Finding the longest Collatz sequence using Go, concurrently
$begingroup$
I am working through this problem on project euler.
Basic purpose of the code:
/*
if n is even n -> n/2
if n is odd n -> 3n + 1
which starting number under 1 million produces the longest sequence?
*/
The trouble with the code is that when I implement concurrency as I have, the program runs considerably more slowly than when I am just returning values directly from the get_seq_len function. Am I implementing concurrency incorrectly? Or is the program just not complex enough that there is speed to be gained from concurrency or something?
package main
import ("fmt";"math";"time")
This function takes a number x and then using the formula described in the above comment block creates a sequence of numbers from it until the number 1 is reached. Originally I had the function simply returning the length of the sequence, but now I have it sending the length to the specified channel.
func get_seq_len(x int, c chan int) {
var count int = 1
for {
if x == 1 {
break
} else if x%2 == 0 {
x = x / 2
} else {
x = 3 * x + 1
}
count++
}
// return count + 1
c <- count + 1
}
Here the main function just step through every number from 1 to whatever the max is and then tests the length of the sequence for that number. I use channel c to pass the sequenced length back to main().
func main() {
start := time.Now()
var max int = int(math.Pow(10,6))
// var max int = int(math.Pow(3,3))
var max_length int = 0
var max_num int = 0
c := make(chan int, 50)
for i:=1; i<max; i++ {
// length := get_seq_len(i, c)
go get_seq_len(i, c)
length := <-c
if length > max_length { max_length = length; max_num = i }
}
fmt.Println(max_num)
t := time.Now()
elapsed := t.Sub(start)
fmt.Println(elapsed)
}
performance programming-challenge go concurrency collatz-sequence
edited 9 mins ago
200_success
130k16153417
asked 1 hour ago
The NightmanThe Nightman
1947
$endgroup$
add a comment |
$begingroup$
I am working through this problem on project euler.
Basic purpose of the code:
/*
if n is even n -> n/2
if n is odd n -> 3n + 1
which starting number under 1 million produces the longest sequence?
*/
The trouble with the code is that when I implement concurrency as I have, the program runs considerably more slowly than when I am just returning values directly from the get_seq_len function. Am I implementing concurrency incorrectly? Or is the program just not complex enough that there is speed to be gained from concurrency or something?
package main
import ("fmt";"math";"time")
This function takes a number x and then using the formula described in the above comment block creates a sequence of numbers from it until the number 1 is reached. Originally I had the function simply returning the length of the sequence, but now I have it sending the length to the specified channel.
func get_seq_len(x int, c chan int) {
var count int = 1
for {
if x == 1 {
break
} else if x%2 == 0 {
x = x / 2
} else {
x = 3 * x + 1
}
count++
}
// return count + 1
c <- count + 1
}
Here the main function just step through every number from 1 to whatever the max is and then tests the length of the sequence for that number. I use channel c to pass the sequenced length back to main().
func main() {
start := time.Now()
var max int = int(math.Pow(10,6))
// var max int = int(math.Pow(3,3))
var max_length int = 0
var max_num int = 0
c := make(chan int, 50)
for i:=1; i<max; i++ {
// length := get_seq_len(i, c)
go get_seq_len(i, c)
length := <-c
if length > max_length { max_length = length; max_num = i }
}
fmt.Println(max_num)
t := time.Now()
elapsed := t.Sub(start)
fmt.Println(elapsed)
}
performance programming-challenge go concurrency collatz-sequence
edited 9 mins ago
200_success
130k16153417
asked 1 hour ago
The NightmanThe Nightman
1947
$endgroup$
add a comment |
$begingroup$
I am working through this problem on project euler.
Basic purpose of the code:
/*
if n is even n -> n/2
if n is odd n -> 3n + 1
which starting number under 1 million produces the longest sequence?
*/
The trouble with the code is that when I implement concurrency as I have, the program runs considerably more slowly than when I am just returning values directly from the get_seq_len function. Am I implementing concurrency incorrectly? Or is the program just not complex enough that there is speed to be gained from concurrency or something?
package main
import ("fmt";"math";"time")
This function takes a number x and then using the formula described in the above comment block creates a sequence of numbers from it until the number 1 is reached. Originally I had the function simply returning the length of the sequence, but now I have it sending the length to the specified channel.
func get_seq_len(x int, c chan int) {
var count int = 1
for {
if x == 1 {
break
} else if x%2 == 0 {
x = x / 2
} else {
x = 3 * x + 1
}
count++
}
// return count + 1
c <- count + 1
}
Here the main function just step through every number from 1 to whatever the max is and then tests the length of the sequence for that number. I use channel c to pass the sequenced length back to main().
func main() {
start := time.Now()
var max int = int(math.Pow(10,6))
// var max int = int(math.Pow(3,3))
var max_length int = 0
var max_num int = 0
c := make(chan int, 50)
for i:=1; i<max; i++ {
// length := get_seq_len(i, c)
go get_seq_len(i, c)
length := <-c
if length > max_length { max_length = length; max_num = i }
}
fmt.Println(max_num)
t := time.Now()
elapsed := t.Sub(start)
fmt.Println(elapsed)
}
performance programming-challenge go concurrency collatz-sequence
edited 9 mins ago
200_success
130k16153417
asked 1 hour ago
The NightmanThe Nightman
1947
$endgroup$
I am working through this problem on project euler.
Basic purpose of the code:
/*
if n is even n -> n/2
if n is odd n -> 3n + 1
which starting number under 1 million produces the longest sequence?
*/
The trouble with the code is that when I implement concurrency as I have, the program runs considerably more slowly than when I am just returning values directly from the get_seq_len function. Am I implementing concurrency incorrectly? Or is the program just not complex enough that there is speed to be gained from concurrency or something?
package main
import ("fmt";"math";"time")
This function takes a number x and then using the formula described in the above comment block creates a sequence of numbers from it until the number 1 is reached. Originally I had the function simply returning the length of the sequence, but now I have it sending the length to the specified channel.
func get_seq_len(x int, c chan int) {
var count int = 1
for {
if x == 1 {
break
} else if x%2 == 0 {
x = x / 2
} else {
x = 3 * x + 1
}
count++
}
// return count + 1
c <- count + 1
}
Here the main function just step through every number from 1 to whatever the max is and then tests the length of the sequence for that number. I use channel c to pass the sequenced length back to main().
func main() {
start := time.Now()
var max int = int(math.Pow(10,6))
// var max int = int(math.Pow(3,3))
var max_length int = 0
var max_num int = 0
c := make(chan int, 50)
for i:=1; i<max; i++ {
// length := get_seq_len(i, c)
go get_seq_len(i, c)
length := <-c
if length > max_length { max_length = length; max_num = i }
}
fmt.Println(max_num)
t := time.Now()
elapsed := t.Sub(start)
fmt.Println(elapsed)
}
performance programming-challenge go concurrency collatz-sequence
performance programming-challenge go concurrency collatz-sequence
edited 9 mins ago
200_success
130k16153417
asked 1 hour ago
The NightmanThe Nightman
1947
edited 9 mins ago
200_success
130k16153417
asked 1 hour ago
The NightmanThe Nightman
1947
edited 9 mins ago
200_success
130k16153417
edited 9 mins ago
200_success
130k16153417
edited 9 mins ago
200_success
130k16153417
130k16153417
asked 1 hour ago
The NightmanThe Nightman
1947
asked 1 hour ago
The NightmanThe Nightman
1947
asked 1 hour ago
The NightmanThe Nightman
1947
1947
add a comment |
add a comment |
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