How to resolve the algorithm Set consolidation step by step in the Wren programming language
Published on 12 May 2024 09:40 PM
How to resolve the algorithm Set consolidation step by step in the Wren programming language
Table of Contents
Problem Statement
Given two sets of items then if any item is common to any set then the result of applying consolidation to those sets is a set of sets whose contents is: Given N sets of items where N>2 then the result is the same as repeatedly replacing all combinations of two sets by their consolidation until no further consolidation between set pairs is possible. If N<2 then consolidation has no strict meaning and the input can be returned.
See also
Let's start with the solution:
Step by Step solution about How to resolve the algorithm Set consolidation step by step in the Wren programming language
Source code in the wren programming language
import "/set" for Set
var consolidateSets = Fn.new { |sets|
var size = sets.count
var consolidated = List.filled(size, false)
var i = 0
while (i < size - 1) {
if (!consolidated[i]) {
while (true) {
var intersects = 0
for (j in i+1...size) {
if (!consolidated[j]) {
if (!sets[i].intersect(sets[j]).isEmpty) {
sets[i].addAll(sets[j])
consolidated[j] = true
intersects = intersects + 1
}
}
}
if (intersects == 0) break
}
}
i = i + 1
}
return (0...size).where { |i| !consolidated[i] }.map { |i| sets[i] }.toList
}
var unconsolidatedSets = [
[Set.new(["A", "B"]), Set.new(["C", "D"])],
[Set.new(["A", "B"]), Set.new(["B", "D"])],
[Set.new(["A", "B"]), Set.new(["C", "D"]), Set.new(["D", "B"])],
[Set.new(["H", "I", "K"]), Set.new(["A", "B"]), Set.new(["C", "D"]),
Set.new(["D", "B"]), Set.new(["F", "G", "H"])]
]
for (sets in unconsolidatedSets) {
System.print("Unconsolidated: %(sets)")
System.print("Cosolidated : %(consolidateSets.call(sets))\n")
}
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