Date: Mon, 28 Nov 94 17:30:41 GMT
This article explains how BSP (binary space partitioning) trees can be used in a game such as DOOM as part of the rendering pipeline to perform back-face culling, partial Z-ordering and hidden surface removal.
To explain the use of BSP trees, it is best to start with an example. Consider a very simple DOOM level.
A---------------------------------a----------------------------------B
| | | |
| | y | |
d1 | | b1
| | f' | |
| | | |
| C--------------------f-----------------------D |
| | | | | |
| | | f" | | |
| d | | b |
| | | | | |
| | e" e e' g' g g" | |
d2 | | | | b2
| | | | | |
| | | | | |
| | E F | |
| | x | |
| | | |
G---------------------------------c----------------------------------H
----c1---- ----------------------c2-------------------- -----c3-----
The level consists of a room within a room. The player cannot go outside of the area within the square ABHG.
First some definitions (sorry ?
The vertices are marked A-H, the faces are marked a-g.
We define a line by using an ordered pair of vertices, so that
a = (A,B) e = (E,C) f = (C,D) g = (F,D)
We say a point is to the left of a line if it is to the left of the vector between its two vertices, taken in order.
So, in the above example, nothing is to the left of line a; everything is to the right of it. Note that this depends upon our defintion of line a, and if we had defined a = (B,A) then everything would be to the left of line a.
A face is a side of a line which is visible to the player. Wall e above, for example, has two faces (marked e' and e"). Not all walls have two faces - if the player can never see one side of a wall it only has one.
A face is fully defined by an ordered pair of vertices and an ordered pair of faces - a left face and a right face.
The BSP tree for the example above might look like this:
f
/ \
/ \
/ \
a,d1,b1 e
/ \
/ \
/ \
d2,c1 g
/ \
/ \
/ \
c2 c3,b2
Each node contains a line. Everything to the left of that line is in the left subtree, and everything to the right of that line is in the right subtree.
Note that face d is neither completely to the right of nor to the left of face f. To accomodate this, we split it up into two halves, and put one half into the left subtree and one half into the right subtree. Thus, we have to generate new faces in order to build the BSP tree.
I will explain how the BSP tree is created later. Firstly, I will give the algorithm used to render a picture using the tree.
Suppose the player is standing at position 'x', and looking North.
We start at the top of the tree at line f. We are standing to the right of line f, so we go down the LEFT of the tree. This is because we want the furthest polygons first.
We come to the left-hand-most terminating node. We write down the faces here in our notepad. "a,d1,b1".
Since we've come to a terminator, we back up a level. Back to the top, but we have to go down the right subtree yet. Firstly, though, we look at face f - the deciding face for this node. We've got everything behind it in our list, we've yet to look at anything in front of it, but we must put it into our list. Note that face f has two sides - f' and f". Since we already know we're on the right of line f, we know that we can only see its right side - so we write f" in our notepad. It now says a,d1,b1,f".
Note, though, that if we were looking south (i.e. our line-of-sight vector points away from face f) then we could not see either face f or anything on the other side of face f - in this case, we just don't bother going any further down the tree.
Now we go down the subtree and come to node e. We are on the right of e, so we go down the left subtree and get a terminal node - we just write d2,c1 in our notepad.
Back up, decide on which side of e to put in. We decide e'. The notepad now says a,d1,b1,f",d2,c1,e'.
Down the right subtree to node g. We're on the left, so down the right subtree to c3,b2, up, check g (we're on the left = g'), back down to the final node, get c2, up, up, up, and we're done.
The notepad ends up saying:
a d1 b1 f" d2 c1 e' c3 b2 g' c2
If we draw these walls, in this order, then we will get the correct scene. I would recommend using a one-dimensional Z-buffer to get finer granularity than the painter's algorithm provides, before plotting the walls. Note also that some walls are behind you - however, since you need to calculate their z coordinates for the perspective transform, you can merely discard faces with negative z values.
Creating the BSP tree
The BSP tree almost creates itself. The only difficulty is knowing when to stop recursing. Notice that the terminal nodes are just put into the list - so a sufficient condition for a group of faces to form a terminal node is that they can be drawn in a set order without any mistakes occuring in the drawing. That is, if wherever the player can stand, the group of walls will never obscure each other.
So let us begin: Choose face f (the choice is fairly arbitrary - it is best to choose faces which don't split many other faces up. However, in this case it is unavoidable). Split up faces d and b, because they straddle the line f. (The line you are splitting along is known as the nodeline in DOOM-speak).
Then put everything to the left of f in the left subtree, and vice-versa:
f
/ \
/ \
/ \
a,d1,b1 b2,c,d2,e,g
We can terminate the left node - because walls a,d1 and b1 form a convex shape, they can never overlap each other from any point of view. However, on the other side, face e can obscure face d2 from certain viewpoints (our example viewpoint above, for one) so we divide along side e. This causes side c to be split, but side a is not split because it's not in our current list of sides.
The next level is:
f
/ \
/ \
/ \
a,d1,b1 e
/ \
/ \
/ \
d2,c1 b2,c2,g
Now, c1 and d2 never overlap, so we have another terminal node. We next divide along line g, splitting c2 into c2 and c3, and the last nodes are terminals (a node with one face in is always terminal :-).
This is the basic idea behind a BSP tree - to give an example how effective it is, consider standing at point y and looking North. Because you're looking away from face f, you don't bother recursing down the entire left subtree. This then very quickly gives you the ordered list of faces: a,d1,b1.
Refinements
If at each node we define a bounding box for each subtree, such that every line in a subtree is contained by its corresponding bounding box, then we can cut some invisible polygons (ones which lie to the left or right of the screen) out by comparing each bounding box with the cone of vision - if they don't intersect, then you don't go down the whole subtree. DOOM does this, allowing it to store an *entire* level in one huge BSP tree.
Here's some pseudo-code to traverse the tree. The function left() returns TRUE if the second input vector is to the left of the first input vector. This is a simple dot product, and by pre-calculating the slope of the nodeline can be done with one multiply and one subtract.
vector player ; player's map position vector left_sightline ; vector representing a ray cast through ; the left-most pixel of the screen vector right_sightline ; the right-most pixel of the screen structure node { vector vertex1 vector vertex2 node left_subtree node right_subtree face left_face face right_face box bounding_box bool terminal_node face terminal_node_faces[lots] } recurse(node input) if (cone defined by left and right sightlines does not intersect the node's bounding box) return fi if node.terminal_node ; terminal node - add faces to list add(node.terminal_node_faces) return fi if left(vertex2-vertex1,player-vertex1) ; player is to the left of the nodeline if not left(vertex2-vertex1,right_sightline) ; sight points right - we are looking at the face recurse(node.right_subtree) add(node.left_face) fi ; now go down the left subtree recurse(node.left_subtree) else ; player is to the right of the nodeline if left(vertex2-vertex1,left_sightline) ; sight points left - we are looking at the face recurse(node.left_subtree) add(node.right_face) fi ; now go down the right subtree recurse(node.right_subtree) fi return end recurse
This isn't anywhere near a decent implementation - the data structures, for example, leave a *lot* to be desired ?
It should be possible to encode all the functions inline; in fact, it would be feasible to take a BSP tree and hard-code it into some run-time generated code which you just call to recurse the tree ... but I'm just a hacker at heart ?
Anyway, I hope this helps answer some peoples' questions on this subject. If you have any more questions, please don't hesitate to email me.
Catch you later,
Eddie xxx
ee@datcon.co.uk
===========================================================================
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Wolfenstein 3D and proud of it!! : do and forget.
=================================: Something like that, anyway.
ee@datcon.co.uk ==========================================
