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.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?`

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.)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`

?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]`

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. The method displayed above would traverse the sequence (at least, its first

`n`

members)*twice*. Can you do it in a way that only traverses (that part of the) sequence once?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 []`

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])`

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 []`

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 ...`

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 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])`

.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`

.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`

.Write a function

`reverse`

that returns the reverse of a sequence. Thus,`reverse [1, 2, 3, 4]`

should evaluate to`[4, 3, 2, 1]`

.