Published May 2000,February 2011.

- 1 Introduction
- 2 First Person Shooters
- 3 Strategy Games
- 4 Simulation Games
- 5 Conclusion
- 6 Answers
- 7 Interesting Web Pages

The purpose of this article is to have a look at how
mathematics is used in computer games. I'll use examples from
computer games you've probably already played. There are lots of
different types of computer games, and I'll talk about how maths is
used in some of the following examples:

The First Person
Shooter (FPS) is a type of game where you run around 3D
levels carrying a big gun shooting stuff. Examples of this sort of
game include Doom ,
Quake , Half Life , Unreal or Goldeneye . There are other games that
look very similar, but aren't first person shooters, for instance
Zelda: Ocarina of Time or
Mario 64 .

The Strategy games
are divided into two main types, Real Time Strategy (RTS), and
Turn Based Strategy (not
usually called TBS for some reason). These games usually involve
building and managing a city or civilization and also fighting wars
by controlling troops. Examples of real time strategy games are
Age of Empires ,
Command &; Conquer ,
Tiberian Sun . Examples of
turn based strategy games are Civilization and Alpha Centauri .

Simulation games are
games that try to make something as realistic as possible. For
instance, Flight Sims are
computer games which try to realistically simulate flying an
aeroplane or helicopter. Two games of this sort are Microsoft Flight Simulator and
Red Baron . Space sims are like flight sims, but
with spaceships instead of planes. For instance, Wing Commander or X-Wing vs. Tie Fighter . Racing games are games which simulate
driving different sort of cars. For instance, Need for Speed , NASCAR Racing , Gran Turismo or Driver .

There are some exercises which you can do (if you want)
throughout this article. The answers are at the end of the article,
but do have a go at solving them on your own first. If there are
any bits of this article you don't understand, you should either
ask a question on the NRICH web board, or send an email to me
at

.

If you find any of the article patronising, I apologise, my
excuse is that the article is aimed at people of many different
ages so there might well be bits you already know. If you already
understand one bit, you can just skip through until you get to
something more interesting.

The most amazing things about FPS are their incredible graphics. They look almost real, none of this would have been possible without the use of advanced maths. Here are some pictures from the early games (Wolfenstein) to the most recent games (Quake III Arena). All of the following screen shots are from games by iD software.

To begin to explain how these games work, you need to know a bit
about *geometry* , *vectors* and
*transformations* .

*Simple geometric figures*

If you have a line and a plane, you can find the point where the
line cuts through the plane. In fact, sometimes you can't find the
intersection, because they don't meet and sometimes the line is
inside the plane so they meet at every point on the line , but this
doesn't happen in the cases we're interested in. We call this the
intersection of the line
and the plane. Here is a picture of what this looks like.

*Intersection of a line and a plane*

A vector is a mathematical
way of representing a point. A vector is 3 numbers, usually called
$x$, $y$ and $z$. You can think of these numbers as how far you
have to go in 3 different directions to get to a point. For
instance, put one arm out pointing to the right, and the other
pointing straight forward. I can now give you a vector and you'll
be able to find the point I'm talking about. For instance, if I say
$x=3$, $y=1$, $z=5$, you find the point by walking 3 metres in the
direction of your right hand, then 1 metre in the direction of your
left hand, and then getting a ladder and climbing up 5 metres. Here
is a picture of a vector.

*Picture of a vector and directions*

Vectors are written as $(x,y,z)$, for instance $(1,2,3)$ means move
1 in the x-direction, 2 in the y-direction and 3 in the
z-direction.

One confusing thing about vectors is that they are sometimes used
to represent a point, and sometimes they are used to represent a
direction. The vector $(1,0,0)$ can mean both ``the point you get
to if you move 1 unit in the x-direction from the starting point'',
or it can mean ``move 1 unit in the x-direction from where you are
now''.

**Exercise 1** *[Adding Vectors]*

*If I start at the point represented by the vector $(0,0,0)$,
then I travel along the vector $(1,2,3)$, and then I travel along
the vector $(4,5,6)$, where will I be? This vector is called the
sum of the two vectors,
$(1,2,3)+(4,5,6)=(?,?,?)$.*

*Translations and rotations*

**Exercise 2** *[Translations]*

*If I translate the vector $(1,2,3)$ by 5 in the y-direction,
what is the vector afterwards? In general, if we write a
translation as $T$ and a vector as $\mathbf{v}$, then the new
vector if written as $T\mathbf{v}$. For example, if $T$ is
``translate 1 unit in the x-direction'' and $\mathbf{v}=(0,0,0)$,
then $T\mathbf{v}=(1,0,0)$.*

The basic idea of 3D graphics is to turn a mathematical
description of a world into a picture of what that world would look
like to someone inside the world. The mathematical description
could be in the form of a list, for instance: there is a box with centre $(2,4,7)$ and
sides of length 3, the colour of the box is a bluish grey .
To turn this into a picture, we also need to describe where the
person is and what direction they are looking, for instance:
there is a person at $(10,10,10)$
looking directly at the centre of the box . From this we can
construct what the world would look like to that person.

Imagine there is a painter whose eyes are at the point $P$.
Imagine that he has a glass sheet which he is about to paint on. In
the room he is painting, there is a wooden chest. One of the
corners of the chest is at point $A$, and the painter wants to know
where that corner of the chest should be on his glass sheet. The
way he works it out is to draw a line $L$ from his eyes ($P$) to
the corner of the chest ($A$), then he works out where this line
goes through the canvas, $B$. He can do this, because the glass
sheet is a plane, and I mentioned that you can find the
intersection of a line and a plane above. This point $B$ is where
the corner of the chest should be in his painting. He follows this
rule for every bit of the chest, and ends up with a picture which
looks exactly like the chest. Here are two pictures, the first one
shows the painting when he has only painted the one corner of the
chest, the second one shows what it looks like when he has painted
the entire chest.

*Projection on to a plane*

What I just described above is similar to what the computer is
doing (50 times a second!) every time you run around shooting
hideous monsters in Quake
, although the details are slightly different. In computer games
(at the moment) the description of the world is just a list of
triangles and colours. The newest computer games are using more
complicated descriptions of the world, using curved surfaces, NURBS
and other strange sounding things, however in the end it always
reduces to triangles. For instance, a box can be made using
triangles as illustrated below.

*Box made from triangles*

Here is a much more complicated example, using thousands of
triangles. The first picture shows the triangles used, the second
picture is what it looks like with colours put in.

*Viking Helmet*

The reason for using triangles is that they are a very simple shape, and if you make sure that everything is made from only one type of shape, you don't have to write a separate program for each type of shape in the game.

**Exercise 3** *[Making triangles]
Draw a picture of a box with a smaller box stuck to the
top of it, using only triangles.*

Each time the computer draws a picture of the world, it goes through the following steps: Firstly, it transforms the world (by rotating and translating), so that the person is at position $(0,0,0)$ and the centre of the glass sheet (the centre of the screen in computers) is at $(1,0,0)$. This makes the rest of the calculations much easier. Secondly, it removes all the triangles you can't see so that it can forget about them, for instance the triangles that are behind you or the ones that are so far away that you can't see them. Thirdly, for every remaining triangle, it works out what it would look like when painted on the glass sheet (or drawn on the screen in computers). Finally, it puts the picture it has drawn on the screen. Nowadays, computers are so fast that they can draw hundreds of thousands of triangles every second, making the pictures more and more realistic, as you can see from the pictures at the beginning of this section.

Of course, there is a lot more to it than just that: there is lighting, fog, animation, textures and hundreds of other things. Most of these use maths and physics to a large extent, but what I have described so far is the most important part of making 3D graphics look right.

*Graph with directed edges*

To complicate things even further, we sometimes want to add
something called a cost to
each edge. The idea of a cost is that it indicates how much it
would cost to travel down that edge. A simple example of this is
shown below.

*Graph with costs*

In this graph, most of the people in city $A$ want to get to city
$C$, whereas only a few want to get to city $B$. Unfortunately,
both roads are the same size, this means that there are long
traffic jams on the road from $A$ to $C$, it takes about 30 minutes
to get there. To get from $A$ to $B$ is much easier as most people
are going to $C$, so it only takes 5 minutes. The numbers written
next to the edges indicate how long it takes to travel along that
edge. Here is another (much more complicated) example.

*More complicated graph*

**Exercise 4** *[Shortest Path] Work
out the lowest cost way to get from* A to *K* .

Now you know all you need to know to be able to understand
path finding. If you did the last exercise, you'll have already
solved one example of the sort of problem computers have to solve
to guide troops through complicated maps.

How does all this stuff about graphs help the computer guide
troops around levels? It makes a graph where every interesting
point is a node on the graph, and every way of walking from one
node to another is an edge, then it solves the problem you solved
above to guide the troops. There are some complications. For
starters, what are the interesting points? You might think that
every position on the entire level is interesting, but for most
games this would lead to hundreds of thousands of interesting
points, and finding the path would take years. Instead, the people
making the game decide where the interesting points are. For
instance, if there is a wide open expanse (a big field perhaps),
you don't need a node at every point on the field, because the
troops can walk in a straight line across the field. Basically, you
only need nodes around obstacles. Here is an example of a map of a
level seen from above.

Once you have chosen your interesting points, you need to connect them up with edges, you can only connect up those nodes which don't have an obstacle between them.

**Exercise 5** *[Make your own graph]
Place nodes at the interesting points on the example
map above, then connect up the nodes with edges, remembering that
you can only connect up nodes with straight lines, and the straight
lines cannot go over obstacles.*

Once you have created a graph for a given map, the computer has to go through the following steps to guide the troops. Firstly, it has to work out what the nearest node that he can walk to in a straight line. This node is his starting node. Secondly, he has to work out the node which is nearest to his destination is (making sure he can walk from that node to the destination in a straight line of course!). This node is the destination node. Thirdly, he works out the shortest path connecting his starting node to his destination node. Now, all the troops have to do is walk to the starting node, then walk along all the nodes between the starting node and the destination node, along the connecting edges, then they walk from the destination node to the final destination.

To make things a bit more interesting, we can add costs to all of the edges. For instance, if there is a crocodile pit in the space connecting two nodes, the cost of crossing this crocodile pit is very high, so it might be a better idea to go the long way round even though crossing the crocodile pit is shorter. Of course, if the long way round would take the troops 3 days, and crossing the crocodile pit only took 15 minutes, they might decide that it would be better to take the risk and get there before the battle has moved elsewhere. You have to choose the costs carefully to make sure this sort of problem is solved in the best possible way. Here is an example of a map with edges with costs.

I haven't actually talked about how to get a computer to find the best path from one node to another yet. The total cost of getting from $A_1$ to $A_m$ via $A_2$, $A_3$, ..., $A_{m-2}$, $A_{m-1}$ is found by adding up the cost of going from $A_1$ to $A_2$, $A_2$ to $A_3$, ..., $A_{m-1}$ to $A_m$. The problem is to find the right choice of nodes that this total cost is as small as possible. One way of doing it would be to find all possible ways of getting from one node to the other, work out the total costs of each, and choose the smallest one. Unfortunately, this would take even the fastest computers much too long to do. There is a way of working this route out very quickly, but it is a bit complicated to explain here. However, if you're keen, you can look it up on the internet (search for the A* Algorithm - A* is pronounced A star), or you can check out the references at the end of the article.

**Exercise 6** *[Vectors]*

*(a) What is (1,2,3)+(3,2,1)?*

*(b) What is 10x(9,5,2)?*

*(c) What is the length of (4,2,4)?*

Every object has a position , which is a vector
$\mathbf{x}$. It also has a velocity , which is the direction it
is travelling in, it is also a vector $\mathbf{v}$, sometimes
written as $\mathbf{\dot{x}}$. Every object also has something
called an acceleration ,
this is how fast the velocity is changing, and it is also a vector
$\mathbf{a}$, sometimes written as $\mathbf{\ddot{x}}$. Finally,
every object has something called mass , this is basically how heavy the
object is, and is usually written $m$, but this is not a vector. If
an object starts at position $\mathbf{x}$, and has a velocity
$\mathbf{v}$ which doesn't change, then after $t$ seconds, the
position of the object is $\mathbf{x}_t=\mathbf{x}+\mathbf{v}\times
t$. In case you're wondering what $\mathbf{x}_t$ means, it just
means the position at time $t$. Similarly, $\mathbf{v}_t$ means the
velocity at time $t$. If an objects starts with velocity
$\mathbf{v}$ and has an acceleration $\mathbf{a}$ which doesn't
change, then after $t$ seconds, the velocity of the object is
$\mathbf{v}_t=\mathbf{v}+\mathbf{a} \times t$.

Here is an example of how to use this to work something out.
Suppose Billy-Joe Bob is floating in space, his position is
$(100,150,150)$ and he throws a rock with velocity $(5,2,0)$. Where
is the rock after 10 seconds? To work this out, we use the formula
above: $\mathbf{x}_t=\mathbf{x}+\mathbf{v}\times t$. In this case
(after 10 seconds) the rock is at $(100,150,150)+(5,2,0)\times
10=(100,150,150)+(50,20,0)=(150,170,150)$.

**Exercise 7** *[Throwing Rocks in Space]
Billy-Joe's mortal enemy, Arthur Blenkinsop is at
(240,640,143), reading his newspaper, unaware that Billy-Joe is
watching him. Billy-Joe decides to throw his rock at Arthur.
Billy-Joe is slightly obsessed with the number 7, so he wants the
rock to take exactly 7 seconds to reach Arthur, what velocity
should he throw the rock at?*

If Billy-Joe wasn't in space, the problem would be much harder, because every object on Earth is pulled downwards by an acceleration caused by gravity . This acceleration is the same for every object on Earth, and if the y-direction is upwards, then the acceleration is $\mathbf{g}=(0,-9.81,0)$. The reason the problem is now harder is that the velocity changes, so you can't use the equation above to work out the position at different times. If we have a computer, we can work out things like this, but not exactly. The way we do it is by saying that the velocity is only changing very slightly over very small amounts of time. For instance, the velocity only changes by a very small amount in 0.02 seconds. So, we can work out what happens for very small amounts of time, if $s$ is very small, maybe $s = 0.02$, the equation $\mathbf{x}_{t+s}=\mathbf{x}_t+\mathbf{v}_t \times s$ is very close to being correct. In that equation $\mathbf{x}_{t+s}$ is the position at time $t+s$, $\mathbf{x}_t$ is the position at time $t$, $\mathbf{v}_t$ is the velocity at time t which we can work out because the acceleration doesn't change. By doing this repeatedly, we can get a computer to work out where the rock will be at any time, by working out where it will be after 0.02 seconds, then after 0.04 seconds, then after 0.06 seconds, and so on until we get to the time we want.

**Exercise 8** *[Throwing Rocks on Earth]
Billy-Joe is at position (0,0,0), he throws a rock with
velocity (0,4,0), where will the rock be after 1 second? To answer
this question, the acceleration on the rock is (0,-9.81,0), and you
should use* s = 0.2, you will need to use the method above 5
times to work it out. You might like to see if you can work out how
long it takes for the rock to hit him on the head.

How is all this stuff actually used in computer games? Here's
a simple example. Every time you fire a bullet from your plane in
your favourite flight simulator, the computer has to work out its
position using calculations like the ones above with Billy-Joe
throwing rocks around. The computer has to do these calculations 50
times a second for every bullet in the sky, and on top of that it
has to do much harder calculations for the planes as well.
Unfortunately, things are not as easy as working out where
Billy-Joe's rocks are. It is much more complicated, because in the
real world, there are things like wind and friction because of the
air. Here is how you might include wind in the calculations.
Suppose the wind causes the bullet to have an additional
acceleration of $\mathbf{w}$, and the wind is blowing at a constant
rate (the wind speed isn't changing). Now we have the acceleration
on the bullet is $\mathbf{g}+\mathbf{w}$, which isn't that much
more complicated than before. Friction in the air is much more
complicated though, because the amount of friction depends on the
speed of the bullet. In fact, the friction causes an additional
acceleration of $-k\mathbf{v}$, where $k$ is some positive number
and $\mathbf{v}$ is the velocity. So now the acceleration is
$\mathbf{g}+\mathbf{w}-k\mathbf{v}$: in other words, the
acceleration is changing as well as the velocity! This problem can
be solved using the method above, where you assume that things
don't change in small periods of time.

For the plane, things are even more complicated. If you fly
your plane as high as it will go and then go into a nose dive, you
might find that the wings of your plane rip off (if you are playing
a flight simulator with new planes this might not happen, because
planes are better now than they used to be). This happens because
the air pushes the wings very hard, and they are not attached
strongly enough to the rest of the plane, so they just rip
off.

Physics is one of the hardest bits in making computer games.
Few games have accurate physics, most games only have something
about as complicated as what I described above. It is such a hard
problem that people are researching into how to simulate realistic
physics in computer games even now, and they're quite a long way
from the solution. So if you're interested in programming games and
you're interested in physics, then you might even find yourself
working on this problem in years.

. You should also post if you would like to ask me (or the others) about how maths is used in any other part of making computer games.

**Answer to Question 1** *[Adding
Vectors]*

*The answer is $(1,2,3)+(4,5,6)=(5,7,9)$. This is because
$1+4=5$, $2+5=7$ and $3+6=9$. The reason the answer is those sums
is because to find out how far we go in the x-direction, we work
out how far we have gone in the x-direction from the first vector,
and add on how far we have gone in the x-direction from the second
vector. We do the same for the y- and z-directions. In general,
$(x,y,z)+(u,v,w)=(x+u,y+v,z+w)$.*

**Answer to Question 2**
*[Transformations]*

*The answer is $(1,7,3)$ because we add the vector $(0,5,0)$
to move 5 in the y-direction, and $(1,2,3)+(0,5,0)=(1,7,3)$. In
general, a translation $T$ is something like ``add $(u,v,w)$ to the
vector'', to $T(x,y,z)=(x,y,z)+(u,v,w)=(x+u,y+v,z+w)$. If you
haven't done matrices and
vectors at school, ignore the next bit.*

To rotate a vector by an angle $\theta$ about the x-axis, the matrix is

\mathbf{M}_{\theta} = \left( \begin{array}{ccc}
\cos(\theta) & -\sin(\theta) & 0 \\ \sin(\theta) &
\cos(\theta) & 0 \\ 0 & 0 & 1\\ \end{array}
\right)

So $\mathbf{M_{\theta(x,y,z)}}=(x\cos(\theta)-y\sin(\theta),x\sin(\theta)+y\cos(\theta),z)$.

**Answer to Question 3** *[Making triangles]
Here is one way of doing it, but not the only
way:*

**Answer to Question 4** *[Shortest
Path]*

*The shortest way to get from $A$ to $K$, is to go $A B C H G
J K$.*

**Answer to Question 5** *[Make your own graph]
Here is one way to do it, but this isn't necessarily
the only way to do it.*

If you are creating random levels in a game, you can't decide on where the interesting points are, so you need to think of a way to instruct the computer to decide where the interesting points are. There are lots of ways of doing this, think about how you might go about doing this. Some things to consider are (a) can you get to every point on the map by walking in a straight line from some node? You need to make sure you can. (b) Do you have any silly situations, like the one below where you can get from A to B, but only by taking a very silly route.

**Answer to Question 6** *[Vectors]*

*(a)* $(1,2,3)+(3,2,1)=(4,4,4)$.

*(b)* $10\times (9,5,2)=(90,50,20)$.

*(c)*
$|(4,2,4)|=\sqrt{4^2+2^2+4^2}=\sqrt{16+4+16}=\sqrt{36}=6$.

**Answer to Question 7** *[Throwing Rocks in
Space]*

*Write Billy-Joe's position as $\mathbf{x}$ and Arthur's
position as $\mathbf{y}$. If Billy-Joe throws the rock with
velocity $\mathbf{v}$, then after $t$ seconds, the rock will be at
$\mathbf{x}+\mathbf{v}\times t$. He wants it to hit Arthur, so he
wants $\mathbf{x}+\mathbf{v} \times t = \mathbf{y}$. Rearranging
this equation, we get $\mathbf{v}=\frac{\mathbf{y}-\mathbf{x}}{t}$.
So, the velocity should be the position of Arthur minus the
position of Billy-Joe, divided by the time taken, 7 seconds. So
$\mathbf{v}=\frac{(240,640,143)-(100,150,150)}{7}=\frac{(140,490,-7)}{7}=(20,70,-1)$.*

**Answer to Question 8** *[Throwing Rocks on
Earth]*

*Because the x- and z-directions are 0 in all the vectors, we
can ignore them and just look at the y-directions. At time $t=0$,
the rock is at position 0, with velocity 4 (with vectors, the rock
is at position $(0,0,0)$ with velocity $(0,4,0)$). Therefore, at
time $t=0.2$, the rock will be at position $0+4 \times 0.2=0.8$. At
time $t=0.2$, the velocity will now be $4+0.2 \times -9.81 =
2.038$. So, at time $t=0.4$, the rock will be at position
$0.8+2.038 \times 0.2=1.2076$. At time $t=0.4$, the velocity will
be $4+0.4 \times -9.81=0.076$. So at time $t=0.6$, the rock will be
at $1.2076+0.076 \times 0.2 = 1.2228$. At $t=0.6$, the velocity
will be $4+0.6 \times -9.81 = -1.886$, so at $t=0.8$ the position
will be $1.228-1.886 \times 0.2 = 0.8508$. At $t=0.8$ the velocity
will be $4+0.8 \times -9.81 = -3.848$, so at $t=1.0$ the position
will be $0.8508-3.848 \times 0.2 = 0.0812$. So, after 1 second, the
position of the rock will be $(0,0.0812,0)$.*

For the last part of the question, you work out the velocity at time $t=1.0$ is $4-9.81 \times 1.0 = -5.81$, so the position at time $t=1.2$ is $0.0812-5.81\times 0.2=-1.0808$, so somewhere between $t=1.0$ and $t=1.2$, the rock hits Billy-Joe on the head.

For people interested in learning to program, you can download a free C++ compiler called DJGPP at:

There is also a special ``library'' for making games which you might find useful, called Allegro, at:

http://www.talula.demon.co.uk/allegro

Here are some links on programming games:

This is a brilliant page on path finding:

http://www-cs-students.stanford.edu/~amitp/gameprog.html

Here are some very comprehensive notes on 3D graphics (very hard!):

http://www.cc.gatech.edu/gvu/multimedia/nsfmmedia/cware/graphics/toc.html