1. 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?
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`.
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`.
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.)
let
tail match lambda xs. case xs of
[] then [];
_ & xs' then xs'
end
in tail
4. 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|topics/week1_kapulet_advanced#funct-declarations]], which I will continue to use below.
5. 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
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:
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
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`.
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|/topics/week1_kapulet_advanced#guards]].
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:
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
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.
letrec
double xs = case xs of
[] then [];
x' & xs' then (2*x') & double xs'
end
in double
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). case xs of
[] then [];
x' & xs' then (f x') & map (f, xs')
end;
double match lambda xs. map ((lambda x. 2*x), xs)
in ...
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:
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|topics/week1_kapulet_advanced#funct-declarations]]. 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|/topics/week1_kapulet_advanced#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