Develop Reference

How to calculate prime fibonacci numbers?

(This is an extension from my previous question).
How to print a text notification beside prime fibonacci numbers?
Task:
The section of the assignment I am having difficulty with reads as follows:
Determine which of the first 20 Fibonacci numbers are prime numbers.
Put a “this is a prime” text notification in the printout from the Basic challenge.
Store the FibPrimes in an array called FibPrimes.
Problem:
I cannot seem to figure out how to take each individual fibonacci number, determine it is prime, put the prime fibonacci numbers in an array, and print the output. I was able to make a working program previously, but I didn't use code to calculate the primes, I manually entered them into an array (see previous attempt at bottom).
Attempt:
Here is what I have attempted:
The code that is relevant to this question is emphasized with comments.
package fibonaccinumbers;
public class FibonacciNumbers {
public static void main(String[] args) {
// Creation of Fibonacci Numbers Array.
// Let i represent the index number.
int [] FibNums = new int[20];
FibNums[0] = 0;
FibNums[1] = 1;
// Will add on each successive FibNums.
// Will be used to calculate average.
int FibSum = 1;
// RELEVANT TO QUESTION.
// Creation if Fibonacci Primes Array.
// Let n represent the index number.
int [] FibPrimes = new int[7];
int n=0;
// Printing first two fibonacci numbers.
System.out.println(0);
System.out.println(1 + "*");
// Printing remaining fibonacci numbers up to 20th term.
for (int i=2; i<FibNums.length;i++){ // Begin number generation loop.
FibNums[i] = FibNums[i-1] + FibNums[i-2];
// Checks if the fibonacci number is odd.
// A number is not odd if two divides into it evenly.
boolean oddcheck = true;
if (FibNums[i]%2==0){
oddcheck = false;
}
// RELEVANT TO QUESTION.
// A prime number can only be divided by 1 and inself.
// Divide FibNums[i] by every number inbetween.
// If a number divides evenly, it is not a prime (exception: 2).
// Else, the number is a prime.
boolean primecheck;
for (int divisor = 2; divisor < FibNums[i]/2; divisor++){
if (FibNums[i] % divisor == 0){
primecheck = false;
} else {
primecheck = true;
}
// REVELANT TO QUESTION.
// Add FibNums[i] to the FibPrimes[n] array if it is a prime.
if (primecheck == false){
FibPrimes[n] = FibNums[i];
n = n + 1;
}
// RELEVANT TO QUESTION.
// If any element in the FibPrimes array is equal to the FibNums
// array, then the number is a prime.
for (n=0; n<FibPrimes.length; n++){
if (FibNums[i] == FibPrimes[n]){
System.out.print("This is a prime." + " ");
}
}
// Prints odd fibonacci numbers with a star beside it.
// Prints even fibonacci numbers with no star beside it.
if (oddcheck == true){
System.out.println(FibNums[i] + "*");
} else {
System.out.println(FibNums[i]);
}
FibSum = FibSum + FibNums[i];
} // End number generation loop.
System.out.print ( "The average is" + " " + FibSum/20);
}
}
Output:
0
1*
The average is 0The average is 0The average is 0The average is 0This is a prime. 8
This is a prime. 8
The average is 013*
13*
13*
13*
The average is 321*
This is incorrect.
Previous Attempt:
This solution "worked", but in order to not take the "lazy route" I must do the calculations using code. Only relevant snippets are shown:
// Creation if Fibonacci Primes Array.
// Ideally, this should be caulculated.
int [] FibPrimes = new int[7];
FibPrimes[0] = 2;
FibPrimes[1] = 3;
FibPrimes[2] = 5;
FibPrimes[3] = 13;
FibPrimes[4] = 89;
FibPrimes[5] = 233;
FibPrimes[6] = 1597;
// If any element in the FibPrimes array is equal to the FibNums
// array, then the number is a prime.
for (int n=0; n<FibPrimes.length; n++){
if (FibNums[i] == FibPrimes[n]){
System.out.print("This is a prime." + " ");
}
}
Output:
0
1*
1*
This is a prime. 2
This is a prime. 3*
This is a prime. 5*
8
This is a prime. 13*
21*
34
55*
This is a prime. 89*
144
This is a prime. 233*
377*
610
987*
This is a prime. 1597*
2584
4181*
The average is 547
This is the desired output! However, I cannot use this because I must calculate the prime fibonacci numbers using code. This is the "lazy route".
Thank you.

Just 10 mins on google and you will be able to create something simple and fast.
Using the mathematical formulas below :
which you can find more one this paper you can make your prime fibonacci sequence. In addition you will need a way to check which numbers in the sequence are primes so a quick way is through AKS algorithm
Here is a full example of all the above :
public class FibonacciSequence {
public static void main(String[] args) {
for (int i = 1; i <= 20; i++) {
int num = (int) ((getY_1(i) - getY_2(i)) / Math.sqrt(5));
if (isPrime(num)) {
System.out.print("This is a prime : ");
}
System.out.println(num);
}
}
public static double getY_1(int N) {
return Math.pow((1 + Math.sqrt(5.0)) / 2.0, N);
}
public static double getY_2(int N) {
return Math.pow((1 - Math.sqrt(5.0)) / 2.0, N);
}
public static boolean isPrime(int num) {
if (num == 2 || num == 3)
return true;
if (num % 2 == 0 || num % 3 == 0) {
return false;
}
int i = 5;
int s = 2;
while (i * i <= num) {
if (num % i == 0) {
return false;
}
i += s;
s = 6 - s;
}
return true;
}
}
Of course you can exclude the value (1) if you don't feel like identifying it as a prime :P.
Output :
This is a prime : 1
This is a prime : 1
This is a prime : 2
This is a prime : 3
This is a prime : 5
8
This is a prime : 13
21
34
55
This is a prime : 89
144
This is a prime : 233
377
610
987
This is a prime : 1597
2584
4181
6765
P.S : I had free time :P

Counting numbers of even numbers and odd numbers in java array

I have a program that reads a list of integers, and then display the number of even numbers and odd numbers. We assume that the input ends with 0. Here is the sample run of the program.
Input numbers: 1 2 3 4 5 6 7 8 9 0
Odd: 5 Even: 4
However, my result is
Odd: 5 and Even: 5.
The problem is that 0 is counted as an even number. This is my code
public class Q75 {
public static void main(String[] args){
java.util.Scanner input = new java.util.Scanner (System.in);
double [] numbers = new double[10];
System.out.print("Enter numbers: ");
for(int i = 0;i<numbers.length;i++){
numbers[i] = input.nextDouble();
}
int Evens = 0;
int Odd = 0;
for(int i = 0;i<numbers.length;i++){
if(numbers[i]%2 == 0){
Evens++;
}else{
Odd++;
}
}
System.out.println("The number of odd numbers: " + Odd);
System.out.println("The number of even numbers: " + Evens);
}
}

There are two options
A) Adding another branch in your if statements i.e.
if(number[i] > 0) {
if(number[i] % 2 >0)
Odd++;
else
Evens++;
}
NB: changing the else branch to else if(number[i] >0), you can do without the outer if condition.
B) Since the list of number ends with 0 you can put this as a condition in your for loop i.e.
for(int i =0; i < numbers.length && numbers[i] > 0 ; i++)
Also as a rule of thumb variable names in java start with a small letter

Just don't check the last element: Use i < numbers.length - 1
for(int i = 0;i < numbers.length - 1; i++) {
//
}

How to print prime numbers up to the user's entered integer?

Good afternoon everyone,
I'm currently trying to create a program that does the following:
Develop a code that will print all prime numbers up to the user's entered
number. An example of the output:
Enter an integer (2 or above): 19
The prime numbers up to you integer are:
2
3
5
7
11
13
17
19
Unfortunately, there is also a list of "requirements" that need to be met as well:
If the user enters a number below 2, your program should print a message that the number is not valid, and then stop.
A number is a prime number if it is not divisible by any number except 1 and itself.
For this program, in order to test a number to see if it is prime, you should try to divide the number by every value from 2 up to the number-1, to see if it divides evenly or not. For example:
--To see if 5 is a prime number:
5 does not divide evenly by 2
5 does not divide evenly by 3
5 does not divide evenly by 4
therefore 5 is a prime number
--To see if 9 is a prime number:
9 does not divide evenly by 2
9 divides evenly by 3
therefore 9 is not a prime number
This program requires you to write nested loops (that is, a loop inside a loop). One loop will be used to count from 2 up to the user's number so that you can test each of these numbers to see it it is prime. For each of these numbers, x:
A nested loop will check all values from 2 up to x-1 to see if x divides evenly by any of them.
You will need to use a Boolean variable (also referred to as a flag variable) to help you determine whether or not to print a number to the screen
The question above is in regards to my code so far below:
import java.util.*;
public class Something3 {
public static void main(String[] args) {
Scanner kbd = new Scanner(System.in);
//Variable declaration.
int limit, number;
//get input till which prime number to be printed
System.out.print("Enter an integer (2 or above): ");
limit = kbd.nextInt();
kbd.close();
//Will print prime numbers till the limit (user entered integer).
number = 2;
if (limit >=2) {
System.out.println("The prim numbers up to your interger are: " + limit+"\n");
for(int i = 0; i <= limit;){
//print prime numbers only
if(isPrime(number)){
System.out.println(number +"\n");
i++;
}
number = number + 1;
}
}
else
System.out.println("Number is not vaild");
}
//Prime number is not divisible by any number other than 1 and itself
//return true if number is prime.
public static boolean isPrime(int number){
for(int i=2; i==number; i++){
if(number%i == 0){
return false; //Number is divisible, thus not prime.
}
}
return true;//The number is prime.
}
}
Is the limit variable I'm using the issue? Any help would be much appreciated.

your isPrime() method is returning wrong result. for loop condition is wrong i==number it should be i < number
public static boolean isPrime(int number){
for(int i=2; i*i <= number; i++){
if( number % i == 0){
return false; // Number is divisible, thus not prime.
}
}
return true; //The number is prime.
}
To check a number prime or not you don't have to check divisbility from 2 to N, all you need to check upto sqrt(N).
To find prime numbers in a range(N) use Sieve of Eratosthenes

Your actual problem was number variable. Here's your solution. There's no need of variable number. Below is your solution
import java.util.*;
public class Something3 {
public static void main(String[] args) {
Scanner kbd = new Scanner(System.in);
// Variable declaration.
int limit;
// get input till which prime number to be printed
System.out.print("Enter an integer (2 or above): ");
limit = kbd.nextInt();
kbd.close();
if (limit >= 2) {
System.out.println("The prim numbers up to your interger are: "
+ limit + "\n");
for (int i = 1; i <= limit; i++) {
// print prime numbers only
if (isPrime(i)) {
System.out.println(i);
}
}
} else
System.out.println("Number is not vaild");
}
// Prime number is not divisible by any number other than 1 and itself
// return true if number is prime.
public static boolean isPrime(int n) {
if (n % 2 == 0)
// The only even prime is 2.
return (n == 2);
for (int i = 3; i * i <= n; i += 2) {
if (n % i == 0)
return false;
}
return true;
}
}

import java.util.Scanner;
class prime
{
public static void main(String args[])
{
Scanner sc=new Scanner(System.in);
System.out.println("enter the value of n");
int n=sc.nextInt();
for(int i=2;i<=n;i++)
{
int count=0;
for(int j=1;j<=i;j++)
{
if(i%j==0)
{
count++;
}
}
if(count==2)
{
System.out.println(i+" ");
}
}
}
}

Using a while loop to compare to a value in a array

I need the program to run in a loop until 0 is entered. My code will end with 0 entered but when attempting to run the program with numbers entered it still ends the program. instead of running the numbers entered. The while loop is to keep the program running unless a 0 is entered.
import java.util.Scanner;
public class CountCompare {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.print("Enter the integers between 1 and 100 (0 to end, 0 < to exit): ");
int[] counts = new int[100];
// Count occurrence of numbers
count(counts);
while(counts[0] > 0){
// Display results
for (int i = 0; i < counts.length; i++) {
if (counts[i] > 0)
System.out.println((i + 1) + " occurs " + counts[i] +
" time" + (counts[i] > 1 ? "s" : ""));
}
System.out.print("Enter the integers between 1 and 100 : ");
// Count occurrence of numbers
count(counts);
}
System.out.print("\nEnd of run");
}
/** Method count reads integers between 1 and 100
* and counts the occurrences of each */
public static void count(int[] counts){
Scanner input = new Scanner(System.in);
int num; // holds user input
do {
num = input.nextInt();
if (num >= 1 && num <= 100)
counts[num - 1]++;
} while (num != 0);
}
}
I have posted the entire program.
output looks like this
Enter the integers between 1 and 100 (0 to end, <0 to exit):
23 23 4 5 6 7 8
0
4 occurs 1 time
5 occurs 1 time
6 occurs 1 time
7 occurs 1 time
8 occurs 1 time
23 occurs 2 times
Enter the integers between 1 and 100:

Your program still ends because:
int[] counts = new int[100];
You have defined the limit of the counts here. This means your loop will run
for (int i = 0; i < counts.length; i++)// counts.length=100;
So as far as you code suggest you want to end the user input when user input 0. So you might do this:
int x=1;
int y;
Scanner sc= new Scanner(System.in);
while(x!=0){
System.out.println("Enter your values");
y=sc.nextInt();
if(y==0){
x=0;
}
else{
System.out.println("You entered "+y);
}

int count=new Scanner(System.in).nextInt();
int myarray[]=new int[count];
for(int tmp=0; tmp<count;)
myarray[tmp]=++tmp;
while(count != 0){
for(int inc=1; inc<=count; inc++){
System.out.println(inc + "times occur");
}
System.out.println("Enter 0 to exit");
count=new Scanner(System.in).nextInt();
}

You should take a look at the line
while(counts[0] > 0) {
and try to figure out what is the purpose of this while loop in the main method.

using an Array to display user inputed values hav

i have an assignment , i have been working on it all day, i finished it, and it works, but not in the way my teacher wants it, i have been racking my brain for hours without any luck
this is the assignment question
"write a program that inputs five each of which is between 10 and 100 inclusive,as each number i read, display it only if it is not a duplicate number already read, provide for the "worst case," in which all five numbers are different, use the smallest possible array to solve this problem. Display the complete set of unique values input after the user inputs each new value."
this is the code i wrote
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int sid[] = new int[5];
int count = 0;
int x = 0;
int num = 0;
while (x < sid.length)
{
System.out.println("Enter number: ");
num = input.nextInt();
if ((num >= 10) && (num <= 100)) {
boolean digit = false;
x++;
for (int i = 0; i < sid.length; i++)
{ if (sid[i] == num)
digit = true;
}
if (!digit) {
sid[count] = num;
count++;
}
else
System.out.printf("the number was entered before \n");
}
else
System.out.println("number must be between 10 and 100");
for (int i =0; i < x; i++) {
System.out.print(sid[i] +" ");
}
System.out.println();
}
}
}
my problem is in the output the output is like this
{
Enter number:
11
11
Enter number:
21
11 21
Enter number:
34
11 21 34
Enter number:
11
the number was entered before
11 21 34 0
Enter number:
44
11 21 34 44 0
}
But my teacher wants the output too look like this, without the 0's at the end, and without changing the value of the array sid from sid[5] to sid[4]
{
Enter number:
11
11
Enter number:
21
11 21
Enter number:
34
11 21 34
Enter number:
11
the number was entered before
11 21 34
Enter number:
44
11 21 34 44
}
i have been trying for hours now without any luck.
ANY help is appreciated.
Thank you.
cheers.

You only have to add an if condition to your output loop.
for (int i =0; i < x; i++) {
if(sid[i]!=0)
System.out.print(sid[i] +" ");
}

When you print the answer check for '0' and ignore them. or else decrements x when you found out the current number is a duplicate.
for (int i = 0; i < sid.length; i++)
{ if (sid[i] == num)
digit = true;
x--;
}
or
for (int i =0; i < x; i++) {
if(sid[i]!=0)
System.out.print(sid[i] +" ");
}
And if it is not necessary to use arrays you can use Sets. Then It would be much easier.

Put the x increment (x++) in the if (!digit)
if (!digit) {
sid[count] = num;
count++;
x++;
}
remove the x-increment (x++) from the if statement below
if ((num >= 10) && (num <= 100)) {
boolean digit = false;
x++;
}
that way the zeros are eliminated and the program still runs no matter the number of duplicates tried.