# Fibonacci Search

Given a sorted array arr[] of siz n and an element x to be searched in it. Return index of x if it is present in array else return -1.
Examples:

```Input:  arr[] = {2, 3, 4, 10, 40}, x = 10
Output:  3
Element x is present at index 3.

Input:  arr[] = {2, 3, 4, 10, 40}, x = 11
Output:  -1
Element x is not present.
```

Fibonacci Search is a comparison-based technique that uses Fibonacci numbers to search an element in a sorted array.

Similarities with Binary Search:

1. Works for sorted arrays
2. A Divide and Conquer Algorithm.
3. Has Log n time complexity.

Differences with Binary Search:

1. Fibonacci Search divides given array in unequal parts
2. Binary Search uses division operator to divide range. Fibonacci Search doesn’t use /, but uses + and -. The division operator may be costly on some CPUs.
3. Fibonacci Search examines relatively closer elements in subsequent steps. So when input array is big that cannot fit in CPU cache or even in RAM, Fibonacci Search can be useful.

Background:
Fibonacci Numbers are recursively defined as F(n) = F(n-1) + F(n-2), F(0) = 0, F(1) = 1. First few Fibinacci Numbers are 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, …

Observations:
Below observation is used for range elimination, and hence for the O(log(n)) complexity.

```F(n - 2) ≈ (1/3)*F(n) and
F(n - 1) ≈ (2/3)*F(n).```

Algorithm:
Let the searched element be x.

The idea it to first find the smallest Fibonacci number that is greater than or equal to length of given array. Let the found fibonacci number be fib (m’th fibonacci number). We use (m-2)’th Fibonacci number as index (If it is a valid index). Let (m-2)’th Fibonacci Number be i, we compare arr[i] with x, if x is same, we return i. Else if x is greater, we recur for subarray after i, else we recur for subarray before i.

Below is complete algorithm
Let arr[0..n-1] be th input array and element to be searched be x.

1. Find the smallest Fibonacci Number greater than or equal n. Let this number be fibM [m’th Fibonacci Number]. Let the two Fibonacci numbers preceding it be fibMm1 [(m-1)’th Fibonacci Number and fibMm2 [(m-2)’th Fibonacci Number./li>
2. While the array has elements to be inspected:
1. Compare x with the last element of the range covered by fibMm2
2. If x matches, return index
3. Else If x is less than the element, move the three Fibonacci variables two Fibonacci down, indicating elimination of approximately rear two-third of the remaining array.
4. Else x is greater than the element, move the three Fibonacci variables one Fibonacci down. Reset offset to index. Together these indicate elimination of approximately front one-third of the remaining array.
3. Since there might be a single element remaining for comparison, check if fibMm1 is 1. If Yes, compare x with that remaining element. If match, return index.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

```// C program for Fibonacci Search
#include <stdio.h>

// Utility function to find minimum of two elements
int min(int x, int y) { return (x<=y)? x : y; }

/* Returns index of x if present,  else returns -1 */
int fibMonaccianSearch(int arr[], int x, int n)
{
/* Initialize fibonacci numbers */
int fibMMm2 = 0;   // (m-2)'th Fibonacci No.
int fibMMm1 = 1;   // (m-1)'th Fibonacci No.
int fibM = fibMMm2 + fibMMm1;  // m'th Fibonacci

/* fibM is going to store the smallest Fibonacci
Number greater than or equal to n */
while (fibM < n)
{
fibMMm2 = fibMMm1;
fibMMm1 = fibM;
fibM  = fibMMm2 + fibMMm1;
}

// Marks the eliminated range from front
int offset = -1;

/* while there are elements to be inspected. Note that
we compare arr[fibMm2] with x. When fibM becomes 1,
fibMm2 becomes 0 */
while (fibM > 1)
{
// Check if fibMm2 is a valid location
int i = min(offset+fibMMm2, n-1);

/* If x is greater than the value at index fibMm2,
cut the subarray array from offset to i */
if (arr[i] < x)
{
fibM  = fibMMm1;
fibMMm1 = fibMMm2;
fibMMm2 = fibM - fibMMm1;
offset = i;
}

/* If x is greater than the value at index fibMm2,
cut the subarray after i+1  */
else if (arr[i] > x)
{
fibM  = fibMMm2;
fibMMm1 = fibMMm1 - fibMMm2;
fibMMm2 = fibM - fibMMm1;
}

/* element found. return index */
else return i;
}

/* comparing the last element with x */
if(fibMMm1 && arr[offset+1]==x)return offset+1;

return -1;
}

/* driver function */
int main(void)
{
int arr[] = {10, 22, 35, 40, 45, 50, 80, 82,
85, 90, 100};
int n = sizeof(arr)/sizeof(arr[0]);
int x = 85;
printf("Found at index: %d",
fibMonaccianSearch(arr, x, n));
return 0;
}
```

Output:

`Found at index 8`

Illustration:
Let us understand the algorithm with below example:

Illustration assumption: 1-based indexing. Target element x is 85. Length of array n = 11.

Smallest Fibonacci number greate than or equal to 11 is 13. As per our illustration, fibMm2 = 5, fibMm2 = 8, and fibM = 13.

Another implementation detail is the offset variable (zero initialized). It marks the range that has been eliminated, starting from the front. We will update it time to time.

Now since the offset value is an index and all indices inclusive it and below it have been eliminated, it only makes sense to add something to it. Since fibMm2 marks approximately one-third of our array, as well as the indices it marks are sure to be valid ones, we can add fibMm2 to offset and check the element at index i = min(offset + fibMm2, n).

Visualization:

Time Complexity analysis:
The worst case will occur when we have our target in the larger (2/3) fraction of the array, as we proceed finding it. In other words, we are eliminating the smaller (1/3) fraction of the array every time. We call once for n, then for(2/3) n, then for (4/9) n and henceforth.

Consider that:

# GATE CS Corner    Company Wise Coding Practice

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.
3.6 Average Difficulty : 3.6/5.0
Based on 21 vote(s)