X-Git-Url: http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=blobdiff_plain;f=exercises%2Fassignment3_answers.mdwn;h=ae35731b0333982c915d357e9ed1d4bd92225848;hp=7530534f3c4e086c6260aaaface0b1b6b4126809;hb=dd911c78a79577243800cafec55f75ef9d76d63a;hpb=cff9857e52ac59c4905efeb80fcf8a91c2fd1ed8 diff --git a/exercises/assignment3_answers.mdwn b/exercises/assignment3_answers.mdwn index 7530534f..ae35731b 100644 --- a/exercises/assignment3_answers.mdwn +++ b/exercises/assignment3_answers.mdwn @@ -167,7 +167,7 @@ where `one` abbreviates `succ zero`, and `two` abbreviates `succ (succ zero)`. > let leq? = \l r. zero? (sub l r) in > ... - > Here is another solution. Jim crafted this particular implementation, but like a great deal of the CS knowledge he's gained over the past eight years, Oleg Kiselyov pointed the way. + > Here is another solution. Jim crafted this particular implementation, but like a great deal of the CS knowledge he's gained over the past eight years, Oleg Kiselyov pointed the way. > let leq? = (\base build consume. \l r. r consume (l build base) fst) > ; where base is @@ -238,23 +238,6 @@ Reduce the following forms, if possible: Using the mapping specified in this week's notes, translate the following lambda terms into combinatory logic: - -Let's say that for any lambda term T, [T] is the equivalent Combinatory Logic term. Then we define the [.] mapping as follows. - - 1. [a] = a - 2. [(\aX)] = @a[X] - 3. [(XY)] = ([X][Y]) - -Wait, what is that @a ... business? Well, that's another operation on (a variable and) a CL expression, that we can define like this: - - 4. @aa = I - 5. @aX = KX if a is not in X - 6. @a(Xa) = X if a is not in X - 7. @a(XY) = S(@aX)(@aY) - - - -
1. `[\x x] = @x x = I`
2. `[\x y. x] = @x [\y. x] = @x. (@y x) = @x (Kx) = S (@x K) (@x x) = S (KK) I`; in general expressions of this form `S(KM)I` will behave just like `M` for any expression `M` @@ -284,6 +267,12 @@ S (S (KS) (S (KK) (S (KS) K))) (KI); this is the B combinator, whi 25. 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. + This generalization depends on you omitting the translation rule: + + 6. @a(Xa) = X if a is not in X + + > With that shortcut rule omitted, then there turn out to be one `I` in the result corresponding to each occurrence of a bound variable in the original term. + Evaluation strategies in Combinatory Logic ------------------------------------------ @@ -323,8 +312,8 @@ Reduce to beta-normal forms:
1. `(\x. x (\y. y x)) (v w) ~~> v w (\y. y (v w))`
2. `(\x. x (\x. y x)) (v w) ~~> v w (\x. y x)` -
3. `(\x. x (\y. y x)) (v x) ~~> v w (\y. y (v x))` -
4. `(\x. x (\y. y x)) (v y) ~~> v w (\u. u (v y))` +
5. `(\x. x (\y. y x)) (v x) ~~> v x (\y. y (v x))` +
6. `(\x. x (\y. y x)) (v y) ~~> v y (\u. u (v y))`
7. `(\x y. x y y) u v ~~> u v v`
8. `(\x y. y x) (u v) z w ~~> z (u v) w`