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Shortest way between two points by going through all other points from point collection

Given a point collection defined by x and y coordinates.
In this collection I get the start point, the end point and all the other n-2 points.
I have to find the shortest way between the start point and end point by going through all the other points. The shortest way is defined by its value and if possible the crossing point order.
At a first look this seems to be a graph problem, but i am not so sure about that right now, any way i am trying to find this shortest way by using only geometric relations since currently all the information that i have is only the x and y coordinates of the points, and which point is the start point and which is the end point.
My question is, can this way be found by using only geometric relations?
I am trying to implement this in C#, so if some helpful packages are available please let me know.

The simplest heuristic with reasonable performance is 2-opt. Put the points in an array, with the start point first and the end point last, and repeatedly attempt to improve the solution as follows. Choose a starting index i and an ending index j and reverse the subarray from i to j. If the total cost is less, then keep this change, otherwise undo it. Note that the total cost will be less if and only if d(p[i - 1], p[i]) + d(p[j], p[j + 1]) > d(p[i - 1], p[j]) + d(p[i], p[j + 1]), so you can avoid performing the swap unless it's an improvement.
There are a possible number of improvements to this method. 3-opt and k-opt consider more possible moves, resulting in better solution quality. Data structures for geometric search, kd-trees for example, decrease the time to find improving moves. As far as I know, the state of the art in local search algorithms for TSP is Keld Helsgaun's LKH.
Another family of algorithms is branch and bound. These return optimal solutions. Concorde (as far as I know) is the state of the art here.
Here's a Java implementation of the O(n^2 2^n) DP that Niklas described. There are many possible improvements, e.g., cache the distances between points, switch to floats (maybe), reorganize the iteration so that subsets are enumerating in increasing order of size (to allow only the most recent layer of minTable to be retained, resulting in a significant space saving).
class Point {
private final double x, y;
Point(double x, double y) {
this.x = x;
this.y = y;
}
double distanceTo(Point that) {
return Math.hypot(x - that.x, y - that.y);
}
public String toString() {
return x + " " + y;
}
}
public class TSP {
public static int[] minLengthPath(Point[] points) {
if (points.length < 2) {
throw new IllegalArgumentException();
}
int n = points.length - 2;
if ((1 << n) <= 0) {
throw new IllegalArgumentException();
}
byte[][] argMinTable = new byte[1 << n][n];
double[][] minTable = new double[1 << n][n];
for (int s = 0; s < (1 << n); s++) {
for (int i = 0; i < n; i++) {
int sMinusI = s & ~(1 << i);
if (sMinusI == s) {
continue;
}
int argMin = -1;
double min = points[0].distanceTo(points[1 + i]);
for (int j = 0; j < n; j++) {
if ((sMinusI & (1 << j)) == 0) {
continue;
}
double cost =
minTable[sMinusI][j] +
points[1 + j].distanceTo(points[1 + i]);
if (argMin < 0 || cost < min) {
argMin = j;
min = cost;
}
}
argMinTable[s][i] = (byte)argMin;
minTable[s][i] = min;
}
}
int s = (1 << n) - 1;
int argMin = -1;
double min = points[0].distanceTo(points[1 + n]);
for (int i = 0; i < n; i++) {
double cost =
minTable[s][i] +
points[1 + i].distanceTo(points[1 + n]);
if (argMin < 0 || cost < min) {
argMin = i;
min = cost;
}
}
int[] path = new int[1 + n + 1];
path[1 + n] = 1 + n;
int k = n;
while (argMin >= 0) {
path[k] = 1 + argMin;
k--;
int temp = s;
s &= ~(1 << argMin);
argMin = argMinTable[temp][argMin];
}
path[0] = 0;
return path;
}
public static void main(String[] args) {
Point[] points = new Point[20];
for (int i = 0; i < points.length; i++) {
points[i] = new Point(Math.random(), Math.random());
}
int[] path = minLengthPath(points);
for (int i = 0; i < points.length; i++) {
System.out.println(points[path[i]]);
System.err.println(points[i]);
}
}
}

The Euclidean travelling salesman problem can be reduced to this and it's NP-hard. So unless your point set is small or you have a very particular structure, you should probably look out for an approximation. Note that the Wikipedia article mentions the existence of a PTAS for the problem, which could turn out to be quite effective in practice.
UPDATE: Since your instances seem to have only few nodes, you can use a simple exponential-time dynamic programming approach. Let f(S, p) be the minimum cost to connect all the points in the set S, ending at the points p. We have f({start}, start) = 0 and we are looking for f(P, end), where P is the set of all points. To compute f(S, p), we can check all potential predecessors of p in the tour, so we have
f(S, p) = MIN(q in S \ {p}, f(S \ {p}, q) + distance(p, q))
You can represent S as a bitvector to save space (just use an single-word integer for maximum simplicity). Also use memoization to avoid recomputing subproblem results.
The runtime will be O(2^n * n^2) and the algorithm can be implemented with a rather low constant factor, so I predict it to be able to solve instance with n = 25 within seconds a reasonable amount of time.

This can be solved using an evolutionary algorithm.
Look at this: http://johnnewcombe.net/blog/post/23

You might want to look at TPL (Task Parallel Library) to speed up the application.
EDIT
I found this Link which has a Traveling Salesman algorithm:
http://msdn.microsoft.com/en-us/magazine/gg983491.aspx
The Source Code is said to be at:
http://archive.msdn.microsoft.com/mag201104BeeColony

Generating permutations of a set (most efficiently)

I would like to generate all permutations of a set (a collection), like so:
Collection: 1, 2, 3
Permutations: {1, 2, 3}
{1, 3, 2}
{2, 1, 3}
{2, 3, 1}
{3, 1, 2}
{3, 2, 1}
This isn't a question of "how", in general, but more about how most efficiently.
Also, I wouldn't want to generate ALL permutations and return them, but only generating a single permutation, at a time, and continuing only if necessary (much like Iterators - which I've tried as well, but turned out to be less efficient).
I've tested many algorithms and approaches and came up with this code, which is most efficient of those I tried:
public static bool NextPermutation<T>(T[] elements) where T : IComparable<T>
{
// More efficient to have a variable instead of accessing a property
var count = elements.Length;
// Indicates whether this is the last lexicographic permutation
var done = true;
// Go through the array from last to first
for (var i = count - 1; i > 0; i--)
{
var curr = elements[i];
// Check if the current element is less than the one before it
if (curr.CompareTo(elements[i - 1]) < 0)
{
continue;
}
// An element bigger than the one before it has been found,
// so this isn't the last lexicographic permutation.
done = false;
// Save the previous (bigger) element in a variable for more efficiency.
var prev = elements[i - 1];
// Have a variable to hold the index of the element to swap
// with the previous element (the to-swap element would be
// the smallest element that comes after the previous element
// and is bigger than the previous element), initializing it
// as the current index of the current item (curr).
var currIndex = i;
// Go through the array from the element after the current one to last
for (var j = i + 1; j < count; j++)
{
// Save into variable for more efficiency
var tmp = elements[j];
// Check if tmp suits the "next swap" conditions:
// Smallest, but bigger than the "prev" element
if (tmp.CompareTo(curr) < 0 && tmp.CompareTo(prev) > 0)
{
curr = tmp;
currIndex = j;
}
}
// Swap the "prev" with the new "curr" (the swap-with element)
elements[currIndex] = prev;
elements[i - 1] = curr;
// Reverse the order of the tail, in order to reset it's lexicographic order
for (var j = count - 1; j > i; j--, i++)
{
var tmp = elements[j];
elements[j] = elements[i];
elements[i] = tmp;
}
// Break since we have got the next permutation
// The reason to have all the logic inside the loop is
// to prevent the need of an extra variable indicating "i" when
// the next needed swap is found (moving "i" outside the loop is a
// bad practice, and isn't very readable, so I preferred not doing
// that as well).
break;
}
// Return whether this has been the last lexicographic permutation.
return done;
}
It's usage would be sending an array of elements, and getting back a boolean indicating whether this was the last lexicographical permutation or not, as well as having the array altered to the next permutation.
Usage example:
var arr = new[] {1, 2, 3};
PrintArray(arr);
while (!NextPermutation(arr))
{
PrintArray(arr);
}
The thing is that I'm not happy with the speed of the code.
Iterating over all permutations of an array of size 11 takes about 4 seconds.
Although it could be considered impressive, since the amount of possible permutations of a set of size 11 is 11! which is nearly 40 million.
Logically, with an array of size 12 it will take about 12 times more time, since 12! is 11! * 12, and with an array of size 13 it will take about 13 times more time than the time it took with size 12, and so on.
So you can easily understand how with an array of size 12 and more, it really takes a very long time to go through all permutations.
And I have a strong hunch that I can somehow cut that time by a lot (without switching to a language other than C# - because compiler optimization really does optimize pretty nicely, and I doubt I could optimize as good, manually, in Assembly).
Does anyone know any other way to get that done faster?
Do you have any idea as to how to make the current algorithm faster?
Note that I don't want to use an external library or service in order to do that - I want to have the code itself and I want it to be as efficient as humanly possible.

This might be what you're looking for.
private static bool NextPermutation(int[] numList)
{
/*
Knuths
1. Find the largest index j such that a[j] < a[j + 1]. If no such index exists, the permutation is the last permutation.
2. Find the largest index l such that a[j] < a[l]. Since j + 1 is such an index, l is well defined and satisfies j < l.
3. Swap a[j] with a[l].
4. Reverse the sequence from a[j + 1] up to and including the final element a[n].
*/
var largestIndex = -1;
for (var i = numList.Length - 2; i >= 0; i--)
{
if (numList[i] < numList[i + 1]) {
largestIndex = i;
break;
}
}
if (largestIndex < 0) return false;
var largestIndex2 = -1;
for (var i = numList.Length - 1 ; i >= 0; i--) {
if (numList[largestIndex] < numList[i]) {
largestIndex2 = i;
break;
}
}
var tmp = numList[largestIndex];
numList[largestIndex] = numList[largestIndex2];
numList[largestIndex2] = tmp;
for (int i = largestIndex + 1, j = numList.Length - 1; i < j; i++, j--) {
tmp = numList[i];
numList[i] = numList[j];
numList[j] = tmp;
}
return true;
}

Update 2018-05-28:
A new multithreaded version (lot faster) is available below as another answer.
Also an article about permutation: Permutations: Fast implementations and a new indexing algorithm allowing multithreading
A little bit too late...
According to recent tests (updated 2018-05-22)
Fastest is mine BUT not in lexicographic order
For fastest lexicographic order, Sani Singh Huttunen solution seems to be the way to go.
Performance test results for 10 items (10!) in release on my machine (millisecs):
Ouellet : 29
SimpleVar: 95
Erez Robinson : 156
Sani Singh Huttunen : 37
Pengyang : 45047
Performance test results for 13 items (13!) in release on my machine (seconds):
Ouellet : 48.437
SimpleVar: 159.869
Erez Robinson : 327.781
Sani Singh Huttunen : 64.839
Advantages of my solution:
Heap's algorithm (Single swap per permutation)
No multiplication (like some implementations seen on the web)
Inlined swap
Generic
No unsafe code
In place (very low memory usage)
No modulo (only first bit compare)
My implementation of Heap's algorithm:
using System;
using System.Collections.Generic;
using System.Diagnostics;
using System.Linq;
using System.Runtime.CompilerServices;
namespace WpfPermutations
{
/// <summary>
/// EO: 2016-04-14
/// Generator of all permutations of an array of anything.
/// Base on Heap's Algorithm. See: https://en.wikipedia.org/wiki/Heap%27s_algorithm#cite_note-3
/// </summary>
public static class Permutations
{
/// <summary>
/// Heap's algorithm to find all pmermutations. Non recursive, more efficient.
/// </summary>
/// <param name="items">Items to permute in each possible ways</param>
/// <param name="funcExecuteAndTellIfShouldStop"></param>
/// <returns>Return true if cancelled</returns>
public static bool ForAllPermutation<T>(T[] items, Func<T[], bool> funcExecuteAndTellIfShouldStop)
{
int countOfItem = items.Length;
if (countOfItem <= 1)
{
return funcExecuteAndTellIfShouldStop(items);
}
var indexes = new int[countOfItem];
// Unecessary. Thanks to NetManage for the advise
// for (int i = 0; i < countOfItem; i++)
// {
// indexes[i] = 0;
// }
if (funcExecuteAndTellIfShouldStop(items))
{
return true;
}
for (int i = 1; i < countOfItem;)
{
if (indexes[i] < i)
{ // On the web there is an implementation with a multiplication which should be less efficient.
if ((i & 1) == 1) // if (i % 2 == 1) ... more efficient ??? At least the same.
{
Swap(ref items[i], ref items[indexes[i]]);
}
else
{
Swap(ref items[i], ref items[0]);
}
if (funcExecuteAndTellIfShouldStop(items))
{
return true;
}
indexes[i]++;
i = 1;
}
else
{
indexes[i++] = 0;
}
}
return false;
}
/// <summary>
/// This function is to show a linq way but is far less efficient
/// From: StackOverflow user: Pengyang : http://stackoverflow.com/questions/756055/listing-all-permutations-of-a-string-integer
/// </summary>
/// <typeparam name="T"></typeparam>
/// <param name="list"></param>
/// <param name="length"></param>
/// <returns></returns>
static IEnumerable<IEnumerable<T>> GetPermutations<T>(IEnumerable<T> list, int length)
{
if (length == 1) return list.Select(t => new T[] { t });
return GetPermutations(list, length - 1)
.SelectMany(t => list.Where(e => !t.Contains(e)),
(t1, t2) => t1.Concat(new T[] { t2 }));
}
/// <summary>
/// Swap 2 elements of same type
/// </summary>
/// <typeparam name="T"></typeparam>
/// <param name="a"></param>
/// <param name="b"></param>
[MethodImpl(MethodImplOptions.AggressiveInlining)]
static void Swap<T>(ref T a, ref T b)
{
T temp = a;
a = b;
b = temp;
}
/// <summary>
/// Func to show how to call. It does a little test for an array of 4 items.
/// </summary>
public static void Test()
{
ForAllPermutation("123".ToCharArray(), (vals) =>
{
Console.WriteLine(String.Join("", vals));
return false;
});
int[] values = new int[] { 0, 1, 2, 4 };
Console.WriteLine("Ouellet heap's algorithm implementation");
ForAllPermutation(values, (vals) =>
{
Console.WriteLine(String.Join("", vals));
return false;
});
Console.WriteLine("Linq algorithm");
foreach (var v in GetPermutations(values, values.Length))
{
Console.WriteLine(String.Join("", v));
}
// Performance Heap's against Linq version : huge differences
int count = 0;
values = new int[10];
for (int n = 0; n < values.Length; n++)
{
values[n] = n;
}
Stopwatch stopWatch = new Stopwatch();
ForAllPermutation(values, (vals) =>
{
foreach (var v in vals)
{
count++;
}
return false;
});
stopWatch.Stop();
Console.WriteLine($"Ouellet heap's algorithm implementation {count} items in {stopWatch.ElapsedMilliseconds} millisecs");
count = 0;
stopWatch.Reset();
stopWatch.Start();
foreach (var vals in GetPermutations(values, values.Length))
{
foreach (var v in vals)
{
count++;
}
}
stopWatch.Stop();
Console.WriteLine($"Linq {count} items in {stopWatch.ElapsedMilliseconds} millisecs");
}
}
}
An this is my test code:
Task.Run(() =>
{
int[] values = new int[12];
for (int n = 0; n < values.Length; n++)
{
values[n] = n;
}
// Eric Ouellet Algorithm
int count = 0;
var stopwatch = new Stopwatch();
stopwatch.Reset();
stopwatch.Start();
Permutations.ForAllPermutation(values, (vals) =>
{
foreach (var v in vals)
{
count++;
}
return false;
});
stopwatch.Stop();
Console.WriteLine($"This {count} items in {stopwatch.ElapsedMilliseconds} millisecs");
// Simple Plan Algorithm
count = 0;
stopwatch.Reset();
stopwatch.Start();
PermutationsSimpleVar permutations2 = new PermutationsSimpleVar();
permutations2.Permutate(1, values.Length, (int[] vals) =>
{
foreach (var v in vals)
{
count++;
}
});
stopwatch.Stop();
Console.WriteLine($"Simple Plan {count} items in {stopwatch.ElapsedMilliseconds} millisecs");
// ErezRobinson Algorithm
count = 0;
stopwatch.Reset();
stopwatch.Start();
foreach(var vals in PermutationsErezRobinson.QuickPerm(values))
{
foreach (var v in vals)
{
count++;
}
};
stopwatch.Stop();
Console.WriteLine($"Erez Robinson {count} items in {stopwatch.ElapsedMilliseconds} millisecs");
});
Usage examples:
ForAllPermutation("123".ToCharArray(), (vals) =>
{
Console.WriteLine(String.Join("", vals));
return false;
});
int[] values = new int[] { 0, 1, 2, 4 };
ForAllPermutation(values, (vals) =>
{
Console.WriteLine(String.Join("", vals));
return false;
});

Well, if you can handle it in C and then translate to your language of choice, you can't really go much faster than this, because the time will be dominated by print:
void perm(char* s, int n, int i){
if (i >= n-1) print(s);
else {
perm(s, n, i+1);
for (int j = i+1; j<n; j++){
swap(s[i], s[j]);
perm(s, n, i+1);
swap(s[i], s[j]);
}
}
}
perm("ABC", 3, 0);

Update 2018-05-28, a new version, the fastest ... (multi-threaded)
Time taken for fastest algorithms
Need: Sani Singh Huttunen (fastest lexico) solution and my new OuelletLexico3 which support indexing
Indexing has 2 main advantages:
allows to get anyone permutation directly
allows multi-threading (derived from the first advantage)
Article: Permutations: Fast implementations and a new indexing algorithm allowing multithreading
On my machine (6 hyperthread cores : 12 threads) Xeon E5-1660 0 # 3.30Ghz, tests algorithms running with empty stuff to do for 13! items (time in millisecs):
53071: Ouellet (implementation of Heap)
65366: Sani Singh Huttunen (Fastest lexico)
11377: Mix OuelletLexico3 - Sani Singh Huttunen
A side note: using shares properties/variables between threads for permutation action will strongly impact performance if their usage is modification (read / write). Doing so will generate "false sharing" between threads. You will not get expected performance. I got this behavior while testing. My experience showed problems when I try to increase the global variable for the total count of permutation.
Usage:
PermutationMixOuelletSaniSinghHuttunen.ExecuteForEachPermutationMT(
new int[] {1, 2, 3, 4},
p =>
{
Console.WriteLine($"Values: {p[0]}, {p[1]}, p[2]}, {p[3]}");
});
Code:
using System;
using System.Runtime.CompilerServices;
namespace WpfPermutations
{
public class Factorial
{
// ************************************************************************
protected static long[] FactorialTable = new long[21];
// ************************************************************************
static Factorial()
{
FactorialTable[0] = 1; // To prevent divide by 0
long f = 1;
for (int i = 1; i <= 20; i++)
{
f = f * i;
FactorialTable[i] = f;
}
}
// ************************************************************************
[MethodImpl(MethodImplOptions.AggressiveInlining)]
public static long GetFactorial(int val) // a long can only support up to 20!
{
if (val > 20)
{
throw new OverflowException($"{nameof(Factorial)} only support a factorial value <= 20");
}
return FactorialTable[val];
}
// ************************************************************************
}
}
namespace WpfPermutations
{
public class PermutationSaniSinghHuttunen
{
public static bool NextPermutation(int[] numList)
{
/*
Knuths
1. Find the largest index j such that a[j] < a[j + 1]. If no such index exists, the permutation is the last permutation.
2. Find the largest index l such that a[j] < a[l]. Since j + 1 is such an index, l is well defined and satisfies j < l.
3. Swap a[j] with a[l].
4. Reverse the sequence from a[j + 1] up to and including the final element a[n].
*/
var largestIndex = -1;
for (var i = numList.Length - 2; i >= 0; i--)
{
if (numList[i] < numList[i + 1])
{
largestIndex = i;
break;
}
}
if (largestIndex < 0) return false;
var largestIndex2 = -1;
for (var i = numList.Length - 1; i >= 0; i--)
{
if (numList[largestIndex] < numList[i])
{
largestIndex2 = i;
break;
}
}
var tmp = numList[largestIndex];
numList[largestIndex] = numList[largestIndex2];
numList[largestIndex2] = tmp;
for (int i = largestIndex + 1, j = numList.Length - 1; i < j; i++, j--)
{
tmp = numList[i];
numList[i] = numList[j];
numList[j] = tmp;
}
return true;
}
}
}
using System;
namespace WpfPermutations
{
public class PermutationOuelletLexico3<T> // Enable indexing
{
// ************************************************************************
private T[] _sortedValues;
private bool[] _valueUsed;
public readonly long MaxIndex; // long to support 20! or less
// ************************************************************************
public PermutationOuelletLexico3(T[] sortedValues)
{
_sortedValues = sortedValues;
Result = new T[_sortedValues.Length];
_valueUsed = new bool[_sortedValues.Length];
MaxIndex = Factorial.GetFactorial(_sortedValues.Length);
}
// ************************************************************************
public T[] Result { get; private set; }
// ************************************************************************
/// <summary>
/// Sort Index is 0 based and should be less than MaxIndex. Otherwise you get an exception.
/// </summary>
/// <param name="sortIndex"></param>
/// <param name="result">Value is not used as inpu, only as output. Re-use buffer in order to save memory</param>
/// <returns></returns>
public void GetSortedValuesFor(long sortIndex)
{
int size = _sortedValues.Length;
if (sortIndex < 0)
{
throw new ArgumentException("sortIndex should greater or equal to 0.");
}
if (sortIndex >= MaxIndex)
{
throw new ArgumentException("sortIndex should less than factorial(the lenght of items)");
}
for (int n = 0; n < _valueUsed.Length; n++)
{
_valueUsed[n] = false;
}
long factorielLower = MaxIndex;
for (int index = 0; index < size; index++)
{
long factorielBigger = factorielLower;
factorielLower = Factorial.GetFactorial(size - index - 1); // factorielBigger / inverseIndex;
int resultItemIndex = (int)(sortIndex % factorielBigger / factorielLower);
int correctedResultItemIndex = 0;
for(;;)
{
if (! _valueUsed[correctedResultItemIndex])
{
resultItemIndex--;
if (resultItemIndex < 0)
{
break;
}
}
correctedResultItemIndex++;
}
Result[index] = _sortedValues[correctedResultItemIndex];
_valueUsed[correctedResultItemIndex] = true;
}
}
// ************************************************************************
}
}
using System;
using System.Collections.Generic;
using System.Threading.Tasks;
namespace WpfPermutations
{
public class PermutationMixOuelletSaniSinghHuttunen
{
// ************************************************************************
private long _indexFirst;
private long _indexLastExclusive;
private int[] _sortedValues;
// ************************************************************************
public PermutationMixOuelletSaniSinghHuttunen(int[] sortedValues, long indexFirst = -1, long indexLastExclusive = -1)
{
if (indexFirst == -1)
{
indexFirst = 0;
}
if (indexLastExclusive == -1)
{
indexLastExclusive = Factorial.GetFactorial(sortedValues.Length);
}
if (indexFirst >= indexLastExclusive)
{
throw new ArgumentException($"{nameof(indexFirst)} should be less than {nameof(indexLastExclusive)}");
}
_indexFirst = indexFirst;
_indexLastExclusive = indexLastExclusive;
_sortedValues = sortedValues;
}
// ************************************************************************
public void ExecuteForEachPermutation(Action<int[]> action)
{
// Console.WriteLine($"Thread {System.Threading.Thread.CurrentThread.ManagedThreadId} started: {_indexFirst} {_indexLastExclusive}");
long index = _indexFirst;
PermutationOuelletLexico3<int> permutationOuellet = new PermutationOuelletLexico3<int>(_sortedValues);
permutationOuellet.GetSortedValuesFor(index);
action(permutationOuellet.Result);
index++;
int[] values = permutationOuellet.Result;
while (index < _indexLastExclusive)
{
PermutationSaniSinghHuttunen.NextPermutation(values);
action(values);
index++;
}
// Console.WriteLine($"Thread {System.Threading.Thread.CurrentThread.ManagedThreadId} ended: {DateTime.Now.ToString("yyyyMMdd_HHmmss_ffffff")}");
}
// ************************************************************************
public static void ExecuteForEachPermutationMT(int[] sortedValues, Action<int[]> action)
{
int coreCount = Environment.ProcessorCount; // Hyper treading are taken into account (ex: on a 4 cores hyperthreaded = 8)
long itemsFactorial = Factorial.GetFactorial(sortedValues.Length);
long partCount = (long)Math.Ceiling((double)itemsFactorial / (double)coreCount);
long startIndex = 0;
var tasks = new List<Task>();
for (int coreIndex = 0; coreIndex < coreCount; coreIndex++)
{
long stopIndex = Math.Min(startIndex + partCount, itemsFactorial);
PermutationMixOuelletSaniSinghHuttunen mix = new PermutationMixOuelletSaniSinghHuttunen(sortedValues, startIndex, stopIndex);
Task task = Task.Run(() => mix.ExecuteForEachPermutation(action));
tasks.Add(task);
if (stopIndex == itemsFactorial)
{
break;
}
startIndex = startIndex + partCount;
}
Task.WaitAll(tasks.ToArray());
}
// ************************************************************************
}
}

The fastest permutation algorithm that i know of is the QuickPerm algorithm.
Here is the implementation, it uses yield return so you can iterate one at a time like required.
Code:
public static IEnumerable<IEnumerable<T>> QuickPerm<T>(this IEnumerable<T> set)
{
int N = set.Count();
int[] a = new int[N];
int[] p = new int[N];
var yieldRet = new T[N];
List<T> list = new List<T>(set);
int i, j, tmp; // Upper Index i; Lower Index j
for (i = 0; i < N; i++)
{
// initialize arrays; a[N] can be any type
a[i] = i + 1; // a[i] value is not revealed and can be arbitrary
p[i] = 0; // p[i] == i controls iteration and index boundaries for i
}
yield return list;
//display(a, 0, 0); // remove comment to display array a[]
i = 1; // setup first swap points to be 1 and 0 respectively (i & j)
while (i < N)
{
if (p[i] < i)
{
j = i%2*p[i]; // IF i is odd then j = p[i] otherwise j = 0
tmp = a[j]; // swap(a[j], a[i])
a[j] = a[i];
a[i] = tmp;
//MAIN!
for (int x = 0; x < N; x++)
{
yieldRet[x] = list[a[x]-1];
}
yield return yieldRet;
//display(a, j, i); // remove comment to display target array a[]
// MAIN!
p[i]++; // increase index "weight" for i by one
i = 1; // reset index i to 1 (assumed)
}
else
{
// otherwise p[i] == i
p[i] = 0; // reset p[i] to zero
i++; // set new index value for i (increase by one)
} // if (p[i] < i)
} // while(i < N)
}

Here is the fastest implementation I ended up with:
public class Permutations
{
private readonly Mutex _mutex = new Mutex();
private Action<int[]> _action;
private Action<IntPtr> _actionUnsafe;
private unsafe int* _arr;
private IntPtr _arrIntPtr;
private unsafe int* _last;
private unsafe int* _lastPrev;
private unsafe int* _lastPrevPrev;
public int Size { get; private set; }
public bool IsRunning()
{
return this._mutex.SafeWaitHandle.IsClosed;
}
public bool Permutate(int start, int count, Action<int[]> action, bool async = false)
{
return this.Permutate(start, count, action, null, async);
}
public bool Permutate(int start, int count, Action<IntPtr> actionUnsafe, bool async = false)
{
return this.Permutate(start, count, null, actionUnsafe, async);
}
private unsafe bool Permutate(int start, int count, Action<int[]> action, Action<IntPtr> actionUnsafe, bool async = false)
{
if (!this._mutex.WaitOne(0))
{
return false;
}
var x = (Action)(() =>
{
this._actionUnsafe = actionUnsafe;
this._action = action;
this.Size = count;
this._arr = (int*)Marshal.AllocHGlobal(count * sizeof(int));
this._arrIntPtr = new IntPtr(this._arr);
for (var i = 0; i < count - 3; i++)
{
this._arr[i] = start + i;
}
this._last = this._arr + count - 1;
this._lastPrev = this._last - 1;
this._lastPrevPrev = this._lastPrev - 1;
*this._last = count - 1;
*this._lastPrev = count - 2;
*this._lastPrevPrev = count - 3;
this.Permutate(count, this._arr);
});
if (!async)
{
x();
}
else
{
new Thread(() => x()).Start();
}
return true;
}
private unsafe void Permutate(int size, int* start)
{
if (size == 3)
{
this.DoAction();
Swap(this._last, this._lastPrev);
this.DoAction();
Swap(this._last, this._lastPrevPrev);
this.DoAction();
Swap(this._last, this._lastPrev);
this.DoAction();
Swap(this._last, this._lastPrevPrev);
this.DoAction();
Swap(this._last, this._lastPrev);
this.DoAction();
return;
}
var sizeDec = size - 1;
var startNext = start + 1;
var usedStarters = 0;
for (var i = 0; i < sizeDec; i++)
{
this.Permutate(sizeDec, startNext);
usedStarters |= 1 << *start;
for (var j = startNext; j <= this._last; j++)
{
var mask = 1 << *j;
if ((usedStarters & mask) != mask)
{
Swap(start, j);
break;
}
}
}
this.Permutate(sizeDec, startNext);
if (size == this.Size)
{
this._mutex.ReleaseMutex();
}
}
private unsafe void DoAction()
{
if (this._action == null)
{
if (this._actionUnsafe != null)
{
this._actionUnsafe(this._arrIntPtr);
}
return;
}
var result = new int[this.Size];
fixed (int* pt = result)
{
var limit = pt + this.Size;
var resultPtr = pt;
var arrayPtr = this._arr;
while (resultPtr < limit)
{
*resultPtr = *arrayPtr;
resultPtr++;
arrayPtr++;
}
}
this._action(result);
}
private static unsafe void Swap(int* a, int* b)
{
var tmp = *a;
*a = *b;
*b = tmp;
}
}
Usage and testing performance:
var perms = new Permutations();
var sw1 = Stopwatch.StartNew();
perms.Permutate(0,
11,
(Action<int[]>)null); // Comment this line and...
//PrintArr); // Uncomment this line, to print permutations
sw1.Stop();
Console.WriteLine(sw1.Elapsed);
Printing method:
private static void PrintArr(int[] arr)
{
Console.WriteLine(string.Join(",", arr));
}
Going deeper:
I did not even think about this for a very long time, so I can only explain my code so much, but here's the general idea:
Permutations aren't lexicographic - this allows me to practically perform less operations between permutations.
The implementation is recursive, and when the "view" size is 3, it skips the complex logic and just performs 6 swaps to get the 6 permutations (or sub-permutations, if you will).
Because the permutations aren't in a lexicographic order, how can I decide which element to bring to the start of the current "view" (sub permutation)? I keep record of elements that were already used as "starters" in the current sub-permutation recursive call and simply search linearly for one that wasn't used in the tail of my array.
The implementation is for integers only, so to permute over a generic collection of elements you simply use the Permutations class to permute indices instead of your actual collection.
The Mutex is there just to ensure things don't get screwed when the execution is asynchronous (notice that you can pass an UnsafeAction parameter that will in turn get a pointer to the permuted array. You must not change the order of elements in that array (pointer)! If you want to, you should copy the array to a tmp array or just use the safe action parameter which takes care of that for you - the passed array is already a copy).
Note:
I have no idea how good this implementation really is - I haven't touched it in so long.
Test and compare to other implementations on your own, and let me know if you have any feedback!
Enjoy.

Here is a generic permutation finder that will iterate through every permutation of a collection and call an evalution function. If the evalution function returns true (it found the answer it was looking for), the permutation finder stops processing.
public class PermutationFinder<T>
{
private T[] items;
private Predicate<T[]> SuccessFunc;
private bool success = false;
private int itemsCount;
public void Evaluate(T[] items, Predicate<T[]> SuccessFunc)
{
this.items = items;
this.SuccessFunc = SuccessFunc;
this.itemsCount = items.Count();
Recurse(0);
}
private void Recurse(int index)
{
T tmp;
if (index == itemsCount)
success = SuccessFunc(items);
else
{
for (int i = index; i < itemsCount; i++)
{
tmp = items[index];
items[index] = items[i];
items[i] = tmp;
Recurse(index + 1);
if (success)
break;
tmp = items[index];
items[index] = items[i];
items[i] = tmp;
}
}
}
}
Here is a simple implementation:
class Program
{
static void Main(string[] args)
{
new Program().Start();
}
void Start()
{
string[] items = new string[5];
items[0] = "A";
items[1] = "B";
items[2] = "C";
items[3] = "D";
items[4] = "E";
new PermutationFinder<string>().Evaluate(items, Evaluate);
Console.ReadLine();
}
public bool Evaluate(string[] items)
{
Console.WriteLine(string.Format("{0},{1},{2},{3},{4}", items[0], items[1], items[2], items[3], items[4]));
bool someCondition = false;
if (someCondition)
return true; // Tell the permutation finder to stop.
return false;
}
}

Here is a recursive implementation with complexity O(n * n!)1 based on swapping of the elements of an array. The array is initialised with values from 1, 2, ..., n.
using System;
namespace Exercise
{
class Permutations
{
static void Main(string[] args)
{
int setSize = 3;
FindPermutations(setSize);
}
//-----------------------------------------------------------------------------
/* Method: FindPermutations(n) */
private static void FindPermutations(int n)
{
int[] arr = new int[n];
for (int i = 0; i < n; i++)
{
arr[i] = i + 1;
}
int iEnd = arr.Length - 1;
Permute(arr, iEnd);
}
//-----------------------------------------------------------------------------
/* Method: Permute(arr) */
private static void Permute(int[] arr, int iEnd)
{
if (iEnd == 0)
{
PrintArray(arr);
return;
}
Permute(arr, iEnd - 1);
for (int i = 0; i < iEnd; i++)
{
swap(ref arr[i], ref arr[iEnd]);
Permute(arr, iEnd - 1);
swap(ref arr[i], ref arr[iEnd]);
}
}
}
}
On each recursive step we swap the last element with the current element pointed to by the local variable in the for loop and then we indicate the uniqueness of the swapping by: incrementing the local variable of the for loop and decrementing the termination condition of the for loop, which is initially set to the number of the elements in the array, when the latter becomes zero we terminate the recursion.
Here are the helper functions:
//-----------------------------------------------------------------------------
/*
Method: PrintArray()
*/
private static void PrintArray(int[] arr, string label = "")
{
Console.WriteLine(label);
Console.Write("{");
for (int i = 0; i < arr.Length; i++)
{
Console.Write(arr[i]);
if (i < arr.Length - 1)
{
Console.Write(", ");
}
}
Console.WriteLine("}");
}
//-----------------------------------------------------------------------------
/*
Method: swap(ref int a, ref int b)
*/
private static void swap(ref int a, ref int b)
{
int temp = a;
a = b;
b = temp;
}
1. There are n! permutations of n elements to be printed.

I would be surprised if there are really order of magnitude improvements to be found. If there are, then C# needs fundamental improvement. Furthermore doing anything interesting with your permutation will generally take more work than generating it. So the cost of generating is going to be insignificant in the overall scheme of things.
That said, I would suggest trying the following things. You have already tried iterators. But have you tried having a function that takes a closure as input, then then calls that closure for each permutation found? Depending on internal mechanics of C#, this may be faster.
Similarly, have you tried having a function that returns a closure that will iterate over a specific permutation?
With either approach, there are a number of micro-optimizations you can experiment with. For instance you can sort your input array, and after that you always know what order it is in. For example you can have an array of bools indicating whether that element is less than the next one, and rather than do comparisons, you can just look at that array.

There's an accessible introduction to the algorithms and survey of implementations in Steven Skiena's Algorithm Design Manual (chapter 14.4 in the second edition)
Skiena references D. Knuth. The Art of Computer Programming, Volume 4 Fascicle 2: Generating All Tuples and Permutations. Addison Wesley, 2005.

I created an algorithm slightly faster than Knuth's one:
11 elements:
mine: 0.39 seconds
Knuth's: 0.624 seconds
13 elements:
mine: 56.615 seconds
Knuth's: 98.681 seconds
Here's my code in Java:
public static void main(String[] args)
{
int n=11;
int a,b,c,i,tmp;
int end=(int)Math.floor(n/2);
int[][] pos = new int[end+1][2];
int[] perm = new int[n];
for(i=0;i<n;i++) perm[i]=i;
while(true)
{
//this is where you can use the permutations (perm)
i=0;
c=n;
while(pos[i][1]==c-2 && pos[i][0]==c-1)
{
pos[i][0]=0;
pos[i][1]=0;
i++;
c-=2;
}
if(i==end) System.exit(0);
a=(pos[i][0]+1)%c+i;
b=pos[i][0]+i;
tmp=perm[b];
perm[b]=perm[a];
perm[a]=tmp;
if(pos[i][0]==c-1)
{
pos[i][0]=0;
pos[i][1]++;
}
else
{
pos[i][0]++;
}
}
}
The problem is my algorithm only works for odd numbers of elements. I wrote this code quickly so I'm pretty sure there's a better way to implement my idea to get better performance, but I don't really have the time to work on it right now to optimize it and solve the issue when the number of elements is even.
It's one swap for every permutation and it uses a really simple way to know which elements to swap.
I wrote an explanation of the method behind the code on my blog: http://antoinecomeau.blogspot.ca/2015/01/fast-generation-of-all-permutations.html

As the author of this question was asking about an algorithm:
[...] generating a single permutation, at a time, and continuing only if necessary
I would suggest considering Steinhaus–Johnson–Trotter algorithm.
Steinhaus–Johnson–Trotter algorithm on Wikipedia
Beautifully explained here

It's 1 am and I was watching TV and thought of this same question, but with string values.
Given a word find all permutations. You can easily modify this to handle an array, sets, etc.
Took me a bit to work it out, but the solution I came up was this:
string word = "abcd";
List<string> combinations = new List<string>();
for(int i=0; i<word.Length; i++)
{
for (int j = 0; j < word.Length; j++)
{
if (i < j)
combinations.Add(word[i] + word.Substring(j) + word.Substring(0, i) + word.Substring(i + 1, j - (i + 1)));
else if (i > j)
{
if(i== word.Length -1)
combinations.Add(word[i] + word.Substring(0, i));
else
combinations.Add(word[i] + word.Substring(0, i) + word.Substring(i + 1));
}
}
}
Here's the same code as above, but with some comments
string word = "abcd";
List<string> combinations = new List<string>();
//i is the first letter of the new word combination
for(int i=0; i<word.Length; i++)
{
for (int j = 0; j < word.Length; j++)
{
//add the first letter of the word, j is past i so we can get all the letters from j to the end
//then add all the letters from the front to i, then skip over i (since we already added that as the beginning of the word)
//and get the remaining letters from i+1 to right before j.
if (i < j)
combinations.Add(word[i] + word.Substring(j) + word.Substring(0, i) + word.Substring(i + 1, j - (i + 1)));
else if (i > j)
{
//if we're at the very last word no need to get the letters after i
if(i== word.Length -1)
combinations.Add(word[i] + word.Substring(0, i));
//add i as the first letter of the word, then get all the letters up to i, skip i, and then add all the lettes after i
else
combinations.Add(word[i] + word.Substring(0, i) + word.Substring(i + 1));
}
}
}

//+------------------------------------------------------------------+
//| |
//+------------------------------------------------------------------+
/**
* http://marknelson.us/2002/03/01/next-permutation/
* Rearranges the elements into the lexicographically next greater permutation and returns true.
* When there are no more greater permutations left, the function eventually returns false.
*/
// next lexicographical permutation
template <typename T>
bool next_permutation(T &arr[], int firstIndex, int lastIndex)
{
int i = lastIndex;
while (i > firstIndex)
{
int ii = i--;
T curr = arr[i];
if (curr < arr[ii])
{
int j = lastIndex;
while (arr[j] <= curr) j--;
Swap(arr[i], arr[j]);
while (ii < lastIndex)
Swap(arr[ii++], arr[lastIndex--]);
return true;
}
}
return false;
}
//+------------------------------------------------------------------+
//| |
//+------------------------------------------------------------------+
/**
* Swaps two variables or two array elements.
* using references/pointers to speed up swapping.
*/
template<typename T>
void Swap(T &var1, T &var2)
{
T temp;
temp = var1;
var1 = var2;
var2 = temp;
}
//+------------------------------------------------------------------+
//| |
//+------------------------------------------------------------------+
// driver program to test above function
#define N 3
void OnStart()
{
int i, x[N];
for (i = 0; i < N; i++) x[i] = i + 1;
printf("The %i! possible permutations with %i elements:", N, N);
do
{
printf("%s", ArrayToString(x));
} while (next_permutation(x, 0, N - 1));
}
// Output:
// The 3! possible permutations with 3 elements:
// "1,2,3"
// "1,3,2"
// "2,1,3"
// "2,3,1"
// "3,1,2"
// "3,2,1"

// Permutations are the different ordered arrangements of an n-element
// array. An n-element array has exactly n! full-length permutations.
// This iterator object allows to iterate all full length permutations
// one by one of an array of n distinct elements.
// The iterator changes the given array in-place.
// Permutations('ABCD') => ABCD DBAC ACDB DCBA
// BACD BDAC CADB CDBA
// CABD ADBC DACB BDCA
// ACBD DABC ADCB DBCA
// BCAD BADC CDAB CBDA
// CBAD ABDC DCAB BCDA
// count of permutations = n!
// Heap's algorithm (Single swap per permutation)
// http://www.quickperm.org/quickperm.php
// https://stackoverflow.com/a/36634935/4208440
// https://en.wikipedia.org/wiki/Heap%27s_algorithm
// My implementation of Heap's algorithm:
template<typename T>
class PermutationsIterator
{
int b, e, n;
int c[32]; /* control array: mixed radix number in rising factorial base.
the i-th digit has base i, which means that the digit must be
strictly less than i. The first digit is always 0, the second
can be 0 or 1, the third 0, 1 or 2, and so on.
ArrayResize isn't strictly necessary, int c[32] would suffice
for most practical purposes. Also, it is much faster */
public:
PermutationsIterator(T &arr[], int firstIndex, int lastIndex)
{
this.b = firstIndex; // v.begin()
this.e = lastIndex; // v.end()
this.n = e - b + 1;
ArrayInitialize(c, 0);
}
// Rearranges the input array into the next permutation and returns true.
// When there are no more permutations left, the function returns false.
bool next(T &arr[])
{
// find index to update
int i = 1;
// reset all the previous indices that reached the maximum possible values
while (c[i] == i)
{
c[i] = 0;
++i;
}
// no more permutations left
if (i == n)
return false;
// generate next permutation
int j = (i & 1) == 1 ? c[i] : 0; // IF i is odd then j = c[i] otherwise j = 0.
swap(arr[b + j], arr[b + i]); // generate a new permutation from previous permutation using a single swap
// Increment that index
++c[i];
return true;
}
};

I found this algo on rosetta code and it is really the fastest one I tried. http://rosettacode.org/wiki/Permutations#C
/* Boothroyd method; exactly N! swaps, about as fast as it gets */
void boothroyd(int *x, int n, int nn, int callback(int *, int))
{
int c = 0, i, t;
while (1) {
if (n > 2) boothroyd(x, n - 1, nn, callback);
if (c >= n - 1) return;
i = (n & 1) ? 0 : c;
c++;
t = x[n - 1], x[n - 1] = x[i], x[i] = t;
if (callback) callback(x, nn);
}
}
/* entry for Boothroyd method */
void perm2(int *x, int n, int callback(int*, int))
{
if (callback) callback(x, n);
boothroyd(x, n, n, callback);
}

If you just want to calculate the number of possible permutations you can avoid all that hard work above and use something like this (contrived in c#):
public static class ContrivedUtils
{
public static Int64 Permutations(char[] array)
{
if (null == array || array.Length == 0) return 0;
Int64 permutations = array.Length;
for (var pos = permutations; pos > 1; pos--)
permutations *= pos - 1;
return permutations;
}
}
You call it like this:
var permutations = ContrivedUtils.Permutations("1234".ToCharArray());
// output is: 24
var permutations = ContrivedUtils.Permutations("123456789".ToCharArray());
// output is: 362880

Simple C# recursive solution by swapping, for the initial call the index must be 0
static public void Permute<T>(List<T> input, List<List<T>> permutations, int index)
{
if (index == input.Count - 1)
{
permutations.Add(new List<T>(input));
return;
}
Permute(input, permutations, index + 1);
for (int i = index+1 ; i < input.Count; i++)
{
//swap
T temp = input[index];
input[index] = input[i];
input[i] = temp;
Permute(input, permutations, index + 1);
//swap back
temp = input[index];
input[index] = input[i];
input[i] = temp;
}
}

Bubble sort using recursion in C#

I've wrote this simple piece of code. And I have a slight problem with it.
int [] x = [50,70,10,12,129];
sort(x, 0,1);
sort(x, 1,2);
sort(x, 2,3);
sort(x, 3,4);
for(int i = 0; i < 5; i++)
Console.WriteLine(x[i]);
static int [] sort(int [] x, int i, int j)
{
if(j ==x.length)
return x;
else if(x[i]>x[j])
{
int temp = x[i];
x[i] = x[j];
x[j] = temp;
return sort(x, i, j+1);
}
else
return sort(x, i, j+1);
}
I feel that calling sort 4 time isn't the best soultion. I need a way to handle this using sort() also. I also ask you for your advice, suggestion, or tip.
Thanks

Firstly, your sort is restricted to ints, however you can use the IComparable<T> interface to extend it to any comparable type. Alternatively you could have another parameter for a Comparer<T> to allow the user to define how to compare items in the input.
A recursive bubble sort would probably look something like this: (NOTE: not tested...)
public static T[] BubbleSort(T[] input) where T : IComparable<T>
{
return BubbleSort(input, 0, 0);
}
public static T[] BubbleSort(T[] input, int passStartIndex, int currentIndex) where T : IComparable<T>
{
if(passStartIndex == input.Length - 1) return input;
if(currentIndex == input.Length - 1) return BubbleSort(input, passStartIndex+1, passStartIndex+1);
//compare items at current index and current index + 1 and swap if required
int nextIndex = currentIndex + 1;
if(input[currentIndex].CompareTo(input[nextIndex]) > 0)
{
T temp = input[nextIndex];
input[nextIndex] = input[currentIndex];
input[currentIndex] = temp;
}
return BubbleSort(input, passStartIndex, currentIndex + 1);
}
However, an iterative solution would probably be more efficient and easier to understand...

A simple bubblesort shouldn't need recursion. You could do something like this, just passing in the array to sort:
public int[] Sort(int[] sortArray)
{
for (int i = 0; i < sortArray.Length - 1; i++)
{
for (int j = sortArray.Length - 1; j > i; j--)
{
if (sortArray[j] < sortArray[j - 1])
{
int x = sortArray[j];
sortArray[j] = sortArray[j - 1];
sortArray[j - 1] = x;
}
}
}
return sortArray;
}

Nothing wrong with wanting to learn - couple of obvious things.
Firstly you're already aware that there's a length property for the array - so you could use that to create a loop that gets rid of the multiple calls to sort at the start and makes the length of the array a non problem.
Secondly you might want to think about the way the sort works - how about this: you're attempting to bubble a value up to its correct place in the list (or down if you prefer!) - so for a list of n items, remove the first, sort the remaining n - 1 items (that's the recursive bit) then bubble the first item into place.
Been decades since I thought about this, fun!

another one with only 2 params :p yeah :
static void Sort(IList<int> data)
{
Sort(data, 0);
}
static void Sort(IList<int> data, int startIndex)
{
if (startIndex >= data.Count) return;
//find the index of the min value
int minIndex = startIndex;
for (int i = startIndex; i < data.Count; i++)
if (data[i] < data[minIndex])
minIndex = i;
//exchange the values
if (minIndex != startIndex)
{
var temp = data[startIndex];
data[startIndex] = data[minIndex];
data[minIndex] = temp;
}
//recurring to the next
Sort(data, startIndex + 1);
}
Note : This is completly useless in real life because
- its extremely slow
- its recursion iteration is linear meaning that when you have more than 1k items, it will stackoverflow

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);
}
}

a more simple big adder in c#?

I just had the task in school to write a big adder. Meaning a method that can put very large numbers together.
We had 10 minutes and I did complete it on time. The teacher approved it.
I am not too satisfied with the result though, and I thought I perhaps were taking the wrong approach.
Here is my version:
using System;
using System.Text;
namespace kæmpe_adder
{
static class Program
{
static void Main()
{
var x = "1111";
var y = "111111111";
Console.WriteLine(BigAdder(x, y));
Console.ReadLine();
}
public static StringBuilder BigAdder(string x, string y)
{
var a = new StringBuilder(x);
var b = new StringBuilder(y);
return BigAdder(a, b);
}
public static StringBuilder BigAdder(StringBuilder x, StringBuilder y)
{
int biggest;
int carry = 0;
int sum;
var stringSum = new StringBuilder();
if (x.Length > y.Length)
{
y.FillString(x.Length - y.Length);
biggest = x.Length;
}
else if (y.Length > x.Length)
{
x.FillString(y.Length - x.Length);
biggest = y.Length;
}
else
{
biggest = y.Length;
}
for (int i = biggest - 1; i >= 0; i--)
{
sum = Convert.ToInt32(x[i].ToString()) + Convert.ToInt32(y[i].ToString()) + carry;
carry = sum / 10;
stringSum.Insert(0, sum % 10);
}
if (carry != 0)
{
stringSum.Insert(0, carry);
}
return stringSum;
}
public static void FillString(this StringBuilder str, int max)
{
for (int i = 0; i < max; i++)
{
str.Insert(0, "0");
}
}
}
}
When I wrote it, I thought of how you do it with binaries.
Is there a shorter and/or perhaps simpler way to do this?

From the algebraic point of view your code looks correct. From the design point of view, you would definitely prefer to encapsulate each of these big numbers in a class, so that you don't have to reference the string/string builders all the time. I am also not a big fan of this FillString approach, it seems more reasonable to add the digits while both numbers have non-zero values, and then just add the carry to the bigger number until you are done.
Not sure what was the question about binaries? The normal length numbers (32bit and 64bit) are added by the CPU as a single operation.

There are a number of open source implementations you could look to for inspiration.
http://www.codeproject.com/KB/cs/biginteger.aspx
http://biginteger.codeplex.com/
In general, I would recommend using an array of byte or long for best performance, but the conversion from a string to the array would be non-trivial.
Store the numbers in reverse order; this makes finding equivalent places trivial.
This makes it easier to add differently sized strings numbers:
int place = 0;
int carry = 0;
while ( place < shorter.Length ) {
result.Append (AddDigits (longer[place], shorter[place], ref carry));
++place;
}
while ( place < longer.Length ) {
result.Append (AddDigits (longer[place], 0, ref carry));
++place;
}
if ( carry != 0 )
result.Append (carry.ToString ());