[英]Can XOR be expressed using SKI combinators?
I have question about SKI-Combinators. 我对SKI组合器有疑问。
Can XOR (exclusive or) be expressed using S
and K
combinators only? 只能使用S
和K
组合器来表示XOR(异或)吗?
I have 我有
True = Cancel
False = (Swap Cancel)
where 哪里
Cancel x y = K x y = x
Swap: ff x y = S ff x y = ff y x
Booleans 布尔
Your question is a bit unclear on the details, but it seems that what you mean is that you have the following representation of booleans: 您的问题在细节上还不清楚,但是您的意思似乎是您具有以下布尔值表示形式:
T := K
F := S K
This works because it means the following reductions hold: 之所以有效,是因为它意味着以下减少适用:
T t e => t
F t e => e
in other words, bte
can be interpreted as IF b THEN t ELSE e
. 换句话说, bte
可以解释为IF b THEN t ELSE e
。
XOR in terms of IF _ THEN _ ELSE _
根据IF _ THEN _ ELSE _
XOR
So given this framework, how do we implement XOR? 因此,有了这个框架,我们如何实现XOR? We can formulate XOR as an IF
expression: 我们可以将XOR表示为IF
表达式:
xor x y := IF x THEN (not y) ELSE y = (IF x THEN not ELSE id) y
which can be eta-reduced to 可以减少到
XOR x := IF x THEN not ELSE id = x not id
Some function combinators 一些功能组合器
We have id = SKK
as standard, and not
can be expressed as flip
, since flip bte = bet = IF b THEN e ELSE t = IF (not b) THEN t ELSE e
. 我们将id = SKK
作为标准,而not
表示为flip
,因为flip bte = bet = IF b THEN e ELSE t = IF (not b) THEN t ELSE e
。 flip
it self is quite involved but doable as flip
它本身是相当参与但可行的
flip := S (S (K (S (KS) K)) S) (KK)
Now we just need to figure out a way to write a function that takes x
and applies it on the two terms NOT
and ID
. 现在,我们只需要找出一种编写函数的方法,该函数接受x
并将其应用于NOT
和ID
这两个术语。 To get there, we first note that if we set 要到达那里,首先要注意的是
app := id
then 然后
app f x = (id f) x = f x
and so, 所以,
(flip app) x f = f x
We are almost there, since everything so far shows that 我们快到了,因为到目前为止的所有事情都表明
((flip app) id) ((flip app) not x) = ((flip app) not x) id = (x not) id = x not id
The last step is to make that last line point-free on x
. 最后一步是使最后一行在x
无点。 We can do that with a function composition operator: 我们可以使用函数组合运算符来做到这一点:
((flip app) id) ((flip app) not x) = compose ((flip app) id) ((flip app) not) x
where the requirement on compose
is that compose
的要求是
compose f g x = f (g x)
which we can get by setting 我们可以通过设置来获得
compose f g := S (K f) g
Putting it all together 放在一起
To summarize, we got 总而言之,我们得到了
xor := compose ((flip app) id) ((flip app) not)
or, fully expanded: 或完全扩展:
xor = S (K ((flip app) id)) ((flip app) not)
= S (K ((flip app) (SKK))) ((flip app) flip)
= S (K ((flip SKK) (SKK))) ((flip SKK) flip)
= S (K (((S (S (K (S (KS) K)) S) (KK)) SKK) (SKK))) (((S (S (K (S (KS) K)) S) (KK)) SKK) (S (S (K (S (KS) K)) S) (KK)))
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