# Assignment 6 (week 7)

## Evaluation order in Combinatory Logic

1. Give a term that the lazy evaluators (either the Haskell evaluator,
or the lazy version of the OCaml evaluator) do not evaluate all the
way to a normal form, i.e., that contains a redex somewhere inside of
it after it has been reduced.

<!-- reduce3 (FA (K, FA (I, I))) -->


2. One of the 
[[criteria we established for classifying reduction strategies|
topics/week3_evaluation_order]]
strategies is whether they reduce subexpressions hidden under lambdas.
That is, for a term like `(\x y. x z) (\x. x)`, do we reduce to
`\y.(\x.x) z` and stop, or do we reduce further to `\y.z`?  Explain
what the corresponding question would be for CL. Using either the
OCaml CL evaluator or the Haskell evaluator developed in the wiki
notes, prove that the evaluator does reduce expressions inside of
"functional" CL expressions.  Then provide a modified evaluator that
does not perform reductions in those positions.

<!-- just add early no-op cases for Ka and Sab -->

3. In the previous homework, one of the techniques for controlling
evaluation order was wrapping expressions in a `let`: `let x = blah in
foo`, you could be sure that `blah` would be evaluated by the time the
interpreter considered `foo` (unless you did some fancy footwork with
thunks).  That suggests the following way to try to arrive at eager
evaluation in our Haskell evaluator for CL:

    reduce4 t = case t of
      I -> I
      K -> K
      S -> S
      FA a b -> 
        let b' = reduce4 b in
        let a' = reduce4 a in
        let t' = FA a' b' in
          if (is_redex t') then reduce4 (reduce_one_step t')
                           else t'                                

Will this work?  That is, will `reduce4 (FA (FA K I) skomega)` go into
an infinite loop?  Run the code to find out, if you must, but write
down your guess (and your rationale) first.

<!-- Doesn't work: infinite loop. -->