Mathematics (from Greek μάθημα máthēma, “knowledge, study, learning”) is the study of topics such as quantity (numbers), structure, space, and change. There is a range of views among mathematicians and philosophers as to the exact scope and definition of mathematics

1. 1
Game MathematicsGame Mathematics

2.  MatricesMatrices
 VectorsVectors
 Fixed-point Real NumbersFixed-point Real Numbers
 Triangle MathematicsTriangle Mathematics
 Intersection IssuesIntersection Issues
 Euler AnglesEuler Angles
 Angular DisplacementAngular Displacement
 QuaternionQuaternion
 Differential Equation BasicsDifferential Equation Basics
2
Essential Mathematics for Game DevelopmentEssential Mathematics for Game Development

3.  Matrix basicsMatrix basics
– DefinitionDefinition
– TransposeTranspose
– AdditionAddition
3
MatricesMatrices
A = (aij) =
a11 .. a1n
. .
. .
am1 .. amn
C = A T
cij = aji
C = A + B cij = aij + bij

4. – Scalar-matrix multiplicationScalar-matrix multiplication
– Matrix-matrix multiplicationMatrix-matrix multiplication
4
C = αA cij = αaij
C = A B cij = Σaikbkj
k = 1
r
cij = row x column

5.  Transformations inTransformations in MatrixMatrix formform
– A point or a vector is a row matrix (de facto convention)A point or a vector is a row matrix (de facto convention)
5
V = [x y z]
– Using matrix notation, a pointUsing matrix notation, a point VV is transformed underis transformed under
translation, scaling and rotation as :translation, scaling and rotation as :
V’ = V + D
V’ = VS
V’ = VR
where D is a translation vector and
S and R are scaling and rotation matrices

6. 6
– To make translation be a linear transformation, weTo make translation be a linear transformation, we
introduce theintroduce the homogeneous coordinate systemhomogeneous coordinate system
V (x, y, z, w)
where w is always 1
– Translation TransformationTranslation Transformation
x’ = x + Tx
y’ = y + Ty
z’ = z + Tz
V’ = VT
[x’
y’
z’
1] = [x y z 1]
= [x y z 1] T
1 0 0 0
0 1 0 0
0 0 1 0
Tx Ty Tz 1

7. 7
– Scaling TransformationScaling Transformation
x’ = xSx
y’ = ySy
z’ = zSz
V’ = VS
[x’
y’
z’
1] = [x y z 1]
= [x y z 1] S
Sx 0 0 0
0 Sy 0 0
0 0 Sz 0
0 0 0 1
Here Sx, Sy and Sz are scaling factors.

8. 8
– Rotation TransformationsRotation Transformations
1 0 0 0
0 cosθ sinθ 0
0 -sinθ cosθ 0
0 0 0 1
Rx =
Ry =
Rz =
cosθ 0 -sinθ 0
0 1 0 0
sinθ 0 cosθ 0
0 0 0 1
cosθ sinθ 0 0
-sinθ cosθ 0 0
0 0 1 0
0 0 0 1

9. 9
– Net Transformation matrixNet Transformation matrix
– Matrix multiplication areMatrix multiplication are not commutativenot commutative
[x’
y’
z’
1] = [x y z 1] M1
and
[x” y” z” 1] = [x’ y’ z’ 1] M2
then the transformation matrices can be concatenated
M3 = M1 M2
and
[x” y” z” 1] = [x y z 1] M3
M1 M2 = M2 M1

10.  A vector is an entity that possessesA vector is an entity that possesses magnitudemagnitude
andand directiondirection..
 A 3D vector is a triple :A 3D vector is a triple :
– VV = (v= (v11, v, v22, v, v33)), where each component, where each component vvii is a scalar.is a scalar.
 A ray (directed line segment), that possessesA ray (directed line segment), that possesses
positionposition,, magnitudemagnitude andand directiondirection..
10
VectorsVectors
(x1,y1,z1)
(x2,y2,z2)
V = (x2-x1, y2-y1, z2-z1)

11.  Addition of vectorsAddition of vectors
 Length of vectorsLength of vectors
11
X = V + W
= (x1, y1, z1)
= (v1 + w1, v2 + w2, v3 + w3)
V
W
V + W
V
W
V + W
|V| = (v1
2
+ v2
2
+ v3
2
)1/2
U = V / |V|

12.  Cross product of vectorsCross product of vectors
– DefinitionDefinition
– ApplicationApplication
» A normal vector to a polygon is calculated from 3 (non-collinear)A normal vector to a polygon is calculated from 3 (non-collinear)
vertices of the polygon.vertices of the polygon.
12
X = V X W
= (v2w3-v3w2)i + (v3w1-v1w3)j + (v1w2-v2w1)k
where i, j and k are standard unit vectors :
i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1)
Np
V2
V1
polygon defined by 4 points
Np = V1 X V2

13. »Normal vector transformationNormal vector transformation
13
N(X) = detJ J-1T
N(x)
where X = F(x)
J the Jacobian matrix, Ji(x) =
δF(x)
δxi
"Global and Local Deformations of Solid Primitives"
Alan H. Barr
Computer Graphics Volume 18, Number 3 July 1984

14. Normal vector transformation:Normal vector transformation:
ExampleExample
Given plan: S = S(x, 0, z), x in [0, 1], z in [0, 1]
Transform the plane to the curved surface of a half cylinder:
x’ = r – r cos (x/r)
y’ = r sin (x /r)
z’ = z
where r is the radius of the cylinder and r = 1/π. 14

15.  Dot product of vectorsDot product of vectors
– DefinitionDefinition
– ApplicationApplication
15
|X| = V . W
= v1w1 + v2w2 + v3w3
θ
V
W
cosθ =
V . W
|V||W|

16.  Fixed Point Arithmetics : N bits (signed) IntegerFixed Point Arithmetics : N bits (signed) Integer
– Example : N = 16 gives range –32768Example : N = 16 gives range –32768 ≤≤ aaii ≤≤ 3276732767
– We can use fixed scale to get the decimalsWe can use fixed scale to get the decimals
16
Fixed Point Arithmetics (1/2)Fixed Point Arithmetics (1/2)
a = ai / 28
1 1 1
8 integer bits
8 fractional bits
ai = 315, a = 1.2305

17.  Multiplication then Requires RescalingMultiplication then Requires Rescaling
 Addition just Like NormalAddition just Like Normal
17
Fixed Point Arithmetics (2/2)Fixed Point Arithmetics (2/2)
e = a.
c = ai / 28 .
ci / 28
⇒ ei = (ai
.
ci) / 28
e = a+c = ai / 28
+ ci / 28
⇒ ei = ai + ci

18.  Compression for Floating-point Real NumbersCompression for Floating-point Real Numbers
– 4 bytes reduced to 2 bytes4 bytes reduced to 2 bytes
– Lost some accuracy but affordableLost some accuracy but affordable
– Network data transferNetwork data transfer
 Software 3D RenderingSoftware 3D Rendering
18
Fixed Point Arithmetics - ApplicationFixed Point Arithmetics - Application

19. 19
h
ha
hb
hc
Aa
Ac
Ab
h = ha + hb + hc
where A = Aa + Ab + Ac
If (Aa < 0 || Ab < 0 || Ac < 0) then
the point is outside the triangle
“Triangular Coordinates”
Aa Ab Ac
A A A
p
(xa,ya,za)
(xb,yb,zb)
(xc,yc,zc)
Triangular CoordinatesTriangular Coordinates

20. 20
Area of a triangle in 2D
xa ya
A = ½ xb yb
xc yc
xa ya
= ½ (xa*yb + xb*yc + xc*ya – xb*ya – xc*yb – xa*yc)
Triangle Area – 2DTriangle Area – 2D
(xa,ya,za)
(xb,yb,zb)
(xc,yc,zc)

21. 21
Area of a triangle in 3D
Triangle Area – 3DTriangle Area – 3D
(x0,y0,z0)
(x1,y1,z1)
(x2,y2,z2)
u
v
u = p1 – p0
v = p2 – p0
area = ½ |u x v |
How to use triangular coordinates to determine the
location of a point? Area is always positive!
p1
p0
p2

22.  Terrain FollowingTerrain Following
 Hit TestHit Test
 Ray CastRay Cast
 Collision DetectionCollision Detection
22
Triangular Coordinate System - ApplicationTriangular Coordinate System - Application

23.  Ray CastRay Cast
 Containment TestContainment Test
23
IntersectionIntersection

24. 24
Ray Cast – The RayRay Cast – The Ray
x = x0 + (x1 – x0) t
y = y0 + (y1 – y0) t, t = 0,
z = z0 + (z1 – z0) t
{
 Shoot a ray to calculate the intersection betweenShoot a ray to calculate the intersection between
the ray and modelsthe ray and models
 Use a parametric equation to represent a rayUse a parametric equation to represent a ray
8
 The ray emits from (xThe ray emits from (x00,y,y00,z,z00))
 Only the tOnly the t ≥≥ 0 is the answer candidate0 is the answer candidate
 The smallest positive t is the answerThe smallest positive t is the answer

25. 25
Ray Cast – The PlaneRay Cast – The Plane
 Each triangle in the Models has its plane equationEach triangle in the Models has its plane equation
 UseUse ax + by + cz + d = 0ax + by + cz + d = 0 as the plane equationas the plane equation
 (a, b, c)(a, b, c) is the plane normal vectoris the plane normal vector
 |d||d| is the distance of the plane to originis the distance of the plane to origin
 Substitute the ray equation into the planeSubstitute the ray equation into the plane
equationequation
 Solve theSolve the tt to Find the Intersectto Find the Intersect
 Check whether or not the intersect point insiderCheck whether or not the intersect point insider
the trianglethe triangle

26. 26
Intersection = 1, inside
Intersection = 2, outside
Intersection = 0, outside
Trick : Parametric equation for a ray which is parallel to the x-axis
x = x0 + t
y = y0 , t = 0,
{
8
(x0, y0)
2D Containment Test2D Containment Test
“if the No. of intersection is odd, the point is inside,
otherwise, it is outside”

27. 27
3D Containment Test3D Containment Test
“if the No. of intersection is odd, the point is inside,
otherwise, is outside”
 Same as the 2D containment testSame as the 2D containment test

28. Rotation vs OrientationRotation vs Orientation
28
 Orientation: relative to a referenceOrientation: relative to a reference
alignmentalignment
 Rotation:Rotation:
 change object from one orientation tochange object from one orientation to
anotheranother
 Represent an orientation as a rotationRepresent an orientation as a rotation
from the reference alignmentfrom the reference alignment

29. 29
 A rotation is described as a sequence of rotationsA rotation is described as a sequence of rotations
about three mutually orthogonal coordinates axesabout three mutually orthogonal coordinates axes
fixed in space (e.g.fixed in space (e.g. world coordinate systemworld coordinate system))
– X-roll, Y-roll, Z-rollX-roll, Y-roll, Z-roll
 There are 6 possible ways to define a rotationThere are 6 possible ways to define a rotation
– 3!3!
R(θ1, θ2, θ3) represents an x-roll, followed by y-roll, followed by z-roll
R(θ1, θ2, θ3) = c2c3 c2s3 -s2 0
s1s2c3-c1s3 s1s2s3+c1c3 s1c2 0
c1s2c3+s1s3 c1s2s3-s1c3 c1c2 0
0 0 0 1
where si = sinθi and ci = cosθi
Euler AnglesEuler Angles
Left hand system

30. 30
 Interpolation happening on each angleInterpolation happening on each angle
 Multiple routes for interpolationMultiple routes for interpolation
 More keys for constraintsMore keys for constraints
 Can lead to gimbal lockCan lead to gimbal lock
z
x
y
R
z
x
y
R
Euler Angles & InterpolationEuler Angles & Interpolation

31. 31
 RR((θθ,, nn),), nn is the rotation axisis the rotation axis
n
r Rr
θ
n
r
rv
rh
V
θ
rv
V
Rrv
rh = (n.
r)n
rv = r - (n.
r)n , rotate into position Rrv
V = nxrv = nxr
Rrv = (cosθ)rv + (sinθ)V
->
Rr = Rrh + Rrv
= rh + (cosθ)rv + (sinθ)V
= (n.
r)n + (cosθ) (r - (n.
r)n) + (sinθ) nxr
= (cosθ)r + (1-cosθ) n (n.
r) + (sinθ) nxr
Angular DisplacementAngular Displacement

32. 32
 Sir William Hamilton (1843)Sir William Hamilton (1843)
 From Complex numbers (a +From Complex numbers (a + iib),b), ii 22
= -1= -1
 16,October, 1843,16,October, 1843, Broome BridgeBroome Bridge inin DublinDublin
 11 realreal + 3+ 3 imaginaryimaginary = 1= 1 quaternionquaternion
 qq = a + b= a + bii + c+ cjj + d+ dkk
 ii22
== jj22
== kk22
= -1= -1
 ijij == kk && jiji = -= -kk, cyclic permutation, cyclic permutation ii--jj--kk--ii
 qq = (= (ss,, vv), where (), where (ss,, vv) =) = ss ++ vvxxii ++ vvyyjj ++ vvzzkk
QuaternionQuaternion

33. 33
q1 = (s1, v1) and q2 = (s2, v2)
q3 = q1q2 = (s1s2 - v1
.
v2 , s1v2 + s2v1 + v1xv2)
Conjugate of q = (s, v), q = (s, -v)
qq = s2
+ |v|2
= |q|2
A unit quaternion q = (s, v), where qq = 1
A pure quaternion p = (0, v)
Noncommutative
Quaternion AlgebraQuaternion Algebra

34. 34
Take a pure quaternion p = (0, r)
and a unit quaternion q = (s, v) where qq = 1
and define Rq(p) = qpq-1
where q-1
= q for a unit quaternion
Rq(p) = (0, (s2
- v.
v)r + 2v(v.
r) + 2svxr)
Let q = (cosθ, sinθ n), |n| = 1
Rq(p) = (0, (cos2
θ - sin2
θ)r + 2sin2
θ n(n.
r) + 2cosθsinθ nxr)
= (0, cos2θr + (1 - cos2θ)n(n.
r) + sin2θ nxr)
Conclusion :
The act of rotating a vector r by an angular displacement (θ, n)
is the same as taking this displacement, ‘lifting’ it into
quaternion space, by using a unit quaternion (cos(θ/2),
sin(θ/2)n)
Quaternion VS Angular DisplacementQuaternion VS Angular Displacement

35. 35
1-2y2
-2z2
2xy-2wz 2xz+2wy 0
2xy+2wz 1-2x2
-2z2
2yz-2wx 0
2xz-2wy 2yz+2wx 1-2x2
-2y2
0
0 0 0 1
q = (w,x,y,z)
Conversion: Quaternion to MatrixConversion: Quaternion to Matrix

36. 36
M0 M1 M2 0
M3 M4 M5 0
M6 M7 M8 0
0 0 0 1
float tr, s;
tr = m[0] + m[4] + m[8];
if (tr > 0.0f) {
s = (float) sqrt(tr + 1.0f);
q->w = s/2.0f;
s = 0.5f/s;
q->x = (m[7] - m[5])*s;
q->y = (m[2] - m[6])*s;
q->z = (m[3] - m[1])*s;
}
else {
float qq[4];
int i, j, k;
int nxt[3] = {1, 2, 0};
i = 0;
if (m[4] > m[0]) i = 1;
if (m[8] > m[i*3+i]) i = 2;
j = nxt[i]; k = nxt[j];
s = (float) sqrt((m[i*3+i] - (m[j*3+j] + m[k*3+k])) + 1.0f);
qq[i] = s*0.5f;
if (s != 0.0f) s = 0.5f/s;
qq[3] = (m[j+k*3] - m[k+j*3])*s;
qq[j] = (m[i+j*3] + m[j+i*3])*s;
qq[k] = (m[i+k*3] + m[k+i*3])*s;
q->w = qq[3];
q->x = qq[0];
q->y = qq[1];
q->z = qq[2];
}
Conversion: Matrix to QuaternionConversion: Matrix to Quaternion
http://en.wikipedia.org/wiki/Conversi
on_between_quaternions_and_Euler_a
ngles

37. 37
 Spherical linear interpolation,Spherical linear interpolation, slerpslerp
A
B
P
φ
t
slerp(q1, q2, t) = q1 + q2
sin((1 - t)φ)
sinφ sinφ
sin(tφ)
Quaternion InterpolationQuaternion Interpolation

38. 38
 Initial value problemsInitial value problems
 ODEODE
– Ordinary differential equationOrdinary differential equation
 Numerical solutionsNumerical solutions
– Euler’s methodEuler’s method
– The midpoint methodThe midpoint method
– Kunge –Kutta methodsKunge –Kutta methods
Differential Equation BasicsDifferential Equation Basics

39. 39
 An ODEAn ODE
 Vector fieldVector field
 SolutionsSolutions
– Symbolic solutionSymbolic solution
– Numerical solutionNumerical solution
x = f (x, t)
where f is a known function
x is the state of the system, x is the x’s time derivative
x & x are vectors
x(t0) = x0, initial condition
.
Start here Follow the vectors …
Initial Value ProblemsInitial Value Problems
.
.

40. 40
 A numerical solutionA numerical solution
– A simplification fromA simplification from Tayler seriesTayler series
 Discrete time steps starting with initial valueDiscrete time steps starting with initial value
 Simple but not accurateSimple but not accurate
– Bigger steps, bigger errorsBigger steps, bigger errors
– Step error: OStep error: O((h22
)) errorserrors
– Total error: O(h)Total error: O(h)
 Can be unstableCan be unstable
 Not efficientNot efficient
x(t + h) = x(t) + h f(x, t)
Euler’s MethodEuler’s Method

41. 41
Concept : x(t0 + h) = x(t0) + h x(t0) + h2
/2 x(t0) + O(h3
)
Result : x(t0 + h) = x(t0) + h[ f (x0 + h/2 f(x0 , t0), t0 +h/2) ]
Method : a. Compute an Euler step
∆x = h f(x0 , t0)
b. Evaluate f at the midpoint
fmid = f ( x0+∆x/2, t0 +h/2 )
c. Take a step using the midpoint
x( t+ h) = x(t) + h fmid
. ..
a
b
c
Error term
The Midpoint MethodThe Midpoint Method

42. 42
 Midpoint =Midpoint = Runge-KuttaRunge-Kutta method of order 2method of order 2
 Runge-KuttaRunge-Kutta method of order 4method of order 4
– Step error: OStep error: O(h(h55
))
– Total error: O(hTotal error: O(h44
))
k1 = h f(x0, t0)
k2 = h f(x0 + k1/2, t0 + h/2)
k3 = h f(x0 + k2/2, t0 + h/2)
k4 = h f(x0 + k3, t0 + h)
x(t0+h) = x0 + 1/6 k1 + 1/3 k2 + 1/3 k3 + 1/6 k4
TheThe Runge-KuttaRunge-Kutta MethodMethod

43.  DynamicsDynamics
– Particle systemParticle system
 Game FX SystemGame FX System
43
Initial Value Problems - ApplicationInitial Value Problems - Application
http://upload.wikimedia.org/wikipedia/en/4/44/Strand_Emitter.jpg