COMP90048 Declarative Programming

The University of Melbourne School of Computing and Information Systems
Semester 1, 2020 Sample Assessment
COMP90048 Declarative Programming
Sample Answers Included
Reading Time: 15 minutes Total marks for this paper: 100 Writing Time: 2 hours
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Semester 1, 2020 Sample Assessment Declarative Programming
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Declarative Programming Semester 1, 2020 Sample Assessment
Question 1 [12 marks]
For each of the following Haskell expressions, give its type (which may be a function type, may include type variables, and may include type class constraints) or indicate that it represents a type error. You need not write anything other than the type, or that it is an error.
(<):: Orda=>a->a->Bool
(b) map (+3)
Num a => [a] -> [a]
(d) Nothing Maybe a
(e) zip [True,True,False] [a] -> [(Bool, a)]
(f) flip filter “hello”
(Char -> Bool) -> [Char]
Question 2
(Char -> Bool) -> [Char]
(a -> b -> b) -> b -> [a] -> b —or— Foldable t => (a -> b -> b) -> b
-> t a -> b
For each of the following Haskell expressions, give its value, or explain why it will produce an error or fail to terminate. Assume the Data.List library is loaded, which defines the sort function.
(a) map (length < 3) [[1],[1,2,3]] Type error W (b) filter (not.(==3)) [1,2,3] (c) let e = head [] in 3 3W [12 marks] (d) map fst $ filter snd $ zip "abcde" [True,True,False,True] "abd" Page 3 of 8 Semester 1, 2020 Sample Assessment Declarative Programming (e) head $ sort $ zip [3,0,0,2,0] $ reverse [9,0,0,4,8] (0,0) (f) map snd $ sort $ zip "decl" [1..] [3,1,2,4] Page 4 of 8 Declarative Programming Semester 1, 2020 Sample Assessment Question 3 Consider the following Haskell type for ternary trees: data Ttree t = Nil | Node3 t (Ttree t) (Ttree t) (Ttree t) [30 marks] Suppose we have a Ttree of Doubles and we want a function to find the average of the numbers in the tree. Write a Haskell function which performs this task. If the Ttree is empty, your function should return 0.0. Include type declarations for all your functions. To obtain maximum marks, your code should use a single traversal over the tree and have O(N) worst case time complexity. Sample Answer to Question 3 W Solution using custom fold function for Ttrees ttreeAverage1 :: Ttree Double -> Double
ttreeAverage1 Nil = 0.0
ttreeAverage1 tt =
let (sum,count) = foldTtree plusCount (0.0,0.0) tt
in sum / count
plusCount :: Double -> (Double, Double) -> (Double, Double)
plusCount n (sum,count) = (sum+n, count+1)
foldTtree :: (a -> b -> b) -> b -> Ttree a -> b
foldTtree _ acc Nil = acc
foldTtree f acc (Node3 n left mid right)
= let acc1 = foldTtree f acc right
acc2 = foldTtree f acc1 mid
acc3 = foldTtree f acc2 left
in f n acc3
Alternative full-credit solution
ttreeAverage2 :: Ttree Double -> Double
ttreeAverage2 Nil = 0.0
ttreeAverage2 tt =
let (sum,count) = ttAverage¡¯ tt
in sum / count
ttAverage¡¯ :: Ttree Double -> (Double,Double)
ttAverage¡¯ Nil = (0.0,0.0)
ttAverage¡¯ (Node3 n left mid right) =
let (suml,countl) = ttAverage¡¯ left
(summ,countm) = ttAverage¡¯ mid
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Semester 1, 2020 Sample Assessment Declarative Programming
(sumr,countr) = ttAverage¡¯ right
in (n+suml+summ+sumr, 1+countl+countm+countr)
Partial credit solution using two passes
— Partial credit, two-pass solution. This answer would still get
— most of the marks.
ttreeAverage3 :: Ttree Double -> Double
ttreeAverage3 Nil = 0.0
ttreeAverage3 tt = ttSum tt / ttCount tt
ttSum :: Ttree Double -> Double
ttSum Nil = 0.0
ttSum (Node3 n left mid right) =
n + ttSum left + ttSum mid + ttSum right
ttCount :: Ttree Double -> Double
ttCount Nil = 0.0
ttCount (Node3 n left mid right) =
1 + ttCount left + ttCount mid + ttCount right
Question 4
[10 marks]
Give the formal semantics (meaning) of the following Prolog program. Recall that the formal semantics of a logic program is the set of ground unit clauses that would give the same answers to all queries as the program itself. English descriptions of the meanings of the programs will receive no credit.
p(a). p(b).
q(X,Y) :- p(X), p(Y).
r(a,c). r(d,b).
s(X,Y) :- q(X,Y), r(X,_), r(_,Y).
Sample Answer to Question 4
p(a). p(b).
q(a,a). q(a,b).
r(a,c). r(d,b).
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Declarative Programming Semester 1, 2020 Sample Assessment
Question 5 [20 marks]
For this question, we will represent a set of integers as a binary tree in Prolog, using the atom empty to represent an empty tree or node, and tree(L,N,R) to represent a node with label N (an integer), and left and right subtrees L and R. Naturally, we also insist that N be strictly larger than any label in L and strictly smaller than any in R. We do not require that the tree be balanced. For example,
tree(tree(tree(empty, 1, empty),
tree(empty, 3, tree(empty, 4, empty))),
tree(tree(tree(empty,6,empty),
tree(empty,8,empty)),
tree(empty, 10, empty)))
is one possible representation of the set of numbers from 1 to 10. It might be visualized as
Write a predicate intset insert(N, Set0, Set) such that Set is the same as Set0, except thatNisamemberofSet,butmayormaynotbeamemberofSet0. Thatis,eitherNisa member of Set0 and Set = Set0, or N is not a member of Set0 and is a member of Set, and other than that, Set is the same as Set0. This predicate must work as long as N is bound to an integer and Set0 is ground.
Hint: Prolog¡¯s arithmetic comparison operators are <, >, =< (not <=), and >=. You can also use = and \= for equality and disequality.
Sample Answer to Question 5 W intset_insert(N, empty, tree(empty,N,empty)).
intset_insert(N, tree(Left,Val,Right), Result) :-
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Semester 1, 2020 Sample Assessment
Declarative Programming
-> Result = tree(Left,Val,Right)
-> Result = tree(Left1,Val,Right),
intset_insert(N, Left, Left1)
; Result = tree(Left,Val,Right1),
intset_insert(N, Right, Right1)
Question 6
[16 marks]
Following is a definition of a Prolog predicate to compute the sum of a list of numbers.
sumlist([], 0).
sumlist([N|Ns], Sum) :-
sumlist(Ns, Sum0),
Sum is N + Sum0.
Fill in the blanks in the following transformation of this code to be tail recursive.
sumlist(List, Sum) :- sumlist(List, 0, Sum).
sumlist( [], Sum, Sum ). sVumlist([N|Ns], Sum0, Sum) :-
Sum1 is Sum0 + N , sumlist( Ns, Sum1, Sum ).
¡ª End of Exam ¡ª
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