## Finding the closest fibonacci numbers

find next fibonacci number

test the function using junit check whether it is finding the nth fibonacci number

the function will return the index of the fibonacci number that is closest to the number entered

nth fibonacci number java

negative fibonacci number calculator

how to check whether a number is fibonacci or not in c

1000th fibonacci number

I am trying to solve a bigger problem, and I think that an important part of the program is spent on inefficient computations.

I need to compute for a given number N, the interval [P, Q], where P is the biggest fibonacci number that is <= to N, and Q is the smallest fibonacci number that is >= to N.

Currently, I am using a map to record the value of the fibonacci numbers. A query normally involves searching all the fibonacci numbers up to N, and it is not very time efficient, as it involves a big number of comparisons.

This type of queries will occur quite often in my program, and I am interested in ways that I could improve the lookup, preferably with sub-linear complexity.

The Fibonacci numbers are given by Binet's formula

F(n) = ( phi^n - (1-phi)^n ) / \sqrt{5}

where `phi`

is the golden ratio,

phi = (1 + \sqrt{5}) / 2.

This can be implemented straightforwardly (Python example):

<<fibonacci_binet.py>>= phi = (1 + 5**0.5) / 2 def fib(n): return int(round((phi**n - (1-phi)**n) / 5**0.5))

Because of floating-point rounding errors, this will however only give the right result for `n < 70`

.

Binet's formula can be inverted by ignoring the `(1-phi)^n`

term, which disappears for large `n`

. We can therefore define the inverse Fibonacci function that, when given `F(n)`

, returns `n`

(ignoring that `F(1) = F(2)`

):

<<fibonacci_binet.py>>= from math import log def fibinv(f): if f < 2: return f return int(round(log(f * 5**0.5) / log(phi)))

Here rounding is used to our advantage: it removes the error introduced by our modification to Binet's formula. The function will in fact return the right answer when given any Fibonacci number that can be stored as an exact integer in the computer's memory. On the other hand, it does not verify that the given number actually is a Fibonacci number; inputting a large Fibonacci number or any number close to it will give the same result. Therefore you can use this idea to find the Fibonacci number closest to a given number.

The idea, then is to apply the inverse Fibonacci map to find `N`

and `M`

, the two closest Fibonacci numbers on either side, then use the direct Fibonacci map to compute `P = F(N)`

and `Q = F(M)`

. This involves more computation, but less searching.

**Finding the closest fibonacci numbers,** The Fibonacci numbers are given by Binet's formula. F(n) = ( phi^n - (1-phi)^n ) / \sqrt{5}. where phi is the golden ratio, phi = (1 + \sqrt{5}) / 2. This can be Write a program to find the nearest Fibonacci number. The Fibonacci numbers are 1, 1, 2, 3, 5, 8, 13, 21, 34, 55 etc.,. Assume that the input number will be always greater than 3. There is no upper limit to the input number. Input (Integer) Output (Integer) 9.

**Given a number N, how do I efficiently find the closest Fibonacci ,** Binet's formula for the th Fibonacci number is. ,. where. and . Since |, if . Hence. ,. and so is the integer closest to for each. Therefore, given a positive integer , we Therefore you can use this idea to find the Fibonacci number closest to a given number. The idea, then is to apply the inverse Fibonacci map to find N and M, the two closest Fibonacci numbers on either side, then use the direct Fibonacci map to compute P = F(N) and Q = F(M). This involves more computation, but less searching.

You can use the closed-form expression of the fibonacci numbers.

Since the second term in it is very small, you can approximate it with just the first term, so `n`

can be found with base-golden ratio logarithm.

**Find the next fibonacci number,** Approach: The ratio of two adjacent numbers in the Fibonacci series rapidly approaches ((1 + sqrt(5)) / 2). So if N is multiplied by ((1 + sqrt(5)) / 2) and round it, We can use this to derive the following simpler formula for the n-th Fibonacci number F(n): F(n) = round( Phin/ √5 ) providedn ≥ 0. where the roundfunction gives the nearest integerto its argument. Notice how, as n gets larger, the value of Phin/√5 is almost an integer.

Use the closed form formula: http://en.wikipedia.org/wiki/Fibonacci_number#Closed-form_expression

Then binary search

**Fibonacci Number Formula – Mudd Math Fun Facts,** ] / Sqrt[5]. phi = (1 – Sqrt[5]) / 2 is an associated golden number, also equal to (-1 / Phi). But here, we needed to generate all the fibonacci number up to provided fibonacci number. But, there is quick solution for this problem. Let’s see how ! Note that computing log of a number is O(1) operation in most of the platforms [Source : Stackoverflow] Fibonacci number is described as, Fn = 1 / sqrt(5) (pow(a,n) – pow(b,n)) where

I just did a CodeChef puzzle that was this exact problem (http://www.codechef.com/problems/DPC204). I simply calculated the Fibonacci sequence from 0 to the end of the range, and counted how many were after the beginning of the range. My test for whatever their sample inputs were took 2.6M, and 0.00s, so the nieve solution is plenty fast enough.

Basically, I made a big-unsigned-int class made of `unsigned int[333]`

, and calculate two numbers per loop, to avoid swaps.

start with A=0,B=1; A+=B;B+=A; now A==1,B==2, the next two Fib. numbers, with no swaps. A+=B;B+=A; now A==3,B==5, the next two Fib. numbers, with no swaps.

It is slightly complicated by the fact you have to stop and check if the neither, one, or both numbers are in the range, but A

My solution on CodeChef clocked in at 0.00 seconds, so I think this method ought to be fast enough, you just have to write a function that adds one `uint[333]`

to another `uint[333]`

(using all 32 bits, just chars for each decimal digit)

**Fibonacci prime,** 21 whereas daisies can be found with 34, 55 or even 89 petals. nth fibonacci number = round (n-1th Fibonacci number X golden ratio) f n = round (f n-1 *) Till 4th term, the ratio is not much close to golden ratio (as 3/2 = 1.5, 2/1 = 2, …). So, we will consider from 5th term to get next fibonacci number. To find out the 9th fibonacci number f9 (n = 9) :

**What is the Fibonacci Sequence (aka Fibonacci Series)?,** Find the closest Fibonacci number. Fibonachi.cs. using System;. using System.Collections.Generic;. namespace Fibonachi. {. class Program. {. static void Remember, to find any given number in the Fibonacci sequence, you simply add the two previous numbers in the sequence. To create the sequence, you should think of 0 coming before 1 (the first term), so 1 + 0 = 1. 5 Add the first term (1) and the second term (1).

**The Fibonacci Numbers and Golden section in Nature,** How to find formulae for Fibonacci numbers. 6 Finding the Fibonacci index number i for a given n. 6.1 calculator Nearest Fibonacci Number ≤ n Calculator. Using The Golden Ratio to Calculate Fibonacci Numbers. And even more surprising is that we can calculate any Fibonacci Number using the Golden Ratio: x n = φ n − (1−φ) n √5. The answer comes out as a whole number, exactly equal to the addition of the previous two terms.

**Find the closest Fibonacci number · GitHub,** Given a positive integer (n) find the nearest fibonacci number to (n). If there are more than one fibonacci with equal distance to the given Find the closest Fibonacci number. GitHub Gist: instantly share code, notes, and snippets.

##### Comments

- how about using quicksort or bubblesort to reduce the number of comparisons?
- How large does this go? Fibonacci numbers grow exponentially, so I can imagine you could just precompute all of them up to the maximum size and binary search that list.
- Knowing how big N can be is critical to coming up with a good solution. If N can be 1e100 or more, solutions involving floating-point calculations may be reasonable; if it's limited to 32-bit integers, such calculations would be overkill.
- N can be represented at most as an 8-byte integer.
- +1 from me. Again, I'm envious of a more solid maths background
- @PengOne: Thanks for this very useful answer! For fun, I tried to implement it in Ruby (gist.github.com/IRLeif/40c543a5237a6fce587c), but I'm still trying to grok the last two paragraphs of your answer, in order to put it all together.
- Don't feel intimidated by the more mathematical answers. Since the OP is only interested in integers up to 64 bit, she only needs to precompute about 93 Fibonacci numbers, and a binary search involves at most 7 integer comparisons. This is almost certainly faster than computing even a single logarithm. +1 for your answer.
- agree, a java version of this gives about 4 million searches/second on my 2.6 ghz machine.
- it's what I implemented. I do have a feeling that the more math-inclined responses (PengOne/mange) might be more appropriate then binary searching here!
- I just tested mine. It finds all the Fibonacci numbers between 0 and 1e100 (inclusive) in 0.00428814s cold, 0.00222952s hot on my machine. Probably not as fast as other suggestions, but fast enough for any purpose.
- Could you please explain your algorithm further in your post?
- Could you also describe the solution in text?
- This doesn't answer the question. Also it's C code. Here's a (direct translation to c++, or using standard library) which is much less code and also more efficient (running ~5% faster when tested with this 115MiB input)