top | item 44201618 (no title) reuben364 | 8 months ago Since the de Bruijn indices are limited (and presumably still Turing complete), I wonder how limited you can make them and still be Turing complete. discuss order hn newest jorkingit|8 months ago I suspect the answer is 3: SKI combinator calculus is Turing complete and you need 3 de Bruijn indices to define S.Good call! I got rid of all numbers above 2, I can't count that high anyway ;-)
jorkingit|8 months ago I suspect the answer is 3: SKI combinator calculus is Turing complete and you need 3 de Bruijn indices to define S.Good call! I got rid of all numbers above 2, I can't count that high anyway ;-)
jorkingit|8 months ago
Good call! I got rid of all numbers above 2, I can't count that high anyway ;-)