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primes.go
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primes.go
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package main
import (
"fmt"
"math"
)
func main() {
const max = 100
fmt.Println("primes1")
for _, p := range primes1(max) {
fmt.Println(p)
}
fmt.Println("primes2")
for _, p := range primes2(max) {
fmt.Println(p)
}
}
func primes1(max int) []int {
primes := []int{}
for i := 2; i <= max; i++ {
var isComposite bool
for j := 2; j < i; j++ {
if i%j == 0 {
isComposite = true
break
}
}
if !isComposite {
primes = append(primes, i)
}
}
return primes
}
func primes2(max int) []int {
primes := []int{}
outer:
for i := 2; i <= max; i++ {
for j := 2; j < i; j++ {
if i%j == 0 {
continue outer
}
}
primes = append(primes, i)
}
return primes
}
func primes3(max int) []int {
primes := []int{}
outer:
for i := 2; i <= max; i++ {
for _, j := range primes {
if i%j == 0 {
continue outer
}
}
primes = append(primes, i)
}
return primes
}
func primes4(max int) []int {
primes := []int{}
outer:
for i := 2; i <= max; i++ {
sqrt := int(math.Sqrt(float64(i)))
for _, j := range primes {
if j > sqrt {
break
}
if i%j == 0 {
continue outer
}
}
primes = append(primes, i)
}
return primes
}
func primes5(max int) []int {
primes := []int{2}
outer:
for i := 3; i <= max; i += 2 {
sqrt := int(math.Sqrt(float64(i)))
for _, j := range primes {
if j > sqrt {
break
}
if i%j == 0 {
continue outer
}
}
primes = append(primes, i)
}
return primes
}
type primeGenerator6 struct {
primes []int
}
func (pg *primeGenerator6) Next() int {
l := len(pg.primes)
switch l {
case 0:
pg.primes = []int{2}
return 2
case 1:
pg.primes = append(pg.primes, 3)
return 3
default:
i := pg.primes[l-1] // fetch last prime number generated
outer:
i += 2 // next prime candidate is 2 numbers higher; skip even numbers
sqrt := int(math.Sqrt(float64(i)))
for _, j := range pg.primes {
if j > sqrt {
break
}
if i%j == 0 {
goto outer
}
}
pg.primes = append(pg.primes, i)
return i
}
}
type primeGenerator7 struct {
primes []int
next func() int
}
func (pg *primeGenerator7) Next() int {
if pg.next == nil {
pg.next = func() int {
pg.primes = []int{2}
pg.next = func() int {
pg.primes = append(pg.primes, 3)
pg.next = func() int {
i := pg.primes[len(pg.primes)-1] // fetch last prime number generated
outer:
i += 2 // next prime candidate is 2 numbers higher; skip even numbers
sqrt := int(math.Sqrt(float64(i)))
for _, j := range pg.primes {
if j > sqrt {
break
}
if i%j == 0 {
goto outer
}
}
pg.primes = append(pg.primes, i)
return i
}
return 3
}
return 2
}
}
return pg.next()
}
func primes8(max uint) []int {
// Create a bit array to mark known composite numbers. After loop below, a 0
// value means the corresponding natural number is prime, and a 1 value
// means a composite number. Those defaults were chosen to take advantage of
// the zero value for each byte in a slice, assuming all numbers are prime,
// then mark off composite numbers.
//
// While it is not necessarily needed to store whether 0 and 1 are prime,
// reserving bits for those values reduces the number of math instructions
// required to set and check bits. Therefore, the 0th bit is for natural
// number 0. The 1st bit is for natural number 1. And so
// forth. Additionally, no attempt is ever made to set or check the bits
// representing 0 or 1. While they are 0 at the end of the outer loop,
// indicating they would be prime, they remain 0 simply because we never set
// them at the beginning of this function.
b, r := 1+max>>3, max&7
if r > 0 {
b++
}
composites := make([]byte, b)
for i := uint(2); i <= max; i++ {
for j := uint(i << 1); j <= max; j += i {
major, minor := j>>3, j&7
composites[major] |= 1 << minor
}
}
primes := []int{2}
for i := uint(3); i <= max; i += 2 {
major, minor := i>>3, i&7
if composites[major]&(1<<minor) == 0 {
primes = append(primes, int(i))
}
}
return primes
}
func primes9(max uint) []int {
// Create a bit array to mark known composite numbers. After loop below, a 0
// value means the corresponding natural number is prime, and a 1 value
// means a composite number. Those defaults were chosen to take advantage of
// the zero value for each byte in a slice, assuming all numbers are prime,
// then mark off composite numbers.
//
// While it is not necessarily needed to store whether 0 and 1 are prime,
// reserving bits for those values reduces the number of math instructions
// required to set and check bits. Therefore, the 0th bit is for natural
// number 0. The 1st bit is for natural number 1. And so
// forth. Additionally, no attempt is ever made to set or check the bits
// representing 0 or 1. While they are 0 at the end of the outer loop,
// indicating they would be prime, they remain 0 simply because we never set
// them at the beginning of this function.
b, r := 1+max>>3, max&7
if r > 0 {
b++
}
composites := make([]byte, b)
i := uint(2)
outer:
for {
// phase 1: find next prime, leaving it in 'i'.
major, minor := i>>3, i&7
for composites[major]&(1<<minor) != 0 {
if minor++; minor == 8 {
minor = 0
if major++; major == b {
break outer
}
}
i++
}
// phase 2: set bits for all numbers which are multiples of 'i'.
for j := uint(i << 1); j <= max; j += i {
composites[j>>3] |= 1 << (j & 7)
}
if i++; i > max {
break
}
}
primes := []int{2}
for i := uint(3); i <= max; i += 2 {
if composites[i>>3]&(1<<(i&7)) == 0 {
primes = append(primes, int(i))
}
}
return primes
}
func primes10(max uint) []int {
// Create a bit array to mark known composite numbers. After loop below, a 0
// value means the corresponding natural number is prime, and a 1 value
// means a composite number. Those defaults were chosen to take advantage of
// the zero value for each byte in a slice, assuming all numbers are prime,
// then mark off composite numbers.
//
// While it is not necessarily needed to store whether 0 and 1 are prime,
// reserving bits for those values reduces the number of math instructions
// required to set and check bits. Therefore, the 0th bit is for natural
// number 0. The 1st bit is for natural number 1. And so
// forth. Additionally, no attempt is ever made to set or check the bits
// representing 0 or 1. While they are 0 at the end of the outer loop,
// indicating they would be prime, they remain 0 simply because we never set
// them at the beginning of this function.
b, r := 1+max>>3, max&7
if r > 0 {
b++
}
composites := make([]byte, b)
i := uint(2)
outer:
for {
// phase 1: find next prime, leaving it in 'i'.
for composites[i>>3]&(1<<(i&7)) != 0 {
if i++; i > max {
break outer
}
}
// phase 2: set bits for all numbers which are multiples of 'i'.
for j := uint(i << 1); j <= max; j += i {
composites[j>>3] |= 1 << (j & 7)
}
if i++; i > max {
break
}
}
primes := []int{2}
for i := uint(3); i <= max; i += 2 {
if composites[i>>3]&(1<<(i&7)) == 0 {
primes = append(primes, int(i))
}
}
return primes
}
func primes11(max uint) []int {
// Create a bit array to mark known composite numbers. After loop below, a 0
// value means the corresponding natural number is prime, and a 1 value
// means a composite number. Those defaults were chosen to take advantage of
// the zero value for each byte in a slice, assuming all numbers are prime,
// then mark off composite numbers.
//
// While it is not necessarily needed to store whether 0 and 1 are prime,
// reserving bits for those values reduces the number of math instructions
// required to set and check bits. Therefore, the 0th bit is for natural
// number 0. The 1st bit is for natural number 1. And so
// forth. Additionally, no attempt is ever made to set or check the bits
// representing 0 or 1. While they are 0 at the end of the outer loop,
// indicating they would be prime, they remain 0 simply because we never set
// them at the beginning of this function.
b, r := 1+max>>3, max&7
if r > 0 {
b++
}
composites := make([]byte, b)
for i := uint(2); i <= max; i++ {
if composites[i>>3]&(1<<(i&7)) == 0 {
// When i is prime; mark composites for all multiples of 'i'.
for j := uint(i << 1); j <= max; j += i {
composites[j>>3] |= 1 << (j & 7)
}
}
}
primes := []int{2}
for i := uint(3); i <= max; i += 2 {
if composites[i>>3]&(1<<(i&7)) == 0 {
primes = append(primes, int(i))
}
}
return primes
}
func primes12(max uint) []int {
// Create a bit array to mark known composite numbers. After loop below, a 0
// value means the corresponding natural number is prime, and a 1 value
// means a composite number. Those defaults were chosen to take advantage of
// the zero value for each byte in a slice, assuming all numbers are prime,
// then mark off composite numbers.
//
// While it is not necessarily needed to store whether 0 and 1 are prime,
// reserving bits for those values reduces the number of math instructions
// required to set and check bits. Therefore, the 0th bit is for natural
// number 0. The 1st bit is for natural number 1. And so
// forth. Additionally, no attempt is ever made to set or check the bits
// representing 0 or 1. While they are 0 at the end of the outer loop,
// indicating they would be prime, they remain 0 simply because we never set
// them at the beginning of this function.
b, r := 1+max>>3, max&7
if r > 0 {
b++
}
composites := make([]byte, b)
// Special case mark multiples of 2.
for i := uint(4); i <= max; i += 2 {
composites[i>>3] |= 1 << (i & 7)
}
// Starting at 3, check all odd numbers;
for i := uint(3); i <= max; i += 2 {
if composites[i>>3]&(1<<(i&7)) == 0 {
// When i is prime; mark composites for all multiples of 'i'.
for j := uint(i << 1); j <= max; j += i {
composites[j>>3] |= 1 << (j & 7)
}
}
}
primes := []int{2}
for i := uint(3); i <= max; i += 2 {
if composites[i>>3]&(1<<(i&7)) == 0 {
primes = append(primes, int(i))
}
}
return primes
}
func primes13(max uint) []int {
// Create a bit array to mark known composite numbers. After loop below, a 0
// value means the corresponding natural number is prime, and a 1 value
// means a composite number. Those defaults were chosen to take advantage of
// the zero value for each byte in a slice, assuming all numbers are prime,
// then mark off composite numbers.
//
// While it is not necessarily needed to store whether 0 and 1 are prime,
// reserving bits for those values reduces the number of math instructions
// required to set and check bits. Therefore, the 0th bit is for natural
// number 0. The 1st bit is for natural number 1. And so
// forth. Additionally, no attempt is ever made to set or check the bits
// representing 0 or 1. While they are 0 at the end of the outer loop,
// indicating they would be prime, they remain 0 simply because we never set
// them at the beginning of this function.
b, r := 1+max>>3, max&7
if r > 0 {
b++
}
composites := make([]byte, b)
// Special case mark multiples of 2.
for i := uint(4); i <= max; i += 2 {
composites[i>>3] |= 1 << (i & 7)
}
primes := []int{2}
// Starting at 3, check all odd numbers;
for i := uint(3); i <= max; i += 2 {
if composites[i>>3]&(1<<(i&7)) == 0 {
primes = append(primes, int(i))
// When i is prime; mark composites for all multiples of 'i'.
for j := uint(i << 1); j <= max; j += i {
composites[j>>3] |= 1 << (j & 7)
}
}
}
return primes
}
func primes14(max uint) []int {
// Create a bit array to mark known composite numbers. After loop below, a 0
// value means the corresponding natural number is prime, and a 1 value
// means a composite number. Those defaults were chosen to take advantage of
// the zero value for each byte in a slice, assuming all numbers are prime,
// then mark off composite numbers.
//
// While it is not necessarily needed to store whether 0 and 1 are prime,
// reserving bits for those values reduces the number of math instructions
// required to set and check bits. Therefore, the 0th bit is for natural
// number 0. The 1st bit is for natural number 1. And so
// forth. Additionally, no attempt is ever made to set or check the bits
// representing 0 or 1. While they are 0 at the end of the outer loop,
// indicating they would be prime, they remain 0 simply because we never set
// them at the beginning of this function.
b, r := 1+max>>3, max&7
if r > 0 {
b++
}
composites := make([]byte, b)
// Special case mark multiples of 2.
for i := uint(0); i < b; i++ {
composites[i] = 0x55
}
primes := []int{2}
// Starting at 3, check all odd numbers
for i := uint(3); i <= max; i += 2 {
if composites[i>>3]&(1<<(i&7)) == 0 {
primes = append(primes, int(i))
// When i is prime; mark composites for all multiples of 'i'.
for j := uint(i << 1); j <= max; j += i {
composites[j>>3] |= 1 << (j & 7)
}
}
}
return primes
}
func primes15(max uint) []int {
// Create a bit array to mark known composite numbers. After loop below, a 0
// value means the corresponding natural number is prime, and a 1 value
// means a composite number. Those defaults were chosen to take advantage of
// the zero value for each byte in a slice, assuming all numbers are prime,
// then mark off composite numbers.
//
// While it is not necessarily needed to store whether 0 and 1 are prime,
// reserving bits for those values reduces the number of math instructions
// required to set and check bits. Therefore, the 0th bit is for natural
// number 0. The 1st bit is for natural number 1. And so
// forth. Additionally, no attempt is ever made to set or check the bits
// representing 0 or 1. While they are 0 at the end of the outer loop,
// indicating they would be prime, they remain 0 simply because we never set
// them at the beginning of this function.
b := 1 + max>>6
composites := make([]uint64, b)
// Special case mark multiples of 2.
for i := uint(0); i < b; i++ {
composites[i] = 0x5555555555555555
}
primes := []int{2}
// Starting at 3, check all odd numbers
for i := uint(3); i <= max; i += 2 {
if composites[i>>6]&(1<<(i&63)) == 0 {
primes = append(primes, int(i))
// When i is prime; mark composites for all multiples of 'i'.
for j := uint(i << 1); j <= max; j += i {
composites[j>>6] |= 1 << (j & 63)
}
}
}
return primes
}