X-Git-Url: http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=blobdiff_plain;f=assignment1.mdwn;h=ea8f2511992e97ff0caed2f8f002426dd2bc053b;hp=453fa546366dd175a19ca23db676c9a5fd5179db;hb=a57738ceb3c389a252958613dcc6d125a64384e0;hpb=4abee3b6356890888890b4907fb4c7a24bc4656e diff --git a/assignment1.mdwn b/assignment1.mdwn index 453fa546..ea8f2511 100644 --- a/assignment1.mdwn +++ b/assignment1.mdwn @@ -1,158 +1,118 @@ -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 until no more reductions are possible. We'll write `λx` as `\x`. + 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: -
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1. Define a `neg` operator that negates `true` and `false`. + letrec + drop match lambda (n, xs). FILL_IN_THIS_PART + in drop -Expected behavior: + What is the relation between `tail` and `drop`? - (((neg true) 10) 20) +5. Define a function `take` that expects two arguments, in the same form as `drop`, but works like this instead: -evaluates to 20, and + 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] - (((neg false) 10) 20) +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: -evaluates to 10. + 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], []) -
2. Define an `and` operator. + Here's a way to answer this problem making use of your answers to previous questions: -
3. Define an `xor` operator. + 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 -If you haven't seen this term before, here's a truth table: + However, we want you to instead write this function from scratch. - true xor true = false - true xor false = true - false xor true = true - false xor false = false +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: + filter (odd?, [11, 12, 13, 14]) # evaluates to [11, 13] + filter (odd?, [11]) # evaluates to [11] + filter (odd?, [12, 14]) # evaluates to [] -
4. 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: +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: - the-value if-black if-white + 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]) -(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.) +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: -
5. 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.) -
+ double [10, 20, 30] # evaluates to [20, 40, 60] + double [] # evaluates to [] +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: + letrec + map match lambda (f, xs). FILL_IN_THIS_PART; + double match lambda xs. map ((lambda x. 2*x), xs) + in ... -Pairs ------ +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: -Recall our definitions of ordered pairs. + map2 ((lambda (x,y). 10*x + y), [1, 2, 3], [4, 5, 6]) # evaluates to [14, 25, 36] -> the pair **(**x**,**y**)** is defined as `\f. f x y` -To extract the first element of a pair p, you write: +###Extra credit problems### - p (\fst \snd. fst) +* 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 `&`. -Here are some definitions 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: - (define make-pair (lambda (fst) (lambda (snd) (lambda (f) ((f fst) snd))))) - (define get-first (lambda (fst) (lambda (snd) fst))) - (define get-second (lambda (fst) (lambda (snd) snd))) + g z1 # evaluates to [x1, y1] + g z2 # evaluates to [x2, y2] + g z3 # evaluates to [x3, y3] -Now we can write: + Then `unmap2 (g, [z1, z2, z3])` should evaluate to `([x1, x2, x3], [y1, y2, y3])`. - (define p ((make-pair 10) 20)) - (p get-first) ; will evaluate to 10 - (p get-second) ; will evaluate to 20 +* Write a function `takewhile` that expects a `p` argument like `filter`, and also a sequence. The result should behave like this: -If you're bothered by having the pair to the left and the function that -operates on it come second, think about why it's being done this way: the pair -is a package that takes a function for operating 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. (Of course, in the -untyped lambda calculus, absolutely *everything* is a function: functors, -arguments, abstracts, redexes, values---everything.) + takewhile ((lambda x. x < 10), [1, 2, 20, 4, 40]) # evaluates to [1, 2] -If you like, you can disguise what's going on like this: + Note that we stop "taking" once we reach `20`, even though there are still later elements in the list that are less than `10`. - (define lifted-get-first (lambda (p) (p get-first))) - (define lifted-get-second (lambda (p) (p get-second))) +* Write a function `dropwhile` that expects a `p` argument like `filter`, and also a sequence. The result should behave like this: -Now you can write: + dropwhile ((lambda x. x < 10), [1, 2, 20, 4, 40]) # evaluates to [20, 4, 40] - (lifted-get-first p) + Note that we stop "dropping" once we reach `20`, even though there are still later elements in the list that are less than `10`. -instead of: - - (p get-first) - -However, the latter is still what's going on under the hood. - - -
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1. 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 - -Write out the definition of swap in Racket. - - -
2. 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 - -
3. Define a `sixteen` function that makes -sixteen copies of its argument (and stores them in a data structure of -your choice). - -
4. Inspired by our definition of ordered pairs, propose a data structure capable of representing ordered triples. 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 defining the `make-triple` function, you have to show how to extract 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. - -
5. Write a function `second-plus-third` that when given to your triple, returns the result of adding the second and third members of the triple. - -You can help yourself to the following definition: - - (define add (lambda (x) (lambda (y) (+ x y)))) - - - -
+* Write a function `reverse` that returns the reverse of a sequence. Thus, `reverse [1, 2, 3, 4]` should evaluate to `[4, 3, 2, 1]`.