python - Is this an issue with my method of recursion? -

in response this question tried come own solution.

my solution in python 3.x

def efficientalgo(number, x):     print("start number... " + str(number))      # see if number dividable three, if divide 3     if number % 3 == 0:         print("dividing three...")         number /= 3         print(number)         # use recursion see if number can divided again         number += efficientalgo(number, x)         print("returning number division 3 now...")         return number      # divide number 2 if evenly dividable 2    if number % 2 == 0:         print("dividing two...")         number /= 2         print(number)         # use recursion see if number can divided again         number += efficientalgo(number, x)         print("returning number division 2 now...")         print(number)         return number       # if number not one, subtract 1 , call     if number != 1:         print("subtracting 1 now...")         number -= 1         print(number)         # use recursion see if number can divided again         number += efficientalgo(number, x)         print(number)         return number      # if number one, return , finish.     if number == 1:         print("returning 1 now... " + str(number))         return number  print(efficientalgo(100, 1)) 

here "working" pythonfiddle

the output

start number... 100 dividing two... 50.0 start number... 50.0 dividing two... 25.0 start number... 25.0 subtracting 1 now... 24.0 start number... 24.0 dividing three... 8.0 start number... 8.0 dividing two... 4.0 start number... 4.0 dividing two... 2.0 start number... 2.0 dividing two... 1.0 start number... 1.0 returning 1 now... 1.0 -----above line correct output want-----  ---------below line, have no idea going on-------- returning number division 2 now... 2.0 returning number division 2 now... 4.0 returning number division 2 now... 8.0 returning number division 3 now... 40.0 returning number division 2 now... 65.0 returning number division 2 now... 115.0 115.0 

as can see, have implemented recursion incorrectly , getting sort of feedback loop in program gets down correct answer, keeps going after ive returned final number, in case one.

i not understand happening below lines have marked out in output

number should set = efficientalgo(..) not +=:

def efficientalgo(number):     print("start number... " + str(number))     # see if number dividable three, if divide 3     if number % 3 == 0:         print("dividing three...")         number /= 3         print(number)         # use recursion see if number can divided again         number =  efficientalgo(number)     # divide number 2 if evenly dividable 2     if number % 2 == 0:         print("dividing two...")         number /= 2         print(number)         # use recursion see if number can divided again         number = efficientalgo(number)      # if number not one, subtract 1 , call     if number != 1:         print("subtracting 1 now...")         number -= 1         print(number)         # use recursion see if number can divided again         number = efficientalgo(number)    return number 

once expected output:

in [4]: efficientalgo(100) start number... 100 dividing two... 50.0 start number... 50.0 dividing two... 25.0 start number... 25.0 subtracting 1 now... 24.0 start number... 24.0 dividing three... 8.0 start number... 8.0 dividing two... 4.0 start number... 4.0 dividing two... 2.0 start number... 2.0 dividing two... 1.0 start number... 1.0 out[4]: 1.0 

or return:

def efficientalgo(number):     print("start number... " + str(number))     # see if number dividable three, if divide 3     if number % 3 == 0:         print("dividing three...")         number /= 3         print(number)         # use recursion see if number can divided again         return efficientalgo(number)      # divide number 2 if evenly dividable 2     if number % 2 == 0:         print("dividing two...")         number /= 2         print(number)         # use recursion see if number can divided again         return efficientalgo(number)      # if number not one, subtract 1 , call     if number != 1:         print("subtracting 1 now...")         number -= 1         print(number)         # use recursion see if number can divided again         return efficientalgo(number)      return number