Z = \f. (\x. f (\v. x x v)) (\x. f (\v. x x v)) can be much simpler, namely S (S... | Hacker News

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		tromp on Dec 10, 2024 \| parent \| context \| favorite \| on: Tree Calculus Z = \f. (\x. f (\v. x x v)) (\x. f (\v. x x v)) can be much simpler, namely S (S (K (S S (K (S (S K K) (S K K))))) K) (S (K (S S (K (S (S K K) (S K K))))) K)

Trung0246 on Dec 10, 2024 [–]

If you means that it would be like this:

    var Z = S(S(K(S(S)(K(S(S(K)(K))(S(K)(K))))))(K))(S(K(S(S)(K(S(S(K)(K))(S(K)(K))))))(K));

Unfortunately not works with the `count_fn` function from the wikipedia page :(, `too much recursion`.

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