+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?
+
+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 ...
+
+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 match lambda (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 match lambda (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 match lambda (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 match lambda (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 match lambda 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 match lambda (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 match lambda (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
+