Data Analysis by Driven IFS

Exercise Answers

(a) (b) (c)

10. From the graphs of the functions, here are the allowed transitions
(a) 1 -> 1, 1 -> 2, 2 -> 3, 2 -> 4, 3 -> 3, 3 -> 4, 4 -> 1, 4 -> 2
(b) 1 ->1, 1 -> 2, 1 -> 3, 2 -> 3, 2 -> 4, 3 -> 3, 3 -> 4, 4 -> 1, 4 -> 2
(c) 1 -> 1, 1 -> 2, 1 -> 3, 2 -> 4, 3 -> 3, 3 -> 4, 4 -> 1, 4 -> 2
Here are the corresponding occupied addresses
(a) 11, 21, 32, 42, 33, 43, 14, 24
(b) 11, 21, 31, 32, 42, 33, 43, 14, 24
(c) 11, 21, 31, 42, 33, 43, 14, 24

(i) (ii)

So we see function (a) and driven IFS (i) have the same occupied length 2 addresses. Also, function (b) and driven IFS (ii) have the same occupied length 2 addresses. Consequently, (i) and (ii) could be the driven IFS of functions (a) and (b).

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