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1 Introduction
This homework will focus on applications of higher order functions, polymorphism, and userdefined datatypes.
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/05 directory should contain a file named exactly
hw05.pdf containing your written solutions to the homework.
To submit your solutions, run make from the hw/05 directory (that contains a code folder and a file hw05.pdf). This should produce a file hw05.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 hw05.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 (usually either 0.0 or 1.0) corresponding to the “check” section of your latest handin on the “Handin History” page. If this number is 0.0, your submission failed the check script; if it is 1.0, it passed.
Remember that your written solutions must be submitted in PDF format—we do not accept MS Word files or other formats.
Your hw05.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(n1) (* Tests: *)
val 1 = fact 0
val 720 = fact 6
2 Types and Polymorphism
In class we discussed typing rules. In particular:
 A function expression fn x => e has type t > t’ if and only if, by assuming that x
has type t, we can show that e has type t’.
 An application e1 e2 has type t’ if and only if there is a type t such that e1 has type
t > t’ and e2 has type t.
 An expression can be used at any instance of its most general type.
Task 2.1 (4%). Consider the following ML function declaration:
fun all (your, base) =
case your of
0 => base
 => “are belong to us” :: all(your – 1, base)
What is the most general type of all?
Task 2.2 (2%). Consider a different ML function:
fun funny (f, []) = 0
 funny (f, x::xs) = f(x, funny(f, xs))
What is the most general type of funny?
Task 2.3 (2%). Now consider the following function expression:
fn x => (fn y => x)
What is its most general type?
Task 2.4 (2%). Now look at this slightly different expression:
(fn x => (fn y => x)) “Hello, World!”
What is the most general type of this expression?
3 Bases
The digits we use to represent a number depend on the base we use. We usually write numbers in base ten, or using digits from 0 to 9. More generally, numbers in a particular base b can use digits between 0 and b − 1, inclusive. We can notate this base with a subscript, so the number fiftyfour is 5410 . In computer science we also commonly see base 2, or binary numbers, which count with 1s and 0s. For example, 102 = 210 , 112 = 310 , and so on.
We’ll be representing numbers in different bases as an int list of digits with the least
significant digit as the first element of the list, so 11002 is [0, 0, 1, 1].
3.1 Converting to an int
Formally, given a base b and a string of n digits dn dn−1 . . . d1, the numerical value of the string is
n
X bi−1di
i=1
For example, 3410 is 3 ∗ 101 + 4 ∗ 100 = 3 ∗ 10 + 4 ∗ 1 = 34.
We look at the same example in base 2: 1000102 = 1 ∗ 25 + 1 ∗ 21 = 1 ∗ 32 + 1 ∗ 2 = 34. Given
this formula, we can take any string of digits in a base and convert the digits to an int.
Task 3.1 (5%). Define the higherorder function
toInt : int > int list > int
such that for all b > 1 and all L : int list, if L is a list of base b digits, then
toInt b L is the corresponding integer value n. Note that toInt is higherorder, so toInt b should return a function from int list to int. For example,
val base2ToInt = toInt 2 val 2 = base2ToInt [0, 1]
3.2 Converting from an int
For any natural number n and a base b, we can convert n into base b with the following algorithm:
 If n = 0, then stop. You’re done.
 Find the remainder of n divided by b. Prepend this to our current list of digits.
 Do an integer division of n by b.
 Go back to step 1 using n = n div b.
For example, this is the process of converting 4210 to base 5.
42 mod 5 = 2 [42 div 5 = 8]
8 mod 5 = 3 [8 div 5 = 1]
1 mod 5 = 1 [1 div 5 = 0]
Reading from the bottom up, we get 4210 = 1325 , which we can confirm using the formula for converting to base 10. 1 ∗ 52 + 3 ∗ 5 + 2 = 25 + 15 + 2 = 42.
Task 3.2 (5%). Define the higher order function
toBase : int > int > int list
such that for all b > 1, n ≥ 0, toBase b n returns the representation of n in base b. Again, toBase is higherorder, so toBase b should return a function. For example,
val toBase3 = toBase 3 val [2, 1] = toBase3 5
Task 3.3 (5%). Define the higher order function
convert : int * int > int list > int list
such that for all b1, b2 > 1 and for all L : int list such that L is a list of base b1 digits, convert (b1, b2) L changes the representation of the input number from base b1 to base b2. We should have toInt b2 (convert(b1, b2) L) = toInt b1 L hold for convert. You may use toInt and toBase in your solution.
4 HigherOrder Functions
Recently we introduced several new language features: polymorphism, option types, and higherorder functions. In the next problems, you will write some simple functions using these new tools.
You’ll start by writing functions on vectors. Vectors of length n are essentially lists of numbers which indicate a direction and a magnitude in Rn . For example, vectors in R2 are line segments from the origin to the (x, y) point they contain. We will represent vectors in SML as real lists. Mathematically, then, a vector ha1, a2 , …, an i translates to [a1, a2, …, an ].
4.1 Dot product
Recall : To calculate the dot product of vectors ~a and ~b
~a · ~b = ha1 , a2, …, an i · hb1, b2, …, bn i = (a1 ∗ b1 ) + (a2 ∗ b2) + … + (an ∗ bn )
Task 4.1 (4%). Write the function
dotProduct : real list * real list > real
that calculates the dot product of two vectors of the same length. You solution should be nonrecursive and it should instead use higher order functions to accomplish its goal. Hint: You can use
zip: ’a list * ’b list > (’a * ’b) list
4.2 Magnitude
Recall : The magnitude of a vector ~a, denoted by ~a, is
q
~a = ha1, a2 , …, an i =
a2 + a2 + … + a2
1 2 n
Task 4.2 (4%). Write the function
magnitudeOfVector : real list > real
that calculates the magnitude of a given vector. You solution should be nonrecursive and should again use higher order functions. Note that you can use
Math.sqrt : real > real
to calculate the square root of a real number.
4.3 Angle
Now that we can calculate the dot product of two vectors and their magnitudes, we can also calculate the angle between them.
Recall : The angle between two vectors ~a and ~b is
~a · ~b !
θ = cos−1
Task 4.3 (5%). Write the function

~a ∗ ~b
angleBetweenVectors : real list * real list > real
that calculates the angle between given vectors. Please note that you can use
Math.acos : real > real
to calculate the inverse cosine of a real number.
4.4 Extract
Task 4.4 (7%). Write a function
fun extract (p : ’a > bool, l : ’a list) : (’a * ’a list) option =
such that
 If there is some element x of l for which p x = true, then extract(p,l) evaluates to
SOME(x,l’), where l’ is l without that particular x but unchanged otherwise.
 If for every element x of l, p x = false then extract(p,l) evaluates to NONE.
If there is more than one element satisfying the predicate in a particular argument list, it is your choice which to return.
For example:
extract(oddP , [2,3,4]) = SOME (3,[2,4])
extract(oddP , [2,4,6]) = NONE
extract(fn x => String.size x < 2 , [“aaa”,”b”,”bca”])
= SOME (“b”, [“aaa”, “bca”])
extract should be recursive. You should use this function when you implement Blocks
World below.
5 Blocks World
In artificial intelligence, planning is the task of figuring what an agent (a robot, that paperclip in Microsoft Word, your roommate, etc.) should do. One way to solve planning problems is to simulate the circumstances of the agent, so that you can simulate plans, and then search through potential plans for good ones.
A simple planning problem, which is often used to illustrate this idea, is blocks world. The idea is that there are a bunch of blocks on a table:
—  —  — 
A  B  C 
—  —  — 
————————
and a robotic hand. You can pick one block up with the hand:
/\
—
C
—
— —
A B
— —
————————
and place it back on the table or on another block:
—
C
—
— —
A B
— —
————————
Of course, you can’t put a block on one that already has something on it, so in the next two moves we can’t pick up B and then put it on A. A planning problem would be something like “starting with the blocks on the table, make the tower BCA”.
In this problem, you will represent blocks world in ML, so that you can simulate plans
(we won’t ask you to search for plans that achieve specific goals).
At the end of the problem, you’ll be able to interact with Blocks World as in the figure above. We’ve written all the input/output code for you, so you just need to do the interesting bits.
– playBlocks ();
Possible moves:
pickup <block> from table put <block> on table
pickup <block> from <block>
put <block> on <block>
quit
— — —
A B C
— — —
————————
Next move: pickup C from table
/\
—
C
—
— —
A B
— —
———————— Next move: put C on A
—
C
—
— —
A B
— —
————————
Figure 1: Sample Blocks World Interaction
5.1 Rules
We will model Blocks World as follows:
 There are three blocks, A, B, C .
 We will represent the state of the world as a list of facts. There are five kinds of facts:
– Block b is free (available to be picked up)
– Block a is on block b
– Block a is on the table
– The hand is empty
– The hand is holding block b
 At each step, there are four possible moves:
pickup <b> from table put <b> on table pickup <a> from <b> put <a> on <b>
These moves act as follows:
– pickup <a> from table
Before: a is free, and a is on the table, and the hand is empty. After: the hand holds a.
– put <b> on table
Before: the hand holds b.
After: the hand is empty, and b is on the table, and b is free.
– pickup <a> from <b>
Before: a is free, and a is on b, and the hand is empty. After: b is free, and the hand is holding a.
– put <a> on <b>
Before: the hand holds a, and and b free.
After: a is free, the hand is empty, and a is on b.
In these descriptions, the “before” facts must hold about the world for the move to be executed; after executing the move, the “before” facts no longer hold (e.g. after picking up a block, the hand is no longer empty), and the “after” facts holds.
5.2 Tasks
Task 5.1 (5%). First, we will need a function to extract many elements from a list. Write a function
extractMany : (’a * ’a > bool * ’a list * ’a list) > (’a list) option extractMany is polymorphic in the list’s element type, but it needs to test whether two list el
ements are equal. For this reason, extractMany takes an argument function eq:’a * ’a > bool
that can be used to test whether two values of type ’a are equal.
extractMany (eq,toExtract,from) “subtracts” the elements of toExtract from from, checking that all the elements of toExtract are present in from. More formally, if toExtract is a submultiset (according to the definition given in the subsetsum problem on HW 3, but
using eq to determine when an element “appears”) of from, then extractMany(eq,toExtract,from)
returns SOME xs, where xs is from with every element of toExtract removed. If toExtract
is not a submultiset of from, then extractMany(eq,toExtract,from) returns NONE.
This means that the number of times an element occurs matters, but order does not:
extractMany(inteq, [2,1,2], [1,2,3,3,2,4,2]) = SOME [3,3,4,2]
extractMany(inteq, [2,2], [2]) = NONE
You may define this recursively, and should use extract.
Task 5.2 (8%). Define datatypes representing blocks, moves, and facts, according to the above rules:
datatype block = … datatype move = … datatype fact = …
Observe the convention that datatype constructors start with an uppercase letter (e.g. Node
and Empty).
Task 5.3 (2%). Define a state of the world to be a list of facts:
type state = fact list
Fill in
val initial : state = …
to represent the following state: the hand is empty, each of A,B,C is on the table, and each of A,B,C is free.
Task 5.4 (3%). Define a short helper function
consumeAndAdd : (state * fact list * fact list) > state option
consumeAndAdd(s,before,after) subtracts before from s and adds after to the result, checking that every fact in before occurs. More formally, if before is a submultiset of s, then consumeAndAdd(s, before, after) returns SOME s’, where s’ is s with before removed and after added. If before is not a submultiset, consumeAndAdd(s, before, after) returns NONE.
You will need to use the provided function extractManyFacts, which instantiates your
extractMany with an equality operation derived from the fact datatype.
consumeAndAdd should not be recursive.
Task 5.5 (7%). Implement a function
step : (move * state) > state option
If the “before” facts of m hold in s, then step(m,s) must return SOME s’, where s’ is the collection of facts resulting from performing the move m. It should return NONE if the move cannot be applied in that state. This function should not be recursive.
Task 5.6 Optional: In the file blocks_world.sml, fill in your datatype constructors at the spots indicated. You will then be able to play Blocks World interactively as follows:
– use “hw05.sml”;
– use “blocks_world.sml”;
– playBlocks();
This task is optional; do not hand in blocks_world.sml.
Task 5.7 EXTREMELY OPTIONAL CHALLENGE TASK:
If you’re really really bored, here’s something fun you can try.
The textbased interface we made for blocks world works but is kind of bland. Download a graphics library for SML and use it to implement a fancier interface for blocks world. You’ll almost certainly have to make a custom .cm file, so don’t modify the one for this assignment. Make a new one and when you’re done submit it along with the rest of the homework.
If you want to do 2D graphics you can learn about SDL::ML at http://www.hardcoreprocessing.com/Freeware/SDLML.html and if you want to do 3D graphics you can learn about SML3D at http://sml3d.cs.uchicago.edu/.
6 Conflatten
In this question you will prove the correctness for some simple functions on lists. First, consider the declaration for the size function.
fun size [] = 0
 size ([]::R) = 1 + size R
 size ((x::L)::R) = 1 + size (L::R)
Task 6.1 (2%). Describe in a sentence or two what size does. Give the most general type for size.
Now, consider the following functions:
fun flatten [] = []
 flatten (L::R) = L @ flatten R
fun concat [] = []
 concat ([]::R) = concat R
 concat ((x::L)::R) = x :: concat(L::R)
Both of these functions achieve the same end. They take a list of lists and put all the values from each sublist into a single main list. For example,
val [1, 2, 3] = flatten [[1], [], [2, 3]]
val [1, 2, 3] = concat [[1], [], [2, 3]]
val [“a”, “b”, “c”] = flatten [[“a”, “b”], [], [], [“c”], []]
Task 6.2 (12%). Prove Theorem 1 by induction. Think carefully about what variable you induct over, as now you’re inducting over lists of lists instead of just lists. Be sure to cite any lemmas you use in your proof.1
Theorem 1. For all types t and for all L : t list list,
flatten L = concat L
Lemma 1. size is a total function. (ie. for all types t, for all values L of type t list list then size L evaluates to a value.)
Lemma 2. You may assume that
size [] = 0
Lemma 3. For all correctlytyped values R,
size ([] :: R) > size R
Lemma 4. For all correctlytyped values x and R,
size((x :: L) :: R) > size(L :: R)
Lemma 5. For all types t and all values L : t list,
either L = [] or L = x :: L’ for some x : t and L’ : t list.
Lemma 6. For all types t and all values L : t list,
Lemma 7. For all types t and all values L : t list, R : t list, and x : t,
1 It’s interesting to note that we could have stated Theorem 1 a more concisely as
flatten = concat
which is a direct transcription of the intuition of the problem into a formal statement. The statement given is an immediate expansion of this, so we don’t lose anything by being a little bit more verbose.
7 HigherOrder Shrubs
For the next few problems, we’re going to introduce a different treelike data structure called a shrub. Instead of containing data at the nodes, shrubs only have data at the leaves. You’ll be writing and analyzing higher order functions on polymorphic shrubs. Here’s the type definition of shrub:
datatype ’a shrub = Leaf of ’a
 Branch of ’a shrub * ’a shrub
Task 7.1 (5%). Define a higher order function
shrubMap : (’a > ’b) > ’a shrub > ’b shrub
such that for any shrub s : ’a shrub and any total function f : ’a > ’b,
shrubMap f s returns a shrub with f applied to every leaf. For example, to multiply all the leaves by 3 for someShrub : int shrub, we could use shrubMap:
val multThree = shrubMap (fn(n) => n * 3)
val newShrub = multThree someShrub
Task 7.2 (2%). Write a recurrence for WshrubM ap (n) where n is the size of the input shrub. For this and the following problems, assume that the function f given to shrubMap has O(1) work and span, and assume that the input shrub is balanced.
Task 7.3 (1%). Derive an estimate for the bigO of WshrubM ap (n).
Task 7.4 (2%). Write a recurrence for SshrubM ap (n).
Task 7.5 (1%). Derive an estimate for the bigO of SshrubM ap (n).
Task 7.6 (5%). Define a higher order function
shrubCombine : (’a * ’a > ’a) > ’a > ’a shrub > ’a
such that for any shrub s : ’a shrub and any total, associative function f : ’a * ’a > ’a and its corresponding identity i : ’a, shrubCombine f i s returns the result of recursively combining the shrub with f. You can think of shrubCombine like foldl for shrubs. For example, to sum all the leaves in a shrub, we could use shrubCombine:
val sumShrub = shrubCombine (op +) 0 val someSum = sumShrub someShrub
At a leaf, shrubCombine should combine the identity with the value at the leaf (in that order). At a branch, shrubCombine should combine subbranches in lefttoright order.
For f to be associative, for all welltyped values a, b, c we have:
f(a, f(b, c)) = f(f(a, b), c). For example, addition is associative as (1+2)+3 =
1+(2+3), but subtraction is not associative as (1 − 2) − 3 = 1 − (2 − 3). The identity
i of f is a value such that f(i, a) = f(a, i) = a for all welltyped values a.
And now that you’ve defined the shrubbery, you must cut down the mightiest tree in the forest with… a herring!