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Prime Number Counter in Golang
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/* | |
I was able to achieve these speeds on my laptop: Intel Core i7-4500U @ 1.80GHZ | |
2015/08/20 09:28:05 Test 1: 664579 took 47.0074ms | |
2015/08/20 09:28:05 Test 2: 5761455 took 626.9954ms | |
2015/08/20 09:28:13 Test 3: 50847534 took 8.0860009s | |
*/ | |
package main | |
import ( | |
"log" | |
"math" | |
"time" | |
) | |
func main() { | |
for i, v := range []int{10000000, 100000000, 1000000000} { | |
start := time.Now() | |
num := countPrimes(v) | |
elapsed := time.Since(start) | |
log.Printf("Test %d: %v \t took %s", i+1, num, elapsed) | |
} | |
} | |
func countPrimes(limit int) int { | |
// Return if less than 1 | |
if limit <= 1 { | |
return 0 | |
} | |
// Get the sqrt of the limit | |
sqrtLimit := int(math.Sqrt(float64(limit))) | |
// Create array | |
numbers := make([]bool, limit) | |
// Set 1 to prime | |
numbers[0] = true | |
numPrimes := 0 | |
// Count the number of olds | |
if limit%2 == 0 { | |
numPrimes = limit / 2 | |
} else { | |
numPrimes = (limit + 1) / 2 | |
} | |
// Loop through odd numbers | |
for i := 3; i <= sqrtLimit; i += 2 { | |
if !numbers[i] { | |
for j := i*i; j < limit; j += i*2 { | |
if !numbers[j] { | |
numbers[j] = true | |
numPrimes -= 1 | |
} | |
} | |
} | |
} | |
return numPrimes | |
} |
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