Python 2.7
Questo approccio sfrutta le seguenti considerazioni:
Qualsiasi numero intero può essere rappresentato come una somma di potenze di due. Gli esponenti nei poteri di due possono anche essere rappresentati come poteri di due. Per esempio:
8 = 2^3 = 2^(2^1 + 2^0) = 2^(2^(2^0) + 2^0)
Queste espressioni con cui finiamo possono essere rappresentate come set di set (in Python, ho usato il built-in frozenset):
0diventa l'insieme vuoto {}.2^adiventa il set contenente il set che rappresenta a. Ad esempio: 1 = 2^0 -> {{}}e 2 = 2^(2^0) -> {{{}}}.a+bdiventa la concatenazione degli insiemi che rappresentano ae b. Per esempio,3 = 2^(2^0) + 2^0 -> {{{}},{}}
Si scopre che le espressioni del modulo 2^2^...^2possono essere facilmente trasformate nella loro rappresentazione di insieme univoca, anche quando il valore numerico è troppo grande per essere memorizzato come un numero intero.
Per n=20, questo funziona in 8.7s su CPython 2.7.5 sulla mia macchina (un po 'più lento in Python 3 e molto più lento in PyPy):
"""Analyze the expressions given by parenthesizations of 2^2^...^2.
Set representation: s is a set of sets which represents an integer n. n is
given by the sum of all 2^m for the numbers m represented by the sets
contained in s. The empty set stands for the value 0. Each number has
exactly one set representation.
In Python, frozensets are used for set representation.
Definition in Python code:
def numeric_value(s):
n = sum(2**numeric_value(t) for t in s)
return n"""
import itertools
def single_arg_memoize(func):
"""Fast memoization decorator for a function taking a single argument.
The metadata of <func> is *not* preserved."""
class Cache(dict):
def __missing__(self, key):
self[key] = result = func(key)
return result
return Cache().__getitem__
def count_results(num_exponentiations):
"""Return the number of results given by parenthesizations of 2^2^...^2."""
return len(get_results(num_exponentiations))
@single_arg_memoize
def get_results(num_exponentiations):
"""Return a set of all results given by parenthesizations of 2^2^...^2.
<num_exponentiations> is the number of exponentiation operators in the
parenthesized expressions.
The result of each parenthesized expression is given as a set. The
expression evaluates to 2^(2^n), where n is the number represented by the
given set in set representation."""
# The result of the expression "2" (0 exponentiations) is represented by
# the empty set, since 2 = 2^(2^0).
if num_exponentiations == 0:
return {frozenset()}
# Split the expression 2^2^...^2 at each of the first half of
# exponentiation operators and parenthesize each side of the expession.
split_points = xrange(num_exponentiations)
splits = itertools.izip(split_points, reversed(split_points))
splits_half = ((left_part, right_part) for left_part, right_part in splits
if left_part <= right_part)
results = set()
results_add = results.add
for left_part, right_part in splits_half:
for left in get_results(left_part):
for right in get_results(right_part):
results_add(exponentiate(left, right))
results_add(exponentiate(right, left))
return results
def exponentiate(base, exponent):
"""Return the result of the exponentiation of <operands>.
<operands> is a tuple of <base> and <exponent>. The operators are each
given as the set representation of n, where 2^(2^n) is the value the
operator stands for.
The return value is the set representation of r, where 2^(2^r) is the
result of the exponentiation."""
# Where b is the number represented by <base>, e is the number represented
# by <exponent> and r is the number represented by the return value:
# 2^(2^r) = (2^(2^b)) ^ (2^(2^e))
# 2^(2^r) = 2^(2^b * 2^(2^e))
# 2^(2^r) = 2^(2^(b + 2^e))
# r = b + 2^e
# If <exponent> is not in <base>, insert it to arrive at the set with the
# value: b + 2^e. If <exponent> is already in <base>, take it out,
# increment e by 1 and repeat from the start to eventually arrive at:
# b - 2^e + 2^(e+1) =
# b + 2^e
while exponent in base:
base -= {exponent}
exponent = successor(exponent)
return base | {exponent}
@single_arg_memoize
def successor(value):
"""Return the successor of <value> in set representation."""
# Call exponentiate() with <value> as base and the empty set as exponent to
# get the set representing (n being the number represented by <value>):
# n + 2^0
# n + 1
return exponentiate(value, frozenset())
def main():
import timeit
print timeit.timeit(lambda: count_results(20), number=1)
for i in xrange(21):
print '{:.<2}..{:.>9}'.format(i, count_results(i))
if __name__ == '__main__':
main()
(Il concetto del decoratore di memoization è copiato da http://code.activestate.com/recipes/578231-probably-the-fastest-memoization-decorator-in-the-/ .)
Produzione:
8.667753234
0...........1
1...........1
2...........1
3...........2
4...........4
5...........8
6..........17
[...]
19.....688366
20....1619087
Tempi per diversi n:
n time
16 0.240
17 0.592
18 1.426
19 3.559
20 8.668
21 21.402
Qualsiasi nsopra 21 provoca un errore di memoria sulla mia macchina.
Sarei interessato se qualcuno potesse renderlo più veloce traducendolo in un'altra lingua.
Modifica: ottimizzata la get_resultsfunzione. Inoltre, l'utilizzo di Python 2.7.5 anziché 2.7.2 lo ha reso un po 'più veloce.
2^ne quindi non sarebbe necessario tenere traccia di qualsiasi cosa trannen. Cioè, usare solo le regole dell'espiazione sembra saggio. Tuttavia, c'è sicuramente un modo più intelligente e completamente algebrico per farlo.