Your solution should have a form something like this:
- let
+ letrec
drop match lambda (n, xs). FILL_IN_THIS_PART
in drop
Here's a way to answer this problem making use of your answers to previous questions:
- let
+ letrec
drop match ... ; # as in problem 4
take match ... ; # as in problem 5
split match lambda (n, xs). let
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:
- let
+ letrec
map match lambda (f, xs). FILL_IN_THIS_PART;
double match lambda xs. map ((lambda x. 2*x), xs)
in ...
map2 ((lambda (x,y). 10*x + y), [1, 2, 3], [4, 5, 6]) # evaluates to [14, 25, 36]
-EXTRA CREDIT PROBLEMS
+###Extra credit problems###
+
+<OL start=12>
+<li>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 `&`.
+
+<li>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:
+
+ g x1 # evaluates to [y1, z1]
+ g x2 # evaluates to [y2, z2]
+ g x3 # evaluates to [y3, z3]
+
+Then `unmap2 (g, [x1, x2, x3])` should evaluate to `([y1, y2, y3], [z1, z2, z3])`.
+
+<li>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 list that are less than `10`.
+
+<li>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 list that are less than `10`.
-*Will post shortly*
+<li>Write a function `reverse` that returns the reverse of a sequence. Thus, `reverse [1, 2, 3, 4]` should evaluate to `[4, 3, 2, 1]`.
-<!-- take_while, drop_while, split_while -->
+</ol>
-<!-- unmap2 (g, xs) where g x \mapsto (y,z), and unmap2 (g, [x1, x2, x3]) \mapsto ([y1, y2, y3], [z1, z2, z3]) -->