On Tuesday, March 24, 2015 at 6:54:20 AM UTC+8, Terry Reedy wrote:
> On 3/23/2015 2:44 PM, Dave Angel wrote:
>
> > ## Example 2: Using recursion with caching
> > cache = [0, 1]
> > def fib4(n):
> > if len(cache) <= n:
> > value = fib4(n-2) + fib4(n-1)
> > cache.append(value)
> > return cache[n]
> >
> > This one takes less than a millisecond up to n=200 or so.
>
> Iteration with caching, using a mutable default arg to keep the cache
> private and the function self-contained. This should be faster.
>
> def fib(n, _cache=[0,1]):
> '''Return fibonacci(n).
>
> _cache is initialized with base values and augmented as needed.
> '''
> for k in range(len(_cache), n+1):
> _cache.append(_cache[k-2] + _cache[k-1])
> return _cache[n]
>
> print(fib(1), fib(3), fib(6), fib(5))
> # 1 2 8 5
>
> Either way, the pattern works with any matched pair of base value list
> and recurrence relation where f(n), n a count, depends on one or more
> f(k), k < n. 'Matched' means that the base value list is as least as
> long as the maximum value of n - k. For fib, the length and max are both 2.
>
> --
> Terry Jan Reedy
How about adding a new type of
numbers of the form:
(a+sqrt(r0)+ swrt(r1)....)/b with proper operations, and then one can compute
the Fib term directly in a
field.
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