hi, i do not need a fancy elegant code, know this one is ugly and take to much time(i will for now use ctrl+c to break) my problem is that the result are wrong, just test one of the reults printed by this code, and you will see tha the results are wrong. Maybe something i do not understand is happenig wen python round the the cubed root. is this code wrong?
On 5/29/08, Kent Johnson <[EMAIL PROTECTED]> wrote: > On Wed, May 28, 2008 at 8:43 PM, Robert William Hanks > <[EMAIL PROTECTED]> 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 > > Let's see. The inner loop will run 10000000 * 10000000 = > 100000000000000 times. That's a pretty big number. Suppose each > iteration of the loop takes 1 microsecond. (That seems optimistic but > not too much - on my computer iscube(27) takes about 1.3 > microseconds.) Then the entire program will complete in 10000000 * > 10000000 / 1000000 / 60 / 60 / 24 / 365 = 3 years. Other than that it > look OK to me... > > Looks like you need either a better algorithm or a faster computer! > > Kent >
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