How to resolve the algorithm Power set step by step in the Kotlin programming language
How to resolve the algorithm Power set step by step in the Kotlin programming language
Table of Contents
Problem Statement
A set is a collection (container) of certain values, without any particular order, and no repeated values. It corresponds with a finite set in mathematics. A set can be implemented as an associative array (partial mapping) in which the value of each key-value pair is ignored. Given a set S, the power set (or powerset) of S, written P(S), or 2S, is the set of all subsets of S.
By using a library or built-in set type, or by defining a set type with necessary operations, write a function with a set S as input that yields the power set 2S of S.
For example, the power set of {1,2,3,4} is For a set which contains n elements, the corresponding power set has 2n elements, including the edge cases of empty set. The power set of the empty set is the set which contains itself (20 = 1): And the power set of the set which contains only the empty set, has two subsets, the empty set and the set which contains the empty set (21 = 2):
Extra credit: Demonstrate that your language supports these last two powersets.
Let's start with the solution:
Step by Step solution about How to resolve the algorithm Power set step by step in the Kotlin programming language
The provided Kotlin code defines a function named subsets
for a set of elements of type T
. This function computes all the possible subsets (including the empty set) of the given set using recursion and Kotlin's Sequence
s (a.k.a. streams).
Here's how this code works:
-
The
subsets
function is a generic function that operates on a set of elements of typeT
. -
It returns a
Sequence
of sets of typeT
. A sequence is a lazily evaluated collection that can be iterated multiple times without holding the entire collection in memory. -
Inside the function, there's a
when
expression that checks the size of the set (i.e., the number of elements in the set). -
If the size is 0, it means the set is empty. In this case, the function returns a sequence containing a single element: an empty set.
-
If the size of the set is not 0, it indicates that the set contains at least one element. In this case, the code uses recursion to compute the subsets of the set.
-
It retrieves the first element of the set (
head
) and creates a new set (tail
) with all elements except thehead
element. -
The
tail
set is recursively passed to thesubsets
function to compute its subsets (tail.subsets()
). -
The subsets of the
tail
set are then combined with thehead
element to form additional subsets. This is achieved by mapping each subset of thetail
set to a new set containing both thehead
element and the original subset (e.g.,setOf(head) + it
). -
The resulting sequence contains all the subsets of the
tail
set, as well as additional subsets that include thehead
element. -
The
+
operator used insetOf(head) + it
is Kotlin's set union operator. It creates a new set that contains all elements from both thehead
set (containing only thehead
element) and theit
set (which is each subset of thetail
set). -
Finally, the function returns the sequence containing all the subsets of the original set, including the empty set.
In summary, this code generates all the possible subsets of a given set using a recursive approach and Kotlin's Sequence
s. It efficiently computes subsets without the need to explicitly create a complete list in memory, making it suitable for large sets.
Source code in the kotlin programming language
// purely functional & lazy version, leveraging recursion and Sequences (a.k.a. streams)
fun <T> Set<T>.subsets(): Sequence<Set<T>> =
when (size) {
0 -> sequenceOf(emptySet())
else -> {
val head = first()
val tail = this - head
tail.subsets() + tail.subsets().map { setOf(head) + it }
}
}
// if recursion is an issue, you may change it this way:
fun <T> Set<T>.subsets(): Sequence<Set<T>> = sequence {
when (size) {
0 -> yield(emptySet<T>())
else -> {
val head = first()
val tail = this@subsets - head
yieldAll(tail.subsets())
for (subset in tail.subsets()) {
yield(setOf(head) + subset)
}
}
}
}
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