The figures below are results of applying the random IFS alogrithm with the transformations 
T_{3}(x, y) = (x/2, y/2) + (0, 1/2) 
T_{4}(x, y) = (x/2, y/2) + (1/2, 1/2) 
T_{1}(x, y) = (x/2, y/2) 
T_{2}(x, y) = (x/2, y/2) + (1/2, 0) 

with probabilities (p_{1}, p_{2}, p_{3}, p_{4}) 
(0.35,0.30,0.20,0.15) 
(0.50,0.25,0.20,0.05) 
(0.60,0.20,0.15,0.05) 
(0.80,0.10,0.06,0.04) 





Here are the corresponding f(α) curves. 



Note that because each of the log(p_{i})/log(r_{i}) are distinct in all four examples,
we have seen f(alpha_{min}) = f(alpha_{max}) = 0. 
Note that the more uniform the probabilities (the red
curve), the narrower
the domain of the f(α) curve. 
If probabilities are equal, the f(α) curve
collapses to a vertical line segment. 
Recalling the max and min values of α are given by the max and min values of
log(p_{i})/log(r_{i}), we see 
log(.35)/log(.5) ≈ 1.51, log(.15)/log(.5) ≈ 2.74 
log(.50)/log(.5) = 1.00, log(.05)/log(.5) ≈ 4.32 
log(.60)/log(.5) ≈ 0.74, log(.05)/log(.5) ≈ 4.32 
log(.80)/log(.5) ≈ 0.32, log(.04)/log(.5) ≈ 4.64 
