Created
November 19, 2011 20:55
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10553, church encoded as ((4*((3*(((5^3)*7)+4))+1))+1)
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| λs.λz.(λs.λz.s(s(s(s z))))(λz.(λs.λz.s(s(s z)))(λz.(λs.λz.s(s(s z)))(λs.λz.s(s(s(s(s z)))))(λz.s(s(s(s(s(s(s z)))))))(s(s(s(s z)))))(s z))(s z) |
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| These also don't fit in a tweet: | |
| 11491: ((((2^11)*5)+1)+((5^4)*2)) | |
| λs.λz.(λs.λz.s(s(s(s(s(s(s(s(s(s(s z)))))))))))(λs.λz.s(s z))(λz.s(s(s(s(s z)))))(s((λs.λz.s(s(s(s z))))(λs.λz.s(s(s(s(s z)))))(λz.s(s z))z)) | |
| 11514: (((4^5)*11)+((5^3)*2)) | |
| λs.λz.(λs.λz.s(s(s(s(s z)))))(λs.λz.s(s(s(s z))))(λz.s(s(s(s(s(s(s(s(s(s(s z)))))))))))((λs.λz.s(s(s z)))(λs.λz.s(s(s(s(s z)))))(λz.s(s z))z) | |
| 12659: ((7*((7*((4^4)+2))+2))+3) | |
| λs.λz.(λs.λz.s(s(s(s(s(s(s z)))))))(λz.(λs.λz.s(s(s(s(s(s(s z)))))))(λz.(λs.λz.s(s(s(s z))))(λs.λz.s(s(s(s z))))s(s(s z)))(s(s z)))(s(s(s z))) | |
| 13439: (((13^3)*6)+((4^4)+1)) | |
| λs.λz.(λs.λz.s(s(s z)))(λs.λz.s(s(s(s(s(s(s(s(s(s(s(s(s z)))))))))))))(λz.s(s(s(s(s(s z))))))((λs.λz.s(s(s(s z))))(λs.λz.s(s(s(s z))))s(s z)) | |
| 13687: ((7*(((3^2)*((6^3)+1))+2))+2) | |
| λs.λz.(λs.λz.s(s(s(s(s(s(s z)))))))(λz.(λs.λz.s(s z))(λs.λz.s(s(s z)))(λz.(λs.λz.s(s(s z)))(λs.λz.s(s(s(s(s(s z))))))s(s z))(s(s z)))(s(s z)) | |
| 14519: ((7*((8*((4^4)+3))+2))+1) | |
| λs.λz.(λs.λz.s(s(s(s(s(s(s z)))))))(λz.(λs.λz.s(s(s(s(s(s(s(s z))))))))(λz.(λs.λz.s(s(s(s z))))(λs.λz.s(s(s(s z))))s(s(s(s z))))(s(s z)))(s z) | |
| 14734: ((4*(((3^3)+2)*((5^3)+2)))+2) | |
| λs.λz.(λs.λz.s(s(s(s z))))((λs.λz.(λs.λz.s(s(s z)))(λs.λz.s(s(s z)))s(s(s z)))(λz.(λs.λz.s(s(s z)))(λs.λz.s(s(s(s(s z)))))s(s(s z))))(s(s z)) | |
| 14783: ((((7^4)*6)+2)+((5^3)*3)) | |
| λs.λz.(λs.λz.s(s(s(s z))))(λs.λz.s(s(s(s(s(s(s z)))))))(λz.s(s(s(s(s(s z))))))(s(s((λs.λz.s(s(s z)))(λs.λz.s(s(s(s(s z)))))(λz.s(s(s z)))z))) |
Yes, but there's exponentiation, which is free! F.e. (λf.λx.f(f(f x)))(λf.λx.f(f x)) or (3 2) is 2^3.
I found these factorizations:
*Main> (2*3*17)^2+(2*5)^2+7^2
10553
*Main> unchurch $ p(n2(m n2(m(s n2)(p(m n2 n5)(s(s n5))))))(p(n2(m n2 n5))(n2(s(s n5))))
10553
*Main> 2^13+3^7+5^3+7^2
10553
*Main> unchurch $ p(p((p n7(p n3 n3))n2)(n7 n3))(p(n3 n5)(n2 n7))
10553
The second one is promising, it needs only 7 and 2:
*Main> unchurch $ p(p((p n7(s(s(s(s n2)))))n2)(n7(s n2)))(p((s n2)(s(s(s n2))))(n2 n7))
10553
So it would be something like this:
(λs.λz.λp.(λ2.(λ7.p(p((p 7(s(s(s(s 2)))))2)(7(s 2)))(p((s 2)(s(s(s 2))))(2 7)))(s(s(s(s(s 2))))))(s(s z)))(λn.λf.λx.f(n f x))(λf.λx.x)(λa.λb.λf.λx.a f(b f x))
Which is tweetable if I didn't commit any (new) mistakes.
BTW I didn't knew about free exponentiation, sweet!
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Yes. The definition of z and s should be part of the term, so 3 is encoded as:
With this we have no free terms and only λ calculus is required to evaluate it.
Using these rules I don't think it's possible to encode 10553 in a tweet. AFAICS the factorization I used (i.e. "(1+((4_3)5))(1+(4_(1+((2_3)_7))))") is the shortest in terms of operators (only s and m).