MATH 5300 Lecture 3- Summary Date: May 12, 2008 By: Violeta Constantin

Transcription

1 MATH 5300 Lecture 3- Summary Date: May 12, 2008 By: Violeta Constantin Facebook, Blogs and Wiki tools for sharing ideas or presenting work Using Facebook as a tool to ask questions - discussion on GIMP Using Blogs to display assignments/ work read over solutions; have complete sentences Some comments/ advice on Assessment 2: Question (2): A useful tool De Morgan s Laws: NOT(A AND B) = (NOT A) OR (NOT B) NOT( A OR B) = (NOT A) AND (NOT B) Using NOT in the 2 nd formula we obtain: A OR B = NOT((NOT A) AND (NOT B)) It means that we don t need both OR and AND and we could express OR in terms of NOT and AND. Also, NOT = expression with OR and IMPL. {OR, IMPL} generates {OR, IMPL, NOT} and it also generates {OR, NOT, AND, IMPL}. Therefore if we obtain the expressions for OR and IMPL, we will definitely obtain those for NOT and AND. Questions (4) and (5): How to create binary operations when working with ternary inputs. We can try for example: x ~y = x +y (mod3) or x&y = xy +1(mod3) Would they be enough to express all possible two argument functions? Problems that may arise when working in ternary: what do True and False mean? What do AND, OR, NOT mean? More about IMPL binary operator: Example: If I study hard IMPL I will get a good grade T = True F = False Situations x y ximply 1: F F T 2: F T T 3: T F F 4: T T T 2: I did not study F; I got an A T In this case, the sentence is example is T 3: I did study T; I failed F In this case, the sentence in the example is F because it said that if I study then I will get a good grade and I did not. 4: I did study T; I got a good grade T The original statement is true in this case. 1

2 New Topic: COLOR IMAGES How does computer represent color? How do humans see color? We only see a small part of the light spectrum: approximately between 380 and 700 nm. This is called the visible light spectrum. Eye is only sensitive to this much, but camera is more sensitive: radio waves, infrared, etc. Figure 1: The electromagnetic spectrum diagram ( Do animals see more than humans? In human eye there are three types of receptors: red, green, blue (color blindness two types of receptors). 2

3 Figure 2: Absorption of light by red, green, and blue cones in the human eye as a function of wavelength (maxima: 445 nm, 535 nm, 575 nm) obtained by detailed experiments in 1965 ( Figure 3: ( Standardized wavelength values for the three primary colors in 1931 by Commission Internationale de l Eclairage (CIE): Blue = nm Green = nm Red = 700 nm How to create colors? We can describe colors with their coordinates: 0<= x <=1 for Red, 0<= y <=1 for Green The colors of the visible light spectrum [2] color wavelength interval frequency interval red ~ nm ~ THz orange ~ nm ~ THz yellow ~ nm ~ THz green ~ nm ~ THz blue ~ nm ~ THz violet ~ nm ~ THz 3

4 Figure 4: The Chromaticity Diagram ( and Each point inside the triangle determined by Red-Green-Blue can be obtained by using a certain percentage of each of these colors. For example if we want a color that has 70% Red, 20% of Green and 10% of Blue, we can write the equation: (x,y) =.7(x1,y1) +.2 (x2,y2) +.1 (x3,y3) Where (xi,yi) for i=1,2,3 are respectively the coordinates of Red, Green, and Blue and (x,y) are the coordinates of desired color. Note that not all colors can be represented just by using R-G-B! White will be 33% of each R, G, B. The third coordinate is obtained if you fix the energy (luminosity in light). Let X, Y, and Z respectively be the values for the sensors Red, Green, and Blue. Then, x = X/(X + Y + Z) and y = Y/(X + Y + Z). This is the normalization! Vocabulary: Hue dominant wavelength of light Saturation how much white is added to your light Brightness hard to measure; depends on the light around you Luminosity, intensity amount of light energy Each pixel is made of three colors: Red, Green, Blue. If we know the percent at each point, we know the image (color). Not all colors can be represented. Primary colors of light: Red, Green, Blue Primary colors of ink: Yellow, Magenta, Cyan 4

5 Figure5: The additive picture of light which reflects color vs. the additive picture of inks/ pigments which absorb color. ( Light Ink On the computer screen I can represent colors with respect to their coordinates Red, Green, Blue. Their mixture will create White. When we print we can represent color with respect to their coordinates Yellow, Magenta, Cyan. Mixing these inks we obtain a kind of black blackish, therefore for the purpose of printing we will need a fourth color: pure Black. Image is an array of tuples of 3 numbers corresponding for each of Red, Green and Blue components. The percentage of each of these colors will determine the image/ color of the resulting picture. Figure6: Creating Color ( 5

6 For example the color for eyes and mouth of the character is obtained by mixing the colors of eyes and mouth in the three pictures: Red and Blue (no eyes and mouth in the Green component), so we obtain Magenta. The color for the body is White (light) because is a mixture of all three colors. The background is black. 100x100 pixels for each image; each pixel has 8 bits. A Black and White image is an array of 100x100 (1 bit/pixel). CGA card can display only 4 colors (of the 8 available) at the same time: black, cyan, magenta, white or black, red, green, orange. For each pixel are necessary 2 bits. EGA graphics have 16 colors therefore there are necessary 4 bits per pixel. Next VGA It has passed 25 years from Black and White to Plasma TV! Color Red for example can be represented by using three matrices: R for the Red component (all entries are 255), G for the Green component (all entries are 0) and B for the Blue component (all entries 0). When representing color, each entry has to be a number between 0 and 255. In binary 0 is , 255 is , and 256 is Uint8 = short notation for 8 bits. I have an image uniformly the same color. We use 24 bits for each pixel which means 2^24 different uniformly colored pictures. Hue- is the dominant wave length. Saturation the amount of White Intensity =1/3(R+G+B) which is between 0 and 255. (do not use brightness which depends on the surrounding). From: Hue by definition is rotation of colors in color space. What does this mean exactly? Well say you have a red color, and you rotate its hue 90 degrees. This would cause it rotate around the diagonal black-white line, and wind up in the yellow-green area of the color cube. When you change the hue of an image, you are rotating all of its colors by some angle. If you were to rotate an image's hue 180 degrees, then you would end up with every color being shifted to its complimentary (opposite). Red would become cyan, green would become purple, blue would become yellow, and so on. 6

7 Figure 7: The Colour Cube ( Hue Saturation Intensity Space (HSI): From: The HSI color space is very important and attractive color model for image processing applications because it represents color s similarly how the human eye senses colors. The HSI color model represents every color with three components: hue ( H ), saturation ( S ), intensity ( I ). The below figure illustrates how the HIS color space represents colors. The Hue component describes the color itself in the form of an angle between [0,360] degrees. 0 degree mean red, 120 means green 240 means blue. 60 degrees is yellow, 300 degrees is magenta. The Saturation component signals how much the color is polluted with white color. The range of the S component is [0,1]. 7

8 The Intensity range is between [0,1] and 0 means black, 1 means white. As the above figure shows, hue is more meaningful when saturation approaches 1 and less meaningful when saturation approaches 0 or when intensity approaches 0 or 1. Intensity also limits the saturation values. Figure 8: ( Intensity shows how far high and low on the axis we are. Hue will have a point for Red, one for Blue and one for Green. Yellow = (255,255,0) Cyan = (0,255,255) Green = (0,255,0) Blue = (0,0,255) Red = (255,0,0) Magenta = (255,0,255) We need a function that takes those points and sens them into an angle (the hue): Hue= θ for B<=G and 2π θ for B>G 1/ 2[( R G) + ( R B)] θ = arcos 2 1 / 2 [( R G) + ( R B)( G B)] Exercise: Let us calculate hue for Yellow: The coordinates of Yellow: Y=(255,255,0), therefore R=255, G=255, B=0. We substitute into the above formulas: θ = cos -1 1/ 2( ) 1 / 2 [( (255)] = cos -1 (1/2)= 60 ο Hand out: Assignment 3 8