Fast Fibonacci using recursion

Question

I implemented a Fibonacci function in Scala and it works fine however when I enter 50 it takes a long time to compute it because it has to calculate the 2 previous integers each time. I found a function that keeps the 2 previous numbers. However, can somebody tell me how to write this function to make it accept 2 integers instead of 3 and return the last 2 numbers to compute the Fibonacci at a particular index x. Thanks!

    def fastFib(x: Long ): Long = {
      def fast(x:Long , a:Long, b:Long):Long = 
      if (x<=0) a+b 
      else fast(x-1,b,a+b)
      if (x<2) 1 
      else fast(x-2,0,1)
   }

Try searching first. http://stackoverflow.com/questions/7388416/what-is-the-fastest-way-to-write-fibonacci-function-in-scala — psisoyev, Nov 02 '13 at 15:19
Oh well, like in the link posted you can use the explicit formula that doesnt need previous results — Stefan Kunze, Nov 02 '13 at 15:20
memoisation technique can be very helpful. http://blog.tmorris.net/posts/memoisation-with-state-using-scala/ — Shrey, Nov 03 '13 at 08:06
I don't understand the memoisation technique... can somebody explain how my function gets a and b please? I need to understand this code before I use it. Thanks — user2947615, Nov 05 '13 at 21:16

score 0 · Answer 1 · answered Nov 22 '14 at 22:13

You can cache the intermediate results, then you never recompute the same result twice

Here is the code

//this supposed to contains all the value when initialized
//initialized with 0 for all value
val cache = Array [Int] (101);//0 to 100
cache(1)==1;//initial value
cache(2)=1;//initial value

def fibonacciCache(n:Int) : Int = {
  if (n>100)
  {
      println("error");
      return -1;
  }

  if (cache(n)!=0)//means value has been calculated
     return cache(n);
  else
  {
    cache(n)=fibonacciCache(n-1)+fibonacciCache(n-2);
    return cache(n);    
  }
}

Hope that helps

Fast Fibonacci using recursion

1 Answers1