### BSP Trees

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 definition 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 accommodate 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 occurring 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

===========================================================================

Official Archimedes convertor of : Hear and remember, see and understand,

Wolfenstein 3D and proud of it!! : do and forget.

=================================: Something like that, anyway.

ee@datcon.co.uk ==========================================