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+ 2

# I tried asking AI but AI just couldn't get the correct answer to the code below

Why does the code below output 3? def rec(n): n %= 5 if n <= 1: return n else: return rec(n - 1) + rec(n - 2) print(rec(9))

3 Answers

+ 4

The code outputs 3 because the recursion function rec(n) is calculating the Fibonacci sequence modulo 5.
When rec(9) is called, the function will first calculate 9 % 5 = 4. Since n = 4 is not less than or equal to 1, the function will make two recursive calls rec(4 - 1) & rec(4 - 2). For rec(3) [ from the first recursive call ], it calculates 3 % 5 = 3 which is greater than 1, so it again makes two more recursive calls. For rec(2) & rec(1), which then return 2 & 1 respectively. These values are added together in the recursion tree to get the final result: 2 + 1 = 3, which is output as the result of rec(9). Actually the function calculates the Fibonacci sequence modulo 5, leading to the output of 3 for rec(9).

+ 2

In such as situations you can analyze what happend in a code in a debug editor.
Another way if you do not have that is to use the print statement how I present in thi example:
loop=0
def rec(n):
global loop
loop += 1
print(f'loop={loop}')
print(f'input n={n}')
n %= 5
print(f'after % n={n}')
if n <= 1:
print(f'n<=1: n={n}\n')
return n
else:
print(f'else: n={n}\n')
return rec(n - 1) + rec(n - 2)
print(rec(9))
https://sololearn.com/compiler-playground/cBz1F82S0P1o/?ref=app

0

def rec(n):
if n <= 1:
return n
else:
return rec(n - 1) + rec(n - 2)
print(rec(9))