Im stupid, can someone explain to me what the lambda calculus expression for Y combinator: Y = λf·(λx·f (x x)) (λx·f (x x)) means?
Name:
Anonymous2012-01-23 16:42
Y f = f (Y f)
( Y )
Name:
Anonymous2012-01-23 16:46
I get that but how those those parts translate into the lambda expression. I get the start, of it taking the function f as an argument but am confused on what the application (λx·f (x x)) does. I just dont see where the X comes into it or how this exactly finds the fixed point
factr 3
= fact ((λx.fact (xx)) (λx.fact (xx))) 3 [definition of factr]
= (λf.λn.if n = 0 then 1 else n * f(n - 1))
((λx.fact (xx)) (λx.fact (xx))) 3 [definition of fact]
= if 3 = 0 then 1 [β-reduce]
else 3 * ((λx.fact (xx)) (λx.fact (xx))) (3 - 1)
= 3 * ((λx.fact (xx)) (λx.fact (xx))) 2 [3 = 0 is false, if false then T else F is F]
= 3 * fact ((λx.fact (xx)) (λx.fact (xx))) 2 [β-reduce]
Name:
Anonymous2012-01-23 17:31
I really have only a basic idea of lambda calculus, can someone explain to me how (λx·f (x x)) (λx·f (x x)) converts to plain english. And how Y even finds the fixed point
Plain English won't help. If you understand >>2, here's how to get there:
Y g = λf.(λx.f (x x)) (λx.f (x x)) g
replace f with g:
= (λx.g (x x)) (λx.g (x x))
apply the first part to the second:
= g (λx.g (x x)) (λx.g (x x))
the second part is equal to where we started:
= g (Y g)
Name:
Anonymous2012-01-23 19:56
ok thank you I get that now, but what I still dont get is where is the fixed point being returned? I mean if I do Y(g(x)) with g(x)=x^2 where and how along the execution does the solution 1 arises
Name:
Anonymous2012-01-23 19:56
I don't understand any of this gobbledygook, but this helps:
>>7 http://en.wikipedia.org/wiki/Fixed-point_combinator Whereas a fixed-point of a first-order function (a function on "simple" values such as integers) is a first-order value, a fixed point of a higher-order function f is another functionp such that p = f(p). A fixed-point combinator, then, is a function g which produces such a fixed point p for any function f: g(f) = p, where p = f(p)
or, alternatively: g(f) = f(g(f)).
Name:
Anonymous2012-01-23 20:44
Check my dubs
Name:
Anonymous2013-09-01 19:53
In addition to the team aspect, Super Sentai also has a large emphasis on super robots and big scale setpiece battles. Ultraman has big scale fights too, but lacks mecha.
Name:
Anonymous2013-09-01 20:39
no but really I AM GAY YOU FUCKING SHIT AND I DEMAND EVERYONE ACCEPT ME