Implement recursion using one recursive call

Question

Given a function as follow : f(n) = f(n-1) + f(n-3) + f(n-4)

f(0) = 1
f(1) = 2
f(2) = 3
f(3) = 4

I know to implement it using recursion with three recursive calls inside one function. But I want to do it with only one recursion call inside the function. How it can be done ?

To implement using 3 recursive calls here is my code :

def recur(n):
  if n == 0:
     return 1
  elif n == 1:
     return 2
  elif n == 2:
     return 3
  elif n == 3:
     return 4
  else:
     return recur(n-1) + recur(n-3) + recur(n-4) #this breaks the rule because there are 3 calls to recur

score 2 · Answer 1 · edited May 23 '17 at 11:53

2

This answer in C# may give you a hint how to do what you want.

Fibbonacci with one recursive call in Python is as follows:

def main():
  while True:
    n = input("Enter number : ")
    recur(0,1,1,int(n))

def recur(firstNum,secondNum,counter,n):
  if counter==n :
     print (firstNum)
     return
  elif counter < n
      recur (secondNum,secondNum+firstNum,counter+1,n)

edited May 23 '17 at 11:53

Community

1
1

answered Jun 01 '15 at 04:41

George Hanna

382
2
15

I did something like that but getting 14 for n=5 . Here is my code :http://ideone.com/8zk4kI – ms8 Jun 01 '15 at 05:05
I dont want fibonacci number. Please see problem :/ – ms8 Jun 01 '15 at 05:17
Ohh I am so sorry, I misunderstood. – George Hanna Jun 01 '15 at 05:18

sgp · Accepted Answer · 2015-06-01T09:44:56.387

2

Your attempt is in the right direction but it needs a slight change:

def main():
  while True:
    n = input("Enter number : ")
    recur(1,2,3,4,1,int(n))

def recur(firstNum,secondNum,thirdNum,fourthNum,counter,n):  
  if counter==n:
     print (firstNum)
     return
  elif counter < n:
      recur (secondNum,thirdNum,fourthNum,firstNum+secondNum+fourthNum,counter+1,n)

edited Jun 01 '15 at 09:44

answered Jun 01 '15 at 05:45

sgp

1,738
6
17
31

It only gives 1 as output..:/. Reason being 1 is always then then any n so it will always display 1 – ms8 Jun 01 '15 at 09:32
@python_slayer Oops. Check the edit. Is it fine now? – sgp Jun 01 '15 at 09:44

score 0 · Answer 3 · answered Jun 01 '15 at 06:43

At first glance, this looks like a dynamic programming problem. I really like memoization for problems like this because it keeps the code nice and readable, but also gives very good performance. Using python3.2+ you could do something like this (you can do the same thing with older python versions but you'll need to either implement your own lru_cache or install one of the many 3rd party that have similar tools):

import functools

@functools.lru_cache(128)
def recur(n):
  print("computing recur for {}".format(n))
  if n == 0:
     return 1
  elif n == 1:
     return 2
  elif n == 2:
     return 3
  elif n == 3:
     return 4
  else:
     return recur(n-1) + recur(n-3) + recur(n-4)

Notice that the function only gets called once per n:

recur(6)
# computing recur for 6
# computing recur for 5
# computing recur for 4
# computing recur for 3
# computing recur for 1
# computing recur for 0
# computing recur for 2

Implement recursion using one recursive call

3 Answers3

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