in lambda calculus (λ x. λ y. λ s. λ z. x s (y s z)) is used for addition of two Church numerals how can we explain this, is there any good resource the lambda calculus for functional programming ? your help is much appreciated
-
Check the "Related" section at the right side of the page: stackoverflow.com/questions/515413/…, stackoverflow.com/questions/1051033/…, etc.vgru– vgru2009-11-02 17:27:53 +00:00Commented Nov 2, 2009 at 17:27
Add a comment
|
2 Answers
Actually λ f1. λ f2. λ s. λ z. (f1 s (f2 s z)) computes addition because it is in effect substituting (f2 s z), the number represented by f2, to the "zero" inside (f1 s z).
Example: Let's take two for f2, s s z in expanded form. f1 is one: s z. Replace that last z by f2 and you get s s s z, the expanded form for three.
This would be easier with a blackboard and hand-waving, sorry.
Comments
In lambda calculus, you code a datatype in terms of the operations it induces. For instance, a boolean is a just a choice function that takes in input two values a and b and either returns a or b:
true = \a,b.a false = \a,b.b
What is the use of a natural number? Its main computational purpose is to provide a bound to iteration. So, we code a natural number as an operator that takes in input a function f, a value x, and iterate the application of f over x for n times:
n = \f,x.f(f(....(f x)...))
with n occurrences of f.
Now, if you want to iterate n + m times the function f starting from x you must start iterating n times, that is (n f x), and then iterate for m additional times, starting from the previous result, that is
m f (n f x)
Similarly, if you want to iterate n*m times you need to iterate m times the operation of iterating n times f (like in two nested loops), that is
m (n f) x
The previous encoding of datatypes is more formally explained in terms of constructors and corresponding eliminators (the so called Bohm-Berarducci encoding).
