Randomly pick one of the three corners. Call this the current point (Alternative
ID: 3812110 • Letter: R
Question
Randomly pick one of the three corners. Call this the current point (Alternatively, you could also randomly select any point inside the triangle) Randomly pick one of the three corners. this the target point. Calculate the midpoint between the current point and the target point. Draw a single pixel at the midpoint. Set the current point to the midpoint. Co back to step 2 and repeat many times (e.g., 50,000 iterations). It would seem like you should get a random mess of doors since the algorithm randomly picks a target corner each rime. Instead, if this process is repeated many times, you will get a very orderly picture with a repeating structure. To draw a single pixel at coordinate (X, Y) use the drawLine method where the start and endpoints are both (X, Y), To generate a random number x, where 0 xExplanation / Answer
The Sierpinski Triangle, also called Sierpinski Gasket and Sierpinski Sieve.
void drawSierpinski(float x1, float y1, float x2, float y2, float x3, float y3);
//Declaration of the subTriangle function, the coordinates are the 3 corners, and n is the number of recursions.
void subTriangle(int n, float x1, float y1, float x2, float y2, float x3, float y3);
//depth is the number of recursive steps
int depth = 7;
//The main function sets up the screen and then calls the drawSierpinski function
int main(int argc, char *argv[])
{
screen(640, 480, 0, "Sierpinski Triangle");
cls(RGB_White); //Make the background white
drawSierpinski(10, h - 10, w - 10, h - 10, w / 2, 10); //Call the sierpinski function (works with any corners inside the screen)
//After drawing the whole thing, redraw the screen and wait until the any key is pressed
redraw();
sleep();
return(0);
}
//This function will draw only one triangle, the outer triangle (the only not upside down one), and then start the recursive function
void drawSierpinski(float x1, float y1, float x2, float y2, float x3, float y3)
{
//Draw the 3 sides of the triangle as black lines
drawLine(int(x1), int(y1), int(x2), int(y2), RGB_Black);
drawLine(int(x1), int(y1), int(x3), int(y3), RGB_Black);
drawLine(int(x2), int(y2), int(x3), int(y3), RGB_Black);
//Call the recursive function that'll draw all the rest. The 3 corners of it are always the centers of sides, so they're averages
subTriangle
(
1, //This represents the first recursion
(x1 + x2) / 2, //x coordinate of first corner
(y1 + y2) / 2, //y coordinate of first corner
(x1 + x3) / 2, //x coordinate of second corner
(y1 + y3) / 2, //y coordinate of second corner
(x2 + x3) / 2, //x coordinate of third corner
(y2 + y3) / 2 //y coordinate of third corner
);
}
//The recursive function that'll draw all the upside down triangles
void subTriangle(int n, float x1, float y1, float x2, float y2, float x3, float y3)
{
//Draw the 3 sides as black lines
drawLine(int(x1), int(y1), int(x2), int(y2), RGB_Black);
drawLine(int(x1), int(y1), int(x3), int(y3), RGB_Black);
drawLine(int(x2), int(y2), int(x3), int(y3), RGB_Black);
//Calls itself 3 times with new corners, but only if the current number of recursions is smaller than the maximum depth
if(n < depth)
{
//Smaller triangle 1
subTriangle
(
n+1, //Number of recursions for the next call increased with 1
(x1 + x2) / 2 + (x2 - x3) / 2, //x coordinate of first corner
(y1 + y2) / 2 + (y2 - y3) / 2, //y coordinate of first corner
(x1 + x2) / 2 + (x1 - x3) / 2, //x coordinate of second corner
(y1 + y2) / 2 + (y1 - y3) / 2, //y coordinate of second corner
(x1 + x2) / 2, //x coordinate of third corner
(y1 + y2) / 2 //y coordinate of third corner
);
//Smaller triangle 2
subTriangle
(
n+1, //Number of recursions for the next call increased with 1
(x3 + x2) / 2 + (x2 - x1) / 2, //x coordinate of first corner
(y3 + y2) / 2 + (y2 - y1) / 2, //y coordinate of first corner
(x3 + x2) / 2 + (x3 - x1) / 2, //x coordinate of second corner
(y3 + y2) / 2 + (y3 - y1) / 2, //y coordinate of second corner
(x3 + x2) / 2, //x coordinate of third corner
(y3 + y2) / 2 //y coordinate of third corner
);
//Smaller triangle 3
subTriangle
(
n+1, //Number of recursions for the next call increased with 1
(x1 + x3) / 2 + (x3 - x2) / 2, //x coordinate of first corner
(y1 + y3) / 2 + (y3 - y2) / 2, //y coordinate of first corner
(x1 + x3) / 2 + (x1 - x2) / 2, //x coordinate of second corner
(y1 + y3) / 2 + (y1 - y2) / 2, //y coordinate of second corner
(x1 + x3) / 2, //x coordinate of third corner
(y1 + y3) / 2 //y coordinate of third corner
);
}
}
After running the program we will get the above triangle that is given in the problem
This program is to create a window in which a Sierpinski gasket is drawn using multiple calls to draw Triangle. The gasket consists of <depth> sizes of triangles, bounded by the points represented by <a>, <b>, and <c>. Each point is an array of two real numbers, the x-coordinate followed by the y coordinate. Both coordinates will be between 0 and 1.Related Questions
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