CSE 130 Final, Spring 2018
Nadia Polikarpova June 11, 2018
NAME ____________________________________ SID ____________________________________
You have 180 minutes to complete this exam.
Where limits are given, write no more than the amount specified.
You may refer to a double-sided cheat sheet, but no electronic materials.
Questions marked with * are difficult; we recommend solving them last.
Avoid seeing anyone elses work or allowing yours to be seen.
Do not communicate with anyone but an exam proctor.
If you have a question, raise your hand.
Good luck!
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Q1: Lambda Calculus: Sets [20 pts]
In this question you will implement sets of natural numbers in -calculus. Your set data structure has to support the following four operations:
EMPTY INSERT n s HAS s n INTERSECT s1 s2
| |
| |
The empty set
Set that contains the number n
and all elements of the set s
Does set s contain number n?
Set that contains all the elements
common to s1 and s2
You can use any function defined in Appendix I (at the end of the exam). Your implementation must satisfy the following test cases:
let S012 = INSERT ZERO (INSERT ONE (INSERT TWO EMPTY)) let S234 = INSERT TWO (INSERT THREE (INSERT FOUR EMPTY))
eval empty :
HAS EMPTY ZERO
=~> FALSE
eval insert_0 :
HAS S012 ZERO
=~> TRUE
eval insert_1 :
HAS S012 THREE
=~> FALSE
eval intersect_0 :
HAS (INTERSECT S012 S234) TWO
=~> TRUE
eval intersect_1 :
HAS (INTERSECT S012 S234) THREE
=~> FALSE
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1.1 Empty set [5 pts]
let EMPTY = _______________________________________________________ 1.2 Insert an element [5 pts]
let INSERT = ______________________________________________________ 1.3 Membership [5 pts]
let HAS = _________________________________________________________ 1.4 Set intersection [5 pts]
let INTERSECT = ___________________________________________________
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Q2: Datatypes and Recursion: Decision Trees [60 pts]
A binary decision tree (BDT) is an alternative representation of a Boolean formula. In a BDT, each leaf is labeled with True or False, and each internal node is labeled with a variable, and represents branching on the value of that variable.
False
T
xx TF
yy F
zz TF
True
True
True
False
True
False
False
Figure 1: (left) A BDT representation of the formula x (y z). (right) Its evaluation with x = True, y = False, z = True
Figure 1 (left) shows one possible BDT representation of the Boolean formula x (y z). To evaluate a BDT, we start at the root; in each internal node, we descend into the left sub-tree if the nodes variable is True, and into the right sub-tree if the nodes variable is False); the leaf we end up in gives the value of the formula. Figure 1 (right) depicts the evaluation of our example BDT with the following variable values: x = True, y = False, z = True.
In this question, you will implement several Haskell functions that operate on BDTs. We will represent BDTs using the following datatype:
data BDT = Leaf Bool | Node Id BDT BDT
where Id is just a synonym for strings:
type Id = String
We will also use the type Env to represent an environment, i.e. a mapping from variable names to Boolean values:
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type Env = [(Id, Bool)]
Your implementation can rely on the following function to look up the value
of a variable in an environment:
lookup :: Id -> Env -> Bool
Besides lookup, your implementations can use:
any library functions on Booleans; for example: not, (&&), (||), (==) any library functions on strings; for example: (==), (<), (>)
2.1 Evaluation [10 pts]
Implement the function eval, which evaluates a BDT in a given environment. You can assume that the environment contains all variables of the BDT.
Your implementation must satisfy the following test cases, where env = [(x,True), (y,False), (z,True)]:
eval env (Leaf False)
==> False
eval env (Node x (Leaf True) (Leaf False))
==> True
eval env (Node x (Node y (Leaf True) (Leaf False)) (Leaf False))
==> False
eval :: Env -> BDT -> Bool
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2.2 Negation [15 pts]
The cool thing about decision trees is that you can perform logical operations (negation, conjunction, and disjunction) directly on the trees, without having
to convert them back into a traditional formula representation.
Implement the function tNot, which returns a BDT that represents the negation of a given BDT.
Your implementation must satisfy the following test case for any BDT t and any environment env:
eval env (tNot t) == not (eval env t)
==> True
tNot :: BDT -> BDT
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2.3 Conjunction [15 pts]
Implement the function tAnd that computes a conjunction of two BDTs. Your implementation must satisfy the following test case for any two BDTs
tl and tr, and any environment env:
eval env (tAnd tl tr) == (eval env tl && eval env tr)
==> True
Its okay if the resulting BDT has duplicate variables. For example, the
simplest solution will satisfy the following test case (depicted on Figure 2):
let t = Node x (Leaf True) (Leaf False) in tAnd t t
==> Node x (Node x (Leaf True) (Leaf False)) (Leaf False)
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xxx
tAnd ==>
x
True
False
True
False
False
True
False
Figure 2: A test case for tAnd
tAnd :: BDT -> BDT-> BDT
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2.4 Ordered BDTs* [20 pts]
The simple implementation of tAnd from section 3.3 can cause the BDT to have duplicate variables, which makes the tree less compact and slower to evaluate. One way to eliminate this redundancy is to enforce ordering on all the variables in the tree, such that the variable in each node is strictly less
(lexicographically) than all variables in its sub-trees.
For example, the BDT in Figure 1 is ordered, because x < y and y < z both hold. In contrast, the BDT Figure 2 is not ordered, because x < x doesnt hold.Implement the function tAndOrd that computes a conjunction of two ordered BDTs, and returns an ordered BDT.Your implementation must satisfy the following test cases: 7tAndOrd (Node “x” (Leaf True) (Leaf False))(Node “x” (Leaf True) (Leaf False))==> (Node x (Leaf True) (Leaf False))
tAndOrd (Node x (Leaf True) (Leaf False))
(Node y (Leaf True) (Leaf False))
==> (Node x
(Node y (Leaf True) (Leaf False))
(Leaf False))
tAndOrd (Node y (Leaf True) (Leaf False))
(Node x (Leaf True) (Leaf False))
==> (Node x
(Node y (Leaf True) (Leaf False))
(Leaf False))
tAndOrd :: BDT -> BDT -> BDT
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Q3: Higher-Order Functions [40 pts]
Convert each of the given recursive functions into a function that always returns the same result but doesnt directly use recursion. Instead, your function can use the following higher-order functions from the standard library:
map :: (a -> b) -> [a] -> [b]
filter :: (a -> Bool) -> [a] -> [a] foldr ::(a->b->b)->b->[a]->b foldl ::(b->a->b)->b->[a]->b
Apart from these four functions, your implementation can only use the list constructors and library functions on integers (e.g. comparisons). You are allowed to introduce (non-recursive) auxiliary functions.
3.1 List reversal [5 pts]
reverse :: [a] -> [a]
reverse [] = []
reverse (x:xs) = reverse xs ++ [x]
Non-recursive version:
reverse :: [a] -> [a]
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3.2 Absolute values [10 pts]
absValues :: [Int] -> [Int]
absValues [] = []
absValues (x:xs)
|x<0 = (-x):(absValues xs) | otherwise = x :(absValues xs)9Non-recursive version:absValues :: [Int] -> [Int]
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3.3 Remove duplicates [15 pts]
dedup :: [Int] -> [Int]
dedup [] = []
dedup (x:xs) = x:(remove x (dedup xs))
where
remove x [] = []
remove x (y:ys)
| x == y=remove x ys
| otherwise = y:(remove x ys)
Non-recursive version:
dedup :: [Int] -> [Int]
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3.4 Insertion Sort* [20 pts]
sort :: [Int] -> [Int]
sort [] = []
sort (x:xs) = insert x (sort xs)
where
insert x [] = [x] insert x (y:ys) = if x <= ythen x:y:yselse y:(insert x ys)Non-recursive version:sort :: [Int] -> [Int]
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Q4: Semantics and Type Systems [30 pts]
In this question you will use the operational semantics and the type system of Nano2 (given in Appendix II at the end of the exam) to derive some reduction judgments E,e => E,e and typing judgments G |- e :: S.
A complete derivation should satisfy the following conditions:
all judgments in the derivation are complete
every rule (or axiom) application is labeled with the name of the rule all leaves are axioms
Here is an example of a complete reduction derivation:
[Add] –
E, 1 + 2 => E, 3
[Add-L] – E,(1+2)+(4+5) => E,3+(4+5)
4.1 Reduction 1 [10 points]
Complete the following reduction derivation, where
E = [f -> <[], x y -> x + y>]
[______] –
E,=> E,
[______] –
E,=> E,
[______] –
E, f 1 2=> E,
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4.2 Reduction 2 [10 points]
Complete the following reduction derivation (same E as in 4.1):
[______] –
=>
[______] –
E, <[], x y -> x + y> 1 2 =>
4.3 Typing 1 [10 points]
Complete the following typing derivation
[______]- [______]
[______]-
[______]-
[] |- x -> x + 5 ::
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4.4 Typing 2 [10 points]
Complete the following typing derivation where
G = [f : Int -> Int, id : forall a. a -> a]
[______]
G |-
[______] -[______]
G |-G |-
[______]-
G |- id f ::
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Q5: Prolog: Selection sort [30 pts]
In this question, we will implement Selection sort in Prolog. As a reminder, this algorithm sorts a list by repeatedly extracting the minimum element from the input list and appending it to the output list.
Unless otherwise stated, your solution cannot use any library func- tions/predicates or introduce auxiliary predicates.
5.1 Insert [10 points]
Write a Prolog predicate insert(X, Ys, Zs) which is true whenever Zs is the result of inserting the element X into Ys at some position.
Your implementation should satisfy the following test cases
?- insert(1, [2,3], Zs). Zs = [1,2,3];
Zs = [2,1,3];
Zs = [2,3,1];
false.
?- insert(1, Ys, [1,2,1]). Ys = [2,1];
Ys = [1,2];
false.
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5.2 Minimum element [10 points]
Write a Prolog predicate list_min(Acc, Xs, Min) which is true when Min is the minimum between the number Acc and the smallest element of list Xs
(if non-empty).
In this problem, you can use the built-in function min(X,Y) that computes the minimum of two numbers.
Your implementation should satisfy the following test cases
?- list_min(1, [], X). X = 1; false.
?- list_min(1, [3,2], X). X = 1; false.
?- list_min(2, [3,1], X). X = 1; false.
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5.3 Selection Sort [10 points]
Write a Prolog predicate selection_sort(Xs, Ys) which is true when the list Ys contains the same elements as Xs but in ascending order.
Your solution can use the predicates insert and list_min defined in 5.1 and 5.2.
Your implementation should satisfy the following test cases (when queried for the first solution only).
?- selection_sort([3,2,4,1], Ys). Ys = [1,2,3,4].
?- selection_sort([1,2,1,2], Ys). Ys = [1,1,2,2].
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Appendix I: Lambda Calculus Cheat Sheet
Here is a list of definitions you may find useful for Q2
Booleans
letTRUE let FALSE let ITE let NOT let AND let OR
=xy->x
= x y -> y
=bxy-> bxy =bxy-> byx =b1b2-> ITE b1 b2 FALSE =b1b2-> ITE b1 TRUE b2
Pairs
letPAIR =xyb->bxy let FST = p -> p TRUE let SND = p -> p FALSE
Numbers
letZERO =fx->x let ONE = f x -> f let TWO = f x -> f let THREE = f x -> f
x
(f x)
(f (f x))
Arithmetic
let INC let ADD let MUL let ISZ let EQL
=
fx->f(nfx)
=
m->nINCm
=
m->n(ADDm)ZERO =
->n(z->FALSE)TRUE =
m->AND(ISZ(SUBnm))(ISZ(SUBmn))
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Appendix II: Syntax and Semantics of Nano2
Expression syntax:
e ::= n | x | e1 + e2 | let x = e1 in e2 | x -> e | e1 e2 Operational semantics:
[Var] [Add]
[Add-L]
[Add-R]
[Let]
E,x=>E,E[x] ifxindom(E) E,n1+n2=>E,n wheren==n1+n2
E, e1 => E, e1
E, e1 + e2 => E, e1 + e2
E, e2 => E, e2
E, n1 + e2 => E, n1 + e2
E,letx=vine2=>E[x->v],e2 E, e1 => E, e1
[Let-Def]
E, let x = e1 in e2 => E, let x = e1 in e2 [Abs] E, x->e=>E,
[App] E,
E, e1 => E, e1
[App-L] –
E, e1 e2 => E, e1 e2
E, e => E, e
[App-R]
E, v e => E, v e
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Syntax of types:
T ::= Int | T1 -> T2 | a
S ::= T | forall a . S
Typing rules:
[T-Num]G |- n :: Int
G |- e1 :: Int
[T-Add]-
G |- e2 :: Int
G |- e1 + e2 :: Int
[T-Var] G|-x::S ifx:SinG
G, x:T1 |- e :: T2
[T-Abs]
G |- x -> e :: T1 -> T2
G |- e1 :: T1 -> T2 G |- e2 :: T1
[T-App]
G |- e1 e2 :: T2
G |- e1 :: S G, x:S |- e2 :: T
[T-Let]
G |- let x = e1 in e2 :: T
G |- e :: forall a . S
[T-Inst] –
G |- e :: [a / T] S
G |- e :: S
[T-Gen] – if not (a in FTV(G))
G |- e :: forall a . S
Here n N is natural number, v Val is a value, x Var is a variable, e Expr is an expression, E Var Val is an environment, a TVar is a type variable, T Type is a type, S Poly is a type scheme (a poly-type), G Var Poly is a type environment (a context).
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