CS计算机代考程序代写 Lambda Calculus Haskell CSE 130 Midterm, Spring 2018

CSE 130 Midterm, Spring 2018
Nadia Polikarpova May 4, 2018
NAME ____________________________________ SID ____________________________________
• You have 50 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 ma- terials.
• Questions marked with * are difficult; we recommend solving them last.
• Avoid seeing anyone else’s 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!
1

Part I. Lambda Calculus [60 pts] Q1: Reductions [20 pts]
For each λ-term below, circle all valid reductions of that term. It is possible that none, some, or all of the listed reductions are valid. Reminder:
• =a> stands for an α-step (α-renaming)
• =b> stands for a β-step (β-reduction)
• =*> stands for a sequence of zero or more steps, where each step is
either an α-step or a β-step 1.1 [5 pts]
(\x -> x (\x -> x)) (f x)
(A) =b>fx(\x->x)
(B) =b>fx(\x->fx)
(C) =b>fx(\fx->fx)
(D) =a>(\y->y(\x->x))(fy) (E) =a>(\x->x(\y->y))(fx)
1.2 [5 pts]
\x -> (\y z -> x y) (\x -> x)
(A) =a>\x->(\yx->xx)(\x->x) (B) =a>\a->(\yz->ay)(\x->a) (C) =*>\a->(\yz->ay)(\a->a) (D) =b>\xz->x(\x->x)
(E) =b>\yz->(\x->x)y
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1.3 [5 pts]
(\f g x -> f (g x)) (\x -> g x) (\z -> z)
(A) =b>(\fgx->f(gx))(g(\z->z))
(B) =b>(\gx->(\x->gx)(gx))(\z->z) (C) =*>\x->gx
(D) =*>\y->gy
(E) =*>\x->fx
1.4 [5 pts]
(\x y -> \b u v -> b v u) (\x y -> y) (\x y -> y) (\x y -> x) (A) =b>(\y->\buv->bvu)(\xy->y)(\xy->x)
(B) =b>(\y->\buv->bvu)(\xy->y)(\xy->y)
(C) =*>(\buv->bvu)(\xy->x)
(D) =*>\xy->y
(E) =b>\y->(\xy->y)(\xy->y)(\xy->x)
Q2: Lists [40 pts]
We can encode lists in λ-calculus by representing • an empty list as FALSE
• a non-empty list as a PAIR of its head and tail For example, the list [1,2,3] would be represented as:
PAIR ONE (PAIR TWO (PAIR THREE FALSE))
You can find the definitions of all variables used in this question in the
“Lambda Calculus Cheat Sheet” at the end of this exam.
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2.1 Repeat [10 pts]
Implement the function REPEAT, which, given a Church numeral n and any value x, returns a list with n copies of x. You can use any function defined in the “Lambda Calculus Cheat Sheet”.
let REPEAT = ___________________________________________________ The function should satisfy the following test cases:
eval repeat1 :
REPEAT ZERO x
=~> FALSE
eval repeat2 :
REPEAT TWO ONE
=~> PAIR ONE (PAIR ONE FALSE)
2.2 Empty* [20 pts]
Implement the function EMPTY, which, given a list represented as defined above, determines whether the list is empty. You can use any function defined in the “Lambda Calculus Cheat Sheet”.
let EMPTY = ___________________________________________________ The function should satisfy the following test cases:
eval empty1 :
EMPTY FALSE
=~> TRUE
eval repeat2 :
EMPTY (PAIR ZERO FALSE)
=~> FALSE
4

2.3 Length [10 pts]
Recall that recursion can be implemented in λ-calculus using fixpoint combi- nators like this one:
letFIX =\stp->(\x->stp(xx))(\x->stp(xx))
Using the fixpoint combinator, implement the function LEN, which, given a list represented as defined above, computes its length. You can use FIX, EMPTY from 2.2, and any function defined in the “Lambda Calculus Cheat Sheet”.
let LEN = ________________________________________________________ The function should satisfy the following test cases:
eval len1 :
LEN FALSE
=~> ZERO
eval repeat2 :
LEN (PAIR ONE (PAIR TWO FALSE))
=~> TWO
5

Part II. Datatypes and Recursion [50 pts] Q3: Binary Search Trees [50 pts]
Recall that a binary search tree (BST) is a binary tree, where every node stores an integer key x, and the keys are ordered such that all keys in the node’s left sub-tree are strictly less than x, and all keys in the node’s right sub-tree are strictly greater than x. An example BST is depicted below.
Figure 1: Binary Search Tree
In this question, you will implement several Haskell functions that operate on BSTs. We will represent BSTs using the following datatype:
data Tree = Empty | Node Int Tree Tree
In your implementations, you can use any library functions on integers (arith-
metic operators and comparisons), and the append function on lists (++). 3.1 Size [5 pts]
Implement the fuction size that computes the size of a tree. size :: Tree -> Int
_______________________________________________________________
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Your implementation must satisfy the following test cases 6

size Empty
==> 0
size (Node 8 (Node 3 Empty Empty) (Node 10 Empty Empty))
==> 3
3.2 Insert [10 pts]
Implement the fuction insert that inserts a key into a BST. More precisely, insert x t may assume that t is a BST (i.e., elements are ordered as defined above) and must ensure that its result is a BST that contains a key x.
insert :: Int -> Tree -> Tree
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Your implementation must satisfy the following test cases
insert 8 Empty
==> Node 8 Empty Empty
insert 6 (Node 8 (Node 3 Empty Empty) (Node 10 Empty Empty))
==> Node 8
(Node 3 Empty (Node 6 Empty Empty))
(Node 10 Empty Empty)
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3.3 Sort [15 pts]
Implement the helper fuctions fromList and toList so that the sort func- tion below sorts a list of integers (you can assume the input list has no duplcate elements). You implementation can use the function insert from question 3.2.
sort :: [Int] -> [Int]
sort xs = toList (fromList xs)
where
fromList :: [Int] -> Tree
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toList :: Tree -> [Int]
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3.4 Tail-recursive size* [20 pts]
Re-implement the function size from question 3.1 so that it’s tail-recursive. Hint: introduce an auxiliary function with extra arguments that keep track
of what’s already done and what is still left to do.
8

size :: Tree -> Int
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9

Lambda Calculus Cheat Sheet
Here is a list of definitions you may find useful for Q2
— Booleans ——————————–
letTRUE =\xy->x
let FALSE = \x y -> y
letITE =\bxy->bxy
letNOT =\bxy->byx
letAND =\b1b2->ITEb1b2FALSE letOR =\b1b2->ITEb1TRUEb2
— Pairs ———————————–
letPAIR =\xyb->bxy letFST =\p ->pTRUE let SND = \p -> p FALSE
— Numbers ———————————
letZERO =\fx->x
letONE =\fx->fx letTWO =\fx->f(fx) let THREE = \f x -> f (f (f x))
— Arithmetic ——————————
letINC letADD letMUL let ISZ
=\nfx->f(nfx) =\nm->nINCm =\nm->n(ADDm)ZERO =\n->n(\z->FALSE)TRUE
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