assignment1 answers

Define a function zero? that expects a single number as an argument, and returns 'true if that number is 0, else returns 'false.

let
  zero? match lambda x. case x of
                          0 then 'true;
                          y then 'false
                        end
in zero?

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.
```
let
  empty? match lambda xs. case xs of
                            []    then 'true;
                            _ & _ then 'false
                          end
in empty?
```
The second case clause could also just be _ then 'false.
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.)
```
let
  tail match lambda xs. case xs of
                          []      then [];
                          _ & xs' then xs'
                        end
in tail
```

Define a function drop that expects two arguments, in the form (number, sequence), and works 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 []

letrec
  drop match lambda (n, xs). case (n, xs) of
                               (0, _)       then xs;
                               (_, [])      then [];
                               (_, _ & xs') then drop (n-1, xs')
                             end
in drop

What is the relation between tail and drop?

let
  tail xs = drop (1, xs)
in ...

That uses the shorthand explained here, which I will continue to use below.

Define a function take that expects two arguments, in the same form as drop, but works like this instead:

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]

letrec
  take (n, xs) = case (n, xs) of
                   (0, _)        then [];
                   (_, [])       then [];
                   (_, x' & xs') then x' & take (n-1, xs')
                 end
in take

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:

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], [])

letrec
  split (n, xs) = case (n, xs) of
                   (0, _)        then ([], xs);
                   (_, [])       then ([], []);
                   (_, x' & xs') then let
                                        (ys, zs) match split (n-1, xs')
                                      in (x' & ys, zs)
                  end
in split

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.
```
letrec
  filter (p, xs) = case xs of
                     [] then [];
                     x' & xs' when p x' then x' & filter (p, xs');
                     _  & xs'           then      filter (p, xs')
                   end
in filter
```
The above solution uses pattern guards.

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:

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])

letrec
  partition (p, xs) = case xs of
                        []       then ([], []);
                        x' & xs' then let
                                        (ys, zs) match partition (p, xs')
                                      in if p x' then (x' & ys, zs) else (ys, x' & zs)
                      end
in partition

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.
```
letrec
  double xs = case xs of
                []       then [];
                x' & xs' then (2*x') & double xs'
              end
in double
```
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). case xs of
                              []       then [];
                              x' & xs' then (f x') & map (f, xs')
                            end;
  double match lambda xs. map ((lambda x. 2*x), xs)
in ...
```
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:
```
map2 ((lambda (x,y). 10*x + y), [1, 2, 3], [4, 5, 6])  # evaluates to [14, 25, 36]
```
```
letrec
  map2 (f, xs, ys) = case (xs, ys) of
                       ([], _)              then [];
                       (_, [])              then [];
                       (x' & xs', y' & ys') then (f x' y') & map2 (f, xs', ys')
                     end
in map2
```

Extra credit problems

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 &.
```
letrec
  xs && ys = case xs of
               []       then ys;
               x' & xs' then x' & (xs' && ys)
             end
in (&&)
```
This solution is using a variation of the shorthand explained here. We didn't expect you'd know how to deal with the special syntax of &&. You might have just defined this using a regular name, like append.
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 sequence, and the second into a different sequence. Then unmap2 should return those two sequences. Thus if:
```
g z1  # evaluates to (x1, y1)
g z2  # evaluates to (x2, y2)
g z3  # evaluates to (x3, y3)
```
Then unmap2 (g, [z1, z2, z3]) should evaluate to ([x1, x2, x3], [y1, y2, y3]).
```
letrec
  unmap2 (g, zs) = case zs of
                     []       then ([], []);
                     z' & zs' then let
                                     (x, y)   match g z';
                                     (xs, ys) match unmap2 (g, zs')
                                   in (x & xs, y & ys)
                   end
in unmap2
```

Write a function takewhile that expects a p argument like filter, and also a sequence. The result should behave like this:

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 sequence that are less than 10.

letrec
  takewhile (p, xs) = case xs of
                        []       then [];
                        x' & xs' then if p x' then x' & takewhile (p, xs')
                                              else []
                      end
in takewhile

Write a function dropwhile that expects a p argument like filter, and also a sequence. The result should behave like this:
```
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 sequence that are less than 10.
```
letrec
  dropwhile (p, xs) = case xs of
                        x' & xs' when p x' then dropwhile (p, xs');
                        _ & _              then xs;
                        []                 then []
                      end
in dropwhile
```
Unlike the previous solution, this one uses pattern guards, merely for variety. (In this solution the last two case clauses could also be replaced by the single clause _ then xs.)

Write a function reverse that returns the reverse of a sequence. Thus, reverse [1, 2, 3, 4] should evaluate to [4, 3, 2, 1].

letrec
  aux (ys, xs) = case xs of
                   []       then ys;
                   x' & xs' then aux (x' & ys, xs')
                 end;
  reverse xs = aux ([], xs)
in reverse