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-*Reduction*
+1. Define a function `zero?` that expects a single number as an argument, and returns `'true` if that number is `0`, else returns `'false`. Your solution should have a form something like this:
-Find "normal forms" for the following (that is, reduce them as far as it's possible to reduce
-them):
+ let
+ zero? match lambda x. FILL_IN_THIS_PART
+ in zero?
- 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))
+ You can use the `if...then...else` construction if you like, but it will make it easier to generalize to later problems if you use the `case EXPRESSION of PATTERN1 then RESULT1; PATTERN2 then RESULT2; ... end` construction instead.
+2. Define a function `empty?` that expects a sequence of values as an argument (doesn't matter what type of values), and returns `'true` if that sequence is the empty sequence `[]`, else returns `'false`. Here your solution should have a form something like this:
-*Booleans*
+ let
+ empty? match lambda xs. case xs of
+ FILL_IN_THIS_PART
+ end
+ in empty?
-Recall our definitions of true and false.
+3. Define a function `tail` that expects a sequence of values as an argument (doesn't matter what type of values), and returns that sequence with the first element (if any) stripped away. (Applying `tail` to the empty sequence `[]` can just give us back the empty sequence.)
- "true" defined to be `\t \f. t`
- "false" defined to be `\t \f. f`
+4. Define a function `drop` that expects two arguments, in the form (*number*, *sequence*), and works like this:
-In Racket, these can be defined like this:
+ drop (0, [10, 20, 30]) # evaluates to [10, 20, 30]
+ drop (1, [10, 20, 30]) # evaluates to [20, 30]
+ drop (2, [10, 20, 30]) # evaluates to [30]
+ drop (3, [10, 20, 30]) # evaluates to []
+ drop (4, [10, 20, 30]) # evaluates to []
- (define true (lambda (t) (lambda (f) t)))
- (define false (lambda (t) (lambda (f) f)))
+ Your solution should have a form something like this:
+ letrec
+ drop match lambda (n, xs). FILL_IN_THIS_PART
+ in drop
-8. Define a "neg" operator that negates "true" and "false".
-Expeceted behavior: (((neg true) 10) 20) evaluates to 20,
-(((neg false) 10) 20) evaluates to 10.
+ What is the relation between `tail` and `drop`?
-9. Define an "and" operator.
+5. Define a function `take` that expects two arguments, in the same form as `drop`, but works like this instead:
-10. 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
-)
+ take (0, [10, 20, 30]) # evaluates to []
+ take (1, [10, 20, 30]) # evaluates to [10]
+ take (2, [10, 20, 30]) # evaluates to [10, 20]
+ take (3, [10, 20, 30]) # evaluates to [10, 20, 30]
+ take (4, [10, 20, 30]) # evaluates to [10, 20, 30]
-11. Inspired by our definition of boolean values, propose a data structure
-capable of representing one of the two values "black" or "white". If we have
-one of those values, call it a black-or-white-value, we should be able to
-write:
+6. Define a function `split` that expects two arguments, in the same form as `drop` and `take`, but this time evaluates to a pair of results. It works like this:
- the-black-or-white-value if-black if-white
-(where if-black and if-white are anything), and get back one of if-black or
-if-white, depending on which of the black-or-white values we started with. Give
-a definition for each of "black" and "white". (Do it in both lambda calculus
-and also in Racket.)
+ split (0, [10, 20, 30]) # evaluates to ([], [10, 20, 30])
+ split (1, [10, 20, 30]) # evaluates to ([10], [20, 30])
+ split (2, [10, 20, 30]) # evaluates to ([10, 20], [30])
+ split (3, [10, 20, 30]) # evaluates to ([10, 20, 30], [])
+ split (4, [10, 20, 30]) # evaluates to ([10, 20, 30], [])
-12. Now propose a data structure capable of representing one of the three values
-"red" "green" or "blue," based on the same model. (Do it in both lambda
-calculus and also in Racket.)
+ Here's a way to answer this problem making use of your answers to previous questions:
+ letrec
+ drop match ... ; # as in problem 4
+ take match ... ; # as in problem 5
+ split match lambda (n, xs). let
+ ys = take (n, xs);
+ zs = drop (n, xs)
+ in (ys, zs)
+ in split
+ However, we want you to instead write this function from scratch.
-Pairs
------
+7. Write a function `filter` that expects two arguments. The second argument will be a sequence `xs` with elements of some type *t*, for example numbers. The first argument will be a function `p` that itself expects arguments of type *t* and returns `'true` or `'false`. What `filter` should return is a sequence that contains exactly those members of `xs` for which `p` returned `'true`. For example, helping ourself to a function `odd?` that works as you'd expect:
-Recall our definitions of ordered pairs.
+ filter (odd?, [11, 12, 13, 14]) # evaluates to [11, 13]
+ filter (odd?, [11]) # evaluates to [11]
+ filter (odd?, [12, 14]) # evaluates to []
- the pair (x,y) is defined as `\f. f x y`
+8. Write a function `partition` that expects two arguments, in the same form as `filter`, but this time evaluates to a pair of results. It works like this:
-To extract the first element of a pair p, you write:
+ partition (odd?, [11, 12, 13, 14]) # evaluates to ([11, 13], [12, 14])
+ partition (odd?, [11]) # evaluates to ([11], [])
+ partition (odd?, [12, 14]) # evaluates to ([], [12, 14])
- p (\fst \snd. fst)
+9. Write a function `double` that expects one argument which is a sequence of numbers, and returns a sequence of the same length with the corresponding elements each being twice the value of the original element. For example:
-Here are some defintions in Racket:
+ double [10, 20, 30] # evaluates to [20, 40, 60]
+ double [] # evaluates to []
- (define make-pair (lambda (fst) (lambda (snd) (lambda (f) ((f fst) snd)))))
- (define get-first (lamda (fst) (lambda (snd) fst)))
- (define get-second (lamda (fst) (lambda (snd) snd)))
+10. Write a function `map` that generalizes `double`. This function expects a pair of arguments, the second being a sequence `xs` with elements of some type *t*, for example numbers. The first argument will be a function `f` that itself expects arguments of type *t* and returns some type *t'* of result. What `map` should return is a sequence of the results, in the same order as the corresponding original elements. The result should be that we could say:
-Now we can write:
- (define p ((make-pair 10) 20))
- (p get-first) ; will evaluate to 10
- (p get-second) ; will evaluate to 20
+ letrec
+ map match lambda (f, xs). FILL_IN_THIS_PART;
+ double match lambda xs. map ((lambda x. 2*x), xs)
+ in ...
-If you're bothered by having the pair to the left and the function that operates on it come seco\
-nd, think about why it's being done this way: the pair is a package that takes a function for op\
-erating on its elements as an argument, and returns the result of operating on its elemens with \
-that function. In other words, the pair is also a function.
+11. Write a function `map2` that generalizes `map`. This function expects a triple of arguments: the first being a function `f` as for `map`, and the second and third being two sequences. In this case `f` is a function that expects *two* arguments, one from the first of the sequences and the other from the corresponding position in the other sequence. The result should behave like this:
-If you like, you can disguise what's going on like this:
- (define lifted-get-first (lambda (p) (p get-first)))
- (define lifted-get-second (lambda (p) (p get-second)))
+ map2 ((lambda (x,y). 10*x + y), [1, 2, 3], [4, 5, 6]) # evaluates to [14, 25, 36]
-Now you can write:
- (lifted-get-first p)
-instead of:
- (p get-first)
-However, the latter is still what's going on under the hood.
+###Extra credit problems###
-13. Define a "swap" function that reverses the elements of a pair.
-Expected behavior:
- (define p ((make-pair 10) 20))
- ((p swap) get-first) ; evaluates to 20
- ((p swap) get-second) ; evaluates to 10
+* In class I mentioned a function `&&` which occupied the position *between* its arguments, rather than coming before them (this is called an "infix" function). The way that it works is that `[1, 2, 3] && [4, 5]` evaluates to `[1, 2, 3, 4, 5]`. Define this function, making use of `letrec` and the simpler infix operation `&`.
-Write out the definition of swap in Racket.
+* Write a function `unmap2` that is something like the inverse of `map2`. This function expects two arguments, the second being a sequence of elements of some type *t*. The first is a function `g` that expects a single argument of type *t* and returns a *pair* of results, rather than just one result. We want to collate these results, the first into one list, and the second into a different list. Then `unmap2` should return those two lists. Thus if:
+ g z1 # evaluates to [x1, y1]
+ g z2 # evaluates to [x2, y2]
+ g z3 # evaluates to [x3, y3]
-14. Define a "dup" function that duplicates its argument to form a pair
-whose elements are the same.
-Expected behavior:
- ((dup 10) get-first) ; evaluates to 10
- ((dup 10) get-second) ; evaluates to 10
-15. Define a "sixteen" function that makes
-sixteen copies of its argument (and stores them in a data structure of
-your choice).
+ Then `unmap2 (g, [z1, z2, z3])` should evaluate to `([x1, x2, x3], [y1, y2, y3])`.
-16. Inspired by our definition of ordered pairs, propose a data structure capable of representin\
-g ordered tripes. That is,
- (((make-triple M) N) P)
-should return an object that behaves in a reasonable way to serve as a triple. In addition to de\
-fining the make-triple function, you have to show how to extraxt elements of your triple. Write \
-a get-first-of-triple function, that does for triples what get-first does for pairs. Also write \
-get-second-of-triple and get-third-of-triple functions.
+* Write a function `takewhile` that expects a `p` argument like `filter`, and also a sequence. The result should behave like this:
-> I expect some to come back with the lovely
-> (\f. f first second third)
-> and others, schooled in a certain mathematical perversion, to come back
-> with:
-> (\f. f first (\g. g second third))
+ takewhile ((lambda x. x < 10), [1, 2, 20, 4, 40]) # evaluates to [1, 2]
+ Note that we stop "taking" once we reach `20`, even though there are still later elements in the list that are less than `10`.
-17. Write a function second-plus-third that when given to your triple, returns the result of add\
-ing the second and third members of the triple.
+* Write a function `dropwhile` that expects a `p` argument like `filter`, and also a sequence. The result should behave like this:
-You can help yourself to the following definition:
- (define add (lambda (x) (lambda (y) (+ x y))))
+ dropwhile ((lambda x. x < 10), [1, 2, 20, 4, 40]) # evaluates to [20, 4, 40]
+
+ Note that we stop "dropping" once we reach `20`, even though there are still later elements in the list that are less than `10`.
+
+* Write a function `reverse` that returns the reverse of a sequence. Thus, `reverse [1, 2, 3, 4]` should evaluate to `[4, 3, 2, 1]`.