4 ms·

This is what S and K (and I) are: S := λx.λy.λz.x z (y z) K := λx.λy.x I := λx.x There's nothing really more to it. What's weird/amazing is that j

• mikewarot 6y ago
  I still don't understand the way this breaks down... How many parameters do S,K,or I take? If x=11, y=12, z=13 what are S,K? I assume I = 11
  • dvt 6y ago
    This is more of a λ-Calculus question (unrelated to combinators, really). But S takes 3, K takes 2, and I takes 1. This isn't strictly correct because S/K/I isn't a function, but rather a chain of lambdas with no free variables (and we get the number of "parameters the function takes" by counting the lambda symbols). See §1 in [1]. In your specific case, S 11 12 = 12 K 11 12 13 = 11 13 (12 13) I 11 = 11 Keep in mind that the literals "11," "12," and "13" in λ-Calculus aren't numbers. [1] https://personal.utdallas.edu/~gupta/courses/apl/lambda.pdf https://personal.utdallas.edu/~gupta/courses/apl/lambda.pdf
    • mikewarot 6y ago
      If I understand this right... S takes 2 values, and returns 1 K takes 3 values, and returns the 1st, 3rd, and (2nd applied to 3rd) I returns the value it is given (like a no-op)
      • dvt 6y ago
        > K takes 3 values, and returns the 1st, 3rd, and (2nd applied to 3rd) Everything applies: so, e.g., when we see something like ABC, we actually omit the parentheses; if we include them, we have (((A)B)C). This can be confusing because applications associate to the left, but lambdas associate to the right[1]. So 11 13 (12 13) is actually (((11)13)((12)13)). In roundabout English, 11 applies to 13 which then applies to the result of 12 applies to 13. [1] https://www.cs.cmu.edu/~anupamg/251-notes/lambda_calculus.pdf https://www.cs.cmu.edu/~anupamg/251-notes/lambda_calculus.pd...
        • mikewarot 6y ago
          The impedance mismatch is too hard, given the nature of this medium... thanks for trying to help. Maybe next time?
          • carapace 6y ago
            Look here mate yer over thinking it innit? In Python: S = lambda x: lambda y: lambda z: x(z)(y(z)) K = lambda x: lambda y: x I = lambda x: x That there is runnable code and with it you can implement numbers, math, logic, etc. That's all the SKI combinators are: those functions. You apply them to themselves in various patterns and they model numbers and arithmetic and programming languages. In [1]: S = lambda x: lambda y: lambda z: x(z)(y(z)) In [2]: K = lambda x: lambda y: x In [3]: I = lambda x: x In [4]: K(I) Out[4]: <function __main__.<lambda>.<locals>.<lambda>(y)> In [5]: _(S) is I Out[5]: True In [6]: K(I)(S) is I Out[6]: True Hope that helps.
            • mikewarot 6y ago
              Yes, having code that I can play with did help... but I can't figure out S, no matter what I give it, I always get some sort of error. TypeError: 'int' object is not callable
              • dvt 6y ago
                Hint: if you pass it x, y, z where they are functions, it won't complain :) Python is (weakly) typed, so in your case, it tries to call 1(something) and that won't work. In (untyped) Lambda Calculus, everything is a function.
                • mikewarot 6y ago
                  I got S(I)(I)(I)(4) to return 4, everything else is errors.
                  • 6y ago