C# - for Loop Freezes at strange intervals - c#

I am working on Problem 14 on Project Euler, and my code seems to freeze at random intervals for no apparent reason.
static void Main()
{
int maxNum = 0;
int maxLength = 0;
for (int x = 2; x < 1000000; ++x)
{
int num = x;
int length = 0;
while (num != 1)
{
if (num % 2 == 0)
{
num /= 2;
length++;
}
else
{
num = (3 * num) + 1;
length++;
}
}
if (length > maxLength)
{
maxLength = length;
maxNum = x;
}
}
Console.WriteLine(maxNum);
Console.ReadLine();
The number that the program hangs at is different each time I run it and doesn't seem to follow any set patterns. Any ideas on why it would be hanging like this? Thanks in advance.

I've solved it in another way, by caching the result for each step, and I've found your problem. I doubt your program ever stops.
The statement num = (3 * num) + 1 may overflow over Int32.MaxValue and result in a negative number and an infinite loop(?).
In this case, you can solve the problem by using long for your x.

If goes into infinite loop in while (num != 1).

I bet that this version doesn't freeze, there's no reason it should do that.
The Collatz sequences before you hit 1 in the inner while loop are too short to lead to a noticeable delay, and if they would that should always happen at the same numbers.
If you add console output inside the loop then this may allocate memory, and the pauses you see could be due to garbage collection.

Related

how many numbers between 1 to 10 billion contains 14

i tried this code but it takes so long and I can not get the result
public long getCounter([FromBody]object req)
{
JObject param = Utility.GetRequestParameter(req);
long input = long.Parse(param["input"].ToString());
long counter = 0;
for (long i = 14; i <= input; i++)
{
string s = i.ToString();
if (s.Contains("14"))
{
counter += 1;
}
}
return counter;
}
please help
We can examine all non-negative numbers < 10^10. Every such number can be represented with the sequence of 10 digits (with leading zeroes allowed).
How many numbers include 14
Dynamic programming solution. Let's find the number of sequences of a specific length that ends with the specific digit and contains (or not) subsequence 14:
F(len, digit, 0) is the number of sequences of length len that ends with digit and do not contain 14, F(len, digit, 1) is the number of such sequences that contain 14. Initially F(0, 0, 0) = 1. The result is the sum of all F(10, digit, 1).
C++ code to play with: https://ideone.com/2aS17v. The answer seems to be 872348501.
How many times the numbers include 14
First, let's place 14 at the end of the sequence:
????????14
Every '?' can be replaced with any digit from 0 to 9. Thus, there are 10^8 numbers in the interval that contains 14 at the end. Then consider ???????14?, ??????14??, ..., 14???????? numbers. There are 9 possible locations of 14 sequence. The answer is 10^8 * 9 = 90000000.
[Added by Matthew Watson]
Here's the C# version of the C++ implementation; it runs in less than 100ms:
using System;
namespace Demo
{
public static class Program
{
public static void Main(string[] args)
{
const int M = 10;
int[,,] f = new int [M + 1, 10, 2];
f[0, 0, 0] = 1;
for (int len = 1; len <= M; ++len)
{
for (int d = 0; d <= 9; ++d)
{
for (int j = 0; j <= 9; ++j)
{
f[len,d,0] += f[len - 1,j,0];
f[len,d,1] += f[len - 1,j,1];
}
}
f[len,4,0] -= f[len - 1,1,0];
f[len,4,1] += f[len - 1,1,0];
}
int sum = 0;
for (int i = 0; i <= 9; ++i)
sum += f[M,i,1];
Console.WriteLine(sum); // 872,348,501
}
}
}
If you want a brute force solution it could be something like this (please, notice, that we should avoid time consuming string operations like ToString, Contains):
int count = 0;
// Let's use all CPU's cores: Parallel.For
Parallel.For(0L, 10000000000L, (v) => {
for (long x = v; x > 10; x /= 10) {
// Get rid of ToString and Contains here
if (x % 100 == 14) {
Interlocked.Increment(ref count); // We want an atomic (thread safe) operation
break;
}
}
});
Console.Write(count);
It returns 872348501 within 6 min (Core i7 with 4 cores at 3.2GHz)
UPDATE
My parallel code calculated the result as 872,348,501 in 9 minutes on my 8- processor-core Intel Core I7 PC.
(There is a much better solution above that takes less than 100ms, but I shall leave this answer here since it provides corroborating evidence for the fast answer.)
You can use multiple threads (one per processor core) to reduce the calculation time.
At first I thought that I could use AsParallel() to speed this up - however, it turns out that you can't use AsParallel() on sequences with more than 2^31 items.
(For completeness I'm including my faulty implementation using AsParallel at the end of this answer).
Instead, I've written some custom code to break the problem down into a number of chunks equal to the number of processors:
using System;
using System.Linq;
using System.Threading.Tasks;
namespace Demo
{
class Program
{
static void Main()
{
int numProcessors = Environment.ProcessorCount;
Task<long>[] results = new Task<long>[numProcessors];
long count = 10000000000;
long elementsPerProcessor = count / numProcessors;
for (int i = 0; i < numProcessors; ++i)
{
long end;
long start = i * elementsPerProcessor;
if (i != (numProcessors - 1))
end = start + elementsPerProcessor;
else // Last thread - go right up to the last element.
end = count;
results[i] = Task.Run(() => processElements(start, end));
}
long sum = results.Select(r => r.Result).Sum();
Console.WriteLine(sum);
}
static long processElements(long inclusiveStart, long exclusiveEnd)
{
long total = 0;
for (long i = inclusiveStart; i < exclusiveEnd; ++i)
if (i.ToString().Contains("14"))
++total;
return total;
}
}
}
The following code does NOT work because AsParallel() doesn't work on sequences with more than 2^31 items.
static void Main(string[] args)
{
var numbersContaining14 =
from number in numbers(0, 100000000000).AsParallel()
where number.ToString().Contains("14")
select number;
Console.WriteLine(numbersContaining14.LongCount());
}
static IEnumerable<long> numbers(long first, long count)
{
for (long i = first, last = first + count; i < last; ++i)
yield return i;
}
You compute the count of numbers of a given length ending in 1, 4 or something else that don't contain 14. Then you can extend the length by 1.
Then the count of numbers that do contain 14 is the count of all numbers minus those that don't contain a 14.
private static long Count(int len) {
long e1=0, e4=0, eo=1;
long N=1;
for (int n=0; n<len; n++) {
long ne1 = e4+e1+eo, ne4 = e4+eo, neo = 8*(e1+e4+eo);
e1 = ne1; e4 = ne4; eo = neo;
N *= 10;
}
return N - e1 - e4 - eo;
}
You can reduce this code a little, noting that eo = 8*e1 except for the first iteration, and then avoiding the local variables.
private static long Count(int len) {
long e1=1, e4=1, N=10;
for (int n=1; n<len; n++) {
e4 += 8*e1;
e1 += e4;
N *= 10;
}
return N - 9*e1 - e4;
}
For both of these, Count(10) returns 872348501.
One easy way to calculate the answer is,
You can fix 14 at a place and count the combination of the remaining numbers right to it,
and do this for all the possible positions where 14 can be place such that the number is still less than 10000000000,Lets take a example,
***14*****,
Here the '*' before 14 can be filled by 900 ways and the * after 14 can be filled by 10^5 ways so total occurrence will be 10^5*(900),
Similarly you can fix 14 at other positions to calculate the result and this solution will be very fast O(10) or simply O(1), while the previous solution was O(N), where N is 10000000000
You can use the fact that in each 1000 (that is from 1 to 1000 and from 1001 to 2000 etc)
the 14 is found: 19 times so when you receive your input divide it by 1000 for example you received 1200 so 1200/1000
the result is 1 and remainder 200, so we have 1 * 19 "14"s and then you can loop over the 200.
you can extend for 10000 (that is count how many "14"s there are in 10000 and fix it to a global variable) and start dividing by 10000 then and apply the equation above, then you divide the remainder by 1000 and apply the equation and add the two results.
You can extend it as the fact that for all hundreds (that is from 1 to 100 and from 201 to 300) the "14" is found only 1 except for the second hundred (101 to 200).

When arrays go awry?

I'm trying to learn C# by solving mathematical problems. For example, I'm working on finding the sum of factors of 3 or 5 in the first 1000 positive numbers. I have the basic shell of the code laid out, but it isn't behaving how I'm expecting it to.
Right now, instead of getting a single output of 23, I am instead getting 1,1,3,3,5,5,7,7,9,9. I imagine I messed up the truncate function somehow. Its a bloody mess, but its the only way I can think of checking for factors. Second, I think that the output is writing during the loop, instead of patiently waiting for the for() loop to finish.
using System;
namespace Problem1
{
class Problem1
{
public static void Main()
{
//create a 1000 number array
int[] numberPool = new int[10];
//use for loop to assign the first 1000 positive numbers to the array
for (int i = 0; i < numberPool.Length; i++)
{
numberPool[i] = i + 1;
}
//check for factors of 3 or 5 using if/then statment
foreach (int i in numberPool)
if ((i / 3) == Math.Truncate((((decimal)(i / 3)))) || ((i / 5) == Math.Truncate(((decimal)(i / 5)))))
{
numberPool[i] = i;
}
else
{
numberPool[i] = 0;
}
//throw the 0s and factors together and get the sum!
int sum = 0;
for (int x = 0;x < numberPool.Length;x++)
{
sum = sum + numberPool[x];
}
Console.WriteLine(sum);
Console.ReadLine();
//uncomment above if running in vbs
}
}
}
The foreach loop has a few errors.
If you want to modify the array you are looping through use a for loop. Also, use modulus when checking remainders.
for (int i = 0; i < numberPool.Length; i++)
{
if (numberPool[i] % 3 == 0 || numberPool[i] % 5 == 0)
{
// Do nothing
}
else
{
numberPool[i] = 0;
}
}
Modulus (%) will give the remainder when dividing two integers.
Another useful shortcut, variable = variable + x can be replaced with variable += x
Please note that there are more concise ways of doing this but since you are learning the language I will leave that for you to find.
#kailanjian gave some great advice for you but here is another way your initial logic can be simplified for understanding:
//the sum of factors
int sum = 0;
//the maximum number we will test for
int maxNum = 1000;
//iterate from 1 to our max number
for (int i = 1; i <= maxNum; i++)
{
//the number is a factor of 3 or 5
if (i % 3 == 0 || i % 5 == 0)
{
sum += i;
}
}
//output our sum
Console.WriteLine(sum);
You also stated:
Second, I think that the output is writing during the loop, instead of patiently waiting for the for() loop to finish.
Your program logic will execute in the order that you list it and won't move on to the next given command until it is complete with the last. So your sum output will only be printed once it has completed our for loop iteration.

Why is Console.WriteLine speeding up my application?

Ok so this is kind of weird. I have an algorithm to find the highest possible numerical palindrome that is a multiple of two factors who each have K digits.
The method I'm using to find the highest valid palindrome is to look at the highest possible palindrome for the number set (i.e. if k=3, the highest possible is 999999, then 998899, etc). Then I check if that palindrome has two factors with K digits.
For debugging, I thought it would be a good idea to print to the console each of the palindromes I was checking (to make sure I was getting them all. To my surprise, adding
Console.WriteLine(palindrome.ToString());
to each iteration of finding a palindrome dropped my runtime a whopping 10 seconds from ~24 to ~14.
To verify, I ran the program several times, then commented out the Console command and ran that several times, and every time it was shorter with the Console command.
This just seems weird, any ideas?
Here's the source if anyone wants to take a whack at it:
static double GetHighestPalindromeBench(int k)
{
//Because the result of k == 1 is a known quantity, and results in aberrant behavior in the algorithm, handle as separate case
if (k == 1)
{
return 9;
}
/////////////////////////////////////
//These variables will be used in HasKDigitFactors(), no need to reprocess them each time the function is called
double kTotalSpace = 10;
for (int i = 1; i < k; i++)
{
kTotalSpace *= 10;
}
double digitConstant = kTotalSpace; //digitConstant is used in HasKDigits() to determine if a factor has the right number of digits
double kFloor = kTotalSpace / 10; //kFloor is the lowest number that has k digits (e.g. k = 5, kFloor = 10000)
double digitConstantFloor = kFloor - digitConstant; //also used in HasKDigits()
kTotalSpace--; //kTotalSpace is the highest number that has k digits (e.g. k = 5, kTotalSpace = 99999)
/////////////////////////////////////////
double totalSpace = 10;
double halfSpace = 10;
int reversionConstant = k;
for (int i = 1; i < k * 2; i++)
{
totalSpace *= 10;
}
double floor = totalSpace / 100;
totalSpace--;
for (int i = 1; i < k; i++)
{
halfSpace *= 10;
}
double halfSpaceFloor = halfSpace / 10; //10000
double halfSpaceStart = halfSpace - 1; //99999
for (double i = halfSpaceStart; i > halfSpaceFloor; i--)
{
double value = i;
double palindrome = i;
//First generate the full palindrome
for (int j = 0; j < reversionConstant; j++)
{
int digit = (int)value % 10;
palindrome = palindrome * 10 + digit;
value = value / 10;
}
Console.WriteLine(palindrome.ToString());
//palindrome should be ready
//Now we check the factors of the palindrome to see if they match k
//We only need to check possible factors between our k floor and ceiling, other factors do not solve
if (HasKDigitFactors(palindrome, kTotalSpace, digitConstant, kFloor, digitConstantFloor))
{
return palindrome;
}
}
return 0;
}
static bool HasKDigitFactors(double palindrome, double totalSpace, double digitConstant, double floor, double digitConstantFloor)
{
for (double i = floor; i <= totalSpace; i++)
{
if (palindrome % i == 0)
{
double factor = palindrome / i;
if (HasKDigits(factor, digitConstant, digitConstantFloor))
{
return true;
}
}
}
return false;
}
static bool HasKDigits(double value, double digitConstant, double digitConstantFloor)
{
//if (Math.Floor(Math.Log10(value) + 1) == k)
//{
// return true;
//}
if (value - digitConstant > digitConstantFloor && value - digitConstant < 0)
{
return true;
}
return false;
}
Note that I have the Math.Floor operation in HasKDigits commented out. This all started when I was trying to determine if my digit check operation was faster than the Math.Floor operation. Thanks!
EDIT: Function call
I'm using StopWatch to measure processing time. I also used a physical stopwatch to verify the results of StopWatch.
Stopwatch stopWatch = new Stopwatch();
stopWatch.Start();
double palindrome = GetHighestPalindromeBench(6);
stopWatch.Stop();
TimeSpan ts = stopWatch.Elapsed;
string elapsedTime = String.Format("{0:00}:{1:00}:{2:00}:{3:00}", ts.Hours, ts.Minutes, ts.Seconds, ts.Milliseconds / 10);
Console.WriteLine();
Console.WriteLine(palindrome.ToString());
Console.WriteLine();
Console.WriteLine(elapsedTime);
I have tested your code. My system is an i7-3770 3.40 GHz, quad-core with hyperthreading, so 8 cores available.
Debug build, with and without the console Writeline statement (commented out or not), in debug mode or not, the times vary from about 8.7 to 9.8 sec. As a Release build it comes down to about 6.8-7.0 sec either way. Th figures were the same inside VS and from the command line. So your observation is not reproduced.
On performance monitor with no console output I see one core at 100%, but it switches between cores 1,4,5 and 8. Without console output there is activity on other cores. Max CPU usage never exceeds 18%.
In my judgment your figure with console output is probably consistent with mine, and represents the true value. So your question should read: why is your system so slow when it's not doing console output?
The answer is: because there is something different about your computer or your project which we don't know about. I've never seen this before, but something is soaking up cycles and you should be able to find out what it is.
I've written this as an answer although it isn't really an answer. If you get more facts and update your question, hopefully I can provide a better answer.

Is my program taking too much time to execute?

I wanted to solve a question from project euleur about finding the largest prime number of a big number. I run my code on a virtual machine on Visual studio 2012, and the code seems froze. When I step into the loop, the code works well, but when I execute it, the console is always there. It is as if the program is still running. Could it be that the program takes time to execute?
My Code
static void Main(string[] args)
{
long number = 5;
for (long i = 1; i < 600851475143; i++)
{
if (i % 2 != 0 && i % 1 == 0 && i % i == 0)
number = i;
}
}
I ran this bit of code and it does take a while to run, but it does seem to progress (i does increment). Try this to determine whether i is a prime:
public static bool IsPrime(long candidate)
{
// Test whether the parameter is a prime number.
if ((candidate & 1) == 0)
{
return candidate == 2;
}
// Note:
// ... This version was changed to test the square.
// ... Original version tested against the square root.
// ... Also we exclude 1 at the very end.
for (int i = 3; (i * i) <= candidate; i += 2)
{
if ((candidate % i) == 0)
{
return false;
}
}
return candidate != 1;
}
I can't claim credit for this. It is from http://www.dotnetperls.com/prime.
Add some Console.WriteLines to your main method to its progress:
static void Main(string[] args)
{
long number = 5;
for (long i = 1; i < 600851475143; i++)
{
if (IsPrime(i))
{
Console.WriteLine(i);
number = i;
}
}
}
There's other resources out there for these algorithms too:
http://csharpinoneroom.blogspot.com/2008/03/find-prime-numer-at-fastest-speed.html
Good luck!
Your algorithm is incorrect. Here is a simple way to find the prime factors of a composite number, suitable for Project Euler:
function factors(n)
f := 2
while f * f <= n
if n % f == 0
output f
n := n / f
else
f := f + 1
output n
This works by dividing n by each f in succession, reducing n and outputting f whenever a prime factor is found. The final factor is the remaining n when f is greater than the square root of n, at which point n must be prime.
There are other ways to factor integers. When you are ready for more, I modestly recommend this essay at my blog.
Ultimately, it doesn't matter if you write the fastest code in the world if it doesn't work correctly. Yours doesn't, speed aside.

Program to find prime numbers

I want to find the prime number between 0 and a long variable but I am not able to get any output.
The program is
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace ConsoleApplication16
{
class Program
{
void prime_num(long num)
{
bool isPrime = true;
for (int i = 0; i <= num; i++)
{
for (int j = 2; j <= num; j++)
{
if (i != j && i % j == 0)
{
isPrime = false;
break;
}
}
if (isPrime)
{
Console.WriteLine ( "Prime:" + i );
}
isPrime = true;
}
}
static void Main(string[] args)
{
Program p = new Program();
p.prime_num (999999999999999L);
Console.ReadLine();
}
}
}
Can any one help me out and find what is the possible error in the program?
You can do this faster using a nearly optimal trial division sieve in one (long) line like this:
Enumerable.Range(0, Math.Floor(2.52*Math.Sqrt(num)/Math.Log(num))).Aggregate(
Enumerable.Range(2, num-1).ToList(),
(result, index) => {
var bp = result[index]; var sqr = bp * bp;
result.RemoveAll(i => i >= sqr && i % bp == 0);
return result;
}
);
The approximation formula for number of primes used here is π(x) < 1.26 x / ln(x). We only need to test by primes not greater than x = sqrt(num).
Note that the sieve of Eratosthenes has much better run time complexity than trial division (should run much faster for bigger num values, when properly implemented).
Try this:
void prime_num(long num)
{
// bool isPrime = true;
for (long i = 0; i <= num; i++)
{
bool isPrime = true; // Move initialization to here
for (long j = 2; j < i; j++) // you actually only need to check up to sqrt(i)
{
if (i % j == 0) // you don't need the first condition
{
isPrime = false;
break;
}
}
if (isPrime)
{
Console.WriteLine ( "Prime:" + i );
}
// isPrime = true;
}
}
You only need to check odd divisors up to the square root of the number. In other words your inner loop needs to start:
for (int j = 3; j <= Math.Sqrt(i); j+=2) { ... }
You can also break out of the function as soon as you find the number is not prime, you don't need to check any more divisors (I see you're already doing that!).
This will only work if num is bigger than two.
No Sqrt
You can avoid the Sqrt altogether by keeping a running sum. For example:
int square_sum=1;
for (int j=3; square_sum<i; square_sum+=4*(j++-1)) {...}
This is because the sum of numbers 1+(3+5)+(7+9) will give you a sequence of odd squares (1,9,25 etc). And hence j represents the square root of square_sum. As long as square_sum is less than i then j is less than the square root.
People have mentioned a couple of the building blocks toward doing this efficiently, but nobody's really put the pieces together. The sieve of Eratosthenes is a good start, but with it you'll run out of memory long before you reach the limit you've set. That doesn't mean it's useless though -- when you're doing your loop, what you really care about are prime divisors. As such, you can start by using the sieve to create a base of prime divisors, then use those in the loop to test numbers for primacy.
When you write the loop, however, you really do NOT want to us sqrt(i) in the loop condition as a couple of answers have suggested. You and I know that the sqrt is a "pure" function that always gives the same answer if given the same input parameter. Unfortunately, the compiler does NOT know that, so if use something like '<=Math.sqrt(x)' in the loop condition, it'll re-compute the sqrt of the number every iteration of the loop.
You can avoid that a couple of different ways. You can either pre-compute the sqrt before the loop, and use the pre-computed value in the loop condition, or you can work in the other direction, and change i<Math.sqrt(x) to i*i<x. Personally, I'd pre-compute the square root though -- I think it's clearer and probably a bit faster--but that depends on the number of iterations of the loop (the i*i means it's still doing a multiplication in the loop). With only a few iterations, i*i will typically be faster. With enough iterations, the loss from i*i every iteration outweighs the time for executing sqrt once outside the loop.
That's probably adequate for the size of numbers you're dealing with -- a 15 digit limit means the square root is 7 or 8 digits, which fits in a pretty reasonable amount of memory. On the other hand, if you want to deal with numbers in this range a lot, you might want to look at some of the more sophisticated prime-checking algorithms, such as Pollard's or Brent's algorithms. These are more complex (to put it mildly) but a lot faster for large numbers.
There are other algorithms for even bigger numbers (quadratic sieve, general number field sieve) but we won't get into them for the moment -- they're a lot more complex, and really only useful for dealing with really big numbers (the GNFS starts to be useful in the 100+ digit range).
First step: write an extension method to find out if an input is prime
public static bool isPrime(this int number ) {
for (int i = 2; i < number; i++) {
if (number % i == 0) {
return false;
}
}
return true;
}
2 step: write the method that will print all prime numbers that are between 0 and the number input
public static void getAllPrimes(int number)
{
for (int i = 0; i < number; i++)
{
if (i.isPrime()) Console.WriteLine(i);
}
}
It may just be my opinion, but there's another serious error in your program (setting aside the given 'prime number' question, which has been thoroughly answered).
Like the rest of the responders, I'm assuming this is homework, which indicates you want to become a developer (presumably).
You need to learn to compartmentalize your code. It's not something you'll always need to do in a project, but it's good to know how to do it.
Your method prime_num(long num) could stand a better, more descriptive name. And if it is supposed to find all prime numbers less than a given number, it should return them as a list. This makes it easier to seperate your display and your functionality.
If it simply returned an IList containing prime numbers you could then display them in your main function (perhaps calling another outside function to pretty print them) or use them in further calculations down the line.
So my best recommendation to you is to do something like this:
public void main(string args[])
{
//Get the number you want to use as input
long x = number;//'number' can be hard coded or retrieved from ReadLine() or from the given arguments
IList<long> primes = FindSmallerPrimes(number);
DisplayPrimes(primes);
}
public IList<long> FindSmallerPrimes(long largestNumber)
{
List<long> returnList = new List<long>();
//Find the primes, using a method as described by another answer, add them to returnList
return returnList;
}
public void DisplayPrimes(IList<long> primes)
{
foreach(long l in primes)
{
Console.WriteLine ( "Prime:" + l.ToString() );
}
}
Even if you end up working somewhere where speration like this isn't needed, it's good to know how to do it.
EDIT_ADD: If Will Ness is correct that the question's purpose is just to output a continuous stream of primes for as long as the program is run (pressing Pause/Break to pause and any key to start again) with no serious hope of every getting to that upper limit, then the code should be written with no upper limit argument and a range check of "true" for the first 'i' for loop. On the other hand, if the question wanted to actually print the primes up to a limit, then the following code will do the job much more efficiently using Trial Division only for odd numbers, with the advantage that it doesn't use memory at all (it could also be converted to a continuous loop as per the above):
static void primesttt(ulong top_number) {
Console.WriteLine("Prime: 2");
for (var i = 3UL; i <= top_number; i += 2) {
var isPrime = true;
for (uint j = 3u, lim = (uint)Math.Sqrt((double)i); j <= lim; j += 2) {
if (i % j == 0) {
isPrime = false;
break;
}
}
if (isPrime) Console.WriteLine("Prime: {0} ", i);
}
}
First, the question code produces no output because of that its loop variables are integers and the limit tested is a huge long integer, meaning that it is impossible for the loop to reach the limit producing an inner loop EDITED: whereby the variable 'j' loops back around to negative numbers; when the 'j' variable comes back around to -1, the tested number fails the prime test because all numbers are evenly divisible by -1 END_EDIT. Even if this were corrected, the question code produces very slow output because it gets bound up doing 64-bit divisions of very large quantities of composite numbers (all the even numbers plus the odd composites) by the whole range of numbers up to that top number of ten raised to the sixteenth power for each prime that it can possibly produce. The above code works because it limits the computation to only the odd numbers and only does modulo divisions up to the square root of the current number being tested.
This takes an hour or so to display the primes up to a billion, so one can imagine the amount of time it would take to show all the primes to ten thousand trillion (10 raised to the sixteenth power), especially as the calculation gets slower with increasing range. END_EDIT_ADD
Although the one liner (kind of) answer by #SLaks using Linq works, it isn't really the Sieve of Eratosthenes as it is just an unoptimised version of Trial Division, unoptimised in that it does not eliminate odd primes, doesn't start at the square of the found base prime, and doesn't stop culling for base primes larger than the square root of the top number to sieve. It is also quite slow due to the multiple nested enumeration operations.
It is actually an abuse of the Linq Aggregate method and doesn't effectively use the first of the two Linq Range's generated. It can become an optimized Trial Division with less enumeration overhead as follows:
static IEnumerable<int> primes(uint top_number) {
var cullbf = Enumerable.Range(2, (int)top_number).ToList();
for (int i = 0; i < cullbf.Count; i++) {
var bp = cullbf[i]; var sqr = bp * bp; if (sqr > top_number) break;
cullbf.RemoveAll(c => c >= sqr && c % bp == 0);
} return cullbf; }
which runs many times faster than the SLaks answer. However, it is still slow and memory intensive due to the List generation and the multiple enumerations as well as the multiple divide (implied by the modulo) operations.
The following true Sieve of Eratosthenes implementation runs about 30 times faster and takes much less memory as it only uses a one bit representation per number sieved and limits its enumeration to the final iterator sequence output, as well having the optimisations of only treating odd composites, and only culling from the squares of the base primes for base primes up to the square root of the maximum number, as follows:
static IEnumerable<uint> primes(uint top_number) {
if (top_number < 2u) yield break;
yield return 2u; if (top_number < 3u) yield break;
var BFLMT = (top_number - 3u) / 2u;
var SQRTLMT = ((uint)(Math.Sqrt((double)top_number)) - 3u) / 2u;
var buf = new BitArray((int)BFLMT + 1,true);
for (var i = 0u; i <= BFLMT; ++i) if (buf[(int)i]) {
var p = 3u + i + i; if (i <= SQRTLMT) {
for (var j = (p * p - 3u) / 2u; j <= BFLMT; j += p)
buf[(int)j] = false; } yield return p; } }
The above code calculates all the primes to ten million range in about 77 milliseconds on an Intel i7-2700K (3.5 GHz).
Either of the two static methods can be called and tested with the using statements and with the static Main method as follows:
using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;
static void Main(string[] args) {
Console.WriteLine("This program generates prime sequences.\r\n");
var n = 10000000u;
var elpsd = -DateTime.Now.Ticks;
var count = 0; var lastp = 0u;
foreach (var p in primes(n)) { if (p > n) break; ++count; lastp = (uint)p; }
elpsd += DateTime.Now.Ticks;
Console.WriteLine(
"{0} primes found <= {1}; the last one is {2} in {3} milliseconds.",
count, n, lastp,elpsd / 10000);
Console.Write("\r\nPress any key to exit:");
Console.ReadKey(true);
Console.WriteLine();
}
which will show the number of primes in the sequence up to the limit, the last prime found, and the time expended in enumerating that far.
EDIT_ADD: However, in order to produce an enumeration of the number of primes less than ten thousand trillion (ten to the sixteenth power) as the question asks, a segmented paged approach using multi-core processing is required but even with C++ and the very highly optimized PrimeSieve, this would require something over 400 hours to just produce the number of primes found, and tens of times that long to enumerate all of them so over a year to do what the question asks. To do it using the un-optimized Trial Division algorithm attempted, it will take super eons and a very very long time even using an optimized Trial Division algorithm as in something like ten to the two millionth power years (that's two million zeros years!!!).
It isn't much wonder that his desktop machine just sat and stalled when he tried it!!!! If he had tried a smaller range such as one million, he still would have found it takes in the range of seconds as implemented.
The solutions I post here won't cut it either as even the last Sieve of Eratosthenes one will require about 640 Terabytes of memory for that range.
That is why only a page segmented approach such as that of PrimeSieve can handle this sort of problem for the range as specified at all, and even that requires a very long time, as in weeks to years unless one has access to a super computer with hundreds of thousands of cores. END_EDIT_ADD
Smells like more homework. My very very old graphing calculator had a is prime program like this. Technnically the inner devision checking loop only needs to run to i^(1/2). Do you need to find "all" prime numbers between 0 and L ? The other major problem is that your loop variables are "int" while your input data is "long", this will be causing an overflow making your loops fail to execute even once. Fix the loop variables.
One line code in C# :-
Console.WriteLine(String.Join(Environment.NewLine,
Enumerable.Range(2, 300)
.Where(n => Enumerable.Range(2, (int)Math.Sqrt(n) - 1)
.All(nn => n % nn != 0)).ToArray()));
The Sieve of Eratosthenes answer above is not quite correct. As written it will find all the primes between 1 and 1000000. To find all the primes between 1 and num use:
private static IEnumerable Primes01(int num)
{
return Enumerable.Range(1, Convert.ToInt32(Math.Floor(Math.Sqrt(num))))
.Aggregate(Enumerable.Range(1, num).ToList(),
(result, index) =>
{
result.RemoveAll(i => i > result[index] && i%result[index] == 0);
return result;
}
);
}
The seed of the Aggregate should be range 1 to num since this list will contain the final list of primes. The Enumerable.Range(1, Convert.ToInt32(Math.Floor(Math.Sqrt(num)))) is the number of times the seed is purged.
ExchangeCore Forums have a good console application listed that looks to write found primes to a file, it looks like you can also use that same file as a starting point so you don't have to restart finding primes from 2 and they provide a download of that file with all found primes up to 100 million so it would be a good start.
The algorithm on the page also takes a couple shortcuts (odd numbers and only checks up to the square root) which makes it extremely efficient and it will allow you to calculate long numbers.
so this is basically just two typos, one, the most unfortunate, for (int j = 2; j <= num; j++) which is the reason for the unproductive testing of 1%2,1%3 ... 1%(10^15-1) which goes on for very long time so the OP didn't get "any output". It should've been j < i; instead. The other, minor one in comparison, is that i should start from 2, not from 0:
for( i=2; i <= num; i++ )
{
for( j=2; j < i; j++ ) // j <= sqrt(i) is really enough
....
Surely it can't be reasonably expected of a console print-out of 28 trillion primes or so to be completed in any reasonable time-frame. So, the original intent of the problem was obviously to print out a steady stream of primes, indefinitely. Hence all the solutions proposing simple use of sieve of Eratosthenes are totally without merit here, because simple sieve of Eratosthenes is bounded - a limit must be set in advance.
What could work here is the optimized trial division which would save the primes as it finds them, and test against the primes, not just all numbers below the candidate.
Second alternative, with much better complexity (i.e. much faster) is to use a segmented sieve of Eratosthenes. Which is incremental and unbounded.
Both these schemes would use double-staged production of primes: one would produce and save the primes, to be used by the other stage in testing (or sieving), much above the limit of the first stage (below its square of course - automatically extending the first stage, as the second stage would go further and further up).
To be quite frank, some of the suggested solutions are really slow, and therefore are bad suggestions. For testing a single number to be prime you need some dividing/modulo operator, but for calculating a range you don't have to.
Basically you just exclude numbers that are multiples of earlier found primes, as the are (by definition) not primes themselves.
I will not give the full implementation, as that would be to easy, this is the approach in pseudo code. (On my machine, the actual implementation calculates all primes in an Sytem.Int32 (2 bilion) within 8 seconds.
public IEnumerable<long> GetPrimes(long max)
{
// we safe the result set in an array of bytes.
var buffer = new byte[long >> 4];
// 1 is not a prime.
buffer[0] = 1;
var iMax = (long)Math.Sqrt(max);
for(long i = 3; i <= iMax; i +=2 )
{
// find the index in the buffer
var index = i >> 4;
// find the bit of the buffer.
var bit = (i >> 1) & 7;
// A not set bit means: prime
if((buffer[index] & (1 << bit)) == 0)
{
var step = i << 2;
while(step < max)
{
// find position in the buffer to write bits that represent number that are not prime.
}
}
// 2 is not in the buffer.
yield return 2;
// loop through buffer and yield return odd primes too.
}
}
The solution requires a good understanding of bitwise operations. But it ways, and ways faster. You also can safe the result of the outcome on disc, if you need them for later use. The result of 17 * 10^9 numbers can be safed with 1 GB, and the calculation of that result set takes about 2 minutes max.
I know this is quiet old question, but after reading here:
Sieve of Eratosthenes Wiki
This is the way i wrote it from understanding the algorithm:
void SieveOfEratosthenes(int n)
{
bool[] primes = new bool[n + 1];
for (int i = 0; i < n; i++)
primes[i] = true;
for (int i = 2; i * i <= n; i++)
if (primes[i])
for (int j = i * 2; j <= n; j += i)
primes[j] = false;
for (int i = 2; i <= n; i++)
if (primes[i]) Console.Write(i + " ");
}
In the first loop we fill the array of booleans with true.
Second for loop will start from 2 since 1 is not a prime number and will check if prime number is still not changed and then assign false to the index of j.
last loop we just printing when it is prime.
Very similar - from an exercise to implement Sieve of Eratosthenes in C#:
public class PrimeFinder
{
readonly List<long> _primes = new List<long>();
public PrimeFinder(long seed)
{
CalcPrimes(seed);
}
public List<long> Primes { get { return _primes; } }
private void CalcPrimes(long maxValue)
{
for (int checkValue = 3; checkValue <= maxValue; checkValue += 2)
{
if (IsPrime(checkValue))
{
_primes.Add(checkValue);
}
}
}
private bool IsPrime(long checkValue)
{
bool isPrime = true;
foreach (long prime in _primes)
{
if ((checkValue % prime) == 0 && prime <= Math.Sqrt(checkValue))
{
isPrime = false;
break;
}
}
return isPrime;
}
}
Prime Helper very fast calculation
public static class PrimeHelper
{
public static IEnumerable<Int32> FindPrimes(Int32 maxNumber)
{
return (new PrimesInt32(maxNumber));
}
public static IEnumerable<Int32> FindPrimes(Int32 minNumber, Int32 maxNumber)
{
return FindPrimes(maxNumber).Where(pn => pn >= minNumber);
}
public static bool IsPrime(this Int64 number)
{
if (number < 2)
return false;
else if (number < 4 )
return true;
var limit = (Int32)System.Math.Sqrt(number) + 1;
var foundPrimes = new PrimesInt32(limit);
return !foundPrimes.IsDivisible(number);
}
public static bool IsPrime(this Int32 number)
{
return IsPrime(Convert.ToInt64(number));
}
public static bool IsPrime(this Int16 number)
{
return IsPrime(Convert.ToInt64(number));
}
public static bool IsPrime(this byte number)
{
return IsPrime(Convert.ToInt64(number));
}
}
public class PrimesInt32 : IEnumerable<Int32>
{
private Int32 limit;
private BitArray numbers;
public PrimesInt32(Int32 limit)
{
if (limit < 2)
throw new Exception("Prime numbers not found.");
startTime = DateTime.Now;
calculateTime = startTime - startTime;
this.limit = limit;
try { findPrimes(); } catch{/*Overflows or Out of Memory*/}
calculateTime = DateTime.Now - startTime;
}
private void findPrimes()
{
/*
The Sieve Algorithm
http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes
*/
numbers = new BitArray(limit, true);
for (Int32 i = 2; i < limit; i++)
if (numbers[i])
for (Int32 j = i * 2; j < limit; j += i)
numbers[j] = false;
}
public IEnumerator<Int32> GetEnumerator()
{
for (Int32 i = 2; i < 3; i++)
if (numbers[i])
yield return i;
if (limit > 2)
for (Int32 i = 3; i < limit; i += 2)
if (numbers[i])
yield return i;
}
IEnumerator IEnumerable.GetEnumerator()
{
return GetEnumerator();
}
// Extended for Int64
public bool IsDivisible(Int64 number)
{
var sqrt = System.Math.Sqrt(number);
foreach (var prime in this)
{
if (prime > sqrt)
break;
if (number % prime == 0)
{
DivisibleBy = prime;
return true;
}
}
return false;
}
private static DateTime startTime;
private static TimeSpan calculateTime;
public static TimeSpan CalculateTime { get { return calculateTime; } }
public Int32 DivisibleBy { get; set; }
}
public static void Main()
{
Console.WriteLine("enter the number");
int i = int.Parse(Console.ReadLine());
for (int j = 2; j <= i; j++)
{
for (int k = 2; k <= i; k++)
{
if (j == k)
{
Console.WriteLine("{0}is prime", j);
break;
}
else if (j % k == 0)
{
break;
}
}
}
Console.ReadLine();
}
static void Main(string[] args)
{ int i,j;
Console.WriteLine("prime no between 1 to 100");
for (i = 2; i <= 100; i++)
{
int count = 0;
for (j = 1; j <= i; j++)
{
if (i % j == 0)
{ count=count+1; }
}
if ( count <= 2)
{ Console.WriteLine(i); }
}
Console.ReadKey();
}
U can use the normal prime number concept must only two factors (one and itself).
So do like this,easy way
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace PrimeNUmber
{
class Program
{
static void FindPrimeNumber(long num)
{
for (long i = 1; i <= num; i++)
{
int totalFactors = 0;
for (int j = 1; j <= i; j++)
{
if (i % j == 0)
{
totalFactors = totalFactors + 1;
}
}
if (totalFactors == 2)
{
Console.WriteLine(i);
}
}
}
static void Main(string[] args)
{
long num;
Console.WriteLine("Enter any value");
num = Convert.ToInt64(Console.ReadLine());
FindPrimeNumber(num);
Console.ReadLine();
}
}
}
This solution displays all prime numbers between 0 and 100.
int counter = 0;
for (int c = 0; c <= 100; c++)
{
counter = 0;
for (int i = 1; i <= c; i++)
{
if (c % i == 0)
{ counter++; }
}
if (counter == 2)
{ Console.Write(c + " "); }
}
This is the fastest way to calculate prime numbers in C#.
void PrimeNumber(long number)
{
bool IsprimeNumber = true;
long value = Convert.ToInt32(Math.Sqrt(number));
if (number % 2 == 0)
{
IsprimeNumber = false;
}
for (long i = 3; i <= value; i=i+2)
{
if (number % i == 0)
{
// MessageBox.Show("It is divisible by" + i);
IsprimeNumber = false;
break;
}
}
if (IsprimeNumber)
{
MessageBox.Show("Yes Prime Number");
}
else
{
MessageBox.Show("No It is not a Prime NUmber");
}
}
class CheckIfPrime
{
static void Main()
{
while (true)
{
Console.Write("Enter a number: ");
decimal a = decimal.Parse(Console.ReadLine());
decimal[] k = new decimal[int.Parse(a.ToString())];
decimal p = 0;
for (int i = 2; i < a; i++)
{
if (a % i != 0)
{
p += i;
k[i] = i;
}
else
p += i;
}
if (p == k.Sum())
{ Console.WriteLine ("{0} is prime!", a);}
else
{Console.WriteLine("{0} is NOT prime", a);}
}
}
}
There are some very optimal ways to implement the algorithm. But if you don't know much about maths and you simply follow the definition of prime as the requirement:
a number that is only divisible by 1 and by itself (and nothing else), here's a simple to understand code for positive numbers.
public bool IsPrime(int candidateNumber)
{
int fromNumber = 2;
int toNumber = candidateNumber - 1;
while(fromNumber <= toNumber)
{
bool isDivisible = candidateNumber % fromNumber == 0;
if (isDivisible)
{
return false;
}
fromNumber++;
}
return true;
}
Since every number is divisible by 1 and by itself, we start checking from 2 onwards until the number immediately before itself. That's the basic reasoning.
You can do also this:
class Program
{
static void Main(string[] args)
{
long numberToTest = 350124;
bool isPrime = NumberIsPrime(numberToTest);
Console.WriteLine(string.Format("Number {0} is prime? {1}", numberToTest, isPrime));
Console.ReadLine();
}
private static bool NumberIsPrime(long n)
{
bool retVal = true;
if (n <= 3)
{
retVal = n > 1;
} else if (n % 2 == 0 || n % 3 == 0)
{
retVal = false;
}
int i = 5;
while (i * i <= n)
{
if (n % i == 0 || n % (i + 2) == 0)
{
retVal = false;
}
i += 6;
}
return retVal;
}
}
An easier approach , what i did is check if a number have exactly two division factors which is the essence of prime numbers .
List<int> factorList = new List<int>();
int[] numArray = new int[] { 1, 0, 6, 9, 7, 5, 3, 6, 0, 8, 1 };
foreach (int item in numArray)
{
for (int x = 1; x <= item; x++)
{
//check for the remainder after dividing for each number less that number
if (item % x == 0)
{
factorList.Add(x);
}
}
if (factorList.Count == 2) // has only 2 division factors ; prime number
{
Console.WriteLine(item + " is a prime number ");
}
else
{Console.WriteLine(item + " is not a prime number ");}
factorList = new List<int>(); // reinitialize list
}
Here is a solution with unit test:
The solution:
public class PrimeNumbersKata
{
public int CountPrimeNumbers(int n)
{
if (n < 0) throw new ArgumentException("Not valide numbre");
if (n == 0 || n == 1) return 0;
int cpt = 0;
for (int i = 2; i <= n; i++)
{
if (IsPrimaire(i)) cpt++;
}
return cpt;
}
private bool IsPrimaire(int number)
{
for (int i = 2; i <= number / 2; i++)
{
if (number % i == 0) return false;
}
return true;
}
}
The tests:
[TestFixture]
class PrimeNumbersKataTest
{
private PrimeNumbersKata primeNumbersKata;
[SetUp]
public void Init()
{
primeNumbersKata = new PrimeNumbersKata();
}
[TestCase(1,0)]
[TestCase(0,0)]
[TestCase(2,1)]
[TestCase(3,2)]
[TestCase(5,3)]
[TestCase(7,4)]
[TestCase(9,4)]
[TestCase(11,5)]
[TestCase(13,6)]
public void CountPrimeNumbers_N_AsArgument_returnCountPrimes(int n, int expected)
{
//arrange
//act
var actual = primeNumbersKata.CountPrimeNumbers(n);
//assert
Assert.AreEqual(expected,actual);
}
[Test]
public void CountPrimairs_N_IsNegative_RaiseAnException()
{
var ex = Assert.Throws<ArgumentException>(()=> { primeNumbersKata.CountPrimeNumbers(-1); });
//Assert.That(ex.Message == "Not valide numbre");
Assert.That(ex.Message, Is.EqualTo("Not valide numbre"));
}
}
in the university it was necessary to count prime numbers up to 10,000 did so, the teacher was a little surprised, but I passed the test. Lang c#
void Main()
{
int number=1;
for(long i=2;i<10000;i++)
{
if(PrimeTest(i))
{
Console.WriteLine(number+++" " +i);
}
}
}
List<long> KnownPrime = new List<long>();
private bool PrimeTest(long i)
{
if (i == 1) return false;
if (i == 2)
{
KnownPrime.Add(i);
return true;
}
foreach(int k in KnownPrime)
{
if(i%k==0)
return false;
}
KnownPrime.Add(i);
return true;
}
for (int i = 2; i < 100; i++)
{
bool isPrimeNumber = true;
for (int j = 2; j <= i && j <= 100; j++)
{
if (i != j && i % j == 0)
{
isPrimeNumber = false; break;
}
}
if (isPrimeNumber)
{
Console.WriteLine(i);
}
}

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