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src/algorithms/math/linear-prime-sieve/README.md

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+# Linear Prime Sieve
+
+Linear Prime Sieve is an algorithm for finding all prime numbers up to some limit `n` in linear time.
+
+
+
+## How it works
+
+1. Create an array `leastPrimeFactor` of `n + 1` positions filled with `0`(to represent the least prime factors of numbers `0` through `n`)
+2. Start at position `i = 2` (the first prime number)
+3. If the least prime factor of `i` is `0` it means `i` is a prime number, so we add it to our list of `primes`.
+4. We start setting the least prime factor for numbers `x`, where x = `p` * `i`, `p` is `x`'s least prime factor and `p` <= `leastPrimeFactor(i)`, we do 
+   this by traversing the list of primes accumulated till now.
+5. Increment the value of `i`.
+6. We repeat steps 3 through 5 until `i` exceeds `n`.   
+
+
+When the algorithm terminates, all the prime numbers from `0` through `n` will be added to our primes list
+
+
+
+## Complexity
+
+The algorithm has a complexity of `O(n)`.
+
+
+
+## Proof Of Correctness
+
+In order to prove the time complexity, we need to show that the alogrithm sets all leastPrimeFactor[] correctly, and that every value will be set only once.
+If so the algorithm will have linear time complexity, since all the remaining work of the algorithm is of `O(n)`.
+We Observe that every number `i` has exactly one representation of form:
+ `i = leastPrimeFactor[i] * x`, where leastPrimeFactor[i] <= leastPrimeFactor[x].
+ 
+In our Alogrith, for every `x`, it goes through all prime numbers it could be multiplied by, i.e. all prime numbers up to leastPrimeFactor[x] inclusive,
+in order to get the numbers in the form given above.
+Hence, the algorithm will go through every composite number exactly once, setting the correct values for leastPrimeFactor[] there.
+
+
+
+## References
+
+- David Gries, Jayadev Misra. A Linear Sieve Algorithm for Finding Prime Numbers [1978].
+- [CP-Algorithms/Linear-Sieve](https://cp-algorithms.com/algebra/prime-sieve-linear.html).

src/algorithms/math/linear-prime-sieve/__test__/linearPrimeSieve.test.js

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+import linearPrimeSieve from '../linearPrimeSieve';
+
+describe('linearPrimeSieve', () => {
+  it('should find all primes less than or equal to n', () => {
+    expect(linearPrimeSieve(5)).toEqual([2, 3, 5]);
+    expect(linearPrimeSieve(10)).toEqual([2, 3, 5, 7]);
+    expect(linearPrimeSieve(100)).toEqual([
+      2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41,
+      43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97,
+    ]);
+  });
+});

src/algorithms/math/linear-prime-sieve/linearPrimeSieve.js

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+/**
+ * @param {number} maxNumber
+ * @return {number[]}
+ */
+export default function linearPrimeSieve(maxNumber) {
+  const primes = [];
+  // leastPrimeFactor[i] gives us the least prime factor of 'i'
+  const leastPrimeFactor = new Array(maxNumber + 1).fill(0);
+
+  for (let i = 2; i <= maxNumber; i += 1) {
+    if (!leastPrimeFactor[i]) {
+      /* leastPrimeFactor[i] = 0 means 'i' itself is its least prime factor,
+       * i.e 'i' is prime
+       */
+      leastPrimeFactor[i] = i;
+      primes.push(i);
+    }
+
+    /*
+     *  start setting leastPrimeFactor[] for numbers 'x', where x = p * i, p is x's
+     *  least prime factor and p <= leastPrimeFactor[i]. x = p*i, this representation will
+     *  be unique for any number, therefore leastPrimeFactor[x] will be
+     *  set only once.
+     */
+
+    let j = 0;
+    while (j < primes.length) {
+      if ((primes[j] * i > maxNumber) || (primes[j] > leastPrimeFactor[i])) { break; }
+      leastPrimeFactor[primes[j] * i] = primes[j];
+      j += 1;
+    }
+  }
+
+  return primes;
+}