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I am trying to "Find the 10001st prime number" as part of the Project Euler challenges and i have no idea why my code doesn't work. When I tested my isPrime() function, it succeeded in finding whether a number was prime, but my program returns 10200 as the 10001st prime. Why is this?
Here is My Code:
using System;
using System.Collections.Generic;
namespace Challenge_7
{
class Program
{
static void Main(string[] args)
{
Console.WriteLine("Solution to Project Euler");
Console.WriteLine("Challenge 7 - Find the 10001st prime");
Console.WriteLine("\nProject Start:");
List<int> primes = new List<int>();
int number = 1;
while (primes.Count != 10001)
{
if (isPrime(number))
{
primes.Add(number);
Console.WriteLine(number);
}
number++;
}
Console.WriteLine("The 10001st prime is: {0}", primes[10000]);
Console.ReadLine();
}
private static bool isPrime(int n)
{
bool prime = true;
for (int i = 1; i <= Math.Ceiling(Math.Sqrt(n)); i++)
{
for (int j = 1; j <= Math.Ceiling(Math.Sqrt(n)); j++)
{
if (i * j == n)
{
prime = false;
}
}
}
return prime;
}
}
}
Here's a hint::
Imagine a number that is the product of 3 primes.
Lets say 3, 5, and 7 (or) 105;
sqrt(105) == 10.2 so ceiling is 11
There aren't two numbers less than 11 that multiply to 105.
So your algorithm would falsely return true!
Try Again! :-D
The problem is in you loops. Math.Ceiling(Math.Sqrt(10200)) is 101 and you need to check 102 * 100 = 10200 but your loops never reaches 102 and returns 10200 as prime!
You can use the code below for isPrime. It is exists in this link and I changed to C# for you:
private static bool isPrime(int n)
{
if (n <= 1)
return false;
else if (n <= 3)
return true;
else if (n % 2 == 0 || n % 3 == 0)
return false;
int i = 5;
while (i * i <= n)
{
if (n % i == 0 || n % (i + 2) == 0)
return false;
i += 6;
}
return true;
}
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace nthPrimeNumber
{
class Program
{
static void Main(string[] args)
{
ulong starting_number = 1;
ulong last_number = 200000; //only this value needs adjustment
ulong nth_primenumber = 0;
ulong a;
ulong b;
ulong c;
for (a = starting_number; a <= last_number; a++)
{
b = Convert.ToUInt64(Math.Ceiling(Math.Sqrt(a)));
for (c = 2; c <= b; c++)
{
if (a == 1 || (a % c == 0 && c != b))
{
break;
}
else
{
if (c == b)
{
nth_primenumber = nth_primenumber + 1;
break;
}
}
}
if (nth_primenumber == 10001)
{
break;
}
}
Console.WriteLine(nth_primenumber + "st" + "prime number is " + a);
Console.ReadKey();
}
}
}
The program above generates prime numbers between 1 and 200000. The program counts the generated prime numbers and checks if the generated prime numbers are already 10001. The program prints the 10001st prime number. If the program doesn't show 10001st (or shows below 10001) in the console, it is because the value of the last_number is small -- then try to adjust the value of the last_number (make it bigger). In this program, I have already adjusted the value of the last_number and will print the 10001st prime number.
I am trying to fasten my isPrime function but when I add a condition that if the number is divisible by 2 just return false instead of doing the whole process to find if the number is prime or not, but when I do this, it skips me a number for example the 6th prime is 13, without the condition if it's divisble by 2 I get 13, but when I add it I get 17.
static bool isPrime(long n)
{
bool prime = false;
int div = 0;
if (n % 2 == 0)
return false;
else
for (long i = 1; i < n + 1; i++)
{
if (n % i == 0)
div++;
if (div == 2)
prime = true;
else
prime = false;
}
return prime;
}
You need to check for the special case of 2 first, it's even but prime.
As an additional optimization, you can improve the bounds you're looping up to; there's no need to go as high as n + 1.
If you want to speed up the solution you can do something like that:
static bool isPrime(long n) {
// all integers less than 1 (1 is included) are not prime
if (n <= 1)
return false;
// Error in your code: 2 is prime, even if other even numbers aren't
if (n % 2 == 0)
return (n == 2);
// there's no need to loop up to n: sqrt(n) is quite enough
long max = (long) (Math.Sqrt(n) + 0.1);
// skip even numbers when looping: i +=2
for (long i = 3; i <= max; i += 2) {
// the early return the better
if (n % i == 0)
return false;
}
return true;
}
if (n % 2 == 0)
return (n == 2);
Modulus operation is slow in most architectures if you compare it to bit level checking. You can choose to only check the least significant bit of each number and get the answer if it's even or odd number.
Just do a AND operation with 1 and the value. The result tells you if it's even.
Example:
100101011101 (the value to check)
000000000001 (bitmask to AND with)
000000000001 (the result after AND, still 1 ---> odd number (true in boolean))
So:
if(!(value & 1))
{
//even
return false;
};
At first user gives a number (n) to program, for example 5.
the program must find the smallest number that can be divided to n (5).
and this number can only consist of digits 0 and 9 not any other digits.
for example if user gives 5 to program.
numbers that can be divided to 5 are:
5, 10, 15, 20, 25, 30, ..., 85, 90, 95, ...
but 90 here is the smallest number that can be divided to 5 and also consist of digits (0 , 9). so answer for 5 must be 90.
and answer for 9 is 9, because it can be divided to 9 and consist of digit (9).
my code
string a = txtNumber.Text;
Int64 x = Convert.ToInt64(a);
Int64 i ,j=1,y=x;
bool t = false;
for (i = x + 1; t == false; i++)
{
if (i % 9 == 0 && i % 10 == 0 && i % x == 0)
{
j = i;
for (; (i /= 10) != 0; )
{
i /= 10;
if (i == 0)
t = true;
continue;
}
}
}
lblAnswer.Text = Convert.ToString(j);
If you're happy to go purely functional then this works:
Func<IEnumerable<long>> generate = () =>
{
Func<long, IEnumerable<long>> extend =
x => new [] { x * 10, x * 10 + 9 };
Func<IEnumerable<long>, IEnumerable<long>> generate2 = null;
generate2 = ns =>
{
var clean = ns.Where(n => n > 0).ToArray();
return clean.Any()
? clean.Concat(generate2(clean.SelectMany(extend)))
: Enumerable.Empty<long>();
};
return generate2(new[] { 9L, });
};
Func<long, long?> f = n =>
generate()
.Where(x => x % n == 0L)
.Cast<long?>()
.FirstOrDefault();
So rather than iterate through all possible values and test for 0 & 9 and divisibility, this just generates only numbers with 0 & 9 and then only tests for visibility. It is much faster this way.
I can call it like this:
var result = f(5L); // 90L
result = f(23L); //990909L
result = f(123L); //99999L
result = f(12321L); //90900999009L
result = f(123212L); //99909990090000900L
result = f(117238L); //990990990099990990L
result = f(1172438L); //null == No answer
These results are super fast. f(117238L) returns a result on my computer in 138ms.
You can try this way :
string a = txtNumber.Text;
Int64 x = Convert.ToInt64(a);
int counter;
for (counter = 1; !isValid(x * counter); counter++)
{
}
lblAnswer.Text = Convert.ToString(counter*x);
code above works by searching multiple of x incrementally until result that satisfy criteria : "consist of only 0 and or 9 digits" found. By searching only multiple of x, it is guaranteed to be divisible by x. So the rest is checking validity of result candidate, in this case using following isValid() function :
private static bool isValid(int number)
{
var lastDigit = number%10;
//last digit is invalid, return false
if (lastDigit != 0 & lastDigit != 9) return false;
//last digit is valid, but there is other digit(s)
if(number/10 >= 1)
{
//check validity of digit(s) before the last
return isValid(number/10);
}
//last digit is valid, and there is no other digit. return true
return true;
}
About strange empty for loop in snippet above, it is just syntactic sugar, to make the code a bit shorter. It is equal to following while loop :
counter = 1;
while(!isValid(input*counter))
{
counter++;
}
Use this simple code
int inputNumber = 5/*Or every other number, you can get this number from input.*/;
int result=1;
for (int i = 1; !IsOk(result,inputNumber); i++)
{
result = i*inputNumber;
}
Print(result);
IsOk method is here:
bool IsOk(int result, int inputNumber)
{
if(result%inputNumber!=0)
return false;
if(result.ToString().Replace("9",string.Empty).Replace("0",string.Empty).Length!=0)
return false;
return true;
}
My first solution has very bad performance, because of converting a number to string and looking for characters '9' and '0'.
New solution:
My new solution has very good performance and is a technical approach since of using Breadth-first search(BFS).
Algorithm of this solution:
For every input number, test 9, if it is answer print it, else add 2 child numbers (90 & 99) to queue, and continue till finding answer.
int inputNumber = 5;/*Or every other number, you can get this number from input.*/
long result;
var q = new Queue<long>();
q.Enqueue(9);
while (true)
{
result = q.Dequeue();
if (result%inputNumber == 0)
{
Print(result);
break;
}
q.Enqueue(result*10);
q.Enqueue(result*10 + 9);
}
Trace of number creation:
9
90,99
900,909,990,999
9000,9009,9090,9099,9900,9909,9990,9999
.
.
.
I wrote this code for console, and i used goto command however it is not prefered but i could not write it with only for.
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace main
{
class Program
{
static void Main(string[] args)
{
Console.WriteLine("Enter your number");
Int64 x = Convert.ToInt64(Console.ReadLine());
Int64 y, j, i, k, z = x;
x = x + 5;
loop:
x++;
for (i = 0, y = x; y != 0; i++)
y /= 10;
for (j = x, k = i; k != 0; j /= 10, k--)
{
if (j % 10 != 9)
if (j % 10 != 0)
goto loop;
}
if (x % z != 0)
goto loop;
Console.WriteLine("answer:{0}",x);
Console.ReadKey();
}
}
}
I'm a 17 year old student currently in software engineering and web development and im having trouble right now with some of my coding. I need to make a project that will alow the user to input a number anywherefrom 0 to 999 and tell whether it is a prime number or not. The code i have so far is....
using System;
using System.Collections.Generic;
using System.Linq;
using System.Web;
using System.Web.UI;
using System.Web.UI.WebControls;
public partial class _Default : System.Web.UI.Page
{
protected void Page_Load(object sender, EventArgs e)
{
}
public void primeNumber()
{
int primeNumber1 = int.Parse(Request.Form["Text4"]);
if (primeNumber1 % 1 == 0 & ! (primeNumber1 % 2 == 0 &
primeNumber1 % 3 == 0 &
primeNumber1 % 4 == 0 &
primeNumber1 % 5 == 0 &
primeNumber1 % 6 == 0 &
primeNumber1 % 7 == 0 &
primeNumber1 % 8 == 0 &
primeNumber1 % 9 == 0))
{
Response.Write(" This is a prime number! ");
}
else
{
Response.Write(" This is not a prime Number! ");
}
}
}
... but i cannot get this program to display the correct answer. Any help would be greatly appreciated. Thanks!
You have got the concept of prime numbers wrong. Your code would for example report that 3 is not a prime number, because you check if the number divides evenly in three even if the number entered is three.
The simplest solution would be to loop from 2 and up to primeNumber1 - 1 and check if any of those divides evenly with the number. As you are using a loop, you also need a variable to hold what the result was, as you don't have a single expression that returns the result.
Something like:
bool prime = true;
for (int i = 2; i <= primeNumber1 - 1; i++) {
if (primeNumber1 % i == 0) {
prime = false;
}
}
This is of course the simplest possible solution that solves the problem, for reasonably small numbers. You can for example improve on the solution by exiting out of the loop as soon as you know that it's not a prime number.
You also don't need to loop all the way to primeNumber1 - 1, but only as high as the square root of the number, but you can find out about that if you read up on methods for checking prime numbers.
You need to handle the special cases of 1 and 2 also. By definition 1 is not a prime number, but 2 is.
http://en.wikipedia.org/wiki/Prime_number
bool IsPrime(int number) {
if (number == 1) return false;
if (number == 2) return true;
for (int i = 2; i < number; ++i) {
if (number % i == 0) return false;
}
return true;
}
A little google-fu or a little navel-gazing about prime numbers in general, will lead you to the naive algorithm:
For all n such that 0 < n:
There are two "special case" prime numbers, 1 and 2.
All even numbers > 2 are non-prime, by definition
If you think about the nature of factoring, the largest possible factor you have to consider is the square root of n, since above that point, the factors are reflexive (i.e., the possible factorizations of 100 are 1*100 , 2*50 , 4*25 , 5*20 , 10*10 , 20*5 , 25*4, 50*2 and 100*1 — and the square root of 100 is...10).
That should lead you to an implementation that looks something like this:
static bool IsPrime( int n )
{
if ( n < 1 ) throw new ArgumentOutOfRangeException("n") ;
bool isPrime = true ;
if ( n > 2 )
{
isPrime = ( 0 != n & 0x00000001 ) ; // eliminate all even numbers
if ( isPrime )
{
int limit = (int) Math.Sqrt(n) ;
for ( int i = 3 ; i <= limit && isPrime ; i += 2 )
{
isPrime = ( 0 != n % i ) ;
}
}
}
return isPrime ;
}
Anytime you find yourself in programming repeating a test on a sequential range of numbers you're doing the wrong thing. A better construct for this is a loop. This will give you the range of numbers in an identifier which can then be used to write the repetive code one time. For example I could rewrite this code
primeNumber1 % 2 == 0 &
primeNumber1 % 3 == 0 &
primeNumber1 % 4 == 0 &
primeNumber1 % 5 == 0 &
primeNumber1 % 6 == 0 &
primeNumber1 % 7 == 0 &
primeNumber1 % 8 == 0 &
primeNumber1 % 9 == 0))
As follows
bool anyFactors = false;
for (int i = 2; i <= 9; i++) {
if (primeNumber1 % i != 0) {
anyFactors = true;
break;
}
}
At this point I can now substitute the value allTrue for the original condition you wrote.
if (primeNumber1 % 1 == 0 && !anyFactors)
I can also expand the number of values tested here by substiting a different number for the conditional check of the loop. If I wanted to check 999 values I would instead write
for (int i = 2; i <= 999; i++) {
...
}
Additionally you don't want to use & in this scenario. That is for bit level and operations. You are looking for the logical and operator &&
Try the code below:
bool isPrimeNubmer(int n)
{
if (n >=0 && n < 4) //1, 2, 3 are prime numbers
return true;
else if (n % 2 == 0) //even numbers are not prime numbers
return false;
else
{
int j = 3;
int k = (n + 1) / 2 ;
while (j <= k)
{
if (n % j == 0)
return false;
j = j + 2;
}
return true;
}
}
I am trying to write a prime number function in C# and I am wondering if the follow code will work. It "appears" to work with the first 50 numbers or so. I just want to make sure it will work no matter how big the number is:
static bool IsPrime(int number)
{
if ((number == 2) || (number == 3) || (number == 5) || (number == 7) || (number == 9))
return true;
if ((number % 2 != 0) && (number % 3 != 0) && (number % 5 != 0) &&
(number % 7 != 0) && (number % 9 != 0) && (number % 4 != 0) &&
(number % 6 != 0))
return true;
return false;
}
No it won't work! Try 121 = 11 * 11 for example which obviously isn't a prime.
For any number given to your function, that is a product of the prime numbers X1, X2, ..., Xn(where n >= 2) with all of them being greater or equal to 11, your function will return true. (And also, as already said, 9 isn't a prime).
From wikipedia you can see that:
In mathematics, a prime number (or a prime) is a natural number that has exactly two distinct natural number divisors: 1 and itself.
so a very simple and naive algorithm on checking whether a number is prime could be:
public bool CalcIsPrime(int number) {
if (number == 1) return false;
if (number == 2) return true;
if (number % 2 == 0) return false; // Even number
for (int i = 2; i < number; i++) { // Advance from two to include correct calculation for '4'
if (number % i == 0) return false;
}
return true;
}
For better algorithms check here: Primality Test
If you want to check your code, do inlcude a test, here's a test case written in xunit.
[Theory]
[MemberData(nameof(PrimeNumberTestData))]
public void CalcIsPrimeTest(int number, bool expected) {
Assert.Equal(expected, CalcIsPrime(number));
}
public static IEnumerable<object[]> PrimeNumberTestData() {
yield return new object[] { 0, false };
yield return new object[] { 1, false };
yield return new object[] { 2, true };
yield return new object[] { 3, true };
yield return new object[] { 4, false };
yield return new object[] { 5, true };
yield return new object[] { 6, false };
yield return new object[] { 7, true };
yield return new object[] { 8, false };
yield return new object[] { 9, false };
yield return new object[] { 10, false };
yield return new object[] { 11, true };
yield return new object[] { 23, true };
yield return new object[] { 31, true };
yield return new object[] { 571, true };
yield return new object[] { 853, true };
yield return new object[] { 854, false };
yield return new object[] { 997, true };
yield return new object[] { 999, false };
}
It had to be done...
public static bool IsPrime(this int number)
{
return (Enumerable.Range(1,number).Where(x => number % x == 0).Count() == 2);
}
This approach definitely won't work, unless your if statement explicitly enumerates all the prime numbers between 0 and sqrt(INT_MAX) (or the C# equivalent).
To properly check for primality, you basically need to attempt to divide your number by every prime number less than its square root. The Sieve of Eratosthenes algorithm is your best bet.
You are apparently writing from a contrafactual dimension where 9 is a prime number, so I guess that our answers might not work for you. Two things though:
Prime number generating functions are a non-trivial but exiting matter, the Wikipedia page is a good starter (http://en.wikipedia.org/wiki/Formula_for_primes)
from (number%2!=0) it follows (number%4!=0). If you can't divide by 10, then you can't divide by 100 either.
Primality testing is the way to go, but in case you want a quick and dirty hack, here's something.
If it's not working fast enough, you can build a class around it and store the PrimeNumbers collection from call to call, rather than repopulating it for each call.
public bool IsPrime(int val)
{
Collection<int> PrimeNumbers = new Collection<int>();
int CheckNumber = 5;
bool divisible = true;
PrimeNumbers.Add(2);
PrimeNumbers.Add(3);
// Populating the Prime Number Collection
while (CheckNumber < val)
{
foreach (int i in PrimeNumbers)
{
if (CheckNumber % i == 0)
{
divisible = false;
break;
}
if (i * i > CheckNumber) { break; }
}
if (divisible == true) { PrimeNumbers.Add(CheckNumber); }
else { divisible = true; }
CheckNumber += 2;
}
foreach (int i in PrimeNumbers)
{
if (CheckNumber % i == 0)
{
divisible = false;
break;
}
if (i * i > CheckNumber) { break; }
}
if (divisible == true) { PrimeNumbers.Add(CheckNumber); }
else { divisible = true; }
// Use the Prime Number Collection to determine if val is prime
foreach (int i in PrimeNumbers)
{
if (val % i == 0) { return false; }
if (i * i > val) { return true; }
}
// Shouldn't ever get here, but needed to build properly.
return true;
}
There are some basic rules you can follow to check if a number is prime
Even numbers are out. If x % 2 = 0, then it is not prime
All non-prime numbers have prime factors. Therefore, you only need test a number against primes to see if it factors
The highest possible factor any number has is it's square root. You only need to check if values <= sqrt(number_to_check) are even divisible.
Using that set of logic, the following formula calculates 1,000,000 Primes Generated in: 134.4164416 secs in C# in a single thread.
public IEnumerable<long> GetPrimes(int numberPrimes)
{
List<long> primes = new List<long> { 1, 2, 3 };
long startTest = 3;
while (primes.Count() < numberPrimes)
{
startTest += 2;
bool prime = true;
for (int pos = 2; pos < primes.Count() && primes[pos] <= Math.Sqrt(startTest); pos++)
{
if (startTest % primes[pos] == 0)
{
prime = false;
}
}
if (prime)
primes.Add(startTest);
}
return primes;
}
Bear in mind, there is lots of room for optimization in the algorithm. For example, the algorithm could be parallelized. If you have a prime number (let's say 51), you can test all the numbers up to it's square (2601) for primeness in seperate threads as all it's possible prime factors are stored in the list.
static List<long> PrimeNumbers = new List<long>();
static void Main(string[] args)
{
PrimeNumbers.Add(2);
PrimeNumbers.Add(3);
PrimeNumbers.Add(5);
PrimeNumbers.Add(7);
for (long i = 11; i < 10000000; i += 2)
{
if (i % 5 != 0)
if (IsPrime(i))
PrimeNumbers.Add(i);
}
}
static bool IsPrime(long number)
{
foreach (long i in PrimeNumbers)
{
if (i <= Math.Sqrt(number))
{
if (number % i == 0)
return false;
}
else
break;
}
return true;
}
this is a simple one
only odd numbers are prime....so
static bool IsPrime(int number)
{
int i;
if(number==2)
return true; //if number is 2 then it will return prime
for(i=3,i<number/2;i=i+2) //i<number/2 since a number cannot be
{ //divided by more then its half
if(number%i==0) //if number is divisible by i, then its not a prime
return false;
}
return true; //the code will only reach here if control
} //is not returned false in the for loop
This is a simple code for find prime number depend on your input.
static void Main(string[] args)
{
String input = Console.ReadLine();
long num = Convert.ToInt32(input);
long a, b, c;
c = 2;
for(long i=3; i<=num; i++){
b = 0;
for (long j = 2; j < i ; j++) {
a = i % j;
if (a != 0) {
b = b+1;
}
else {
break;
}
}
if(b == i-2){
Console.WriteLine("{0}",i);
}
}
Console.ReadLine();
}
ExchangeCore Forums have a good bit of code that will pretty much let you generate any ulong number for primes. But basically here's the gist:
int primesToFind = 1000;
int[] primes = new int[primesToFind];
int primesFound = 1;
primes[0] = 2;
for(int i = 3; i < int.MaxValue() && primesFound < primesToFind; i++)
{
bool isPrime = true;
double sqrt = Math.sqrt(i);
for(int j = 0; j<primesFound && primes[j] <= sqrt; j++)
{
if(i%primes[j] == 0)
{
isPrime = false;
break;
}
}
if(isPrime)
primes[primesFound++] = i;
}
Once this code has finished running your primes will all be found in the primes array variable.
https://www.khanacademy.org/computing/computer-science/cryptography/comp-number-theory/a/trial-division
public static bool isPrime(int number)
{
for (int k = 2; k <= Math.Ceiling(Math.Sqrt(number)); k++)
{
if (number > k && number % k == 0)
break;
if (k >= Math.Ceiling(Math.Sqrt(number)) || number == k)
{
return true;
}
}
return false;
}
Prime Numbers from 0 - 1 Million in less than two tenths of a second
Just finished it. Last test was 0.017 seconds.
Regular HP Laptop. 2.1 GHz
It takes longer when it gets larger. For primes 1 - 1 billion , my last test was 28.6897 seconds. It might be faster in your program because I was casting class objects to get parameter values in mine.
Method Info
Uses the Sieve of Eratosthenes
Accepts floor and ceiling as arguments
Uses arrays instead of lists for fast performance
Size of array is initialized according to Rosser-Schoenfeld upper bound
Skips multiples of 2, 5, and 7 when assigning values
Max range is between 0 and 2,147,483,646 (1 min 44.499 s)
Heavily commented
Using
using System;
using System.Diagnostics;
using System.Collections;
Method
private static int[] GetPrimeArray(int floor, int ceiling)
{
// Validate arguments.
if (floor > int.MaxValue - 1)
throw new ArgumentException("Floor is too high. Max: 2,147,483,646");
else if (ceiling > int.MaxValue - 1)
throw new ArgumentException("Ceiling is too high. Max: 2,147,483,646");
else if (floor < 0)
throw new ArgumentException("Floor must be a positive integer.");
else if (ceiling < 0)
throw new ArgumentException("Ceiling must be a positve integer.");
else if (ceiling < floor)
throw new ArgumentException("Ceiling cannot be less than floor.");
// This region is only useful when testing performance.
#region Performance
Stopwatch sw = new Stopwatch();
sw.Start();
#endregion
// Variables:
int stoppingPoint = (int)Math.Sqrt(ceiling);
double rosserBound = (1.25506 * (ceiling + 1)) / Math.Log(ceiling + 1, Math.E);
int[] primeArray = new int[(int)rosserBound];
int primeIndex = 0;
int bitIndex = 4;
int innerIndex = 3;
// Handle single digit prime ranges.
if (ceiling < 11)
{
if (floor <= 2 && ceiling >= 2) // Range includes 2.
primeArray[primeIndex++] = 2;
if (floor <= 3 && ceiling >= 3) // Range includes 3.
primeArray[primeIndex++] = 3;
if (floor <= 5 && ceiling >= 5) // Range includes 5.
primeArray[primeIndex++] = 5;
return primeArray;
}
// Begin Sieve of Eratosthenes. All values initialized as true.
BitArray primeBits = new BitArray(ceiling + 1, true);
primeBits.Set(0, false); // Zero is not prime.
primeBits.Set(1, false); // One is not prime.
checked // Check overflow.
{
try
{
// Set even numbers, excluding 2, to false.
for (bitIndex = 4; bitIndex < ceiling; bitIndex += 2)
primeBits[bitIndex] = false;
}
catch { } // Break for() if overflow occurs.
}
// Iterate by steps of two in order to skip even values.
for (bitIndex = 3; bitIndex <= stoppingPoint; bitIndex += 2)
{
if (primeBits[bitIndex] == true) // Is prime.
{
// First position to unset is always the squared value.
innerIndex = bitIndex * bitIndex;
primeBits[innerIndex] = false;
checked // Check overflow.
{
try
{
// Set multiples of i, which are odd, to false.
innerIndex += bitIndex + bitIndex;
while (innerIndex <= ceiling)
{
primeBits[innerIndex] = false;
innerIndex += bitIndex + bitIndex;
}
}
catch { continue; } // Break while() if overflow occurs.
}
}
}
// Set initial array values.
if (floor <= 2)
{
// Range includes 2 - 5.
primeArray[primeIndex++] = 2;
primeArray[primeIndex++] = 3;
primeArray[primeIndex++] = 5;
}
else if (floor <= 3)
{
// Range includes 3 - 5.
primeArray[primeIndex++] = 3;
primeArray[primeIndex++] = 5;
}
else if (floor <= 5)
{
// Range includes 5.
primeArray[primeIndex++] = 5;
}
// Increment values that skip multiples of 2, 3, and 5.
int[] increment = { 6, 4, 2, 4, 2, 4, 6, 2 };
int indexModulus = -1;
int moduloSkipAmount = (int)Math.Floor((double)(floor / 30));
// Set bit index to increment range which includes the floor.
bitIndex = moduloSkipAmount * 30 + 1;
// Increase bit and increment indicies until the floor is reached.
for (int i = 0; i < increment.Length; i++)
{
if (bitIndex >= floor)
break; // Floor reached.
// Increment, skipping multiples of 2, 3, and 5.
bitIndex += increment[++indexModulus];
}
// Initialize values of return array.
while (bitIndex <= ceiling)
{
// Add bit index to prime array, if true.
if (primeBits[bitIndex])
primeArray[primeIndex++] = bitIndex;
checked // Check overflow.
{
try
{
// Increment. Skip multiples of 2, 3, and 5.
indexModulus = ++indexModulus % 8;
bitIndex += increment[indexModulus];
}
catch { break; } // Break if overflow occurs.
}
}
// Resize array. Rosser-Schoenfeld upper bound of π(x) is not an equality.
Array.Resize(ref primeArray, primeIndex);
// This region is only useful when testing performance.
#region Performance
sw.Stop();
if (primeArray.Length == 0)
Console.WriteLine("There are no prime numbers between {0} and {1}",
floor, ceiling);
else
{
Console.WriteLine(Environment.NewLine);
for (int i = 0; i < primeArray.Length; i++)
Console.WriteLine("{0,10}:\t\t{1,15:#,###,###,###}", i + 1, primeArray[i]);
}
Console.WriteLine();
Console.WriteLine("Calculation time:\t{0}", sw.Elapsed.ToString());
#endregion
return primeArray;
}
Comments are welcome! Especially improvements.
Here we must have to consider the square root factor. A prime number can be verified if it is not divisible by any number less than the value of square root of any near number.
static bool isPrime(long number)
{
if (number == 1) return false;
if (number == 2) return true;
if (number % 2 == 0) return false; //Even number
long nn= (long) Math.Abs(Math.Sqrt(number));
for (long i = 3; i < nn; i += 2) {
if (number % i == 0) return false;
}
return true;
}