1.G. The Random IFS Algorithm

Fixed Points

A fixed point of a transformation T(x,y) is a point left unchanged by the transformation. That is,

T(x, y) = (x, y)

If T(x,y) is a contraction, then it has exactly one fixed point.

Example 1 T(x, y) = (x/2, y/2) has fixed point (0, 0). (x, y) = T(x, y) = (x/2, y/2). So x = x/2 and y = y/2, hence x = 0 and y = 0. Example 2 T(x, y) = (x/2, y/2) + (1/2, 0) has fixed point (1, 0). (x, y) = T(x, y) = (x/2, y/2) + (1/2, 0). So x = x/2 + 1/2 and y = y/2, hence x = 1 and y = 0. Example 3 T(x, y) = (a⋅x, b⋅y) + (c, d) has fixed point (c/(1-a), d/(1-b)). (x, y) = T(x, y) = (a⋅x, b⋅y) + (c, d). So x = a⋅x + c and y = b⋅y + d, hence x = c/(1-a) and y = d/(1-b). We have assumed neither a nor b is 1. What happens if either of these is true?

Return to Definition and Illustration of the Random IFS Algorithm.