content="Data-structures and algorithms in Standard ML. Notes from a lecture course given by Michael Fourman at the University of West Australia, Spring 1994." /> rel="start" type="text/html" href="/mfourman/index.html" title="Michael Fourman" /> Sample Mid-term; Solutions

Sample Mid-term; Solutions

Michael P. Fourman

October 31, 2006

1 Introduction

This document contains solutions to the sample questions given in Lecture Note 10

  1. 5 marks
    Give the responses of the ML system to the following sequence of declarations
    val a = 1;  
    val b = 2;  
    fun f a = a + b;  
    val b = 3;  
    f b;

    the responses of the system are as follows:

    val a = 1 : int  
    val b = 2 : int  
    val f = fn : int -> int  
    val b = 3 : int  
    val it = 5 : int

    The point here is to check that you understand, and can apply, the scoping rules, applied to the bindings of a and b. The exam on Thursday may include let and local declarations, in a similar question.

  2. Long Question 10 marks
    The following datatype can be used to represent trees whose nodes can have an arbitrary number of children.
    datatype  ’a  Tree  = Tree  of ’a  * ’a  Tree  list
    1. What tree does the following expression denote (i.e draw a picture):
      Tree(1, [Tree(2, [ ]), Tree(3, [Tree(4,[ ])])])
    2. Define a function to calculate the number of leaves in such a tree.
      fun sum []       = 0  
        | sum (h :: t) = h + sum t  
      fun leaves (Tree(_,[])) = 1  
        | leaves (Tree(_,ts)) = sum (map leaves ts)


      fun leaves (Tree(_,[])) = 1  
        | leaves (Tree(x,t :: ts)) = leaves t + leaves (Tree(x, ts))

    3. We can assign a level to each node in a tree as follows. The node at the root is at level 1. Its children are at level 2. Their children are at level 3 and so on.

      Suppose we are interested in trees where an internal node at level n always has exactly n children. Define a function check : ’a Tree ->bool that checks whether a given tree has this property.

      The recursion is not straightforward: to check the property for a tree, we must check a slightly different property for its subtrees. We therefor introduce an auxiliary function, checkk, with an extra parameter; checkk k checks that the appropriate property holds for a subtree rooted at level k:

      fun length [] = 0  
        | length (_::t) = 1 + length t  
      fun andl [] = true (* and over a list of booleans *)  
        | andl (h :: t) = h andalso andl t  
      fun checkk k (Tree(_,[])) = true (* nothing to check for a leaf *)  
        | checkk k (Tree(_,ts)) = ((length ts) = k)  
                                  (* check there are k children *)  
                                  andl (map (checkk (k+1)) ts)  
                                  (* subtrees are at level (k+1) *)  
      fun check  t           = checkk 1 t

  3. Long Question 10 marks
    The EQueue signature is like the signature Queue, but is extended with an additional operation multiple enqueue, menq:(Item  list * Queue) -> Queue, intended to add a number of items (in an arbitrary order) to the queue in a single operation.
    signature EQueue =  
        type Item  
        type Queue  
        val empty : Queue  
        val enq : (Item * Queue) -> Queue  
        val deq : Queue -> (Item * Queue)  
        val menq: (Item list * Queue) -> Queue  

    An implementation of a stack, including this operation, uses the type declaration

    type Queue = Item list list

    the operations empty and menq are implemented as follows:

    val empty = []  
    fun menq(items, q) = items :: q

    1. Complete the following declarations of the functions enq and deq for this implementation
      fun enq(item,       []) = [[item]]  
        | enq(item, (h :: t)) = (item :: h) :: t  
          (* or, alternatively, [item] :: h :: t *)  
      fun deq((h :: t) :: r)  = (h, t :: r)  
        | deq([] :: r)        = deq r  
        | deq []              = raise Deq

      The point here is to take care with the types. Since a stack is being represented as a list of lists, we need to make a list, [[item]], whose only member is the singleton list, [item], to represent a stack with one entry. When adding an item to a non-empty stack, we have a choice: we can either add the item to the list at the head of the list of lists, or we can form a new singleton list and add this to the list of lists.

    2. What is the complexity of the three operations
      1. enq, O(1)
      2. deq, O(1)
      3. menq O(1)

      for this implementation?

      Notice that, for a conventional stack implementation we would have to implement menq using multiple calls of enq. The complexity would be O(n), where n is the number of items being added in one go.

  4. Long Question 10 marks
    An implementation of sets of integers is designed to represent a set by a list without repetitions, kept in increasing order. Here is the function union : Set*Set -> Set from this implementation
    fun union(a, []) = a  
      | union([], b) = b  
      | union(ah :: at, bh :: bt) =  
              if ah < bh then ah :: union(at, bh :: bt)  
              else if ah = bh then ah :: union(at, bt)  
              else bh :: union(ah :: at, bt)

    1. What is the complexity of this implementation of union?

      O(n), where n is the sum of the sizes of the sets; there is at most one recursive call for each of these elements.

    2. Give an implementation of the operation insert : (int*Set) ->Set compatible with this representation.
      fun insert (x,[]) = [x]  
        | insert (x,h :: t) = if x < h then x :: h :: t  
                              else if h < x then h :: insert(x,t)  
                              else (* x = h *) h :: t

      This is book-work: a similar definition was given in the notes to implement a priority queue.

    3. Give an O(n) implementation of the operation intersect : Set*Set -> Set, compatible with this representation.
      fun intersect(a, []) = []  
        | intersect([], b) = []  
        | intersect(ah :: at, bh :: bt) =  
                if ah < bh then intersect(at, bh :: bt)  
                else if ah = bh then ah :: intersect(at, bt)  
                else intersect(ah :: at, bt)

      This follows the pattern given in the declaration of union.

©Michael Fourman 1994-2006