这里的递归如何工作？

Question

Code 1:

public static int fibonacci (int n){ 
    if (n == 0 || n == 1) { 
        return 1; 
    } else { 
        return fibonacci (n-1) + fibonacci (n-2); 
    }        
} 

How can you use fibonacci if you haven't gotten done explaining what it is yet? I've been able to understand using recursion in other cases like this:

Code 2:

class two 
{
    public static void two (int n) 
    {
        if (n>0) 
        {
            System.out.println (n) ;
            two (n-1) ;
        }
        else
        {
            return ;
        }
    } 

    public static void main (String[] arg) 
    {
        two (12) ;
    }
}

In the case of code 2, though, n will eventually reach a point at which it doesn't satisfy n>0 and the method will stop calling itself recursively. In the case of code 2, though, I don't see how it would be able to get itself from 1 if n=1 was the starting point to 2 and 3 and 5 and so on. Also, I don't see how the line return fibonacci (n-1) + fibonacci (n-2) would work since fibonacci (n-2) has to contain in some sense fibonacci (n-1) in order to work, but it isn't there yet.

The book I'm looking at says it will work. How does it work?

Answer 1

Well, putting aside what a compiler actually does to your code (it's horrible, yet beautiful) and what how a CPU actually interprets your code (likewise), there's a fairly simple solution.

Consider these text instructions:

To sort numbered blocks:

1. pick a random block.
2. if it is the only block, stop.
3. move the blocks with lower numbers to the left side, higher numbers to the right.
4. sort the lower-numbered blocks.
5. sort the higher-numbered blocks.

When you get to instructions 4 and 5, you are being asked to start the whole process over again. However, this isn't a problem, because you still know how to start the process, and when it all works out in the end, you've got a bunch of sorted blocks. You could cover the instructions with slips of paper and they wouldn't be any harder to follow.

Answer 2

In the case of code 2 though n will eventualy reach a point at which it doesnt satisfy n>0 and the method will stop calling itself recursivly

to make it look similar you can replace condition if (n == 0 || n == 1) with if (n < 2)

Also i don't see how the line `return fibonacci (n-1) + fibonacci (n-2) would work since fibbonacci n-2 has to contain in some sense fibonacci n-1 in order to wrok but it isn't there yet.

I suspect you wanted to write: " since fibbonacci n-1 has to contain in some sense fibonacci n-2 "
If I'm right, then you will see from the example below, that actually fibonacci (n-2) will be called twice for every recursion level ( fibonacci(1) in the example):
1. when executing fibonacci (n-2) on the current step
2. when executing fibonacci ((n-1)-1) on the next step

(Also take a closer look at the Spike's comment )

Suppose you call fibonacci(3) , then call stack for fibonacci will be like this:
(Veer provided more detailed explanation )

n=3. fibonacci(3)  
n=3. fibonacci(2) // call to fibonacci(n-1)
   n=2. fibonacci(1) // call to fibonacci(n-1)
      n=1. returns 1
   n=2. fibonacci(0) // call to fibonacci(n-2)
      n=0. returns 1
   n=2. add up, returns 2
n=3. fibonacci(1) //call to fibonacci(n-2)
   n=1. returns 1
n=3. add up, returns 2 + 1

Note, that adding up in fibonacci(n) takes place only after all functions for smaller args return (ie fibonacci(n-1) , fibonacci(n-2) ... fibonacci(2) , fibonacci(1) , fibonacci(0) )

To see what is going on with call stack for bigger numbers you could run this code.

public static String doIndent( int tabCount ){
    String one_tab = new String("   ");
    String result = new String("");
    for( int i=0; i < tabCount; ++i )
       result += one_tab;
    return result;
}

public static int fibonacci( int n, int recursion_level )
{
    String prefix = doIndent(recursion_level) + "n=" + n + ". ";

    if (n == 0 || n == 1){
        System.out.println( prefix + "bottommost level, returning 1" );
        return 1;
    }
    else{
        System.out.println( prefix + "left fibonacci(" + (n-1) + ")" );
        int n_1 = fibonacci( n-1, recursion_level + 1 );

        System.out.println( prefix + "right fibonacci(" + (n-2) + ")" );
        int n_2 = fibonacci( n-2, recursion_level + 1 );

        System.out.println( prefix + "returning " + (n_1 + n_2) );
        return n_1 + n_2;
    }
}

public static void main( String[] args )
{
    fibonacci(5, 0);
}

Answer 3

The trick is that the first call to fibonacci() doesn't return until its calls to fibonacci() have returned.

You end up with call after call to fibonacci() on the stack, none of which return, until you get to the base case of n == 0 || n == 1. At this point the (potentially huge) stack of fibonacci() calls starts to unwind back towards the first call.

Once you get your mind around it, it's kind of beautiful, until your stack overflows.

Answer 4

"How can you use Fibonacci if you haven't gotten done explaining what it is yet?"

This is an interesting way to question recursion. Here's part of an answer: While you're defining Fibonacci, it hasn't been defined yet , but it has been declared . The compiler knows that there is a thing called Fibonacci, and that it will be a function of type int -> int and that it will be defined whenever the program runs.

In fact, this is how all identifiers in C programs work, not just recursive ones. The compiler determines what things have been declared, and then goes through the program pointing uses of those things to where the things actually are (gross oversimplification).

Answer 5

Let me walkthrough the execution considering n=3. Hope it helps.

When n=3 => if condition fails and else executes

return fibonacci (2) + fibonacci (1);  

Split the statement:

1. Find the value of fibonacci(2)
2. Find the value of fibonacci(1)
   // Note that it is not fib(n-2) and it is not going to require fib(n-1) for its execution. It is independent. This applies to step 1 also.
3. Add both values
4. return the summed up value

The way it gets executed(Expanding the above four steps):

1. Find the value of fibonacci(2)

   1. if fails, else executes
   2. fibonacci(1)
      1. if executes
      2. value '1' is returned to step 1.2. and the control goes to step 1.3.
   3. fibonacci(0)
      1. if executes
      2. value '1' is returned to step 1.3. and the control goes to step 1.4.
   4. Add both
      1. sum=1+1=2 //from steps 1.2.2. and 1.3.2.
   5. return sum // value '2' is returned to step 1. and the control goes to step 2
      

2. Find the value of fibonacci(1)

   1. if executes
   2. value '1' is returned

3. Add both values

   1. sum=2+1 //from steps 1.5. and 2.2.
4. return the summed up value //sum=3

Answer 6

Try to draw an illustration yourself, you will eventually see how it works. Just be clear that when a function call is made, it will fetch its return value first. Simple.

Answer 7

尝试调试并使用watch来了解变量的状态

Answer 8

Understanding recursion requires also knowing how the call stack works ie how functions call each other.
If the function didn't have the condition to stop if n==0 or n==1, then the function would call itself recursively forever. It works because eventually, the function is going to petter out and return 1. at that point, the return fibonacci (n-1) + fibonacci (n-2) will also return with a value, and the call stack gets cleaned up really quickly.

Answer 9

I'll explain what your PC is doing when executing that piece of code with an example:

Imagine you're standing in a very big room. In the room next to this room you have massive amounts of paper, pens and tables. Now we're going to calculate fibonacci(3):

We take a table and put it somewhere in the room. On the table we place a paper and we write "n=3" on it. We then ask ourselves "hmm, is 3 equal to 0 or 1?". The answer is no, so we will do "return fibonacci (n-1) + fibonacci (n-2);".

There's a problem however, we have no idea what "fibonacci (n-1)" and "fibonacci (n-2)" actually do. Hence, we take two more tables and place them to the left and right of our original table with a paper on both of them, saying "n=2" and "n=1".

We start with the left table, and wonder "is 2 equal to 0 or 1?". Of course, the answer is no, so we will once again place two tables next to this table, with "n=1" and "n=0" on them.

Still following? This is what the room looks like:

n=1

n=2 n=3 n=1

n=0

We start with the table with "n=1", and hey, 1 is equal to 1, so we can actually return something useful! We write "1" on another paper and go back to the table with "n=2" on it. We place the paper on the table and go to the other table, because we still don't know what we're going to do with that other table.

"n=0" of course returns 1 as well, so we write that on a paper, go back to the n=2 table and put the paper there. At this point, there are two papers on this table with the return values of the tables with "n=1" and "n=0" on them, so we can compute that the result of this method call is actually 2, so we write it on a paper and put it on the table with "n=3" on it.

We then go to the table with "n=1" on it all the way to the right, and we can immediately write 1 on a paper and put it back on the table with "n=3" on it. After that, we finally have enough information to say that fibonacci(3) returns 3.

---

It's important to know that the code you are writing is nothing more than a recipe. All the compiler does is transform that recipe in another recipe your PC can understand. If the code is completely bogus, like this:

    public static int NotUseful()
    {
        return NotUseful();
    }

will simply loop endlessly, or as in my example, you'll keep on placing more and more tables without actually getting anywhere useful. Your compiler doesn't care what fibonacci(n-1) or fibonacci(n-2) actually do.