2019-06-09

Land of some other order

We start with

#lang racket

#lang lazy

And make some natural and/or unnatural numbers:

(struct zero () #:transparent)
(struct succ (pred) #:transparent)

(define one (succ (zero)))
(define two (succ one))
(define (inf) (succ (inf)))

And a less-than-or-equal-to function:

(define (<= a b)
  (cond
    [(zero? a) #t]
    [(zero? b) #f]
    [else (<= (succ-pred a) (succ-pred b))]))

(<= is a function with parameters a and b: If a is zero then true else if b is zero then false else try with one-less-than-a and one-less-than-b instead.)

Okay. We can try to apply the function to some arguments. The following are are fine and evaluate to #t and #f.

(<= one two)

(<= two one)

If we started with #lang racket the next two will run forever and we won’t get values back. If we started with #lang lazy they’re fine and evaluate to #t and #f.

(<= two (inf))

(<= (inf) two)

The next one will run forever in both #lang racket and #lang lazy.

(<= (inf) (inf))

(Forever means until we are out of memory or something.)

Anyway. We can choose:

If we want as many expressions as possible to give back values, we might prefer #lang lazy
If we want as many applications as possible of <= to give back values, we might prefer #lang racket

(In #lang racket the forever happens when evaluating the (inf)-arguments before the <=-function is applied. In #lang lazy we might have to decide: Is it the (inf) or the <= that causes forever? Is it reasonably to expect there to be a base case?)