a) Briefly explain the difference between an SML (also called algebraic) datatype and an abstract datatype.
Algebraic datatypes are one of several kinds of types in SML. (Others include record, tuple, and function types). Although there are several standard datatypes already defined by SML (for example, list), the user can write new datatype declarations; here is one of several examples we have seen so far in the course:
datatype color = Red | Black datatype rbtree = Leaf | Branch of value*color*rbtree*rbtree
A datatype declaration concretely specifies the representation of a type as a one of a set of constructors; these constructors and the implicit destructors that are invoked when pattern matching on a datatype are the operations of the algebraic datatype.
Abstract data types (ADTs) are a programming methodology rather than a language feature, though this methodology is supported by SML's ability to declare abstract types in module signatures. From the client view, an ADT is an abstract type whose representation is hidden, plus some operations whose implementation is also hidden. An implementer has the flexibility to change the representation and implementation of these types because the user does not depend on them.
[This answer is longer and more complete than we expected yours to be!]
b) Why is unit testing needed even if integration testing succeeds? Explain briefly.
Integration testing tests an entire program, but not the individual modules of the program. The modules may contain bugs that are difficult or impossible to reproduce by integration testing. Unit testing gives more assurance that the program works correctly. Even if integration testing completely tests program functionality, bugs that would be caught by unit testing may otherwise show up as program code evolves to use the module differently.
c) Explain briefly how hash tables implement the set membership operation in O(1) time.
A hash table containing n elements hashes an element to be tested for membership into one of O(n) buckets using O(1) time. Assuming that the hash function has a uniform distribution over the buckets, each bucket contains O(1) elements on average, so the lookup within that bucket is constant time. Therefore the expected time to perform the whole operation is constant, i.e. O(1).
A value is an expression that does not need further evaluation; if
typed to the SML prompt, exactly the same thing is reported as a result. For
example, it may be a integer constant, a
fn expression, a tuple of
values, or a datatype constructor applied to a value. Complete each of the
following let-expressions by giving a value that can replace
???" and cause the whole thing to evaluate to the
let fun zardoz(x: string*string): int = case Int.fromString((#2 x)^(#1 x)) of NONE => 41 | SOME n => n*2 in zardoz("1",???) end
let fun zardoz (x:string, (y:(int*int) option, z: int)):int = case y of SOME (7,x) => x*z-1 | SOME (x,z) => x*z | NONE => size(x)*z+1 in zardoz(???) end
let datatype sa = Done of int | Function of sa->int fun zardoz(f:sa):int = case f of Done(n) => n | Function(g) => g(f) in zardoz(Function(???)) end
DataMiners Inc is designing an application to track stock portfolios and
needs code to analyze a history of stock prices and find the maximum profit that
could have been obtained by buying and selling at just the right moments. The
stock price history is stored as a list of ints ordered in time; the maximum
profit is the largest value y-x
where x, y
are integers in the list and y occurs in
the list after x. For example, for
[4,6,7,2]the maximum profit would be 7-4=3.
Note that it would not be 7-2=5 because 7 does not occur after 2 in
the list. The maximum profit can even be negative if the stock price always
declines in value throughout the history.
The programmers at DataMiners have written the following analysis code; they have hired you to figure out whether it works and analyze its performance.
fun maxDiff_x(data: int list, x: int, currentMax: int) = foldl (fn(y,max) => Int.max(y-x,max)) currentMax data
fun mpLoop (data: int list, currentMax:int) : int = case data of  => currentMax | x::xs => mpLoop(xs, maxDiff_x (xs, x, currentMax)) fun maxProfit (data: int list) : int option = case data of x::y::xs => SOME (mpLoop (data, y-x)) | _ => NONE
a) Give a specification of the function
maxDiff_x(data, x, currentMax) is the larger of either currentMax or the largest difference between some element of data and x. Note that data is allowed to be empty.
b) Give a specification of the function
mpLoop(data, currentMax) is the larger of either currentMax or the largest difference y-x between two elements x and y in data, where y strictly follows x in the list. Note that data may be empty.
c) Give a tight asymptotic bound on the running time of
(no justification needed).
Q(n) where n is the length of data. This is because maxDiff_x simply calls foldl with an O(1) anonymous function.
d) Derive a recurrence relation for the worst-case running time of
as a function of the length of the list
The time to call mpLoop on a list of size n includes:
Therefore the recurrence is:
T(0) = k1
T(n) = k1 + k2n + T(n-1)
e) Use your recurrence relation to determine the worst-case time bound
for this code. Justify this bound by using the recurrence relation.
Two possible solutions. The first is to expand the terms of the recurrence. (We could replace k1 and k2 with 1 here and get the same answer, but just to show we don't have to, here's how it works out:
T(n) = k1 + k2n + T(n-1)
T(n) = k1 + k2n + k1 + k2(n-1) + T(n-2)
T(n) = k1 + k2n + k1 + k2(n-1) + k1 + k2n + k1 + k2(n-2) + T(n-3)
T(n) = (n+1)k1+ k2 · Si = 1 to n i
T(n) = (n+1)k1 + k2(½n(n+1)) = ½k2n2 + n(½k2 + k1) + k1
which is clearly Q(n2) because for sufficiently large n it is bounded above by, say, k2n2 , and below by, say, ¼k2n2.
The other solution is a proof by induction. Here we replace k2
with 1 and discard the O(1) k1
P(n) : T(n) £ cn2
Claim: P(n) holds for all n
Proof by induction on Base n=0. T(0) = 0 £ c·02 = 0
Inductive Step: Assume P(k) is true, prove P(k+1) is true.
T(k+1) = k + T(k) [from recurrence]
£ k + ck2 [by inductive hypothesis]
£ ck2 + 2ck + c
The following is a simplified version of the SML signature for strings, with only some operations specified fully:
signature STRING = sig (* A string is a finite sequence of characters. For example, the * string "abd" is the characters #"a", #"b", #"d" in sequence. *) type string (* size(s) is the number of characters in the string. *) val size : string -> int (* sub(s, i) is the character in the string at index i, * where indices start at zero in the usual manner. *) val sub : (string * int) -> char (* ^ is ordinary string concatenation *) val ^ : (string * string) -> string (* implode(lst) is the string containing the characters in lst * in the same order as in lst, e.g. implode[#"a",#"b"] is "ab" * and implode nil is the empty string "". *) val implode : char list -> string (* explode is the inverse of implode *) val explode : string -> char list ... end
a) (2 pts) Explain briefly what is wrong with the specification for
sub and give a corrected specification.
It doesn't specify what happens when the index i is out of bounds. We need to add a requires (or checks) clause:
Requires: 0 <= i < size(s)
While the default SML implementation of this signature is adequate for a broad range of applications, there are cases where a different implementation is better. For example, if the program is doing a lot of string concatenations (^), the standard
String implementation may not be fast enough because
it does string concatenation in time O(n)
in the length of the string. An alternate implementation of strings looks like
structure Rope :> STRING = struct datatype string = Empty | Single of char | Concat of string * string * int (* Abstraction function: * Empty represents the empty string * Single(c) represents the single-character string containing c * Concat(s1,s2,n) represents the concatenation of * s1 with s2, i.e., s1^s2. * Representation invariant: ??? *) fun size(s) = case s of Empty => 0 | Single(c) => 1 | Concat(s1,s2,n) => n fun op^(s1, s2) = Concat(s1, s2, size(s1) + size(s2)) fun sub(s, i:int) = case s of Empty => raise Subscript | Single(c) => if i = 0 then c else raise Subscript | Concat(s1,s2,n) => let val n1 = size(s1) in if i < n1 then sub(s1,i) else sub(s2, ???) end fun implode(cl) = let fun implode_n(cl: char list, n: int): string = case cl of nil => Empty | c::nil => Single(c) | c::cl' => Concat(Single(c), implode_n(cl', n-1), n) in implode_n(cl, List.length(cl)) end fun explode(s) = ??? end
In the rest of this problem, the type name
string refers to the
b) (4 pts) The representation invariant is conspicuously
missing from the
Rope implementation. Give a representation
invariant. Hint: consider what rep invariant is needed in order to make the size
For the representation
Concat(s1,s2,n), n is equal to
the sum of the sizes of s1 and s2. It is also the length of the string. We also
might include ``no
Empty nodes in non-empty strings'' in the
representation invariant — we do not need it for correctness, but without it
we won't be able to prove any running time bounds. But then we'd have to make
the code for ^ a slightly more complicated.
s2of length at most n, what is the asymptotic worst-case performance of the concatenation expression
s1^s2as a function of n? Justify in one sentence.
d) (4 pts) The sub function has three question marks in place of an
expression of type
O(1) time, because
size itself is
int. What expression should replace the
We know that the size of
s2 has index
s1^s2 must have
s2. Therefore we need
??? = i-n1
sub. Describe two different representations of the same string where one representation is much better than the other.
f) (5 pts) Given a
sub has worst-case
O(height) run time. The tree
representing a string of
n characters can have height up to
if the left-hand children are always
Singles. It can also have
lg(n) if it is balanced.
stringof length n, what is the asymptotic worst-case performance of
subas a function of n? Justify in 1-2 sentences.
g) (7 pts) Give code that implements
O(n) by the argument above.
explode(your implementation need not be asymptotically efficient).
Here are two implementations. The second one is asymptotically efficient but more complicated.
fun explode(s: string) = case s of Empty =>  | Single(c) => [c] | Concat(s1,s2,n) => explode(s1) @ explode(s2)
fun explode(s) = let fun explodeApp(s, lst) = case s of Empty => lst | Single(c) => c::lst | Concat(s1,s2,n) => explodeApp(s1,explodeApp(s2, lst)) in explodeApp(s, ) end
h) (EXTRA CREDIT: 2 pts -- save till you are done with everything
else!) Give a better implementation of
implode that improves
the worst-case asymptotic efficiency of
sub on strings that it
fun implode(cl): string = let fun implode_n(cl: char list, n: int): string = case n of 0 => Empty | 1 => Single(hd cl) | n => let val m = n div 2 in Concat(implode_n(cl, m), implode_n(List.drop(cl, m), n-m), n) end in implode_n(cl, List.length(cl)) end
Recall the definition of
fun foldl (f: ('a*'b) -> 'b) (base:'b) (x:'a list):'b = case x of nil => base | hd::tl => foldl f (f (hd,base)) tl
For each of the following tasks, show how
foldl can be used to
reimplement the function in 1–2 lines of functional code, without using
version should always produce the same result as the reference code. In your
code, include the full type signature for the function provided as the argument
and remember to check your code to make sure that that function returns the type
it needs to.
a) (5 pts) Given a list of integers, find the sum of all elements prior to the first 0. Reference code to be rewritten:
fun SumPriorZero(a: int list) = case a of 0::t => 0 | h::t => h + SumPriorZero(t) | nil => 0Answer: fun SumPriorZero(a: int list) = #1 (foldl (fn (e, (x, b)) => if b orelse e=0 then (x,true) else (e+x,false)) (0, false) a)
b) (6 pts) Return a list containing every nth element in a list. Reference code:
fun everyNth(l:'a list, n:int):'a list = let fun iter(l:'a list, counter:int):'a list = case (l,counter) of (,_) =>  |(x::xs,1) => x::(iter(xs,n)) |(x::xs,c) => iter(xs,c-1) in iter(l,n) endAnswer: fun everyNth(l: 'a list, n:int):'a list = List.rev (#2(foldl (fn (elt,(c,lst))=>if (c=1) then (n,elt::lst) else (c-1,lst)) (n,) l))