Reputation: 474
I have written two recursive functions, which work almost identically. I was trying to get the recursion right and then I accidentally stumbled upon the answer, but the syntax confuses me:
def fac(N):
"""
Factorial of N.
"""
################# This makes no goddamn sense to me #################
if N == 1:
return N
####### because N == 1, yet 'return N' is the correct result ########
elif N == 0:
return 1
return N*fac(N-1)
How can N == 1 be true as the exit condition but also store the result of fac(N)? Same thing with a function prod()
, which is the analog to sum()
.
def prod(List):
"""
Product of all numbers in a list.
"""
if len(List) == 1:
return List[-1]
return List[-1]*prod(List[:-1])
I don't get how the end result is stored in List[-1]
. Does the python interpreter understand return arg*func(arg)
in a special way?
Upvotes: 1
Views: 54
Reputation: 2569
Nothing special, but in case like this, use print and explore.
def fac(N):
""" Factorial of N. """
if N == 1:
return N
elif N == 0:
return 1
return N*fac(N-1)
Let see how it's work for fac(3)
# fac(3)
# fac(3) => 3 * fac(3-1)
# fac(3) => 3 * fac(3-1) => 2 * fac(2-1)
# fac(3) => 3 * fac(3-1) => 2 * fac(2-1) => return 1
# fac(3) => 3 * fac(3-1) <= 2 * 1
# fac(3) <= 3 * 2 * 1
# 6
Upvotes: 1
Reputation: 1167
Consider the loops required to calculate fac(4)
:
1: fac(4) -> 4 * fac(3) # It then has to calculate fac(3)
2: fac(3) -> 3 * fac(2) # It then has to calculate fac(2)
3: fac(2) -> 2 * fac(1) # It then has to calculate fac(1)
4: fac(1) -> 1 # Finally we've returned a value - now back up through the loops
3: fac(2) -> 2 * fac(1) == 2 * 1 == 2
2: fac(3) -> 3 * fac(2) == 3 * 2 == 6
1: fac(4) -> 4 * fac(3) == 4 * 6 == 24
The second part is effectively the same thing - recurse down until you get a value, then keep plugging that in all the way back up to your original request.
Upvotes: 1