && tiles[0,4] != blue) //pseudocode obviously
 {
       walls[0,3,0,4] = true;
 }

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seandolan · May 19, 2018 at 08:48 PM 0

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Thanks for your comment. I am not sure if it's just that I misunderstand but wouldn't something like this still generate 2 walls on the same points. For example wall[0,3,0,4] = true but also wall[0,4,0,3] = true. So 2 walls would be placed if running through the code to generate the walls?

Cherno seandolan · May 19, 2018 at 09:27 PM 0

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It doesn't matter because you would just check both cases at the same time instead of each one separately.

 if(walls[0,3,0,4] == true || walls[0,4,0,3] == true) 
 {
        Generate wall
 }

Also, since in your loop, you only count upwards, you can't get this situation anyway. The first coordinate value will always be lower than the second.

 for(int x = 0; x < walls.GetLength(0) - 1; x++) 
 {
      for(int y = 0; y < walls.GetLength(1) - 1; y++) 
      { 
             if(x == 0)
              {
                    Generate west border wall
              }
             if(y == 0)
              {
                    Generate south border wall (Make America Great Again!)
              }
              if(walls[x,y,x,y+1] == true)
              {
                    Generate north wall
              }
              if(walls[x,y,x + 1,y] == true)
              {
                    Generate east wall
              }
      }
 }

seandolan Cherno · May 19, 2018 at 09:31 PM 0

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I am generating all the walls in one hit. So when looping through the wall generation in your example it looks like I am back at the same problem. If I loop to the point of reaching [0,3] I could check wall [0,3,0,4] and [0,4,0,3] and generate a wall. Then when I get to [0,4] I would then be checking again both [0,3,0,4] and [0,4,0,3] and generating a 2nd wall. Maybe I am missing something. I appreciate your help on this. Thank you so much.

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Answer 2

Answer by abedathman8 · May 20, 2018 at 02:25 AM

In the tile-based approach, each visual element is broken down into smaller pieces, called tiles, of a standard size. These tiles will be arranged to form the game world according to pre-determined level data - usually a 2D array. Tutuapp 9Apps Aptoide

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Answer 3

Answer by Lardalot · May 20, 2018 at 07:39 AM

You could just only check to places walls on the bottom or right as you scan your 2D array. That way you wont get duplicates.

Also always place walls on the top of row 0 or left of column 0.

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How do you calculate walls between tiles in a 3d environment?

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