Question

OK, our professor explained (kinda) this problem, but it still doesn’t make much sense.

Question: Implement the function knice(f,a,b,k) that will return 1 if for some integer a <= x <= b and some integer n <= k, n applications of f on x will be x, (e.g. f(f(f...(f(x)))) = x) and 0 if not.

What the professor provided was:

def knice(f,a,b,k):
    f(f(f(...(f(x)))) = x
    for i = a to b:
        y = f(i)
        if y = i break
    for j = z to k:
        y = f(y)
        if y = i break

Personally, that example makes no sense to me, so looking to see if I can get clarification.

OP EDIT 1/19/2012 3:03pm CST

This is the final function that was figured out with the help of the GTA:

def f(x):
    return 2*x-3

def knice(f,a,b,k):
x = a
while x <= b:
    n = 1
    y = f(x)
    if y == x:
        return 1
    while n <= k:
        y = f(y)
        n=n+1
        if y == x:
            return 1
    x=x+1   
return 0

Answer 1

Ignore his code; you should write whatever you feel comfortable with and work out the kinks later.

You want to work out whether

• f(a) = a, or f(f(a)) = a, or …, or f^n(a) = a, or,
• f(a+1) = a+1, or f(f(a+1)) = a+1, or …, or f^n(a+1) = a+1, or,
• …
• f(b) = b, or f(f(b)) = b, or …, or f^n(b) = b.

An obvious algorithm should come to mind immediately: try all these values one-by-one! You will need two (nested) loops, because you are iterating over a rectangle of values. Can you now see what to do?