On 3/23/2015 7:12 PM, Dave Angel wrote:
On 03/23/2015 06:53 PM, Terry Reedy wrote:

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.


I almost used a default value as a cache, but didn't want to confuse the
OP.  Also, your present code does not use recursion, so probably
wouldn't make the prof happy.

I did not read the initial posts with that requirement. Iteration is easily converted to tail recursion, which is definitely the way to calculate recurrences with more than one term without a full cache.

On the other hand, my loop makes some non-obvious assumptions, like that
append is always the right place to put a new value.  I knew it'd  work,
since fib calls itself with n-1.  But it wouldn't directly work if the
function had recursed on n-2 and n-3.  Yours would.

I prefer iteration, but I still figure this is an assignment.

def fib(n, _cache=[0,1]):
    '''Return fibonacci(n).

    _cache is initialized with base values and augmented as needed.
    '''
    k = len(_cache)  # min n without a value in cache
    if k <= n:
        _cache.append(_cache[k-2] + _cache[k-1])
        return fib(n)
    else:
        return _cache[n]

print(fib(1), fib(3), fib(6), fib(5))
# 1 2 8 5

If one does not like the append as a statement, and prefer it as part of the return expression, this works too.

def fib(n, _dummy=None, _cache=[0,1]):
    k = len(_cache)
    return (fib(n, _cache.append(_cache[k-2] + _cache[k-1]), _cache)
               if k <= n else _cache[n])

However, I am not a fan of puritanical functionalism.

--
Terry Jan Reedy

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