[프로세싱] 재귀 함수 예제 : 몬드리안 작품

아래의 소스코드는 프로세싱으로 재귀 함수를 이용해 몬드리안의 작품과 유사한 그림을 그리는 코드이다.

// Draw a Mondrian-inspired image using recursion
void piet(int x0, int y0, int x1, int y1, int N) {
 if (N == 0) {
    // Base case -- draw a colorful rectangle with a thick black border
    int sw = 3; //this is the stroke width for the rectangle's border
    color c[] = { #ff0000, #00ff00, #0000ff, #ffff00, #ffffff}; //Mondrian color palatte
    fill(c[int(random(c.length))]);
    strokeWeight(sw);
    rect (x0,y0,x1-x0-sw,y1-y0-sw);
 } else {
    //Recursive step -- break the current rectangle into 4 new random rectangles
    int i = int(random(x0,x1));
    int j = int(random(y0,y1));
    piet(x0,y0,i,j,N-1); // upper left rectangle
    piet(i,y0,x1,j,N-1); // upper right rectangle
    piet(x0,j,i,y1,N-1); // lower left rectangle
    piet(i,j,x1,y1,N-1); // lower right rectangle
 }
}




The function accepts 5 arguments: the coordinates of the upper left hand corner of a rectangle (x0 and y0), the coordinates of the lower right hand corner (x1 and x2), and the current recursive depth (N). The function does two basic things. First, it tests to see if it’s reached the “base case,” when N = 0. If so, then all it does is draw a rectangle that is randomly colored from the basic Mondrian palette of colors (red, greed, blue, yellow, and white). If it’s not the base case, then it selects a random point (i,j) inside the current rectangle, and then calls piet() 4 more times, passing in the various coordinates that form the 4 new interior rectangles. This “divide and conquer” approach of having a function that calls itself is the hallmark of recursion.