Algorithms for Visual Design Using the Processing Language by Kostas Terzidis

By Kostas Terzidis

Because the first e-book to proportion the required algorithms for growing code to scan with layout difficulties within the processing language, this e-book bargains a chain of time-honored methods which can functionality as development blocks and encourages you to then use these development blocks to test, discover, and channel your suggestions, principles, and rules into strength strategies. The booklet covers such subject matters as dependent shapes, strong geometry, networking and databases, actual computing, photo processing, photo person interfaces, and extra.

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For example: noStroke(); rect(30,30,50,20); rect(40,40,20,30); will draw the shape in Figure 1-16. Figure 1-16: Two rectangles with no strokes ■■ strokeWeight() will increase the width of the stroke. It takes an integer number that specifies the number of pixels of the stroke’s width. For example, strokeWeight(4); rect(30,30,50,20); rect(40,40,20,30); will draw the shape in Figure 1-17. Figure 1-17: Two rectangles with thick strokes 19 20 Chapter 1 ■■ n Elements of the Language background() specifies the gray value or color of the display background.

Chapter 1 n Elements of the Language Table 1-7: Repetition Patterns Formula Result x = i/3; 00011122233344455566 x = i/4; 00001111222233334444 x = ($i+1)/2; 011223344556677889910 x = ($i+2)/2; 1122334455667788991010 x = i%2; 01010101010101010101 x = i%3; 01201201201201201201 x = i%4; 01230123012301230123 x = (i+1)%4; 12301230123012301230 x = (i+2)%4; 23012301230123012301 x = (i/2)%2; 00110011001100110011 x = (i/3)%2; 00011100011100011100 x = (i/4)%2; 00112233001122330011 These patterns can be classified into three categories.

Table 1-2: Arithmetic Operations Operator Use Description + op1 + op2 Adds op1 and op2 - op1 - op2 Subtracts op2 from op1 * op1 * op2 Multiplies op1 by op2 / op1 / op2 Divides op1 by op2 % op1 % op2 Computes the remainder of dividing op1 by op2 For example, to get the sum of two numbers, you can write: int sum; sum = 5 + 6; // not initialized because we do not know how much // now sum is 11 Note that the addition operation occurs on the right side of the equal sign, and the result is assigned to the variable on the left side of the equal sign.

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