1. Computer Graphics in Java CR325

2. What is Computer Graphics? A kind of Data processing Voice and Signal Processing = 1D data processing Image Processing = 2D data processing Computer Graphics = 3D data in with 2D data out. Geometry is transformed into images.

3. What is Computer Animation? Like Computer Graphics with a difference. Animation=movement. This is simulated with image sequences. Output is an image sequence. Input is 3D geometry.

4. Why study Computer Graphics? Entertainment Visualization of data (analysis) Simulation can be visualized with CG. CAD (Computer Aided Design)

5. What do people study in CG? How to do 2D drawing –How to draw a line –How to draw a circle, etc. How to do 3D drawing –How to draw a 3D line –How to draw a sphere, etc.

6. Where do we get input? Pick device –Gives us a location –Mouse, light-pen, arrow-keys, keyboard, –Joystick, tablet, scanner, image –video camera with a cooperative target. –Geometries –Dynamics, kinematics (math).

7. Introduction to 2D Graphics Draw some 2D geometries. Input is 2D geometry Ouput is image.

8. How do we represent an image? Image Processing answer: 2D array of pixels Computer Answer: BREP:vectors to the right is a bitmap vector vs. bitmap

9. What is an image? A 2D array of PIXELS.

10. What is a PIXEL A picture element. Typically a pixel can be described as an intensity for light at a given location. –Location (200,20), I = 255; // gray scale –Location (34,50), R=255, G= 0, B =0; //red pixel. Looks like a dot!

11. What is an Image? a 2D array of pixels a pixel is a scalar quantity that represents the intensity of light. It smallest area within the given image.

12. How do we represent a pixel? how do we represent light?

13. Trading an eye for an ear

14. Spectra of the Human Visual System ranges from 400 nm wavelength to about 800 nm, 1 nm = 10 ** -9 meters d = rate * time rate = c = 3 * 10 ** 8 meters / second d = wavelength = 400 * 10 **-9 meters T = lambda/c, f = 1/T = c / lambda f = 3 * 10 ^ 8 / 4 * 10 ^ -7 =750.*10^12?

15. An eye is a Multi-mega pixel camera It has a lens (adjustable zoom) It has an automatic f-stop (iris 2-8 mm) It has a sensor plane (100 million pixels) The sensor has a transfer function senstive to mesopic range; 380 to about 700 nm

16. The eyes have a USB2 data rate! 250,000 neurons in the optic nerve variable voltage output on EACH nerve 17.5 million neural samples per second 12.8 bits per sample 224 Mbps, per eye (a 1/2 G bps system!). Compression using lateral inhibition between the retinal neurons

17. Response curves Eye has Gaussian response to light. Gives rise to biologically motivated image processing

18. Quantization of an Image Computers use digital cameras -> quantization

19. Quantization Error is visible

20. Displays Color Monitors are made for people physiological arguments are used for display design

21. What is light? any electromagnetic energy visible light is typically visible to the human visual system (eye).

22. How do we represent a pixel assume that the pixel is for humans. You need to know the dynamic range of the eye! 1 part in a million is needed for a pixel that displays on a monitor for people. Typically we use 24 bits per pixel, 8 bits per color.

23. peter packed a pickled pixel its color was that of hash. And every system that displayed it would crash!

24. Displays are designed for human visual system (eye) Painting with LIGHT (additive synthesis) display This is not painting with MUD! (subtractive synthesis) printer

25. Overview public int filterRGB (int rgb) { int r = (rgb & 0xff0000) >>16; int g = (rgb & 0xff00) >> 8; int b = rgb & 0xff; int gr = (r + g + b) / 3; int p = 0xff000000 | (gr << 16) | (gr << 8) | gr; return p // p = 0xffgrgrgr;

26. How would do you pack RGB? int pack(int r, int g, int b) { return 0xff000000 | (r << 16) | (g << 8) | b; }

27. How do I represent the following in decimal int I = 1; int k = ~I+1; how much is k? Ans: -1

28. Intro to bitwise ops and - & - bitwise (not && is logical) or - | - bitwise (not || is logical) not - ~ - bitwise (not !, is logical) xor - ^ - bitwise (has no logical correspondence)

29. Intro to ANDing int I = 0; int j = 1; I&I is 0 I&J is 0 J&I is 0 J&J is 1

30. Intro to Oring int i=0xF0; S={0,1,2,3,4,5,6,7,8,9,a,b,c,d,e,f} int j =0x0F; i|I is 240 = 0xF0 i|j is 255 in decimal = 0xFF 0xF0 = (in base 2) 11110000 0x0F = (in base 2) 00001111 0x0F | 0xF0 = 11111111

31. Using hex for ANDing int i=0x7a; int j =0x0F; how much is i&j ? 01111100 & 00001111 = 00001100 = 0x0c

32. using hex for not int i=0x7a; int j =0x0F; how much is ~j? 0xfffffff0 nibble is 4 bits, byte is 8 bits

33. using hex for xor int i=0x7a; int j =0x0F; how much is i^j? 01111100 = I^ 00001111 = j 01110011 = i^j = 0x73

34. Common Radicies in java, you can use: binary (base 2) Radix 2 octal (base 8) Radix 8 decimal (base 10) Radix 10 hexidecimal (base 16) Radix 16

35. How come they call it base 2? You can only have 0 and 1, so why call it base 2? Why not call it base 1? Cardinality of the set of symbols is the radix S = {0, 1}, card(s)=2

36. Why call it a base? The base of the exponent! 1101, base 2 is, in decimal 1*2^3 + 1 * 2^2 + 0 * 2 ^ 1 + 1 = 8 + 4 + 0 + 1 = 13 base 10. This is called positional notation.

37. Doing the positional notation in hex 0x28 = 2 * 16 ^ 1 + 8 * 16 ^ 0 = 2 * 16 + 8 = 32 + 8 = 40 base 10