ECS 140A:  Homework Assignment 3

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