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Commit 65e4a7c

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Adding a simple cascading solution to generate a Power Set (#975)
* Add a simple cascading version of generating a PowerSet. * Update README. * Update README. * Update README.
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‎README.md‎

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@@ -99,7 +99,7 @@ a set of rules that precisely define a sequence of operations.
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* **Sets**
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* `B` [Cartesian Product](src/algorithms/sets/cartesian-product) - product of multiple sets
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* `B` [Fisher–Yates Shuffle](src/algorithms/sets/fisher-yates) - random permutation of a finite sequence
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* `A` [Power Set](src/algorithms/sets/power-set) - all subsets of a set (bitwiseand backtracking solutions)
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* `A` [Power Set](src/algorithms/sets/power-set) - all subsets of a set (bitwise, backtracking, and cascading solutions)
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* `A` [Permutations](src/algorithms/sets/permutations) (with and without repetitions)
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* `A` [Combinations](src/algorithms/sets/combinations) (with and without repetitions)
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* `A` [Longest Common Subsequence](src/algorithms/sets/longest-common-subsequence) (LCS)

‎src/algorithms/sets/power-set/README.md‎

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# Power Set
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Power set of a set `S` is the set of all of the subsets of `S`, including the
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empty set and `S` itself. Power set of set `S` is denoted as `P(S)`.
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empty set and `S` itself. Power set of set `S` is denoted as `P(S)`.
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For example for `{x, y, z}`, the subsets
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are:
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![Power Set](https://www.mathsisfun.com/sets/images/power-set.svg)
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Here is how we may illustrate the elements of the power set of the set `{x, y, z}` ordered with respect to
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Here is how we may illustrate the elements of the power set of the set `{x, y, z}` ordered with respect to
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inclusion:
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![](https://upload.wikimedia.org/wikipedia/commons/e/ea/Hasse_diagram_of_powerset_of_3.svg)
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**Number of Subsets**
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If `S` is a finite set with `|S| = n` elements, then the number of subsets
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of `S` is `|P(S)| = 2^n`. This fact, which is the motivation for the
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If `S` is a finite set with `|S| = n` elements, then the number of subsets
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of `S` is `|P(S)| = 2^n`. This fact, which is the motivation for the
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notation `2^S`, may be demonstrated simply as follows:
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> First, order the elements of `S` in any manner. We write any subset of `S` in
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the format `{γ1, γ2, ..., γn}` where `γi , 1 ≤ i ≤ n`, can take the value
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> First, order the elements of `S` in any manner. We write any subset of `S` in
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the format `{γ1, γ2, ..., γn}` where `γi , 1 ≤ i ≤ n`, can take the value
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of `0` or `1`. If `γi = 1`, the `i`-th element of `S` is in the subset;
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otherwise, the `i`-th element is not in the subset. Clearly the number of
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otherwise, the `i`-th element is not in the subset. Clearly the number of
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distinct subsets that can be constructed this way is `2^n` as `γi ∈ {0, 1}`.
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## Algorithms
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### Bitwise Solution
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Each number in binary representation in a range from `0` to `2^n` does exactly
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what we need: it shows by its bits (`0` or `1`) whether to include related
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element from the set or not. For example, for the set `{1, 2, 3}` the binary
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Each number in binary representation in a range from `0` to `2^n` does exactly
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what we need: it shows by its bits (`0` or `1`) whether to include related
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element from the set or not. For example, for the set `{1, 2, 3}` the binary
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number of `0b010` would mean that we need to include only `2` to the current set.
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| | `abc` | Subset |
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| :---: | :---: | :-----------: |
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| `0` | `000` | `{}` |
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| `1` | `001` | `{c}` |
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| `2` | `010` | `{b}` |
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| `2` | `010` | `{b}` |
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| `3` | `011` | `{c, b}` |
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| `4` | `100` | `{a}` |
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| `5` | `101` | `{a, c}` |
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> See [btPowerSet.js](./btPowerSet.js) file for backtracking solution.
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### Cascading Solution
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This is, arguably, the simplest solution to generate a Power Set.
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We start with an empty set:
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```text
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powerSets = [[]]
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```
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Now, let's say:
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```text
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originalSet = [1, 2, 3]
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```
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Let's add the 1st element from the originalSet to all existing sets:
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```text
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[[]] ← 1 = [[], [1]]
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```
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Adding the 2nd element to all existing sets:
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```text
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[[], [1]] ← 2 = [[], [1], [2], [1, 2]]
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```
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Adding the 3nd element to all existing sets:
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```
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[[], [1], [2], [1, 2]] ← 3 = [[], [1], [2], [1, 2], [3], [1, 3], [2, 3], [1, 2, 3]]
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```
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And so on, for the rest of the elements from the `originalSet`. On every iteration the number of sets is doubled, so we'll get `2^n` sets.
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> See [caPowerSet.js](./caPowerSet.js) file for cascading solution.
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## References
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* [Wikipedia](https://en.wikipedia.org/wiki/Power_set)
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import caPowerSet from '../caPowerSet';
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describe('caPowerSet', () => {
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it('should calculate power set of given set using cascading approach', () => {
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expect(caPowerSet([1])).toEqual([
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[],
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[1],
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]);
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expect(caPowerSet([1, 2])).toEqual([
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[],
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[1],
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[2],
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[1, 2],
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]);
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expect(caPowerSet([1, 2, 3])).toEqual([
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[],
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[1],
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[2],
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[1, 2],
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[3],
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[1, 3],
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[2, 3],
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[1, 2, 3],
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]);
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});
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});
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/**
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* Find power-set of a set using CASCADING approach.
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*
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* @param {*[]} originalSet
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* @return {*[][]}
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*/
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export default function caPowerSet(originalSet) {
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// Let's start with an empty set.
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const sets = [[]];
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/*
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Now, let's say:
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originalSet = [1, 2, 3].
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Let's add the first element from the originalSet to all existing sets:
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[[]] ← 1 = [[], [1]]
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Adding the 2nd element to all existing sets:
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[[], [1]] ← 2 = [[], [1], [2], [1, 2]]
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Adding the 3nd element to all existing sets:
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[[], [1], [2], [1, 2]] ← 3 = [[], [1], [2], [1, 2], [3], [1, 3], [2, 3], [1, 2, 3]]
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And so on for the rest of the elements from originalSet.
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On every iteration the number of sets is doubled, so we'll get 2^n sets.
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*/
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for (let numIdx = 0; numIdx < originalSet.length; numIdx += 1) {
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const existingSetsNum = sets.length;
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for (let setIdx = 0; setIdx < existingSetsNum; setIdx += 1) {
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const set = [...sets[setIdx], originalSet[numIdx]];
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sets.push(set);
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}
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}
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return sets;
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}

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