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# Mathematically Identifying Tile Clicked

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I have come up with a way to identify a clicked tile in an isomertic scene without using a mousemap. It works almost perfectly in a tile scrolled world, it errors when the world is scrolled by pixel for reasons that are related to how this method works. I said earlier that in a tiled scrolled world this method works almost perfectly, this is because I draw with a slight offset, half a tile up and left to get rid of the "jagged" effect of the drawing method i use. Consider the following isometric scene:
___________
| /\  /\ |
|/  \/  \|
|\  /\  /|
| \/  \/ | /____ Down half a tile
| /\  /\ | \
|/  \/  \|
|\  /\  /|
|_\/__\/_|       
Each tile is in the shape, for lack of a better word, of a diamond. The drawing method increments across by the tile width and down half the tile height. This fact coupled with the "diamond" nature of the tiles makes identifying which tile was clicked a bit difficult. Each isometric tile is encased in a rectangular bitmap, and if the same method one would use to detect a clicked tile in a normal rectangular map (TileX = X / SpriteWidth) was used the diamond nature of the tile would cause the wrong tile to be chosen as clicked. However with a bit of manuevering this method can be used in an isometric scene. The first thing is to use the rectangular method to find the clicked tile. This will ignore the inbetween tiles (the tiles with a + going through them). I then multiply the TileY by 2 since the drawing method goes down half a tile thus twice as many tiles are drawn than in a normal rectangular method, and add 1 (a weak hack against the drawing offset to get rid of jaggedness).

__________
| /\ | /\ |
|/  \|/  \|          Isometric Tiles split into their
|\  /|\  /|    /____ rectangular casing.
|_\/_|_\/_|    \
| /\ | /\ |
|/  \|/  \|
|\  /|\  /|
|_\/_|_\/_|    
I use 128x64 tiles for the ground, thus the ratio of the tiles is 2:1. The slope of each side of the tile is .5, so for each line on the sides it should be y = .5x or, x = 2y. This, however, is not the case since the sides of the tiles are not always moving in the same direction as a normal x = 2y graph would.
 /\ /___ m = .5
/  \\
\  /
\/

y = .5x
6|     .
|    /
4|   /.
|  /
2| /.
|/__________
0 1 2 3 4 5 
So I divided the tile into 4 sections and used different offset methods to extract the x for each.
_______
|a/|\b|
|/_|_\|
|\ | /|
|c\|/d|         
Section a: x = -2y + 64 Section b: x = 2y + 64 Section c: x = 2y - 64 Section d: x = -2y + 192 I then compare the mouseX with this solved x value. If it is in section a and it is less than the solved x i go up and left, section b, if it is greater, go up and right etc. It works almost perfectly, and would, were it not for the fact that since I start drawing from -64, -32 and not 0,0 the rectangular grid is not entirely accurate. I said earlier that i add 1 to tileY to try to alleviate this but even with this, the accuaracy is only about 90%. If anyone can help, it would be much appreciated.

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I don''t know whether you are using a square or diamond shaped map? I used a diamond shaped one, with the coordinate axes laid out like this:
            + (xmax,y0)           / \          /   \         /     \(x0,y0) +       + (xmax,ymax)         \     /          \   /           \ /            + (ymax,x0)

In order to translate between the map coordinates and the screen coordinates, I used the following procedures:
OffsetX and OffsetY specify the screen position of point (0,0) on the map.
The tiles I used were diamond shaped, 32 pixels high and 64 wide.
The second function is basically derived from the first, i.e., the equations were rearranged.

procedure TForm1.MapToScreen(MapX, MapY: Double; out ScreenX, ScreenY: Integer);begin  ScreenX := Trunc(32 *(MapY + MapX))- OffsetX;  ScreenY := Trunc(16 *(MapY - MapX))- OffsetY;end;procedure TForm1.ScreenToMap(ScreenX, ScreenY: Integer; out MapX, MapY: Double);var  t1, t2: Integer;begin  t1 := ScreenX + OffsetX;  t2 := 2 *(ScreenY + OffsetY);  MapX := (t1 - t2)/ 64;  MapY := (t1 + t2)/ 64;end;

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Unfortunately that is not what I was looking for, you misunderstood me. I was looking for mouse to world or screen (without a mousemap) and not map to screen and vice versa.

Nonetheless, in case I had misunderstood you and you had meant mouse coordinates by screenX and Y, I tested the code and it did not work.

And to answer your question, I am using the left to right, top to bottom (rectangular) method, hence the jaggedness, to draw the isometric tiles.

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quote:
So I divided the tile into 4 sections and used different offset methods to extract the x for each.
_______|a/|\b||/_|_\||\ | /|   |c\|/d|

Section a: x = -2y + 64
Section b: x = 2y + 64
Section c: x = 2y - 64
Section d: x = -2y + 192

I posted specific equations which only work in an unscrolled world on a tile at (0,0). These are more generic:

Section a: x = 2(vertex_y - mouseY)
Section b: x = 2(mouseY - vertex_y) + 64
Section c: x = 2(mouseY - vertex_y)
Section d: x = 2(vertex_y - mouseY) + 64

This is intepretated as If in section A and less than solved x, up and left else do nothing etc. Also I found that it is section a and c which are giving the trouble...so i guessed correctly, about 88% accuracy.

[edited by - Daerax on November 27, 2003 10:43:24 AM]

[edited by - Daerax on November 27, 2003 10:46:06 AM]

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To calculate which tile has been clicked in a "typical" isometric game (ie. Diablo-style flat, diamond-shaped tiles) is rather simple.

Assuming diamond shaped tiles, half as high as they are wide.

Assuming the tilemap is stored in a two-dimensional array, such that if the origin of the tilemap were centred on the screen, the first row of the tilemap (ie. tilemap[0,0], tilemap[0,1], ..., tilemap[0,n]) would extend from the centre down to the lower right. The first column of the tilemap (ie. tilemap[0,0], tilemap[1,0], ..., tilemap[m,0]) would extend from the centre down to the lower-left.

Given are:
h, w - the height and width of the tile bitmap. h = w / 2

mx, my - the co-ordinates of the mouse cursor with respect to the upper-left corner of the screen.

scrollx, scrolly - the screen-space co-ordinates of the origin of the tilemap with respect to the upper-left corner of the screen..

We need to calculate:
sx, sy - the screen-space co-ordinates of the mouse cursor with respect to the origin of the tilemap (ie, the top point of the diamond that represents the tile at tilemap[0,0]).

tx, ty - the tile-space co-ordinates of the mouse cursor (ie. the tile the cursor is currently over).

So:

sx = mx - scrollx
sy = my - scrolly

And basic geometry means that:

tx = (sy / h) + (sx / w)
ty = (sy / h) - (sx / w)

Obviously you could rewrite the above to exclude sx,sy.

Also, scrollx, scrolly can be calculated from the player''s location (assuming the player''s feet ar always the centre of the screen) quite simply using the same logic:

(px, py) - player''s location in tile-space.

scrollx = ((px - py) * w * 0.5) + (screenWidth * 0.5)
scrolly = ((px + py) * h * 0.5) + (screenHeight * 0.5)

So, given the player''s location on the tilemap and the co-ordinate''s of the mouse click you can calculate the tile that was clicked.

This may be the same as Useless Hacker''s post, I didn''t compare, sorry.

-hotstone

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Hotstone:
I''m working on an isometric game, and I cant find the tile that the mouse is above it.
I saw your answer above, but my tile(0,0) is top left, and tile(n,n) is bottom right...
how can I find the tile?

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Ah, thank you for the reply hotstone, however i couldn't get your code to work. Perhaps it is because I scroll the world in a different manner (player characters are not always in the center). Nonetheless I have remedied the problem by orienting around the 0 and 2 vertex only , instead of vertex 0 and 1, 1 and 2 and 0,3 and 3,2.

   0 1/\3  \/  2//this requies a 2:1 ratio, i.e. 64x32, 128x64x = mouseX / width //truncated...y = mouseY / Heightif mouseX < x * width + width / 2    if mouseY < y * height + heigt/ 2         vertex_y = y * Height          sx = 2 * (vertex_y - mouseY)      if mouseX < sx          upperleft   else      vertex_y = y * Height + height      sx = 2 * (mouseY - vertex_y)      if mouseX < sx          lowerleftelse   if mouseY < y * height + heigt/ 2      vertex_y = y * Height      sx = 2 * (mouseY - vertex_y) + 64      if mouseX > sx          upperright   else      vertex_y = y * Height + Height      sx = 2 * (vertex_y - mouseY) + 64      if mouseX > sx          lowerright

[edited by - Daerax on November 29, 2003 11:28:09 AM]

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I haven''t read all of the posts above, sorry if I''m repeating someone...

In the editor I''m working on I have went a bit overkill concerning the mouse tile selection but I thought I could explain it anyway

My tiles are in ordinary isometric 2:1 ratio. I’m also using masking to render the tiles since they overlap each other. The tile in the upper left are used to create the lower map.

First I just pick the tile using a bounding box (the dark square bounding one tile in the picture) using mouse/screen coordinates. I store this value in two variables, something like selectedTileX and selectedTileY. Notice that one single bounding box actually overlaps eight different tiles. To find out which tile the mouse actually is over I create eight polygons and perform a point within triangle test to determine which one was selected. The first polygon is made from the upper left tile the second polygon is made from the upper right tile and so on, until I have checked all of the polygons/tiles. I had to break up each tile in 3 shapes to be able to check the polygon, 2 triangles and one square in the middle like so:

When I have found out which polygon the mouse is above I just offset the selectedTileX and selectedTileY variables according to the polygon’s position relative to the parent tile (polygon #5 in pic 1). This offset is different depending on if I am on an even or an uneven tile row. For instance if I am on an uneven row and I clicked the second polygon (polygon #1 when staring at zero) I would add selectedTileX by 1 and substract selectedTileY by 3.

Everyone got that, right? =D This method is pretty fast I think and it doesn’t consume a lot of memory (like a big colour scheme or something).

-----------------------------
www.OddGames.com

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  01 /\ 3   \/  2

Actually it had not to do with the vertex i chose to work around (for the left side I use vertex 1, and the right i use 0 and 2) but that I forgot to translate the solved x back to the original screen location as it moves it to 0,0 to solve. So it would get 48 instead of 48 + 128 or 176, or 56 instead of 56 + 640 , 696 .

This line fixed it: sx += width * tilex, where tilex = mouseX / width. Now it truly works with 100% accuracy.
tilex = mouseX / width //truncated...tiley = mouseY / Heightif mouseX < tilex  * width + width / 2    vertex_y = y * Height + Height / 2  if mouseY < tiley  * height + height/ 2                sx = 2 * (vertex_y - mouseY)      sx += width * tilex      if mouseX < sx          upperleft   else         sx = 2 * (mouseY - vertex_y )      sx += width * tilex      if mouseX < sx          lowerleftelse   if mouseY < y * height + heigt/ 2      vertex_y = tiley  * Height      sx = 2 * (mouseY - vertex_y) + 64      sx += width * tilex      if mouseX > sx          upperright   else      vertex_y = tiley  * Height + Height      sx = 2 * (vertex_y - mouseY) + 64      sx += width * tilex      if mouseX > sx          lowerright

[edited by - Daerax on November 29, 2003 8:57:30 PM]

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ElectronicSandClock:
Do you mean that you store the tiles such that the first row (Tilemap[0,0], Tilemap[1,0], ... Tilemap[n,0]) would go horizontally across the screen?

That''s how I initially did it actually, but I found it complicated virtually every aspect of the program. I later separated the concept of the game-world from the screen''s representation of it by imagining that the world is represented by a piece of grid paper, which is then depicted on the screen in an isometric view.

This made life much much easier - the co-ordinates are more intuitive for AI, etc., and I found some simple formulae (mentioned above) could translate between screen-space and tile-space. Also, even though I was making an iso game, I was basing my game on Moria (similar to Nethack, etc.) so it made sense to have the world as a simple two-dimensional grid.

In short, if you are storing your tiles that way then I don''t know how you''d find the tile co-ord''s unfortunately; probably some fiddly like mousemaps etc. I''m afraid.

-hotstone

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