Python Homework - Not making sense

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, (eg 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?

Answer 2

Yeah, I can see why that might be confusing.

Was f(f(f(...(f(x)))) = x wrapped in triple-double-quotes? That's a function documentation string, sort of like commenting your code. It shouldn't have been stand-alone without something protecting it.

Imagine f was called increment_by_one.

Calling increment_by_one 10 times like that on an x of 2 would give 12. No matter how many times you increment, you never seem to get back 2.

Now imagine f was called multiply_by_one.

Calling multiply_by_one 5 times like that on an x of 3 would give 3. Sweet.

So, some example outputs you can test against (you have to write the functions)

knice(increment_by_one, 1, 3, 5) would return 0.

knice(multiply_by_one, 1, 3, 5) would return 1.

As another hint, indentation is important in python.

Answer 3

Here's a concrete example. Start small, and suppose you called knice(f, a=1, b=2, k=1) . For k==1 , we don't have to worry about iterating the function. The only values of x to consider are 1 and 2, so knice can return 1 (ie, True) if f(1)==1 or f(2)==2 .

Now suppose you called knice(f, a=1, b=2, k=2) . You'll have to check f(f(1)) and f(f(2)) as well.

As k gets bigger, you'll have to call f more. And as the range between a and b gets bigger, you'll have to try more values of x as an argument to f .