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## Description

Introduction

This homework will focus on writing functions on lists and proving properties of them. This homework is longer and harder than the previous two: start early!

1.1 Getting The Homework Assignment

The starter files for the homework assignment have been distributed through our git repos- itory, as usual.

1.2 Submitting The Homework Assignment

Submissions will be handled through Autolab, at

https://autolab.cs.cmu.edu

In preparation for submission, your hw/03 directory should contain a file named exactly

hw03.pdf containing your written solutions to the homework.

To submit your solutions, run make from the hw/03 directory (that contains a code folder and a file hw03.pdf). This should produce a file hw03.tar, containing the files that should be handed in for this homework assignment. Open the Autolab web site, find the page for this assignment, and submit your hw03.tar file via the “Handin your work” link.

The Autolab handin script does some basic checks on your submission: making sure that the file names are correct; making sure that no files are missing; making sure that your code compiles cleanly. Note that the handin script is not a grading script—a timely submission that passes the handin script will be graded, but will not necessarily receive full credit. You can view the results of the handin script by clicking the number corresponding to the “check” section of your latest handin on the “Handin History” page. If this number is −10.0, your submission failed the check script; if it is 0.0, it passed.

Remember that your written solutions must be submitted in PDF format—we do not accept MS Word files or other formats.

Your hw03.sml file must contain all the code that you want to have graded for this assignment, and must compile cleanly. If you have a function that happens to be named the same as one of the required functions but does not have the required type, it will not be graded.

1.4 Methodology

You must use the five step methodology discussed in class for writing functions, for every

function you write in this assignment. Recall the five step methodology:

- In the first line of comments, write the name and type of the function.

- In the second line of comments, specify via a REQUIRES clause any assumptions about the arguments passed to the function.

- In the third line of comments, specify via an ENSURES clause what the function com- putes (what it returns).

- Implement the function.

- Provide testcases, generally in the format

val <return value> = <function> <argument value>. For example, for the factorial function presented in lecture:

(* fact : int -> int

* REQUIRES: n >= 0

* ENSURES: fact(n) ==> n!

*)

fun fact (0 : int) : int = 1

| fact (n : int) : int = n * fact(n-1) (* Tests: *)

val 1 = fact 0

val 720 = fact 6

2 Zippidy Doo Da

It’s often convenient to take a pair of lists and make one list of pairs from it. For instance, if we have the lists

[“a”, “b”] and [5, 1, 2, 1]

we might be interested in the list

[(“a”, 5), (“b”, 1)]

Task 2.1 (5%). Write the function

zip : string list * int list -> (string * int) list

that performs the transformation of pairing the nth element from the first list with the nth element of the second list. If your function is applied to a pair of lists of different lengths, the length of the returned list should be the minimum of the lengths of the argument lists.

You should ensure that zip is a total function (but you do not need to formally prove this fact).

Some terminology:

A function f : t -> t’ is total iff the expression f(x) reduces to a value for all values x :

t.

Task 2.2 (5%). Write the function

unzip : (string * int) list -> string list * int list

unzip does the opposite of zip, in the sense that it takes a list of pairs and returns a pair of lists, where the first list in the pair is the list of first elements and the second list is the list of second elements. You should ensure that unzip is a total function. You may assume this fact in your proofs but should cite it if used.

Task 2.3 (15%). Prove the following Theorem by induction on length of lists. 1.

Theorem 1. For all values l : (string * int) list, zip(unzip l) = l.

Hint: Look at the way we reasoned about eval decimal n = n in Lecture 3 and 4 notes.

Task 2.4 (5%). Prove or disprove Theorem 2.

Theorem 2. For all values l1 : string list and l2 : int list,

unzip(zip (l1,l2)) = (l1,l2)

3 Look And Say

3.1 Definition

If l is any list of integers, the look-and-say list of l is obtained by reading off adjacent groups of identical elements in l. For example, the look-and-say list of l = [2, 2, 2] would be read as “three twos”. For our implementation of look-and-say, this would be represented as [(3, 2)].

Similarly, the look-and-say list of

l = [1, 1, 2] is [(2, 1), (1, 2)]

because l is exactly “two ones, then one two.”

We will use the term run to mean a maximal length sublist of a list with all equal elements. For example,

are both runs of the list but

[1, 1, 1] and [5] [1, 1, 1, 5, 2]

[1, 1] and [5, 2] and [1, 2]

are not: [1, 1] is not maximal, [5, 2] has unequal elements, and [1, 2] is not a sublist.

You will now define a function lookSay that computes the look-and-say sequence of its argument using a helper function and a new pattern of recursion.

3.2 Implementation

Before defining the lookSay function, you will write a helper function runWith. runWith

will remove a “run” from the front of a list.

Task 3.1 (5%). Write the function runWith that satisfies the following spec:

runWith: int*int list -> int list * int list runWith (x, L) = (L1, L2)

where L = L1@L2

and every element of L1 is equal to x and L2 does not begin with x.

Note that L2 is either [ ] or has the form y::R where x = y. For example,

runWith(1,[1,2,3]) = ([1], [2,3]) runWith(1,[1,1,2,3]) = ([1,1], [2,3]) runWith(3,[1,2,3]) = ([], [1,2,3])

Note that you can use the function inteq in hw03.sml to compare integers for equality.

Task 3.2 (10%). Now, write the function lookSay : int list -> (int * int) list using this helper function.1

Task 3.3 (3%). Write the function

flatten : (int * int) list -> int list

that will “flatten out” the list of pairs into a flat list of integers.

For example,

flatten([(1, 2)]) = [1,2]

flatten([(1, 2), (3, 4), (5, 6)]) = [1,2,3,4,5,6]

3.3 Cultural Aside

Repeated applications of our “look and say” function, followed by our “flatten” function, results in a special sequence.

Below are the first 5 steps in the instance of this sequence beginning with [1]:

[1] [1,1] [2,1] [1,2,1,1]

[1,1,1,2,2,1] [3,1,2,2,1,1]

This sequence is noted by Conway’s theorem, which states that any element of this sequence will “decay” (by repeated applications of lookSay and flatten) into a “compound” made up of combinations of “primitive elements” (there are 92 of them, plus 2 infinite families) in

24 steps. If you are interested in this sequence, you may wish to consult [Conway(1987)] or other papers about the “look and say” operation.

1 Hint: The recursive call in the inductive case of lookSay will sometimes be on a list that is more than one element shorter.

4 Prefix-Sum

The prefix-sum of a list l is a list s where the ith index element of s is the sum of the first

i + 1 elements of l. For example,

prefixSum [] = []

prefixSum [1,2,3] = [1,3,6]

prefixSum [5,3,1] = [5,8,9]

Note that the first element of list is regarded as position 0.

Task 4.1 (5%). Implement the function

prefixSum : int list -> int list

that computes the prefix-sum. You must use the addToEach function provided, which adds an integer to each element of a list, and your solution must be in O(n2) but not in O(n). This implementation will be simple, but inefficient.

Task 4.2 (5%). Write a recurrence for the work of prefixSum, WprefixSum(n), where n is the length of the input list. Give a closed form for this recurrence. Argue that your closed form does indeed indicate that WprefixSum(n) is O(n2).

You may use variables k0, k1, . . . for constants. You should assume that addToEach is a linear time function: addToEach l evaluates to a value in kn steps where n is the length of l and k is some constant; your recurrence should involve the constant k.

In order to compute the prefix sum operation in linear time, we will use the technique of adding an additional argument: harder problems can be easier.

Task 4.3 (10%). Write the prefixSumHelp function that uses an additional argument to compute the prefix sum operation in linear time. You must determine what the additional argument should be. Once you have defined prefixSumHelp, use it to define the function

prefixSumFast : int list -> int list

that computes the prefix sum.

Task 4.4 (5%). Write a recurrence for the work of prefixSumFast, WprefixSumFast(n), where n is the length of the input list. Give a closed form for this recurrence. Argue that your closed form does indeed indicate that prefixSumFast is in O(n).

5 Sublist

When programming with lists, we often need to work with a segment of a larger list. For example, one might need to access only the last three elements of a list or only the middle element. Any such segment is called a sublist.

More formally: if L is any list, we say that S is a sublist of L starting at i if and only if there exist l1 and l2 such that

and

length l1 = i

For example, [1, 2] is a sublist of [1, 2, 3] starting at 0 because

[ ]@[1, 2]@[3] = [1, 2, 3] and length [ ] = 0

Task 5.1 (3%). The spec for a function sublist that computes sublists as defined above will have the form:

For all l:int list, i:int, k:int, if __ __ then there exists an S such that S is the sublist of l starting at i, and

length S = k and sublist(i, k, l) = S

The blank is called the preconditions, and represents assumptions about the input. Fill in the blank to complete this spec correctly. Note that this formal spec should very closely resemble any REQUIRES and ENSURES clauses on your function.

Task 5.2 (7%). Implement a function

sublist : int * int * int list -> int list

that meets the spec you gave above.

Because the spec has the form of an implication, in the body of sublist you should assume that whatever preconditions you required in Task 5.1 are met: if they are not, your function can do anything you want and still meet its spec!

Note that the definition above implies that we index lists from zero, so

sublist (0, 3, [1,2,3,4]) = [1,2,3]

The spec that you completed above is good because it closely mirrors the abstract notion of a sublist, but bad because it’s very stringent: any code calling sublist must ensure that the assumptions about the input hold or else it will fail. Since the exact mode of failure is not documented in the type or in the spec, this can produce behaviour that’s very hard to debug.

Sometimes, the caller will be able to prove that these assumptions hold because of other specification-level information. Other times, the information available at compile-time will not be enough to ensure that these assumptions are met. In these circumstances, you can use a run-time check to bridge the gap, which is something we will be able to implement once we see exceptions.

6 Subset sum

A multiset is a slight generalization of a set where elements can appear more than once. A submultiset of a multiset M is a multiset, all of whose elements are elements of M . To avoid too many awkward sentences, we will use the term subset to mean submultiset.

It follows from the definition that if U is a sub(multi)set of M , and some element x appears in U k times, then x appears in M at least k times. If M is any finite multiset of integers, the sum of M is

X x

x∈M

With these definitions, the multiset subset sum problem is answering the following question.

Let M be a finite multiset of integers and n a target value. Does there exists any subset U of M such that the sum of U is exactly n?

Consider the subset sum problem given by

M = {1, 2, 1, −6, 10} n = 4

The answer is “yes” because there exists a subset of M that sums to 4, specifically

U1 = {1, 1, 2}

It’s also yes because

U1 = {−6, 10}

sums to 4 and is a subset of M . However,

U3 = {2, 2}

is not a witness to the solution to this instance. While U3 sums to 4 and each of its elements occurs in M , it is not a subset of M because 2 occurs only once in M but twice in U2.

Representation You’ll implement three solutions to the subset sum problem. In all three, we represent multisets of integers as SML values of type int list, where the integers may be negative. You should think of these lists as just an enumeration of the elements of a particular multiset. The order that the elements appear in the list is not important.

6.1 Basic solution

Task 6.1 (10%). Write the function

subsetSum : int list * int -> bool

that returns true if and only if the input list has a subset that sums to the target number. As a convention, the empty list [ ] has a sum of 0. Start from the following useful fact: each element of the set is in the subset, or it isn’t.2

6.2 NP-completeness and certificates

Subset sum is an interesting problem because it is NP-complete. NP-completeness has to do with the time-complexity of algorithms, and is covered in more detail in courses like 15-251, but here’s the basic idea:

- A problem is in P if there is a polynomial-time algorithm for it—that is, an algorithm one whose work is in O(n), or O(n2), or O(n14 ), etc.
- A problem is in NP if an affirmative answer can be verified in polynomial time. Subset sum is in NP. Suppose that you’re presented with a multiset M , another multiset

U , and an integer n. You can easily check that the sum of U is actually n and that U is a

subset of M in polynomial time. This is exactly what the definition of NP requires.

This means we can write an implementation of subset sum which produces a certificate on affirmative instances of the problem—an easily-checked witness that the computed answer is correct. Negative instances of the problem—when there is no subset that sums to n—are not so easily checked.

You will now prove that subsetSum is in NP by implementing a certificate-generating version.

Task 6.2 (7%). Write the function

subsetSumCert : int list * int -> bool * int list

such that for all values M:int list and n:int, if M has a subset that sums to n,

subsetSumCert (M, n) = (true, U) where U is a subset of M which sums to n.

If no such subset exists, subsetSumCert (M, n) = (false, nil). 3

Task 6.3 (Extra Credit).The P = NP problem, one of the biggest open problems in computer science, asks whether there are polynomial-time algorithms for all of the problems in NP. Right now, there are problems in NP, such as subset sum, for which only exponential-time algorithms are known. However, it is known that subset sum is NP-complete, which means that if you could solve it in polynomial time, then you could solve all problems in NP in polynomial time, so P = NP. So, for extra credit, several million dollars, and a PhD, define in SML a function that solves the subset sum problem and has polynomial time work.

2 Hint: It’s easy to produce correct and unnecessarily complicated functions to compute subset sums. It’s almost certain that your solution will have O(2n ) work, so don’t try to optimize your code too much. There is a very clean way to write this in a few (5-10ish) elegant lines.

3 You’ll note that the empty list returned when a qualifying subset does not exist is superfluous; soon, we’ll cover a better way to handle these kinds of situations, called option types.

References

[Conway(1987)] J. Conway. The weird and wonderful chemistry of audioactive decay. In T. Cover and B. Gopinath, editors, Open Problems in Communication and Computation, pages 173–188. Springer-Verlag, 1987.