-## Comprehensions
+** *Work In Progress* **
-3. Using either Kapulet's or Haskell's list comprehension syntax, write an expression that transforms `[3, 1, 0, 2]` into `[3, 3, 3, 1, 2, 2]`.
+## Comprehensions
-*Here is a hint*
+1. In Kapulet, what does `[ [x, 2*x] | x from [1, 2, 3] ]` evaluate to?
-Define a function `dup (n, x)` that creates a list of *n* copies of `x`. Then use list comprehensions to transform `[3, 1, 0, 2]` into `[[3, 3, 3], [1], [], [2, 2]]`. Then use `join` to "flatten" the result.
+2. What does `[ 10*x + y | y from [4], x from [1, 2, 3] ]` evalaute to?
+3. Using either Kapulet's or Haskell's list comprehension syntax, write an expression that transforms `[3, 1, 0, 2]` into `[3, 3, 3, 1, 2, 2]`. [[Here is a hint|assignment3 hints]], if you need it.
## Lists
-7. Continuing to encode lists in terms of their left-folds, how should we write `head`? This is challenging.
+4. Last week you defined `head` in the Lambda Calculus, using our proposed encoding for lists. Now define `empty?` (It should require only small modifications to your solution for `head`.)
+
+5. If we encode lists in terms of their *left*-folds, instead, `[a, b, c]` would be encoded as `\f z. f (f (f z a) b) c`. The empty list `[]` would still be encoded as `\f z. z`. What should `cons` be, for this encoding?
+
+6. Continuing to encode lists in terms of their left-folds, what should `last` be, where `last [a, b, c]` should result in `c`. Let `last []` result in whatever `err` is bound to.
+
+7. Continuing to encode lists in terms of their left-folds, how should we write `head`? This is challenging. [[Here is a hint|assignment3 hints]], if you need it.
+
+
+## Numbers
+
+8. Recall our proposed encoding for the numbers, called "Church's encoding". As we explained last week, it's similar to our proposed encoding of lists in terms of their folds. In last week's homework, you defined `succ` for numbers so encoded. Can you now define `pred` in the Lambca Calculus? Let `pred 0` result in whatever `err` is bound to. This is challenging. For some time theorists weren't sure it could be done. (Here is [some interesting commentary](http://okmij.org/ftp/Computation/lambda-calc.html#predecessor).) However, in this week's notes we examined one strategy for defining `tail` for our chosen encodings of lists, and given the similarities we explained between lists and numbers, perhaps that will give you some guidance in defining `pred` for numbers.
+
+9. Define `leq` for numbers (that is, ≤) in the Lambda Calculus. Here is the expected behavior,
+where `one` abbreviates `succ zero`, and `two` abbreviates `succ (succ zero)`.
+
+ leq zero zero ~~> true
+ leq zero one ~~> true
+ leq zero two ~~> true
+ leq one zero ~~> false
+ leq one one ~~> true
+ leq one two ~~> true
+ leq two zero ~~> false
+ leq two one ~~> false
+ leq two two ~~> true
+ ...
+
+ You'll need to make use of the predecessor function, but it's not essential to understanding this problem that you have successfully implemented it yet. You can treat it as a black box.
+
+## Combinatorial Logic
+
+Reduce the following forms, if possible:
+
+10. `Kxy`
+11. `KKxy`
+12. `KKKxy`
+13. `SKKxy`
+14. `SIII`
+15. `SII(SII)`
+
+<!-- -->
+
+16. Give Combinatorial Logic combinators (that is, expressed in terms of `S`, `K`, and `I`) that behave like our boolean functions. You'll need combinators for `true`, `false`, `neg`, `and`, `or`, and `xor`.
+
+Using the mapping specified in this week's notes, translate the following lambda terms into combinatory logic:
+
+17. `\x x`
+18. `\x y. x`
+19. `\x y. y`
+20. `\x y. y x`
+21. `\x. x x`
+22. `\x y z. x (y z)`
+
+<!-- -->
+
+23. For each of the above translations, how many `I`s are there? Give a rule for describing what each `I` corresponds to in the original lambda term.
+
+More Lambda Practice
+--------------------
+
+Reduce to beta-normal forms:
-*Here is a hint*
+<OL start=24>
+<LI>`(\x. x (\y. y x)) (v w)`
+<LI>`(\x. x (\x. y x)) (v w)`
+<LI>`(\x. x (\y. y x)) (v x)`
+<LI>`(\x. x (\y. y x)) (v y)`
+<LI>`(\x y. x y y) u v`
+<LI>`(\x y. y x) (u v) z w`
+<LI>`(\x y. x) (\u u)`
+<LI>`(\x y z. x z (y z)) (\u v. u)`
+</OL>