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Monday, September 7, 2015

Scala note 2: Functional Sets

Mathematically, we call the function which takes an integer as argument and which returns a boolean indicating whether the given integer belongs to a set, the characteristic function of the set. For example, we can characterize the set of negative integers by the characteristic function (x: Int) => x < 0.
Therefore, we choose to represent a set by its characterisitc function and define a type alias for this representation:
type Set = Int => Boolean
Using this representation, we define a function that tests for the presence of a value in a set:
def contains(s: Set, elem: Int): Boolean = s(elem)

2.1 Basic Functions on Sets

Let’s start by implementing basic functions on sets.
  1. Define a function which creates a singleton set from one integer value: the set represents the set of the one given element. Its signature is as follows:
    def singletonSet(elem: Int): Set
    Now that we have a way to create singleton sets, we want to define a function that allow us to build bigger sets from smaller ones.
  2. Define the functions unionintersect, and diff, which takes two sets, and return, respectively, their union, intersection and differences.diff(s, t) returns a set which contains all the elements of the set s that are not in the set t. These functions have the following signatures:
    def union(s: Set, t: Set): Set
    def intersect(s: Set, t: Set): Set
    def diff(s: Set, t: Set): Set
  3. Define the function filter which selects only the elements of a set that are accepted by a given predicate p. The filtered elements are returned as a new set. The signature of filter is as follows:
    def filter(s: Set, p: Int => Boolean): Set

2.2 Queries and Transformations on Sets

In this part, we are interested in functions used to make requests on elements of a set. The first function tests whether a given predicate is true for all elements of the set. This forall function has the following signature:
def forall(s: Set, p: Int => Boolean): Boolean
Note that there is no direct way to find which elements are in a set. contains only allows to know whether a given element is included. Thus, if we wish to do something to all elements of a set, then we have to iterate over all integers, testing each time whether it is included in the set, and if so, to do something with it. Here, we consider that an integer x has the property -1000 <= x <= 1000 in order to limit the search space.
  1. Implement forall using linear recursion. For this, use a helper function nested in forall. Its structure is as follows (replace the ???):
    def forall(s: Set, p: Int => Boolean): Boolean = {
     def iter(a: Int): Boolean = {
       if (???) ???
       else if (???) ???
       else iter(???)
     }
     iter(???)
    }
  2. Using forall, implement a function exists which tests whether a set contains at least one element for which the given predicate is true. Note that the functions forall and exists behave like the universal and existential quantifiers of first-order logic.
    def exists(s: Set, p: Int => Boolean): Boolean
  3. Finally, write a function map which transforms a given set into another one by applying to each of its elements the given function. map has the following signature:
    def map(s: Set, f: Int => Int): Set


Coursera Scala course week 2 homework. The original question can be found here.

The question asks as two define an object Set, which has a characteristic function that projects an integer to a boolean.

1. define Singleton
This question requires us to create a set with one integer, i.e., given an integer, projects it to a boolean to claim the set contains the integer:

def singletonSet(elem: Int): Set = (x: Int) => x == elem

This means projects an integer x to if x equals given parameter elem. It can also be considered as given any integer, if x equals elem, then set contains x, which defines the singleton set that it only contains elem.

2. define Union, Intersect and Difference

  def union(s: Set, t: Set): Set = (x: Int) => s(x) || t(x)
  def intersect(s: Set, t: Set): Set = (x: Int) => s(x) && t(x)
  def diff(s: Set, t: Set): Set = (x: Int) => s(x) && !t(x)

s(x) indicates contains. Thus, for union, it means either for any x, if it satisfies the condition that either s contains x or t contains x, then x is in union. For intersect, the condition becomes both s and t should contains x. For difference, it means s should contain x but t should not.

3. define filter

def filter(s: Set, p: Int => Boolean): Set = (x: Int) => s(x) && p(x)

filter is a set that for any x in filter, it should satisfies that s contains x and x satisfies predicate p.

4. forall (∀ )

val bound = 1000

  def forall(s: Set, p: Int => Boolean): Boolean = {
    def iter(a: Int): Boolean = {
      if (a > bound) true
      else if (s(a) && !p(a)) false
      else iter(a+1)
    }
    iter(-bound)
  }

This is nothing special. Starts from -bound, if any a in s doesn't satisfy p, then return false. Iterate until bound, then return true.

5. exists(∃ )

def exists(s: Set, p: Int => Boolean): Boolean = !forall(s, x => !p(x))

This one is tricky. The solution means not all elements in s satisfies !p, which in turn indicates there is at least one element in s satisfies p.

6. map

def map(s: Set, f: Int => Int): Set = (y: Int) => exists(s, x => f(x) == y)

This is similar as how we define a singleton set: For any y, if there exists an element x in s that satisfies the condition f(x) equals y, then y is in new Set map.


Source on git.

7 comments:

  1. Hi, I am just a beginner in Scala. Your code helped me a lot. Thanks for sharing

    ReplyDelete
  2. Shirley, You are welcome to give as many hints as possible, but you should not give away the solutions! That's breaking 'Code of Honor' that all attendees of Coursera sign up to.

    ReplyDelete
  3. Awesome intro.....please keep up the great work.

    ReplyDelete
  4. http://islamguzelahlaktir.blogspot.com/

    ReplyDelete
  5. Thank you so much for this blog. I'm auditing the Scala course and I'd be lost without you.

    ReplyDelete
  6. I have a feeling that 99.99% of the people that complete the Scala course broke the honor code. That course is brutal difficult. I'm grateful that the solutions are out there.

    ReplyDelete