03-tailrecursion.student

03-tailrecursion.student - Last Time: * The stack: calling...

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Unformatted text preview: Last Time: * The stack: calling multiple funcitons * Recursion: base case, inductive step * Recursive functions: factorial, numOnes Today: * The recursion "space problem": * what it is, and how to fix it * Tail-recursion: a "small" recursive twist * Testing programs ++++++++++++++++++++++++++++++++++++++++++++++++++++++++ So far, we've introduced the notion of recursive problems and, correspondingly, recursive functions. Something is recursive if it refers to itself. A problem is recursive if it has: * One or more trivial base cases * Other cases defined in terms of smaller problem instances. (these are "inductive steps") A function is recursive if it calls itself. We then gave a few examples of how to map from recursive definitions to recursive functions: identify the base case, and write that down. Then, write the inductive step as a recursive call. Last time, I left you with the following question: Let's consider the resource costs of the recursive version of factorial: int factorial(int x) { if (x == 0) { return 1; } else { return x * factorial(x-1); } } Compare this to a version with a loop: int fact_iter(int x) { int result = 1; while (x) { result *= x; x--; } return result; } Questions for the class, small groups ~5 minutes: * How many multiplications does each version perform? * How much space does each one require? The recursive version is much less efficient in terms of space! Happily, there is a way to re-write the recursive version to use (approximately) the same amount of space as is required by the iterative version. Consider the following implementation: int fact_helper(int n, int result) // REQUIRES: n >= 0 // EFFECTS: returns result * n! { if (n == 0) { return result; } else { return fact_helper(n-1, result * n); } } int factorial(int num) // REQUIRES: n >= 0 // EFFECTS: returns num! { return fact_helper(num, 1); } This function is equivalent to the original factorial---we'll prove this by induction. There are two steps to any indutive proof. First, prove the base case, and second, the inductive step. First, let's state the hypothesis: fact_helper (n, r) = n! * r Naturally, if this is true, then fact_helper(n, 1) would be n!, and factorial() would be correct. So, it's a pretty useful hypothesis. ;-) First, the base case: if n=0, the fact_helper(0,r) = r = 1 * r = (0!) * r so, the hypothesis holds for the base case. next, let's prove that if the hypothesis holds for step n-1, it also holds for n: we know that if n > 0: fact_helper(n, result) = fact_helper(n-1,result*n) by our hypothesis, this is = (n-1)! * (result*n) = ((n-1)! * n) * result : = n! * result so, given that our hypothesis holds for n-1, it also holds for n....
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03-tailrecursion.student - Last Time: * The stack: calling...

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