Robert William Hanks wrote:
Need ti find out whem a number o this form i**3+j**3+1 is acube.
tryed a simple brute force code but, why this not work?
def iscube(n):
cubed_root = n**(1/3.0)
if round(cubed_root)**3 == n:
return True
else:
return False
for i in range(1,10000000):
for j in range(1,10000000):
soma= i**3 +j**3 +1
if isCube(soma):
print i
print j
print soma
Assuming you fixed the problem, do you know of any cases for which i**3
+j**3 +1 is a cube?
Also note this program will run for a LONG time.
You can shorten the run time:
for i in range(1,10000000):
for j in range(j,10000000): # test each combination once
soma = i**3 +j**3 +1
cubed_root = soma**(0.333) # function call and division take
extra time and there is no need for either
if abs(cubed_root - round(cubed_root)) < .01: # this is a
guess at close enough, and cheaper than cubing?
print i
print j
print soma
--
Bob Gailer
919-636-4239 Chapel Hill, NC
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