Computer Graphics... Flashcards

Question

line and sphere intersection point

Answer 1

sphere : (x-Cx)^2 + (y-Cy)^2 + (z- Cz)^2 - r^2 = 0 line : p(t) = p0 + t(p1-p0) == p(t) = s + td intersect is (sx + tdx - Cx)^2 + (sy + tdy - Cy)^2 + (sz + tdz - Cz) ^2 - r^2 = 0

Answer 2

Covering a surface using polygon Drawing diagonals between non adjacenet vertices and diagonals dont intersect

Answer 3

if any straight line joining 2 points of the polygon is contained within is convex otherwise concave

Answer 4

Sum 𝑖=0 𝑛−1 𝐶𝑖𝐶𝑛−𝑖−1

Answer 5

vertex-vertex meshes Stores each vertex together with list of verticies its connected to adv- storage fficient dis - must do computational work to generate faces and edges face-vertex meshes consist ordered list of vertices and faces. for each vertex, list the faces it velongs to Half -edge meshes Each edge defines two half-edges, one in each direction. For each half-edge we store its twin, the next half edge around the polygon, the face it belongs to, the previous half edge.

Answer 6

For every point p - If p is outside or on the circumcircle → triangle is Delaunay. - If any point lies inside the circumcircle → triangle is not Delaunay.

Answer 7

For degree n, we begin n+1 control points p0... pn The first 2 points define start and end t range is 0 and 1 ( some point on line) Linear interpolate (1-t)A + tB

Answer 8

Interpolate P0 and P1 Q0 = (1-t)P0 + tP1 Interpolate P1 and P2 Q1 = (1-t)P1 + tP2 Interpolate between Q0 and Q1 and plug in t (1-t)^2P0 + 2t(1-t)P1 + t^2P2

Answer 9

sum of k = 0 to k = n (NCK) x^k where n is the power n! / k!(n-k)!

Answer 10

Sum from k = 0 to k = n (nCk) t^k (1-t)^n-k bk

Answer 11

Direct illumination - Light Arrives straight from light sources Indirect illumination- Light arrives after interacting with rest of scene Ambient light - Uniform from all directions. Consistent amount of light applied to every point Diffuse Light - Light comes from specific direction. Light reflected with equal intensity in all directions Specular light - highlights on shiny objects. Positiion of viewer important

Answer 12

Compute lighting at each vertex 𝐼 = 𝑘𝑎𝑚𝑏*𝐼𝑎 + 𝑘𝑑𝑖𝑓𝑓*𝐼𝐿(𝑙∙ 𝑛) + 𝑘𝑠𝑝𝑒𝑐𝐼𝐿(𝑟⃗ ∙ 𝑣⃗)^𝛼 kamb - constant for ambient reflectivity of matreial Ia is ambient light intensity kdiff is a constant for diffuse reflectivity of material IL is intensity of light source kspec is constant for specular reflectivity of material Vector to light source l Vector to viewer (camera) v Normal vector n (Mirror/perfect) reflection vector r

Answer 13

r = l + 2u this equals l + 2(n-l) this equals 2(n . l)n-l

Answer 14

𝐼𝑎𝑚𝑏 = 𝑘𝑎𝑚𝑏*𝐼𝑎

Answer 15

𝐼𝑑𝑖𝑓𝑓 = 𝑘𝑑𝑖𝑓𝑓*𝐼𝐿 * 𝑙 ∙ 𝑛 = 𝑘𝑑𝑖𝑓𝑓*𝐼𝐿 cos x

Answer 16

𝐼𝑠𝑝𝑒𝑐 = 𝑘𝑠𝑝𝑒𝑐*𝐼𝐿( 𝑟 ∙ 𝑣)^a = 𝑘𝑠𝑝𝑒𝑐*𝐼𝐿 (cos) 𝛾^𝛼 a is shininess coefficient

Answer 17

C^0 -> function is continuous C^1 -> its derivative is also continuous C^i = its ith derivative exist and continous

Answer 18

For a quadratic Bézier at the end: Tangent∝P2−P1 For a linear Bézier at the start: Tangent∝𝑄1−𝑄0 if there equal then yes

Answer 19

Flat Shading- Each vertex of each polygon has the same color and color whole

Answer 20

Verticies have differnet colors. Interpolate those colors over surface of a triangle

Answer 21

T has verticies V1, V2, V3 Then Wi = 1/distance(Vi,P) Then color c(P) = W1C(P1) + W2C(P2) + W3C(P3) / w1 + w2 + w3

Answer 22

Imagine we are interpolating the triangle with three points A = 𝑋1, 𝑌1 , 𝐵 = 𝑋2, 𝑌2 and C = X3, Y3 : Just remember this equation A+𝑊2( 𝐵 − 𝐴) + 𝑊3(𝐶 − 𝐴)

Answer 23

Phong shading interpolates the normal Phong Shading: Lighting equation used at each pixel Interpolate normal of the two verticies N= y-y1/y2-y1 N1 + y2-y/y2-y1 N2

Answer 24

trace the rays of light to model the interaction of light with objects. Use backward ray tracing to trace rays from camera to the object

Answer 25

Primary ray -ray that leaves eye and travels out to scene Secondary ray- ray that doesent leave eye

Answer 26

Shadow rays- from surface point towards light source Reflection rays -from surface point in mirror direction Refraction ray- from surface point in refracted direction

Answer 27

r1sin(x) = r2sin(y)

Answer 28

t = -an + v where a is a scalar and v is perpendicular to n and n is the normal vector

Answer 29

must be closed under scalar multiplication and vector addition Z^2 is not a vector space as (1,0) x 0.5 = (0.5,0) not integer

Answer 30

W is a subset of vectors in V and V is a vector space. if w is also a vector space then its a subspace of v.

Answer 31

Function taking vector from 1 vector space as input and outputting a vector in another Map from vector space to itself

Answer 32

F(λv) = λf(v) : Homogeneity f(u + v) = f(u) + f(v): addictivity i guess u and v and just be anything e.g u = (x,y) v = (z,w)

Answer 33

L : X -> Y X = domain of L, Y is range Image of L is set of vectors I (subset) Y which l maps some vertex of x Kernel of L is set of vectors k (Subset) X which are mapped by L to zero vector

Answer 34

k = {(x,y) : x = 0} I = {(x,2x) : x E R}

Answer 35

nullity of L Rank of L Dim(X) = Rank(L) + nullity(L)

Answer 36

Equal to number of linearly indepepndednt columns of M The number of non-zero rows in the row reduced matrix is the rank of the matrix.

Answer 37

nxn matrix is invertible if it has rank n

Answer 38

if determinant = 0 a11a22 - a12a21 a11C11 + a12C12 + ...A1nC1n Find Cofactor Cij = (-1)^i+j Mij Find Mij = remove row i and col j and take determinant of matrix remaining

Answer 39

For matrix M and shape T New area = det(M) x original area

Answer 40

adjoint A/ |A| adjoint A is transpose of cofactor matrix of A

Answer 41

X perserves y Cord |1 s | | x | = |x + sy| |0 1 | |y| = | y | Y perserves x cord |1 0| |x| = | x | |s 1| |y| = | sx + y|

Answer 42

|0 1| |x| = |y| |1 0| |y| = |x| 0 −1 −1 0 −1 0 0 1 1 0 0 −1

Answer 43

Answer 44

orthographic projection of a onto b tells us the component of a that travels in the same direction as b. a . b ------- b b . b Cos(θ)=||proj(a)|| / ||a||

Answer 45

Right most tranfsofrmation first Order just do rotate, reflect, shear, scale

Answer 46

x' = tx + x y' = ty + y just add x and y values to original values

Answer 47

1 0 𝑡x 0 1 ty 0 0 1 extend og x,y to (x,y,1) to use 3x3 matrix TM . [x,y,1] = -dot product both to get |x + tx| |y + ty| | 1 |

Answer 48

Add the 001 00 around the matrix for , rotatiton, scaling

Answer 49

|sx 0 0| |0 sy 0| |0 0 sz| Same as 2D with the 001 thing on outside

Answer 50

point p = [x,y,z,1] matrix uses 4x4 matrices [a11, a12, a13, tx] [a21, a22, a23, ty] [a31, a32, a33, tz] [0, 0, 0, 1 ] M . P = [x + tx] [y + ty] [z + tz] [ 1 ]

Answer 51

doubles x coord and thirds y coord

Answer 52

[x - sy] [ y ] [x ] [sx-y]

Answer 53

also [y] [x]

Answer 54

local coordinate system World coordinate system Eye coord system Normalized device coordinates

Answer 55

Modelling Transformation: local coordinates → world coordinates 2. Camera Transformation: world coordinates → eye coordinates 3. Projection Transformation: eye coordinates → NDC 4. Windowing Transformation: NDC → image coordinate syste

Answer 56

𝑎11 𝑎12 𝑎13 𝑡1 𝑎21 𝑎22 𝑎23 𝑡2 𝑎31 𝑎32 𝑎33 𝑡3 0 0 0 1 basically an affine transformation, use 4x4 matrix

Answer 57

Step 1 : Translation 1 0 0 -Cx 0 1 0 -Cy 0 0 1 -Cz 0 0 0 1 // subtract camera position so its postion becomes 0,0,0 step 2: //cam has 3 direction vectors u(right),v(up),w(forward) wx wy wz 0 vx vy vz 0 -ux -uy -uz 0 0 0 0 1 Step 3 Mcamera = Mr Mtr

Answer 58

generate a 2D image by Projecting objs onto view plane Parallel projection (orthographic or othogonal proj) - Maps 3D points to 2D by moving them along a projection direction until they hit an image plane. Perspective Projection = maps 3D points to 2D by projecting them along lines that pass through a single viewpoint until they hit an image plane

Answer 59

orthographic - Specified by a direction of projection - less realistic - parallel lines remain parallel Perspective - Specified by a centre of projection (camera) and the distance from this centre and the image plane -looks realistic -distances and angles not perserved

Answer 60

area in 3D space that is visible from a camera and may appear on the screen. Eliminating the portion of scene that Is outside the viewing frustum is called clipping Discard objects that cant be viewed

Answer 61

For each edge with first point v1, v2 1st Vertex , 2nd Vertex , Output Inside Inside Save 2nd Vertex Inside Outside Save Intersection Outside Outside None Outside Inside Save 2nd vertex and intersection

Answer 62

Frame buffer This is a 2D array which has same size of the screen. The GPU writes the color value into the frame buffer. Then the display reads the color values from frame buffer and puts each value onto the screen

Answer 63

determines if triangle is front facing or backfacing by order of verticies after projected onto screen. if the order is counter-clockwise, the triangle is front facing, if clockwise, it’s back facing.

Answer 64

process of finding and filling the pixels that are inside those primitives. (also scan conversion of a point)

Answer 65

Given some point x,y we need to choose a pixel to place that point in. Option 1. X’ = floor(x), Y’ = floor(y) Option 2. X’=floor(x+0.5), Y = floor(y+0.5)

Answer 66

y= mx + b if m < 1 use formula: Xₖ₊₁ = Xₖ + 1 Yₖ₊₁ = Yₖ + 1 Else use formula Yₖ₊₁ = Yₖ + 1 Xₖ₊₁ = Xₖ + 1/m

Answer 67

find delta(x) find delta(y) Q = 1/max{|delta(x), delta(y)|} Xₙ₊₁ = Xₙ + Q * delta(x) Yₙ₊₁ = Yₙ + Q * delta(y)

Answer 68

Simple and easy to implment Faster than direct use of the line equation since it avoids floating point multiplications. Disadvantages: * Deals with rounding and floating point arithmetic * Slower that breshams line generation

Answer 69

Dda uses floating-point additions and this alg avoids that. Step1: Calculate △ x and △y Step 2: Calculate decision parameters D = 2△y - △x say D = 0 Step 3: Find next point based on D if D < 0 -->D = D + 2dy then next pixel is (x+1, y) if D >=0 -->D= D + 2(dy-dx), next pixel is (x+1, y+1)

Answer 70

Edge equation Scan line equation(look in book) Barycentric coordinates

Answer 71

𝛼+β+𝛾=1, is inside of the triangle if and only if: 0≤𝛼,β,𝛾≤1

Answer 72

c=αc1+βc2+γc3 get that than apply it to each color presented

Answer 73

for each vertices of polygon create edges. e.g AB and determine the equation of that line. now determine its positive half space

Answer 74

Use smaller number of triangles for a model and map a precomputed image onto them Lower rendering time Easily change object appearance

Answer 75

Texture pixel corresponding to points from UV coords if number of texels is same as number of pixels, we can expect sample to be fairly accurate representation of texture if smaller than number of pixels, Than we will color many pixels alike(blurry) if larger than number of pixels, produce less detailed images

Answer 76

Used when texutre image we sample uses less texels than number of pixels in our polygon so our texels are larger than pixels Nearest neighbor mangification- for pixel (5,5) scale x = original width/ output width scale y = original height/ output height src x = xpixel(5) * scale x src y = ypixel(5) * scale y Then round

Answer 77

say x-coordinate odd → green (0,1,0) x-coordinate even → red (1,0,0) find x and y texel xtexel = xpixel * texel width/image width ytexel = ypixel * texture height/image height ID SURRONDING TEXELS xfloor=floor(xtexel),yfloor=floor(ytexel) //BOTTOM LEFT COMPUTE FRACTIONAL DISTANCES fx = xtexel - xfloor fy = ytexel - yfloor than interpolate horizontally C(bottom) = Q11 + fx * (Q21 - Q11) // Q11 is green so (0,1,) C(top) same way with Q12 --> Q22 Than interpolate vertically C=Cbottom+fy⋅(Ctop−Cbottom)

Answer 78

Texture minification is required when the texture image is larger (has more pixels) than the image to be mapped.

Answer 79

Use textures to alter the surface normal but does not change actual shape of surface

Answer 80

process of pre-calculating the brightness of objects in a scene and storing the result in a texture map for later use. provide lighting effects such as global illumination at a relatively low computational cost. lighting of surface cannot be changed in real time/

Answer 81

noise reduction super resolution deblurring

Answer 82

Increase visibility of objects by amplifying difference between colors of objects and colors of their backgrounds Process of adding color information to monochrome photos Framework that matches the visual appearances of images before blending them

Answer 83

Lossy compression technique achieved by compressing a range of values to a single value

Answer 84

Allows to store an image in fewper pixels at cost of lower quality we store the number of pixels in each color.

Answer 85

Sobel filter: creates an image (feature map) emphasising edges. Laplacian filter: used to find areas of rapid change (edges) in images Sharpening Filter: emphasize details and enhance the edges of objects in an image Gaussian filter: image smoothing

Computer Graphics... Flashcards

(112 cards)