Added note on how we compute if a number is a prime.

nelvadas · Dec 16, 2020 · 46bf282 · 46bf282
1 parent 70c6cd1
commit 46bf282
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Showing 2 changed files with 13 additions and 1 deletion.
diff --git a/1/README.md b/1/README.md
@@ -99,7 +99,13 @@ public class PrimesComputer {
             .boxed()
             .collect(Collectors.toList());
     }
-
+
+    /**
+     * If n is not a prime, then n = a * b (some a & b)
+     * Both a and b can't be larger than sqrt(n), or the product of them would be greater than n
+     * One of the factors has to be smaller than sqrt(n).
+     * So, if we don't find any factors by the time we get to sqrt(n), then n must be prime
+     */
     public static boolean isPrime(long n) {
         return LongStream.rangeClosed(2, (long) Math.sqrt(n))
                 .allMatch(i -> n % i != 0);

diff --git a/1/complete/primes/src/main/java/primes/PrimesComputer.java b/1/complete/primes/src/main/java/primes/PrimesComputer.java
@@ -21,6 +21,12 @@ public static List<Long> primeSequence(long min, long max) {
             .collect(Collectors.toList());
     }
 
+    /**
+     * If n is not a prime, then n = a * b (some a & b)
+     * Both a and b can't be larger than sqrt(n), or the product of them would be greater than n
+     * One of the factors has to be smaller than sqrt(n).
+     * So, if we don't find any factors by the time we get to sqrt(n), then n must be prime
+     */
     public static boolean isPrime(long n) {
         return LongStream.rangeClosed(2, (long) Math.sqrt(n))
             .allMatch(i -> n % i != 0);