Could someone help me with this. How do i make a recursive program that returns sum of 1**3 + 2**3 + ... + n**3?
doeman 0 Newbie Poster
Recommended Answers
Jump to PostHere is another way you can skin it:
def sum_cube(n, sum=0): if n > 0: x = n**3 return sum_cube(n-1, sum+x) else: return sum print sum_cube(5) # 225
Jump to PostHere's another recursive way although it is kind of cheating:
>>> sum_cube = lambda N: eval(["0","(N*N*N) + sum_cube(N-1)"][N>0]) >>> sum_cube(10) 3025
All 6 Replies
Gribouillis 1,391 Programming Explorer Team Colleague
A recursive way
def sum_cubes(n):
if n <= 0:
if n < 0:
raise ValueError
else:
return 0
else:
return n ** 3 + sum_cubes(n-1)
if __name__ == "__main__":
print sum_cubes(10)and a non recursive way
def sum_cubes2(n):
if n < 0:
raise ValueError
res = 0
while n:
res += n ** 3
n -= 1
return res
ZZucker 342 Practically a Master Poster
Here is another way you can skin it:
def sum_cube(n, sum=0):
if n > 0:
x = n**3
return sum_cube(n-1, sum+x)
else:
return sum
print sum_cube(5) # 225 doeman 0 Newbie Poster
Here is another way you can skin it:
def sum_cube(n, sum=0): if n > 0: x = n**3 return sum_cube(n-1, sum+x) else: return sum print sum_cube(5) # 225
This is a recursive way right?
ZZucker 342 Practically a Master Poster
Recursive means that the function calls itself, and it does so in line 4.
In Gribouillis's code it does so in line 8.
BearofNH 104 Posting Whiz
Here's another recursive way although it is kind of cheating:
>>> sum_cube = lambda N: eval(["0","(N*N*N) + sum_cube(N-1)"][N>0])
>>> sum_cube(10)
3025 ZZucker 342 Practically a Master Poster
Wow BearofNH, the Auberge Pyrenees-Cevennes of Python.
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