fibonacci

As we all know, the simplest algorithm to generate Fibonacci sequence is as follows: if(n<=0) return 0; else if(n==1) return 1; f(n) = f(n-1) + f(n-2); But this algorithm has some repetitive calculation. For example, if you calculate f(5), it will calculate f(4) and f(3). When you calculate f(4), it will again calculate both f(3) and f(2). Could someone give me a more time-efficient recursive algorithm? One simple way is to calculate it iteratively instead of recursively. This will calculate F(n) in linear time. def fib(n): a,b = 0,1 for i in range(n): a,b = a+b,a return a Hynek -Pichi-

What is an optimal Huffman code for the following characters whose frequencies are the first 8 Fibonacci numbers: a : 1, b : 1, c : 2, d : 3, e : 5, f : 8, g : 13, h : 21? Generalize the case to find an optimal code when the frequencies are the first n Fibonacci numbers. This is one of the assignment problems I have. I'm not asking for a straight answer, just for some resources. Where should I look to put the pieces together to answer the questions? Read - http://en.wikipedia.org/wiki/Huffman_coding - In particular, pay attention to the phrase "A binary tree is generated from left to right

This question already has answers here : How to write the Fibonacci Sequence? (44 answers) Closed 4 years ago . I am new to Python and to these forums. My question is: How can I create a list of n Fibonacci numbers in Python? So far, I have a function that gives the nth Fibonacci number, but I want to have a list of the first n Fib. numbers for future work. For example: fib(8) -> [0,1,1,2,3,5,8,13] Try this, a recursive implementation that returns a list of numbers by first calculating the list of previous values: def fib(n): if n == 0: return [0] elif n == 1: return [0, 1] else: lst = fib(n-1

I'm practicing replacing recursion with while loops, and I'm stuck on the following problem. How many ways can you go up a staircase of length n if you can only take the stairs 1 or 2 at a time? The recursive solution is pretty simple: def stairs(n): if n <= 1: return 1 else: return stairs(n-2) + stairs(n-1) I feel like the structure for the iterative program should go something like this: def stairs_iterative(n): ways = 0 while n > 1: # do something ways +=1 return ways But I don't know what I need to put in the #do something part. Can someone help me? Pseudocode is fine! This amounts to the

I have come up with the following tail-recursive fibonacci generator that works: let { fibo :: Integral x => [x]->x->x->x->[x] fibo l x y 0 = l fibo l x y n = fibo (l ++ [y+x] ++ [y+x+y]) (x+y) (y+x+y) (n-1) } Pardon me for the whole implementation put in one line because i am using the GHCi and haven't quite learnt how to put this in a file and run (i am yet to reach there). What i want to know is how this call: fibo [0, 1] 0 1 5 can be improved. I do not want to pass the initial list with 0 and 1 and then pass 0 and 1 again with the limit. I believe that the implementation can be changed.

I have a method that takes n and returns nth Fibonacci number. Inside the method implementation I use BigDecimal to get the nth Fibonacci number then I use method toBigInteger() to get the number as a BigInteger object and that's surely because I am working with huge numbers in my application. I keep getting correct results until I pass 1475 as an argument for my method. I get NumberFormatException: Infinite or NaN in this case without any clear reason for me. Could you please explain me why am I getting this exception? Here's my method: BigInteger getFib(int n){ double phi = (1 + Math.sqrt(5)