Python 2.7
Bu yaklaşım aşağıdaki hususlardan yararlanır:
Herhangi bir tam sayı, ikisinin güçlerinin toplamı olarak temsil edilebilir. İkisinin gücündeki üsler aynı zamanda ikisinin gücü olarak da temsil edilebilir. Örneğin:
8 = 2^3 = 2^(2^1 + 2^0) = 2^(2^(2^0) + 2^0)
Sonunda bu ifadeler kümeler olarak temsil edilebilir (Python'da yerleşik olarak kullandım frozenset):
0boş küme olur {}.2^atemsil eden seti içeren set haline gelir a. Örneğin: 1 = 2^0 -> {{}}ve 2 = 2^(2^0) -> {{{}}}.a+btemsil eden setlerinin birleştirme olur ave b. Örneğin,3 = 2^(2^0) + 2^0 -> {{{}},{}}
2^2^...^2Sayısal değer bir tamsayı olarak depolanamayacak kadar büyük olsa bile , form ifadelerinin kolayca benzersiz küme gösterimlerine dönüştürülebileceği ortaya çıkmaktadır .
Bu n=20, makinemdeki CPython 2.7.5 üzerinde 8.7s'de çalışıyor (Python 3'te biraz daha yavaş ve PyPy'de çok daha yavaş):
"""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()
(Not dekoratörünün konsepti http://code.activestate.com/recipes/578231-plasılıkla-the-fastest-memoization-decorator-in-the-/ adresinden kopyalanır .)
Çıktı:
8.667753234
0...........1
1...........1
2...........1
3...........2
4...........4
5...........8
6..........17
[...]
19.....688366
20....1619087
Farklı zamanlamalar n:
n time
16 0.240
17 0.592
18 1.426
19 3.559
20 8.668
21 21.402
nYukarıdakilerden herhangi biri makinemde bellek hatasına neden olur.
Başka bir dile çevirerek bunu daha hızlı yapıp yapamayacağımı merak ediyorum.
Düzenle:get_results İşlev optimize edildi . Ayrıca, 2.7.2 yerine Python 2.7.5 kullanmak biraz daha hızlı çalışmasını sağladı.
2^nve bu nedenle dışında herhangi bir şeyi takip etmek gereksiz olacaktırn. Yani, sadece üs alma kurallarını kullanmak akıllıca görünüyor. Ancak, bunu yapmanın kesinlikle daha akıllı ve tamamen cebirsel bir yolu vardır.