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MIT 6.00 Intro Computer Science (OCW)
OpenStudy (anonymous):

Iterative and recursive seem to be the same. Can anyone explain to me, in layman's terms, the difference? Thx.

8 years ago
OpenStudy (anonymous):

A recursive function is a function that calls itself with smaller versions of the original problem until it's in the base case. An iterative program does not call itself.

8 years ago
OpenStudy (anonymous):

I tried the plain English way, but this might work better, at the risk of being more confusing. Honestly, the way I think about it is that whenever I see a function calling itself, I imagine a trapdoor. So in the line: else: return fib(x-1) + fib(x-2) I read as far as else: return fib(x-1) and then I think, OK, the current function stops, and (x-1) falls through to fib(x). Until that's completed, the rest of this function is on hold. That could happen hundreds of times, but as long as x isn't in the base case, it's going to keep falling through that trapdoor into smaller and smaller problems. It might help to use the fibonacci example in the lecture, but expand it to show the flow of control. def fib(x): """Return fibonacci of x, where x is a non-negative int""" if x == 0 or x == 1: return 1 else: return fib(x-1) + fib(x-2) So say we call it with 3 as an argument: 1. if x==0 or x==1: return 1 ##Nope 2. else: return fib(x-1) +fib(x-2) a) call fib with 2 as an argument. This is fib(x-1) in line 2. b) If x ==0 or x ==1: return 1 ##nope c) else: return fib(x-1) + fib(x-2) i) call fib with 1 as an argument.This is fib(x-1) in line c. ii) If x ==0 or x ==1: return 1 ##Yes! we can do this! the fib(x-1) in line c = 1 iii) call fib with 0 as an argument. This is fib(x-2) in line c. iv) If x ==0 or x ==1: return 1 ##Yes! we can do this! the fib(x-2) in line c = 1 c) is now completed. Returns 2 as fib(x-1) in line 2 d) call fib with 1 as an argument. This is fib(x-2) in line 2. i) If x ==0 or x ==1: return 1 ##Yes! we can do this! the fib(x-2) in line 2 == 1 2. is now completed. fib(x-1) = 2, fib (x-2) = 1, so line 2 returns 3.

8 years ago
OpenStudy (anonymous):

Thanks a million! I think I got it - for example - and feel free to correct this: def recur(x): if x == 0: print "This is finally the end of the recursive loop with value x:", x else: print x, "Where x = x - 1:", (x-1) return recur(x-1) I think this is a good example for beginners to understand what happens...and your layman's term explained it nicely. For anything else more complicated than this...I may be at a loss! ;-)

8 years ago
OpenStudy (anonymous):

Exactly. This has also been pretty hard for me to get my head around. I've found that with recursive functions it's important for me to keep in mind what they're actually returning and where they're returning it. In the case of fibonacci, that function is getting an integer whenever it says something like fib(x-1). So if I wanted to understand it without getting that snake-eating-its-own-tail confused feeling, I might think of it as: if x == 0 or x == 1: return 1 else: return [INTEGER] + [INTEGER] and worry about the recursive calls later. it helps me separate the number of recursions from the number of answers being returned (at each level, The number of recursions is based on the size of the initial value--fib(21) recurs way more times than fib(2). But no matter how big the initial value is, you're only ever going to get an integer back from any single function call.

8 years ago
OpenStudy (anonymous):

Sorry, should be: It helps me separate the number of recursions from the number of answers being returned.

8 years ago
OpenStudy (anonymous):

Thanks somnamniac - you've been quite the help. I'll be going over the recursion lectures anyways...they'll make more sense now.

8 years ago
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