fibonacci search

Introduction of Fibonacci search

    Two point search is familiar to everyone , Every time 1/2 Fold check , Then someone thought , If you don't follow 1/2 If you check , Will it be more efficient in some cases ？ So someone thought 0.618( The golden section ), But using this decimal directly is a little less elegant . At this time, people think of Fibonacci series .
   1 1 2 3 5 8 13 21 34 55 89…… We will find that as the Fibonacci sequence increases , The ratio of the two numbers will be closer and closer 0.618, Take advantage of this feature , We can apply the golden ratio to the search technology .

The basic idea

    By using the concept of golden ratio, select the search point in the sequence to find , Improve search efficiency . similarly , Fibonacci search also belongs to an ordered search algorithm .

The total length of the array is F[K]-1,mid The front length is F[K-1]-1, The back length is F[K-2]-1,mid At the golden section .

Algorithm implementation

public class Fibonacci {
    

    public static void main(String[] args) {
    
        int[] arr = {
    1, 8, 10, 89, 1000, 1234};
        System.out.println(fibSearch(arr , 99));
    }

    public static int maxSize = 20;

    public static int[] fib() {
    
        int[] f = new int[maxSize];
        f[0] = 1;
        f[1] = 1;
        for(int i = 2; i < maxSize; i++) {
    
            f[i] = f[i - 1] + f[i - 2];
        }
        return f;
    }

    //  Fibonacci looks for （ In fact, it is the point of each search , Are close to dividing the array into 0.618 ratio , Number of elements on the left / The total number of elements   About equal to  0.618）
    //  I want to search by （ Find the element on the left of the point / The total number of elements ） near 0.618 Get the index of the search element .
    //  Because Fibonacci looked , The ratio of two adjacent items is close 0.618, Then it is used in the search algorithm according to the law of Fibonacci sequence .
    public static int fibSearch(int[] a, int key) {
    
        int low = 0;
        int high = a.length - 1;
        //  A subscript representing the Fibonacci division value 
        int k = 0;
        int mid = 0;
        int f[] = fib();
        //  Get the Fibonacci partition value subscript , According to the length of the lookup array , Decide how many Fibonacci values are needed 
        while(high > f[k] - 1) {
    
            k++;
        }
        //  because f[k] The value may be greater than a The length of , So we need to use Arrays class , Construct a new array , And point to a[]
        //  The insufficient part will be used 0 fill 
        int[] temp = Arrays.copyOf(a, f[k]);
        //  In fact, you need to use a The last number of the array is filled temp
        //  for example ：temp = {1,8,10,89,1000,1234,0,0,0} => {1,8,10,89,1000,1234,1234,1234}
        for(int i = high + 1; i < temp.length; i++) {
    
            temp[i] = a[high];
        }
        //  Use while To cycle through , Find our number key
        while(low <= high) {
    
            mid = low + f[k - 1] - 1;
            //  We should continue to look to the left of the array 
            if(key < temp[mid]) {
    
                high = mid - 1;
                //  Why k--
                //  All elements  =  The front element  +  Back element  f[k] = f[k - 1] + f[k - 2]
                //  Because there are f[k - 1] Elements , So we can continue to split f[k - 1] = f[k - 2] + f[k - 3]
                //  That is to say f[k - 1] Left continue to search k--, Next cycle mid = f[k - 1 - 1] - 1
                k--;
            } else if(key > temp[mid]) {
    
                //  All elements  =  The front element  +  Back element  f[k] = f[k - 1] + f[k - 2]
                //  Look on the right 
                low = mid + 1;
                k -= 2;
            } else {
    
                return Math.min(mid, high);
            }
        }
        return -1;
    }
}