python2.7中递归函数的执行顺序

Question

def fac(n):
    if n==1 or n==2  or n==3:
        print "i am calling fac(",n,")"
        return  n
    else:
        print "i am calling fac(",n,")"
        x=fac(n-1)+fac(n-2)+fac(n-3)
        return  x  

The output of fac(6) is :

fac(6)
i am calling fac( 6 )
i am calling fac( 5 )
i am calling fac( 4 )
i am calling fac( 3 )
i am calling fac( 2 )
i am calling fac( 1 )
i am calling fac( 3 )
i am calling fac( 2 )
i am calling fac( 4 )
i am calling fac( 3 )
i am calling fac( 2 )
i am calling fac( 1 )
i am calling fac( 3 )
20

What is the rule for python2.7 to execute the recursion function?
The result confused me，it can not be analysed from the calculation tree. Why the result is not other forms?

What is the rule for python to deal with recursion calculation?

Answer 1

Python runs each call in the order it encounters the instruction to call it. So, starting at the top of fac with n=6 , it will get to this line:

x=fac(n-1)+fac(n-2)+fac(n-3)

The first thing it will do is to calculate n-1=5 , and run fac(5) - which starts again at the top of the function. It will reach the same place and call fac(4) , which will call fac(3) - which just returns 3. Only now will it calculate n-2=2 and run fac(2) , and then fac(1) and do the addition. Now fac(4) has finished, we're back in fac(5) and we keep going from fac(n-2) .

If you modify your function to keep track of how deeply recursed you are, you can print the calls as a tree structure so you can see what's going on more easily:

def fac(n, level=0):
    print '{}fac({})'.format(level*'\t', n)

    if n==1 or n==2  or n==3:
        return n
    else:
        x = fac(n-1, level+1) + fac(n-2, level+1) + fac(n-3, level+1)
    return x

gives:

fac(6)
   fac(5)
      fac(4)
          fac(3)
          fac(2)
          fac(1)
      fac(3)
      fac(2)
   fac(4)
      fac(3)
      fac(2)
      fac(1)
   fac(3)