Nth Fibonacci number in Go using recursion and concurrency











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The Java Code I'm attempting to translate. I've been trying to implement this java method of getting the nth Fibonacci number in Go but I can't seem to get my code past the number Fibonacci number 35 before it crashes. This method is supposed to be very inefficient but not so inefficient that it doesn't complete.





package main

import (
"fmt"
"time"
)

type Fibonacci struct {
num float64
answer float64
}

func newFibonacci(n float64) *Fibonacci {

f := new(Fibonacci)
f.num = n
c1 := make(chan float64)
c2 := make(chan float64)

if f.num <= 1 {
f.answer = n
} else {
go func() {
fib1 := newFibonacci(n - 1)
c2 <- fib1.answer
}()
go func() {
fib2 := newFibonacci(n - 2)
c1 <- fib2.answer
}()

f.answer = <-c2 + <-c1
}
close(c1)
close(c2)

return f
}

func main() {

numbers := float64{30, 35, 36, 37, 38, 39, 40}
for _, value := range numbers{
start := time.Now()
fmt.Println("getting the ", value, " fibonacci number")
f := newFibonacci(value)
fmt.Println(f.answer)
end := time.Now()
totalTime := end.Sub(start)
fmt.Println("Fibonacci number: ", value, " took ", totalTime, "n")
}

}











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  • You have a combinatorial explodion: cs.toronto.edu/~gfb/csc104/2016W/Lectures/…
    – peterSO
    Nov 5 at 2:24










  • "supposed to be very efficient" and then "recursive fibonacci implementation", these two doesn't go together, period. An efficient fibonacci implementation uses a loop, and isn't recursive. The recursive variant is only good for one thing and that is teaching about the pitfalls of recursion.
    – Lasse Vågsæther Karlsen
    Nov 5 at 12:20












  • "supposed to be very inefficient" and I am using this to teach the pitfalls. I'm hoping to implement a worker pool and limit the amount of workers to the number of available CPUs. I would just like to get this working to show the differences.
    – Swade
    Nov 5 at 17:15















up vote
1
down vote

favorite












The Java Code I'm attempting to translate. I've been trying to implement this java method of getting the nth Fibonacci number in Go but I can't seem to get my code past the number Fibonacci number 35 before it crashes. This method is supposed to be very inefficient but not so inefficient that it doesn't complete.





package main

import (
"fmt"
"time"
)

type Fibonacci struct {
num float64
answer float64
}

func newFibonacci(n float64) *Fibonacci {

f := new(Fibonacci)
f.num = n
c1 := make(chan float64)
c2 := make(chan float64)

if f.num <= 1 {
f.answer = n
} else {
go func() {
fib1 := newFibonacci(n - 1)
c2 <- fib1.answer
}()
go func() {
fib2 := newFibonacci(n - 2)
c1 <- fib2.answer
}()

f.answer = <-c2 + <-c1
}
close(c1)
close(c2)

return f
}

func main() {

numbers := float64{30, 35, 36, 37, 38, 39, 40}
for _, value := range numbers{
start := time.Now()
fmt.Println("getting the ", value, " fibonacci number")
f := newFibonacci(value)
fmt.Println(f.answer)
end := time.Now()
totalTime := end.Sub(start)
fmt.Println("Fibonacci number: ", value, " took ", totalTime, "n")
}

}











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  • You have a combinatorial explodion: cs.toronto.edu/~gfb/csc104/2016W/Lectures/…
    – peterSO
    Nov 5 at 2:24










  • "supposed to be very efficient" and then "recursive fibonacci implementation", these two doesn't go together, period. An efficient fibonacci implementation uses a loop, and isn't recursive. The recursive variant is only good for one thing and that is teaching about the pitfalls of recursion.
    – Lasse Vågsæther Karlsen
    Nov 5 at 12:20












  • "supposed to be very inefficient" and I am using this to teach the pitfalls. I'm hoping to implement a worker pool and limit the amount of workers to the number of available CPUs. I would just like to get this working to show the differences.
    – Swade
    Nov 5 at 17:15













up vote
1
down vote

favorite









up vote
1
down vote

favorite











The Java Code I'm attempting to translate. I've been trying to implement this java method of getting the nth Fibonacci number in Go but I can't seem to get my code past the number Fibonacci number 35 before it crashes. This method is supposed to be very inefficient but not so inefficient that it doesn't complete.





package main

import (
"fmt"
"time"
)

type Fibonacci struct {
num float64
answer float64
}

func newFibonacci(n float64) *Fibonacci {

f := new(Fibonacci)
f.num = n
c1 := make(chan float64)
c2 := make(chan float64)

if f.num <= 1 {
f.answer = n
} else {
go func() {
fib1 := newFibonacci(n - 1)
c2 <- fib1.answer
}()
go func() {
fib2 := newFibonacci(n - 2)
c1 <- fib2.answer
}()

f.answer = <-c2 + <-c1
}
close(c1)
close(c2)

return f
}

func main() {

numbers := float64{30, 35, 36, 37, 38, 39, 40}
for _, value := range numbers{
start := time.Now()
fmt.Println("getting the ", value, " fibonacci number")
f := newFibonacci(value)
fmt.Println(f.answer)
end := time.Now()
totalTime := end.Sub(start)
fmt.Println("Fibonacci number: ", value, " took ", totalTime, "n")
}

}











share|improve this question







New contributor




Swade is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











The Java Code I'm attempting to translate. I've been trying to implement this java method of getting the nth Fibonacci number in Go but I can't seem to get my code past the number Fibonacci number 35 before it crashes. This method is supposed to be very inefficient but not so inefficient that it doesn't complete.





package main

import (
"fmt"
"time"
)

type Fibonacci struct {
num float64
answer float64
}

func newFibonacci(n float64) *Fibonacci {

f := new(Fibonacci)
f.num = n
c1 := make(chan float64)
c2 := make(chan float64)

if f.num <= 1 {
f.answer = n
} else {
go func() {
fib1 := newFibonacci(n - 1)
c2 <- fib1.answer
}()
go func() {
fib2 := newFibonacci(n - 2)
c1 <- fib2.answer
}()

f.answer = <-c2 + <-c1
}
close(c1)
close(c2)

return f
}

func main() {

numbers := float64{30, 35, 36, 37, 38, 39, 40}
for _, value := range numbers{
start := time.Now()
fmt.Println("getting the ", value, " fibonacci number")
f := newFibonacci(value)
fmt.Println(f.answer)
end := time.Now()
totalTime := end.Sub(start)
fmt.Println("Fibonacci number: ", value, " took ", totalTime, "n")
}

}








go recursion concurrency fibonacci goroutine






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asked Nov 5 at 1:36









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New contributor





Swade is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






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Check out our Code of Conduct.












  • You have a combinatorial explodion: cs.toronto.edu/~gfb/csc104/2016W/Lectures/…
    – peterSO
    Nov 5 at 2:24










  • "supposed to be very efficient" and then "recursive fibonacci implementation", these two doesn't go together, period. An efficient fibonacci implementation uses a loop, and isn't recursive. The recursive variant is only good for one thing and that is teaching about the pitfalls of recursion.
    – Lasse Vågsæther Karlsen
    Nov 5 at 12:20












  • "supposed to be very inefficient" and I am using this to teach the pitfalls. I'm hoping to implement a worker pool and limit the amount of workers to the number of available CPUs. I would just like to get this working to show the differences.
    – Swade
    Nov 5 at 17:15


















  • You have a combinatorial explodion: cs.toronto.edu/~gfb/csc104/2016W/Lectures/…
    – peterSO
    Nov 5 at 2:24










  • "supposed to be very efficient" and then "recursive fibonacci implementation", these two doesn't go together, period. An efficient fibonacci implementation uses a loop, and isn't recursive. The recursive variant is only good for one thing and that is teaching about the pitfalls of recursion.
    – Lasse Vågsæther Karlsen
    Nov 5 at 12:20












  • "supposed to be very inefficient" and I am using this to teach the pitfalls. I'm hoping to implement a worker pool and limit the amount of workers to the number of available CPUs. I would just like to get this working to show the differences.
    – Swade
    Nov 5 at 17:15
















You have a combinatorial explodion: cs.toronto.edu/~gfb/csc104/2016W/Lectures/…
– peterSO
Nov 5 at 2:24




You have a combinatorial explodion: cs.toronto.edu/~gfb/csc104/2016W/Lectures/…
– peterSO
Nov 5 at 2:24












"supposed to be very efficient" and then "recursive fibonacci implementation", these two doesn't go together, period. An efficient fibonacci implementation uses a loop, and isn't recursive. The recursive variant is only good for one thing and that is teaching about the pitfalls of recursion.
– Lasse Vågsæther Karlsen
Nov 5 at 12:20






"supposed to be very efficient" and then "recursive fibonacci implementation", these two doesn't go together, period. An efficient fibonacci implementation uses a loop, and isn't recursive. The recursive variant is only good for one thing and that is teaching about the pitfalls of recursion.
– Lasse Vågsæther Karlsen
Nov 5 at 12:20














"supposed to be very inefficient" and I am using this to teach the pitfalls. I'm hoping to implement a worker pool and limit the amount of workers to the number of available CPUs. I would just like to get this working to show the differences.
– Swade
Nov 5 at 17:15




"supposed to be very inefficient" and I am using this to teach the pitfalls. I'm hoping to implement a worker pool and limit the amount of workers to the number of available CPUs. I would just like to get this working to show the differences.
– Swade
Nov 5 at 17:15












2 Answers
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0
down vote













I suggest you do the math behind how many goroutines you're actually launching.



Your first call spawns two goroutines. Each of those spawn two further goroutines. The total number of goroutines spawned will be 2^1 + 2^2 + 2^3 + ... + 2^n. While a lot of these will exit before you reach the number below, it's still a lot of goroutines.



Sum






share|improve this answer




























    up vote
    0
    down vote













    You program create exponential number of go routines. That comes with context switching latency. Try to run it for limited number of go routines. See following code where i run it for limited number of go routines:



    package main

    import (
    "fmt"
    "time"
    )

    type Fibonacci struct {
    num float64
    answer float64
    }

    type goRoutineManager struct {
    goRoutineCnt chan bool
    }

    func (g *goRoutineManager) Run(f func()) {
    select {
    case g.goRoutineCnt <- true:
    go func() {
    f()
    <-g.goRoutineCnt
    }()
    default:
    f()
    }
    }

    func NewGoRoutineManager(goRoutineLimit int) *goRoutineManager {
    return &goRoutineManager{
    goRoutineCnt: make(chan bool, goRoutineLimit),
    }
    }

    func newFibonacci(n float64, gm *goRoutineManager) *Fibonacci {

    f := new(Fibonacci)
    f.num = n
    c1 := make(chan float64, 1)
    c2 := make(chan float64, 1)

    if f.num <= 1 {
    f.answer = n
    } else {

    gm.Run(func() {
    fib1 := newFibonacci(n-1, gm)
    c2 <- fib1.answer
    })

    gm.Run(func() {
    fib2 := newFibonacci(n-2, gm)
    c1 <- fib2.answer
    })

    f.answer = <-c2 + <-c1
    }
    close(c1)
    close(c2)

    return f
    }

    func main() {

    numbers := float64{30, 35, 36, 37, 38, 39, 40} //{30, 35, 36, 37, 38, 39, 40}
    for _, value := range numbers {
    start := time.Now()
    fmt.Println("getting the ", value, " fibonacci number")

    gm := NewGoRoutineManager(3)

    f := newFibonacci(value, gm)
    fmt.Println(f.answer)

    end := time.Now()
    totalTime := end.Sub(start)
    fmt.Println("Fibonacci number: ", value, " took ", totalTime, "n")
    }

    }





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      2 Answers
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      active

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      2 Answers
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      active

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      up vote
      0
      down vote













      I suggest you do the math behind how many goroutines you're actually launching.



      Your first call spawns two goroutines. Each of those spawn two further goroutines. The total number of goroutines spawned will be 2^1 + 2^2 + 2^3 + ... + 2^n. While a lot of these will exit before you reach the number below, it's still a lot of goroutines.



      Sum






      share|improve this answer

























        up vote
        0
        down vote













        I suggest you do the math behind how many goroutines you're actually launching.



        Your first call spawns two goroutines. Each of those spawn two further goroutines. The total number of goroutines spawned will be 2^1 + 2^2 + 2^3 + ... + 2^n. While a lot of these will exit before you reach the number below, it's still a lot of goroutines.



        Sum






        share|improve this answer























          up vote
          0
          down vote










          up vote
          0
          down vote









          I suggest you do the math behind how many goroutines you're actually launching.



          Your first call spawns two goroutines. Each of those spawn two further goroutines. The total number of goroutines spawned will be 2^1 + 2^2 + 2^3 + ... + 2^n. While a lot of these will exit before you reach the number below, it's still a lot of goroutines.



          Sum






          share|improve this answer












          I suggest you do the math behind how many goroutines you're actually launching.



          Your first call spawns two goroutines. Each of those spawn two further goroutines. The total number of goroutines spawned will be 2^1 + 2^2 + 2^3 + ... + 2^n. While a lot of these will exit before you reach the number below, it's still a lot of goroutines.



          Sum







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 5 at 3:55









          Luke Joshua Park

          4,73051630




          4,73051630
























              up vote
              0
              down vote













              You program create exponential number of go routines. That comes with context switching latency. Try to run it for limited number of go routines. See following code where i run it for limited number of go routines:



              package main

              import (
              "fmt"
              "time"
              )

              type Fibonacci struct {
              num float64
              answer float64
              }

              type goRoutineManager struct {
              goRoutineCnt chan bool
              }

              func (g *goRoutineManager) Run(f func()) {
              select {
              case g.goRoutineCnt <- true:
              go func() {
              f()
              <-g.goRoutineCnt
              }()
              default:
              f()
              }
              }

              func NewGoRoutineManager(goRoutineLimit int) *goRoutineManager {
              return &goRoutineManager{
              goRoutineCnt: make(chan bool, goRoutineLimit),
              }
              }

              func newFibonacci(n float64, gm *goRoutineManager) *Fibonacci {

              f := new(Fibonacci)
              f.num = n
              c1 := make(chan float64, 1)
              c2 := make(chan float64, 1)

              if f.num <= 1 {
              f.answer = n
              } else {

              gm.Run(func() {
              fib1 := newFibonacci(n-1, gm)
              c2 <- fib1.answer
              })

              gm.Run(func() {
              fib2 := newFibonacci(n-2, gm)
              c1 <- fib2.answer
              })

              f.answer = <-c2 + <-c1
              }
              close(c1)
              close(c2)

              return f
              }

              func main() {

              numbers := float64{30, 35, 36, 37, 38, 39, 40} //{30, 35, 36, 37, 38, 39, 40}
              for _, value := range numbers {
              start := time.Now()
              fmt.Println("getting the ", value, " fibonacci number")

              gm := NewGoRoutineManager(3)

              f := newFibonacci(value, gm)
              fmt.Println(f.answer)

              end := time.Now()
              totalTime := end.Sub(start)
              fmt.Println("Fibonacci number: ", value, " took ", totalTime, "n")
              }

              }





              share|improve this answer

























                up vote
                0
                down vote













                You program create exponential number of go routines. That comes with context switching latency. Try to run it for limited number of go routines. See following code where i run it for limited number of go routines:



                package main

                import (
                "fmt"
                "time"
                )

                type Fibonacci struct {
                num float64
                answer float64
                }

                type goRoutineManager struct {
                goRoutineCnt chan bool
                }

                func (g *goRoutineManager) Run(f func()) {
                select {
                case g.goRoutineCnt <- true:
                go func() {
                f()
                <-g.goRoutineCnt
                }()
                default:
                f()
                }
                }

                func NewGoRoutineManager(goRoutineLimit int) *goRoutineManager {
                return &goRoutineManager{
                goRoutineCnt: make(chan bool, goRoutineLimit),
                }
                }

                func newFibonacci(n float64, gm *goRoutineManager) *Fibonacci {

                f := new(Fibonacci)
                f.num = n
                c1 := make(chan float64, 1)
                c2 := make(chan float64, 1)

                if f.num <= 1 {
                f.answer = n
                } else {

                gm.Run(func() {
                fib1 := newFibonacci(n-1, gm)
                c2 <- fib1.answer
                })

                gm.Run(func() {
                fib2 := newFibonacci(n-2, gm)
                c1 <- fib2.answer
                })

                f.answer = <-c2 + <-c1
                }
                close(c1)
                close(c2)

                return f
                }

                func main() {

                numbers := float64{30, 35, 36, 37, 38, 39, 40} //{30, 35, 36, 37, 38, 39, 40}
                for _, value := range numbers {
                start := time.Now()
                fmt.Println("getting the ", value, " fibonacci number")

                gm := NewGoRoutineManager(3)

                f := newFibonacci(value, gm)
                fmt.Println(f.answer)

                end := time.Now()
                totalTime := end.Sub(start)
                fmt.Println("Fibonacci number: ", value, " took ", totalTime, "n")
                }

                }





                share|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  You program create exponential number of go routines. That comes with context switching latency. Try to run it for limited number of go routines. See following code where i run it for limited number of go routines:



                  package main

                  import (
                  "fmt"
                  "time"
                  )

                  type Fibonacci struct {
                  num float64
                  answer float64
                  }

                  type goRoutineManager struct {
                  goRoutineCnt chan bool
                  }

                  func (g *goRoutineManager) Run(f func()) {
                  select {
                  case g.goRoutineCnt <- true:
                  go func() {
                  f()
                  <-g.goRoutineCnt
                  }()
                  default:
                  f()
                  }
                  }

                  func NewGoRoutineManager(goRoutineLimit int) *goRoutineManager {
                  return &goRoutineManager{
                  goRoutineCnt: make(chan bool, goRoutineLimit),
                  }
                  }

                  func newFibonacci(n float64, gm *goRoutineManager) *Fibonacci {

                  f := new(Fibonacci)
                  f.num = n
                  c1 := make(chan float64, 1)
                  c2 := make(chan float64, 1)

                  if f.num <= 1 {
                  f.answer = n
                  } else {

                  gm.Run(func() {
                  fib1 := newFibonacci(n-1, gm)
                  c2 <- fib1.answer
                  })

                  gm.Run(func() {
                  fib2 := newFibonacci(n-2, gm)
                  c1 <- fib2.answer
                  })

                  f.answer = <-c2 + <-c1
                  }
                  close(c1)
                  close(c2)

                  return f
                  }

                  func main() {

                  numbers := float64{30, 35, 36, 37, 38, 39, 40} //{30, 35, 36, 37, 38, 39, 40}
                  for _, value := range numbers {
                  start := time.Now()
                  fmt.Println("getting the ", value, " fibonacci number")

                  gm := NewGoRoutineManager(3)

                  f := newFibonacci(value, gm)
                  fmt.Println(f.answer)

                  end := time.Now()
                  totalTime := end.Sub(start)
                  fmt.Println("Fibonacci number: ", value, " took ", totalTime, "n")
                  }

                  }





                  share|improve this answer












                  You program create exponential number of go routines. That comes with context switching latency. Try to run it for limited number of go routines. See following code where i run it for limited number of go routines:



                  package main

                  import (
                  "fmt"
                  "time"
                  )

                  type Fibonacci struct {
                  num float64
                  answer float64
                  }

                  type goRoutineManager struct {
                  goRoutineCnt chan bool
                  }

                  func (g *goRoutineManager) Run(f func()) {
                  select {
                  case g.goRoutineCnt <- true:
                  go func() {
                  f()
                  <-g.goRoutineCnt
                  }()
                  default:
                  f()
                  }
                  }

                  func NewGoRoutineManager(goRoutineLimit int) *goRoutineManager {
                  return &goRoutineManager{
                  goRoutineCnt: make(chan bool, goRoutineLimit),
                  }
                  }

                  func newFibonacci(n float64, gm *goRoutineManager) *Fibonacci {

                  f := new(Fibonacci)
                  f.num = n
                  c1 := make(chan float64, 1)
                  c2 := make(chan float64, 1)

                  if f.num <= 1 {
                  f.answer = n
                  } else {

                  gm.Run(func() {
                  fib1 := newFibonacci(n-1, gm)
                  c2 <- fib1.answer
                  })

                  gm.Run(func() {
                  fib2 := newFibonacci(n-2, gm)
                  c1 <- fib2.answer
                  })

                  f.answer = <-c2 + <-c1
                  }
                  close(c1)
                  close(c2)

                  return f
                  }

                  func main() {

                  numbers := float64{30, 35, 36, 37, 38, 39, 40} //{30, 35, 36, 37, 38, 39, 40}
                  for _, value := range numbers {
                  start := time.Now()
                  fmt.Println("getting the ", value, " fibonacci number")

                  gm := NewGoRoutineManager(3)

                  f := newFibonacci(value, gm)
                  fmt.Println(f.answer)

                  end := time.Now()
                  totalTime := end.Sub(start)
                  fmt.Println("Fibonacci number: ", value, " took ", totalTime, "n")
                  }

                  }






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                  answered Nov 5 at 12:17









                  nightfury1204

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