什么算法会按长度顺序枚举lambda演算的表达式?例如,(λx.x), (λx.(x x)), (λx.(λy.x))
等等?
答案 0 :(得分:2)
作为长度,我会在(无类型)lambda表达式的BNF中选择T
- 扩展(“深度”)的数量:
V ::= x | y
T ::= V |
λV.T |
(T T)
在python中,你可以按照上面给定变量的生成规则和给定的深度定义一个生成器,如下所示:
def lBNF(vars, depth):
if depth == 1:
for var in vars:
yield var
elif depth > 1:
for var in vars:
for lTerm in lBNF(vars,depth-1):
yield 'l%s.%s' % (var,lTerm)
for i in range(1,depth):
for lTerm1 in lBNF(vars,i):
for lTerm2 in lBNF(vars,depth-i):
yield '(%s %s)' % (lTerm1,lTerm2)
现在,您可以枚举/达到给定深度的lambda术语:
vars = ['x','y']
for i in range(1,5):
for lTerm in lBNF(vars,i):
print lTerm
答案 1 :(得分:1)
请参阅第5页的https://arxiv.org/abs/1210.2610。以下是一些示例代码:
from itertools import chain, count
from functools import lru_cache
@lru_cache(maxsize=None)
def terms(size, level=0):
if size == 0:
return tuple(range(level))
else:
abstractions = (
('abs', term)
for term in terms(size - 1, level + 1)
)
applications = (
('app', term1, term2)
for i in range(size)
for term1 in terms(i, level)
for term2 in terms(size - 1 - i, level)
)
return tuple(chain(abstractions, applications))
def string(term):
if isinstance(term, tuple):
if term[0] == 'abs':
return '(λ {})'.format(string(term[1]))
elif term[0] == 'app':
return '({} {})'.format(string(term[1]), string(term[2]))
else:
return term
for size in count():
print('{} terms of size {}'.format(len(terms(size)), size))
for term in terms(size):
pass # input(string(term))
此输出
0 terms of size 0
1 terms of size 1
3 terms of size 2
14 terms of size 3
82 terms of size 4
579 terms of size 5
4741 terms of size 6
43977 terms of size 7
454283 terms of size 8
依此类推(即this sequence)。