+ 1

who can explain this

the answer is seven '1' ... def f(x): if x==0 or x==1: return 1 else: print(f(x-1)) return f(x-1) #print(f(4)) f(4)

17th Jan 2020, 9:29 PM
yq w
2 Respostas
+ 2
Here are a few points to explain a lot of it. Consider each point until you're confident why it is correct before moving on. The #print(f(4)) does nothing because it is a comment but you probably knew that. The function's return value is always 1 because that's what the base case returns and the general case doesn't make any change to the returned value. "general case" = the else section or "x != 0 and x != 1". "base case" = the "x==0 or x==1" case. In every case except the base case, the function makes 2 recursive calls. This means if f gets called with a value of 4, 2^(something related to 4) more calls will be made to f. Here is a little ASCII art to illustrate the recursive branching structure this leads to where the number corresponds with each call's value of x: _4_ / \ 3 3 / \ / \ 2 2 2 2 / \ / \ / \ / \ 1 1 1 1 1 1 1 1 Count the number of values greater than 1 and you get 7 just like you found. That makes sense when each call with a value greater than 1 directly prints 1 one time. How would you describe what the function does? Given a value x, the f prints "1" ( 2 ^ (x - 1) ) - 1 times where ^ indicates exponent.
18th Jan 2020, 1:36 AM
Josh Greig
Josh Greig - avatar
0
Without the print statements you would have 4 function calls. Each function which calls the function goes into a stack. f(4) -> f(3) -> f(2) -> f(1) -> return 1 The stack: f(2) f(3) f(4) f(2) gets 1 from f(1). f(3) gets 1 from f(2). f(4) gets 1 from f(3). Stack is empty. f(4) returns 1. With the print statements you have 15 function calls. Each f(x) with x > 1 prints 1. If you add print(x) at the beginning the function you will see all function calls. If you change the print statement to print(x, f(x - 1)) you will see which function prints 1: f(2) prints 1 f(3) prints 1 f(2) prints 1 f(4) prints 1 f(2) prints 1 f(3) prints 1 f(2) prints 1 The order has to do with the stack. But with the recursion tree from Josh Greig you should be able to understand it.
18th Jan 2020, 3:07 AM
Denise Roßberg
Denise Roßberg - avatar