def power, problem when raising power to decimals
sjmachin at lexicon.net
Thu Apr 17 00:21:18 CEST 2008
skanemupp at yahoo.se wrote:
> how do i solve power(5,1.3)?
Is this a trick question? OK, I'll bite:
>>> 5 ** 1.3
> def power(nbr, po):
> if po==0:
> return 1
> if po>0:
> return nbr*power(nbr, po-1)
> if po<0:
> return 1/power(nbr, -1*po)
> also i found a link which states 0^0 isnt 1 even though every
> calculator ive tried says it is.
> it doesnt say what it is but i presume 0 then.
> but it seems the dude is wrong and it is 1?
Of the possible results of 0 ** 0 (i.e. 1, 0, and NaN), 1 seems to be
the least implausible. It allows X ** 0 to be 1 for all X.
> dont run the code with decimals, it will never leave the function, u
> have to restart the shell(if using the standard python ide)
I presume that by "decimals", you mean "numbers that are not integers".
So you've got it into an infinite loop. Have you tried tracing through
the first 5 or 6 gyrations? This can be done by
(a) putting a debugger breakpoint just after the start of the function
(b) using something like:
print "power(%.6f, %.6f)" % (nbr, po))
junk = raw_input("Press <Enter> to continue -> ")
(c) using a pencil and a piece of scrap paper, write down what is
happening as a typical function call is executed e.g.
power(5, 1.3) => 5 * power(5, 0.3)
power(5, 0.3) => 5 * power(5, -0.7)
power(5, -0.7) => 1 / power (5, 0.7)
Then work out what extra condition you would have to test to stop it
doing that. Then work out how to calculate the return value when that
condition is true.
More information about the Python-list