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diff --git a/exercises/_assignment7.mdwn b/exercises/_assignment7.mdwn
index cd798ab0..d351b1b5 100644
--- a/exercises/_assignment7.mdwn
+++ b/exercises/_assignment7.mdwn
@@ -23,3 +23,28 @@ notes, prove that the evaluator does reduce expressions inside of
 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. -->
+