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`