Insert all the implicit ( )s and λs into the following abbreviated expressions. Don't just insert them freely; rather, provide the official expression, without any notational shortcuts, that is syntactically identical to the form presented.

  1. x x (x x x) x
  2. v w (\x y. v x)
  3. (\x y. x) u v
  4. w (\x y z. x z (y z)) u v

Mark all occurrences of (x y) in the following terms:

(I know the numbering of the homework problems will restart instead of continuing with 5, 6, 7, ... It's too much of a pain to fix it right now. We'll put in a better rendering engine later that will make this work right without laborious work-arounds on our part. Please just renumber the problems appropriately)

  1. (\x y. x y) x y
  2. (\x y. x y) (x y)
  3. \x y. x y (x y)


Find "normal forms" for the following---that is, reduce them until no more reductions are possible. As mentioned in the notes, we'll write λx as \x. If we ever say "reduce" without qualifications, we mean just "beta-reduce" (as opposed to "(beta + eta)-reduce").

  1. (\x \y. y x) z
  2. (\x (x x)) z
  3. (\x (\x x)) z
  4. (\x (\z x)) z
  5. (\x (x (\y y))) (\z (z z))
  6. (\x (x x)) (\x (x x))
  7. (\x (x x x)) (\x (x x x))


For these questions, and the ones on triples below, we're setting them up so as to encourage you to experiment with Racket and to formulate your answer in Scheme/Racket syntax. But you can answer in Lambda Calculus syntax if you prefer.

Recall our definitions of true and false.

true is defined to be \t f. t
false is defined to be \t f. f

In Racket, these functions can be defined like this:

(define true (lambda (t) (lambda (f) t)))
(define false (lambda (t) (lambda (f) f)))

(Note that they are different from Racket's primitive boolean values #t and #f.)

  1. Define a neg operator that negates true and false.

    Expected behavior:

    (((neg true) 10) 20)

    evaluates to 20, and

    (((neg false) 10) 20)

    evaluates to 10.

  2. Define an or operator.

  3. Define an xor operator. If you haven't seen this term before, here's a truth table:

    true   xor  true   == false
    true   xor  false  == true
    false  xor  true   == true
    false  xor  false  == false


Recall our definitions of ordered triples.

the triple (a, b, c) is defined to be \f. f a b c

To extract the first element of a triple t, you write:

t (\fst snd trd. fst)

Here are some definitions in Racket:

(define make-triple  (lambda (fst) (lambda (snd) (lambda (trd) (lambda (f) (((f fst) snd) trd))))))
(define fst_of_three (lambda (fst) (lambda (snd) (lambda (trd) fst))))
(define snd_of_three (lambda (fst) (lambda (snd) (lambda (trd) snd))))

Now we can write:

(define t (((make-triple 10) 20) 30))
(t fst_of_three)  ; will evaluate to 10
(t snd_of_three)  ; will evaluate to 20

If you're puzzled by having the triple to the left and the function that operates on it come second, think about why it's being done this way: the triple is a package that takes a function for operating on its elements as an argument, and returns the result of operating on its elements with that function. In other words, the triple is a higher-order function.

  1. Define the swap12 function that permutes the elements of a triple. Expected behavior:

    (define t (((make-triple 10) 20) 30))
    ((t swap12) fst_of_three) ; evaluates to 20
    ((t swap12) snd_of_three) ; evaluates to 10

    Write out the definition of swap12 in Racket.

  2. Define a dup3 function that duplicates its argument to form a triple whose elements are the same. Expected behavior:

      ((dup3 10) fst_of_three) ; evaluates to 10
      ((dup3 10) snd_of_three) ; evaluates to 10
  3. Define a dup27 function that makes twenty-seven copies of its argument (and stores them in a data structure of your choice).

Folds and Lists

  1. Using Kapulet syntax, define fold_left.

  2. Using Kapulet syntax, define filter (problem 7 in last week's homework) in terms of fold_right and other primitive syntax like lambda, &, and []. Don't use letrec! All the letrec-ing that happens should come from the one inside the definition of fold_right.

  3. Using Kapulet syntax, define && in terms of fold_right. (To avoid trickiness about the infix syntax, just call it append.) As with problem 22 (the previous problem), don't use letrec!

  4. Using Kapulet syntax, define head in terms of fold_right. When applied to a non-empty list, it should give us the first element of that list. When applied to an empty list, let's say it should give us 'err. As with problem 22, don't use letrec!

  5. We mentioned in the Encoding notes that fold_left (flipped_cons, []) xs would give us the elements of xs but in the reverse order. So that's how we can express reverse in terms of fold_left. How would you express reverse in terms of fold_right? As with problem 22, don't use letrec!

    This problem does have an elegant and concise solution, but it may be hard for you to figure it out. We think it will a useful exercise for you to try, anyway. We'll give a hint. Don't look at the hint until you've gotten really worked up about the problem. Before that, it probably will just be baffling. If your mind has really gotten its talons into the problem, though, the hint might be just what you need to break it open.

    There are also other, less cool answers. Perhaps you'll find one of them first.

    Even if you don't get any answer, we think the experience of working on the problem, and then understanding the answer when we reveal it, will be satisfying and worthwhile. It also fits our pedagogical purposes for some of the recurring themes of the class.


  1. Given that we've agreed to Church's encoding of the numbers:

    0 ≡ \f z. z
    1 ≡ \f z. f z
    2 ≡ \f z. f (f z)
    3 ≡ \f z. f (f (f z))

    How would you express the succ function in the Lambda Calculus?