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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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@ -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 (bitwise and 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)
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@ -68,6 +68,44 @@ element.
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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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28
src/algorithms/sets/power-set/__test__/caPowerSet.test.js
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src/algorithms/sets/power-set/__test__/caPowerSet.test.js
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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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src/algorithms/sets/power-set/caPowerSet.js
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src/algorithms/sets/power-set/caPowerSet.js
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@ -0,0 +1,37 @@
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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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