Python 2.7
このアプローチでは、次の考慮事項を活用します。
任意の整数は、2の累乗の合計として表すことができます。2のべき乗の指数も2のべき乗として表すことができます。例えば：
8 = 2^3 = 2^(2^1 + 2^0) = 2^(2^(2^0) + 2^0)
最終的にこれらの式は、セットのセットとして表すことができます（Pythonでは、組み込みのを使用しましたfrozenset）。
0空のセットになり{}ます。2^aを表すセットを含むセットになりaます。例：1 = 2^0 -> {{}}および2 = 2^(2^0) -> {{{}}}。a+b代表組の連結となるaとb。例えば、3 = 2^(2^0) + 2^0 -> {{{}},{}}
2^2^...^2数値が大きすぎて整数として保存できない場合でも、フォームの式を一意のセット表現に簡単に変換できることがわかります。
の場合n=20、これは8.7秒で実行されます私のマシンのCPython 2.7.5でます（Python 3では少し遅く、PyPyでははるかに遅くなります）。
"""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()
（メモ化デコレータのコンセプトは、 http://code.activestate.com/recipes/578231-probably-the-fastest-memoization-decorator-in-the-/ます。）
出力：
8.667753234
0...........1
1...........1
2...........1
3...........2
4...........4
5...........8
6..........17
[...]
19.....688366
20....1619087
異なるタイミングn：
n time
16 0.240
17 0.592
18 1.426
19 3.559
20 8.668
21 21.402
どれか n私のマシン上のメモリエラーに上記の21件の結果。
誰かがそれを別の言語に翻訳することでこれをより速くできるかどうかに興味があります。
編集：get_results機能を最適化しました。また、2.7.2ではなくPython 2.7.5を使用すると、実行速度が少し速くなりました。
2^n、ゆえ以外のものを追跡する必要だろうn。つまり、べき乗の規則を使用するだけで賢明です。ただし、これを行うには、よりスマートで完全に代数的な方法が確実にあります。