12th Mar 2020, 6:02 AM
Amit Biswas
Amit Biswas - avatar
2 Antworten
+ 5
def recur(n): if n == 0: return 0 else: return n + recur(n-1) print(recur(3)) so let's start n is 3 so it will return return 3 + recur(2) we can find the value of recur(2) n = 2 return 2 + recur (1) the above result become return 3 + 2 + recur(1) again for recur(1) return 1 + recur(0) since recur(0) will be 0 because n==0 the whole thing can be written as return 3+2+1+0 which is 6 If you want to stop at 5 you can change n==0 to n==1
12th Mar 2020, 6:07 AM
Utkarsh Sharma
Utkarsh Sharma - avatar
0
thanks a lot
12th Mar 2020, 6:12 AM
Amit Biswas
Amit Biswas - avatar