With all brackets present, the term is λx.(λy.(λz.(xz(yz)))).Ī syntax tree for the term thus looks as follows: λxyz.xz(yz): abstractionĪnd as you also correctly suspected, the application of λxyz.xz to yz would be written as (λxyz.xz)yz. So as you suspect, λxyz.xz(yz) is a three-fold abstraction over xz(yz), which is itself an application of xz to yz, which are again applications of x to z and y to z, repsectively. For full -reduction, we have the following: a.b.(c.c) (d.d) a.b.d.d (x.xxxx) (x.y. bracketing in application terms is left-associative, so xzyz (without brackets around yz) would be read as ((xz)y)z, and to indicate that xz is applied to yz, you need to bracket the term yz. Lambda calculus basics Lambda calculus encodings and Recursion (Lectures 78) Section and Practice Problems Answer: There are no possible steps under CBV and CBN because neither allows reductions inside a lambda term. Somewhat recursive ), but more in-depth definition: A combinator is just a lambda expression with no. In particular, normallyįirst try to form the application of xz to yz, and only then theĪbstractions over x, y and z, rather than first forming the abstraction λxyz.xz and then embedding the resulting term in an applicatn to yz (which is what you would get if abstraction had precedence over application) An Y-combinator is a 'functional' (or a higher-order function a function that operates on other functions) that takes a single argument, which is a function that isn't recursive, and returns a version of the function which is recursive. To be precise, it depends on the exact definition of the syntax of lambda terms, more precisely the rules for ommission of brackets, but there are convention for these rules. The clue here is that this effectively stops the computation at each step until it's actually put into use.Correct. In the same manner (lambda args (apply (g g) args)) is the same as (g g) and you can see that by just applying substitution rules. If you have a function add1 you can make a wrapper (lambda (x) (add1 x)) that you can use instead and it will work. We know (g g) becomes a function so we just give f a function that when applied will generate the actual function and apply it. To salvage that we don't apply (g g) right away. If you look at how this would be applied with an argument you'll notice that Y never returns since before it can apply f in (f (g g)) it needs to evaluate (g g) which in turn evaluates (f (g g)) etc. Well the normal order version looks like this: (define Y So you are asking how the evaluation gets delayed. Notice the implementations is exactly the same and the difference is how the reference to itself is handled. This can be done with Z like this: ((Z (lambda (ackermann) Imagine the Ackermann function: (define ackermann Notice this version uses apply to support multiple argument functions. It's just a little more code that does exactly the same. Now the first thing that happens when this is applied is (g g) and since you can always substitute a whole application with the expansion of it's body the body of the function can get rewritten to: (define Z The applicative version of Y is often called a Z combinator: (define Z In a lazy language like Lazy Racket you can use the normal order version, but not in any of the applicative order programming languages like Scheme.
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |