Question

• Write a JavaFX code that generates Fibonacci number sequence of a certain length (-100 numbers) • The program then uses bucket sort technique to sort the Fibonacci numbers based on the order of their division by the digits 2,3,4,5,6,7,8 and 9 (→ 08 digits) • The sorting order is defined based on the divisibility of numbers starting from 2,3,4 and onwards. As a result of sorting, some of the Fibonacci numbers will be in more than one bucket because of the number being divisible by multiple digits. The program should display a GUI interface with the buckets and their numbers in it.

Answer 1

Given a sorted array arr[] of size 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:

Works for sorted arrays

A Divide and Conquer Algorithm.

Has Log n time complexity.

Differences with Binary Search:

Fibonacci Search divides given array in unequal parts

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.

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 is to first find the smallest Fibonacci number that is greater than or equal to the length of given array. Let the found Fibonacci number be fib (m’th Fibonacci number). We use (m-2)’th Fibonacci number as the 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 the complete algorithm

Let arr[0..n-1] be the input array and element to be searched be x.

Find the smallest Fibonacci Number greater than or equal to 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].

While the array has elements to be inspected:

Compare x with the last element of the range covered by fibMm2

If x matches, return index

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.

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.

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.

// Java program for Fibonacci Search

import java.util.*;

  

class Fibonacci

{   

// Utility function to find minimum  

// of two elements

public static int min(int x, int y)  

{ return (x <= y)? x : y; }

  

/* Returns index of x if present, else returns -1 */

public static 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 less 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 == 1 && arr[offset+1] == x)

return offset+1;

  

/*element not found. return -1 */

return -1;

}

  

// driver code

public static void main(String[] args)

{

int arr[] = {10, 22, 35, 40, 45, 50,  

80, 82, 85, 90, 100};

int n = 11;

int x = 85;

System.out.print ("Found at index: "+

fibMonaccianSearch(arr, x, n));

}

}

Bucket Sort

Bucket sort is mainly useful when input is uniformly distributed over a range. For example, consider the following problem.

Sort a large set of floating point numbers which are in range from 0.0 to 1.0 and are uniformly distributed across the range. How do we sort the numbers efficiently?

A simple way is to apply a comparison based sorting algorithm. The lower bound for Comparison based sorting algorithm (Merge Sort, Heap Sort, Quick-Sort .. etc) is Ω(n Log n), i.e., they cannot do better than nLogn.

Can we sort the array in linear time? Counting sort can not be applied here as we use keys as index in counting sort. Here keys are floating point numbers.  

The idea is to use bucket sort. Following is bucket algorithm.

bucketSort(arr[], n)

1) Create n empty buckets (Or lists).

2) Do following for every array element arr[i].

.......a) Insert arr[i] into bucket[n*array[i]]

3) Sort individual buckets using insertion sort.

4) Concatenate all sorted buckets.

Time Complexity: If we assume that insertion in a bucket takes O(1) time then steps 1 and 2 of the above algorithm clearly take O(n) time. The O(1) is easily possible if we use a linked list to represent a bucket (In the following code, C++ vector is used for simplicity). Step 4 also takes O(n) time as there will be n items in all buckets.

0 0.78 07 10.17 1 0.12 0.17 2 0.39 2 0.21 0.23 0.26 / 3 0.26 3 0.39 4 0.72 4 5 0.94 5 / 6 0.21 6 0.68 7 0.12 7 0.72 0.78 7 80.23 9 0.68 9 0.94 Input Aray Buckets created

The main step to analyze is step 3. This step also takes O(n) time on average if all numbers are uniformly distributed

Following is the implementation of the above algorithm.

// Java program to sort an array

// using bucket sort

import java.util.*;

import java.util.Collections;

  

class GFG {

  

// Function to sort arr[] of size n

// using bucket sort

static void bucketSort(float arr[], int n)

{

if (n <= 0)

return;

  

// 1) Create n empty buckets

@SuppressWarnings("unchecked")

Vector<Float>[] buckets = new Vector[n];

  

for (int i = 0; i < n; i++) {

buckets[i] = new Vector<Float>();

}

  

// 2) Put array elements in different buckets

for (int i = 0; i < n; i++) {

float idx = arr[i] * n;

buckets[(int)idx].add(arr[i]);

}

  

// 3) Sort individual buckets

for (int i = 0; i < n; i++) {

Collections.sort(buckets[i]);

}

  

// 4) Concatenate all buckets into arr[]

int index = 0;

for (int i = 0; i < n; i++) {

for (int j = 0; j < buckets[i].size(); j++) {

arr[index++] = buckets[i].get(j);

}

}

}

  

// Driver code

public static void main(String args[])

{

float arr[] = { (float)0.897, (float)0.565,

(float)0.656, (float)0.1234,

(float)0.665, (float)0.3434 };

  

int n = arr.length;

bucketSort(arr, n);

  

System.out.println("Sorted array is ");

for (float el : arr) {

System.out.print(el + " ");

}

Sorted array is

0.1234 0.3434 0.565 0.656 0.665 0.897