1. Define a function `zero?` that expects a single number as an argument, and returns `'true` if that number is `0`, else returns `'false`. Your solution should have a form something like this:
let
zero? match lambda x. FILL_IN_THIS_PART
in zero?
You can use the `if...then...else` construction if you like, but it will make it easier to generalize to later problems if you use the `case EXPRESSION of PATTERN1 then RESULT1; PATTERN2 then RESULT2; ... end` construction instead.
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`. Here your solution should have a form something like this:
let
empty? match lambda xs. case xs of
FILL_IN_THIS_PART
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.)
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 []
Your solution should have a form something like this:
letrec
drop match lambda (n, xs). FILL_IN_THIS_PART
in drop
What is the relation between `tail` and `drop`?
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]
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], [])
Here's a way to answer this problem making use of your answers to previous questions:
letrec
drop match ... ; # as in problem 4
take match ... ; # as in problem 5
split match lambda (n, xs). let
ys = take (n, xs);
zs = drop (n, xs)
in (ys, zs)
in split
However, we want you to instead write this function from scratch.
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`. For example, helping ourself to a function `odd?` that works as you'd expect:
filter (odd?, [11, 12, 13, 14]) # evaluates to [11, 13]
filter (odd?, [11]) # evaluates to [11]
filter (odd?, [12, 14]) # evaluates to []
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])
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. For example:
double [10, 20, 30] # evaluates to [20, 40, 60]
double [] # evaluates to []
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). FILL_IN_THIS_PART;
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]
###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 `&`.
* 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 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])`.
* 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`.
* 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`.
* Write a function `reverse` that returns the reverse of a sequence. Thus, `reverse [1, 2, 3, 4]` should evaluate to `[4, 3, 2, 1]`.