How to resolve the algorithm Set step by step in the Nanoquery programming language

Published on 12 May 2024 09:40 PM
#Nanoquery

How to resolve the algorithm Set step by step in the Nanoquery programming language

Table of Contents

Problem Statement

A   set  is a collection of elements, without duplicates and without order.

Show each of these set operations:

As an option, show some other set operations. (If A ⊆ B, but A ≠ B, then A is called a true or proper subset of B, written A ⊂ B or A ⊊ B.) As another option, show how to modify a mutable set.

One might implement a set using an associative array (with set elements as array keys and some dummy value as the values). One might also implement a set with a binary search tree, or with a hash table, or with an ordered array of binary bits (operated on with bit-wise binary operators). The basic test, m ∈ S, is O(n) with a sequential list of elements, O(log n) with a balanced binary search tree, or (O(1) average-case, O(n) worst case) with a hash table.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Set step by step in the Nanoquery programming language

Source code in the nanoquery programming language

class set
        declare internal_list

        def set()
                internal_list = list()
        end
        def set(list)
                internal_list = list
        end

        def append(value)
                if not value in internal_list
                        internal_list.append(value)
                end
                return this
        end
        def contains(value)
                return value in internal_list
        end
        def difference(other)
                diff = list()
                for value in this.internal_list
                        diff.append(value)
                end

                for i in range(len(diff) - 1, 0)
                        if diff[i] in other.internal_list
                                diff.remove(i)
                        end
                end

                return new(set, diff)
        end
        def operator=(other)
                for value in other.internal_list
                        if not value in this.internal_list
                                return false
                        end
                end
                return true
        end
        def intersection(other)
                intersect = list()
                for value in this.internal_list
                        if other.contains(value)
                                intersect.append(value)
                        end
                end
                return new(set, intersect)
        end
        def subset(other)
                for value in this.internal_list
                        if not value in other.internal_list
                                return false
                        end
                end
                return true
        end
        def union(other)
                u = list()

                for value in this.internal_list
                        u.append(value)
                end

                for value in other.internal_list
                        if not value in u
                                u.append(value)
                        end
                end

                return new(set, u)
        end

        def toString()
                return str(this.internal_list)
        end
end

import "rosetta-code/set.nq"

a = new(set, {1, 2, 3, 4, 5})
b = new(set, {2, 3, 4, 5, 6, 8})
c = new(set, {2, 3, 4})
d = new(set, {2, 3, 4})

println "a: " + a
println "b: " + b
println "c: " + c
println "d: " + d

println "2 in a: " + a.contains(2)
println "6 in a: " + a.contains(6)

println "a union b: "        + a.union(b)
println "a - b: "            + a.difference(b)
println "c subset of a: "    + c.subset(a)

println "c = d: "            + (c = d)
println "d = c: "            + (d = c)

println "a intersect b: "    + a.intersection(b)
println "add 7 to a: "       + a.append(7)
println "add 2 to a again: " + a.append(2)

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