ECS 140A:

Homework Assignment 3

For each of the following problems, you are to provide two

solutions: one using the pattern-matching techniques discussed

in Episode 18, and one not using those pattern-matching

techniques (i.e., the ones we discussed before talking about

pattern-matching over lists). For example, the two solutions

for myappend might look like:

— without pattern matching

myappend list1 list2

| list1 == [] = list2

| otherwise = (head list1):(myappend (tail list1) list2)

— with pattern matching

myappend_pm [] list2 = list2

myappend_pm (x:xs) list2 = x:(myappend_pm xs list2)

Note that I have ended the name of the pattern-matching solution

with “_pm”. Please do the same for all the pattern-matching

solutions that you submit for this assignment.

To implement the functions below, you may use only the following

list operations — ‘:’, ‘head’, ‘tail’, ‘null’, and ‘elem’ .

(Note that there are pattern matching equivalents for most of

these.) Do not resort to just giving a new name to an existing

Haskell function that already does what we want your function to

do. So, for example

myappend inlist1 inlist2 = inlist1 ++ inlist2

wouldn’t get you any points.

Please make sure you name your functions with the names provided

below (with and without the _pm suffix). Also, include type

declarations with your functions. It will be good practice.

Submit your solutions as a single file named “hw3.hs”.

Grading will be on a 3-point scale for each solution (7 problems

x 2 solutions per problem x 3 points maximum per solution = 42

points maximum).

And now, here are your homework problems:

1) myremoveduplicates

myremoveduplicates “abacad” => “bcad”

myremoveduplicates [3,2,1,3,2,2,1,1] => [3,2,1]

2) myintersection

For this function, the order of the elements in the list

returned by the function doesn’t matter. Also, if the arguments

to the function have duplicate elements in the list, then the

result of the intersection is unspecified.

myintersection “abc” “bcd” => “bc”

myintersection [3,4,2,1] [5,4,1,6,2] => [4,2,1]

myintersection [] [1,2,3] => []

myintersection “abc” “” => “”

3) mynthtail

mynthtail 0 “abcd” => “abcd”

mynthtail 1 “abcd” => “bcd”

mynthtail 2 “abcd” => “cd”

mynthtail 3 “abcd” => “d”

mynthtail 4 “abcd” => “”

mynthtail 2 [1, 2, 3, 4] => [3,4]

mynthtail 4 [1, 2, 3, 4] => []

4) mylast

mylast “” => “”

mylast “b” => “b”

mylast “abcd” => “d”

mylast [1, 2, 3, 4] => [4]

mylast [] => []

5) myreverse

There’s a simple but inefficient solution to this problem, and a

much more efficient solution. For full credit, provide the more

efficient solution.

myreverse “” => “”

myreverse “abc” => “cba”

myreverse [1, 2, 3] => [3, 2, 1]

myreverse [] => []

6) myreplaceall

myreplaceall 3 7 [7,0,7,1,7,2,7] => [3,0,3,1,3,2,3]

myreplaceall ‘x’ ‘a’ “” => “”

myreplaceall ‘x’ ‘a’ “abacad” => “xbxcxd”

7) myordered

myordered [] => True

myordered [1] => True

myordered [1,2] => True

myordered [1,1] => True

myordered [2,1] => False

myordered “abcdefg” => True

myordered “ba” => False