Best search algorithm to find condition within a range

[jonas.thornvall at gmail.com]

Den tisdag 7 april 2015 kl. 21:27:20 UTC+2 skrev Ben Bacarisse:
> Ian Kelly <ian.g.kelly at gmail.com> writes:
> 
> > On Tue, Apr 7, 2015 at 12:55 PM, Terry Reedy <tjreedy at udel.edu> wrote:
> >> On 4/7/2015 1:44 PM, Ian Kelly wrote:
> >>
> >>>>>> def to_base(number, base):
> >>>
> >>> ...     digits = []
> >>> ...     while number > 0:
> >>> ...         digits.append(number % base)
> >>> ...         number //= base
> >>> ...     return digits or [0]
> >>> ...
> >>>>>>
> >>>>>>
> >>>>>> to_base(2932903594368438384328325832983294832483258958495845849584958458435439543858588435856958650865490,
> >>>>>> 429496729)
> >>>
> >>> [27626525, 286159541, 134919277, 305018215, 329341598, 48181777,
> >>> 79384857, 112868646, 221068759, 70871527, 416507001, 31]
> >>> About 15 microseconds.
> >>
> >>
> >> % and probably // call divmod internally and toss one of the results.
> >> Slightly faster (5.7 versus 6.1 microseconds on my machine) is
> >
> > Not on my box.
> >
> > $ python3 -m timeit -s "n = 1000000; x = 42" "n % x; n // x"
> > 10000000 loops, best of 3: 0.105 usec per loop
> > $ python3 -m timeit -s "n = 1000000; x = 42" "divmod(n,x)"
> > 10000000 loops, best of 3: 0.124 usec per loop
> 
> I get similar results, but the times switch over when n is large enough
> to become a bignum.
> 
> -- 
> Ben.

I am not sure you guys realised, that althoug the size of the factors to muliply expands according to base^(exp+1) for each digitplace the number of comparissons needed to reach the digit place (multiple of base^exp+1) is constant with my approach/method.