I'm having an issue with the following code. The code works with no errors but I'm receiving different output values when using a parallel for loop vs a regular for loop. I need to get the parallel for loop working properly because I run this code thousands of times. Does anyone know why my parallel for loop is returning different outputs?
private object _lock = new object();
public double CalculatePredictedRSquared()
{
double press = 0, tss = 0, press2 = 0, press1 = 0;
Vector<double> output = CreateVector.Dense(Enumerable.Range(0, 400).Select(i => Convert.ToDouble(i)).ToArray());
List<double> input1 = new List<double>(Enumerable.Range(0, 400).Select(i => Convert.ToDouble(i)));
List<double> input2 = new List<double>(Enumerable.Range(200, 400).Select(i => Convert.ToDouble(i)));
Parallel.For(0, output.Count, i =>
{
ConcurrentBag<MultipleRegressionInfo> listMRInfoBag = new ConcurrentBag<MultipleRegressionInfo>(listMRInfo);
ConcurrentBag<double> vectorArrayBag = new ConcurrentBag<double>(output);
ConcurrentBag<double[]> matrixList = new ConcurrentBag<double[]>();
lock (_lock)
{
matrixList.Add(input1.Where((v, k) => k != i).ToArray());
matrixList.Add(input2.Where((v, k) => k != i).ToArray());
}
var matrixArray2 = CreateMatrix.DenseOfColumnArrays(matrixList);
var actualResult = vectorArrayBag.ElementAt(i);
var newVectorArray = CreateVector.Dense(vectorArrayBag.Where((v, j) => j != i).ToArray());
var items = FindBestMRSolution(matrixArray2, newVectorArray);
double estimate1 = 0;
if (items != null)
{
lock (_lock)
{
var y = 0d;
var independentCount = matrixArray2.RowCount;
var dependentCount = newVectorArray.Count;
if (independentCount == dependentCount)
{
var populationCount = independentCount;
y = newVectorArray.Average();
for (int l = 0; l < matrixArray2.ColumnCount; l++)
{
var avg = matrixArray2.Column(l).Average();
y -= avg * items[l];
}
}
for (int m = 0; m < 2; m++)
{
var coefficient = items[m];
if (m == 0)
{
estimate1 += input1.ElementAt(i) * coefficient;
}
else
{
estimate1 += input2.ElementAt(i) * coefficient;
}
}
estimate1 += y;
}
}
else
{
lock (_lock)
{
estimate1 = 0;
}
}
lock (_lock)
{
press1 += Math.Pow(actualResult - estimate1, 2);
}
});
for (int i = 0; i < output.Count; i++)
{
List<double[]> matrixList = new List<double[]>();
matrixList.Add(input1.Where((v, k) => k != i).ToArray());
matrixList.Add(input2.Where((v, k) => k != i).ToArray());
var matrixArray = CreateMatrix.DenseOfColumnArrays(matrixList);
var actualResult = output.ElementAt(i);
var newVectorArray = CreateVector.Dense(output.Where((v, j) => j != i).ToArray());
var items = FindBestMRSolution(matrixArray, newVectorArray);
double estimate = 0;
if (items != null)
{
var y = CalculateYIntercept(matrixArray, newVectorArray, items);
for (int m = 0; m < 2; m++)
{
var coefficient = items[m];
if (m == 0)
{
estimate += input1.ElementAt(i) * coefficient;
}
else
{
estimate += input2.ElementAt(i) * coefficient;
}
}
}
else
{
estimate = 0;
}
press2 += Math.Pow(actualResult - estimate, 2);
}
tss = CalculateTotalSumOfSquares(vectorArray.ToList());
var test1 = 1 - (press1 / tss);
var test2 = 1 - (press2 / tss);
}
public Vector<double> CalculateWithQR(Matrix<double> x, Vector<double> y)
{
Vector<double> result = null;
result = MultipleRegression.QR(x, y);
for (int i = 0; i < result.Count; i++)
{
var value = result.ElementAt(i);
if (Double.IsNaN(value) || Double.IsInfinity(value))
{
return null;
}
}
return result;
}
public Vector<double> CalculateWithNormal(Matrix<double> x, Vector<double> y)
{
Vector<double> result = null;
result = MultipleRegression.NormalEquations(x, y);
for (int i = 0; i < result.Count; i++)
{
var value = result.ElementAt(i);
if (Double.IsNaN(value) || Double.IsInfinity(value))
{
return null;
}
}
return result;
}
public Vector<double> CalculateWithSVD(Matrix<double> x, Vector<double> y)
{
Vector<double> result = null;
result = MultipleRegression.Svd(x, y);
for (int i = 0; i < result.Count; i++)
{
var value = result.ElementAt(i);
if (Double.IsNaN(value) || Double.IsInfinity(value))
{
return null;
}
}
return result;
}
public Vector<double> FindBestMRSolution(Matrix<double> x, Vector<double> y)
{
Vector<double> result = null;
result = CalculateWithNormal(x, y);
if (result != null)
{
return result;
}
else
{
result = CalculateWithSVD(x, y);
if (result != null)
{
return result;
}
else
{
result = CalculateWithQR(x, y);
if (result != null)
{
return result;
}
}
}
return result;
}
public double CalculateTotalSumOfSquares(List<double> dependentVariables)
{
double tts = 0;
for (int i = 0; i < dependentVariables.Count; i++)
{
tts += Math.Pow(dependentVariables.ElementAt(i) - dependentVariables.Average(), 2);
}
return tts;
}
Actual Output (Updated results):
test1 = 137431.12889999992 (parallel for loop)
test2 = 7.3770258447689254E- (regular for loop)
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If this runs for the O/P, it will run for other Community Members, whom the O/P has asked for an answer or help.
Try it online!
My new version of the code:
public double CalculatePredictedRSquared()
{
Vector<double> output = CreateVector.Dense(Enumerable.Range(0, 400).Select(i => Convert.ToDouble(i)).ToArray());
List<double> input1 = new List<double>(Enumerable.Range(0, 400).Select(i => Convert.ToDouble(i)));
List<double> input2 = new List<double>(Enumerable.Range(200, 400).Select(i => Convert.ToDouble(i)));
double tss = CalculateTotalSumOfSquares(output.ToList());
IEnumerable<int> range = Enumerable.Range(0, output.Count);
var query = range.Select(i => DoIt(i, output, input1, input2));
var result = 1 - (query.Sum() / tss);
return result;
}
public double DoIt(int i, Vector<double> output, List<double> input1, List<double> input2)
{
List<double[]> matrixList = new List<double[]>
{
input1.Where((v, k) => k != i).ToArray(),
input2.Where((v, k) => k != i).ToArray()
};
var matrixArray = CreateMatrix.DenseOfColumnArrays(matrixList);
var actualResult = output.ElementAt(i);
var newVectorArray = CreateVector.Dense(output.Where((v, j) => j != i).ToArray());
var items = FindBestMRSolution(matrixArray, newVectorArray);
double estimate = 0;
if (items != null)
{
var y = CalculateYIntercept(matrixArray, newVectorArray, items);
for (int m = 0; m < 2; m++)
{
var coefficient = items[m];
if (m == 0)
{
estimate += input1.ElementAt(i) * coefficient;
}
else
{
estimate += input2.ElementAt(i) * coefficient;
}
}
}
else
{
estimate = 0;
}
return Math.Pow(actualResult - estimate, 2);
}
This whole thing is a dog's breakfast; you should abandon that attempt at parallelism entirely.
Start over. Here's what I want you to do. I want you to write a method DoIt that returns double and takes a int i and whatever other state is required to do a single iteration of the loop.
You will then rewrite your method as follows:
public double CalculatePredictedRSquared()
{
Vector<double> output = whatever;
// Whatever other state you need here
IEnumerable<int> range = Enumerable.Range(0, output.Count);
var query = range.Select(i => DoIt(i, whatever_other_state));
return query.Sum();
}
Got it? DoIt is the thing that is inside your loop right now. It must take in i, and output and whatever other vectors you need to pass into it. It must only compute a double -- in this case, the square of the estimate error -- and return that double.
It must be pure: It must not read or write any non-local variable, it must not call any non-pure method, and it must give exactly the same results when given the same inputs, every time. Pure methods are the easiest methods to write, to read, to understand, to test and to parallelize; always try to write pure methods when doing math computations.
Write test cases for DoIt, and test the heck out of it. It's a pure method; you should be able to write lots of test cases. Similarly test any of the pure methods called by DoIt.
Once you are satisfied that DoIt is both correct and pure, then the magic happens. Just change it to:
range.AsParallel().Select...
Then compare the parallel and non-parallel versions. They should produce the same result; if not, then something was impure. Figure out what it was.
Then, verify that the parallel version was faster. If not, then you have failed to do enough work in DoIt to justify parallelism; see https://en.wikipedia.org/wiki/Amdahl%27s_law for details.
A few things:
lock (_lock)
{
matrixList.Add(input1.Where((v, k) => k != i).ToArray());
matrixList.Add(input2.Where((v, k) => k != i).ToArray());
}
You're adding items to a collection that is already thread-safe by design, so no need to lock. While List is not thread-safe, it should be OK to read from it concurrently. From the documentation:
It is safe to perform multiple read operations on a List, but issues can occur if the collection is modified while it’s being read. To ensure thread safety, lock the collection during a read or write operation. To enable a collection to be accessed by multiple threads for reading and writing, you must implement your own synchronization.
Also note that matrixList is stored in a local variable; in this case, the collection can't be called from multiple threads because the entire body of the delegate is guaranteed to run on the same thread - it won't be the case that half of the body of the Parallel.For loop will be run on thread A and that the other half will be run on thread B, for example.
Similarly, there's no reason to lock while making changes to estimate1 because it can't possibly be modified from other threads.
Disclaimer: there's no guarantee as to the degree of parallelism of the Parallel.For loop overall. There's not even a guarantee that it will run in parallel at all.
press1 and press2, however, are not local variables, so you do need to synchronize these somehow. (It would be better if you found some way to avoid locking every time, though, because that'll at least partially kill the point of multithreading).
Perhaps most critically, ConcurrentBag is an unordered collection. You don't show all of the operations you're doing on your matrices, but if you're doing matrix multiplication anywhere this could easily cause wrong results. There is no guarantee that matrix multiplication will commute. While A * B = B * A for integers, this is not true in general for matrices. It's quite possible that your logic is subtly dependent on operations occurring in a particular order (and they won't because ConcurrentBag is unordered).
I have below triangle of numbers which will be sent as parameter to a function
5
9 6
4 6 8
0 7 1 5
Now this will be received as string in below function with the format 5#9#6#4#6#8#0#7#1#5. So far I've tried to ripple only the digits from #
public class Sample
{
public static string validtrianglesum(string input)
{
string sum="0";
foreach(char num in input)
{
if(!num.Equals('#'))
{
Console.PrintLine(num); //here am getting only the nums excluding #
//How to sum up based on each row
}
}
return sum; //return
}
}
how could highest number from each row and sum them and how could I identify the rows to sum it up? Hope to find some help.
Let's break this down as follows:
Firstly, turn the input into an array of numbers:
string input = "5#9#6#4#6#8#0#7#1#5";
var numbers = input.Split('#').Select(int.Parse).ToArray();
Now let's assume we have a MakeTriangular(int[]) method that turns an array of numbers into a sequence of rows with the first row being of length 1, the second of length 2 and so on, so that it returns IEnumerable<IEnumerable<int>>.
Then we can use that along with Linq to calculate the sum of the maximum value in each row as follows:
int sum = MakeTriangular(numbers).Sum(row => row.Max());
Which gives the answer.
The implementation of MakeTriangular() could look like this:
public static IEnumerable<IEnumerable<int>> MakeTriangular(int[] numbers)
{
for (int i = 0, len = 1; i < numbers.Length; i += len, ++len)
yield return new ArraySegment<int>(numbers, i, len);
}
Putting it all together into a compilable Console app:
using System;
using System.Collections.Generic;
using System.Linq;
namespace Demo
{
class Program
{
public static void Main()
{
string input = "5#9#6#4#6#8#0#7#1#5";
var numbers = input.Split('#').Select(int.Parse).ToArray();
int sum = MakeTriangular(numbers).Sum(row => row.Max());
Console.WriteLine(sum);
}
public static IEnumerable<IEnumerable<int>> MakeTriangular(int[] numbers)
{
for (int i = 0, len = 1; i < numbers.Length; i += len, ++len)
yield return new ArraySegment<int>(numbers, i, len);
}
}
}
Summing up all values in each row:
private static IEnumerable<int> Sum(string input)
{
int i = 0, s = 0, z = 1;
foreach (var v in input.Split('#').Select(int.Parse))
{
s += v;
if (++i != z) continue;
z++;
yield return s;
s = i = 0;
}
}
The same in one line:
private static IEnumerable<int> Sum(string input) => new Func<int, int, IEnumerable<int>>((i, z) => input.Split('#').Select(int.Parse).GroupBy(e => i++ == z && (i = 1) != null ? ++z : z, e => e).Select(e => e.Sum()))(0, 1);
Summing up all the highest values in each row:
private static int Sum(string input)
{
int i = 0, s = 0, z = 1, m = 0;
foreach (var v in input.Split('#').Select(int.Parse))
{
if (v > m) m = v;
if (++i != z) continue;
z++;
s += m;
i = m = 0;
}
return s;
}
Same in one line:
private static int Sum(string input) => new Func<int, int, int>((i, z) => input.Split('#').Select(int.Parse).GroupBy(e => i++ == z && (i = 1) != null ? ++z : z, e => e).Select(e => e.Max()).Sum())(0, 1);
I am returning the sums as IEnumerable<int> and with the yield return. If you just want to print out the answers change the return type to void and remove the yield return s; line.
One way to solve this is to determine the size of the triangle. By size I mean the height/width. E.g, the provided triangle has a size of 4.
If the size is n then the number of elements in the triangle will be n(n + 1)/2. When the number of elements in the input is known this can be solved to determine n (the size) by solving a second degree polynomial and picking the positive solution (the expression below involving a square root):
var triangle = "5#9#6#4#6#8#0#7#1#5";
var values = triangle.Split('#').Select(Int32.Parse).ToList();
var sizeAsDouble = (-1 + Math.Sqrt(1 + 8*values.Count))/2;
var size = (Int32) sizeAsDouble;
if (sizeAsDouble != size)
throw new ArgumentException("Input data is not a triangle.");
So with the provided input size will be 4. You can then use the size to select each row in the triangle and perform the desired arithmetic:
var maxValues = Enumerable
.Range(0, size)
.Select(i => new { Start = i*(i + 1)/2, Count = i + 1 })
.Select(x => values.Skip(x.Start).Take(x.Count))
.Select(v => v.Max());
The first Select will compute the necessary indices to correctly slice the array of values which is done in the second Select. Again the formula n(n + 1)/2 is used. If you want to you can merge some of these Select operations but I think spliting them up makes it clearer what is going on.
The output of this will be the numbers 5, 9, 8, 7. If you want to sum these you can do it like this:
return maxValues.Sum();
You can use LINQ:
string input = "5#9#6#4#6#8#0#7#1#5";
var nums = input.Split('#').Select(s => Int32.Parse(s));
var res = Enumerable.Range(0, nums.Count())
.Select(n => nums.Skip(Enumerable.Range(0, n).Sum()).Take(n));
.Where(x => x.Any()); // here you have IEnumerable<int> for every row
.Select(arr => arr.Max());
Please give credit to Widi :) but this is your request
var rows = Sum("5#9#6#4#6#8#0#7#1#5");
var total = rows.Sum();
private static IEnumerable<int> Sum(string inp)
{
int i = 0, s = 0, z = 1;
foreach (var v in inp.Split('#').Select(int.Parse))
{
s = Math.Max(s, v);
if (++i == z)
{
z++;
yield return s;
s = i = 0;
}
}
}
I would use 2 functions:
1st one to convert the string into a tree representation:
List<List<int>> GetTree(string data)
{
List<List<int>> CompleteTree = new List<List<int>>();
List<int> ValuesInLine = new List<int>();
int partsinrow = 1;
int counter = 0;
foreach (string part in data.Split('#'))
{
int value = int.Parse(part);
ValuesInLine.Add(value);
if (++counter == partsinrow)
{
CompleteTree.Add(ValuesInLine);
ValuesInLine = new List<int>();
counter = 0;
partsinrow++;
}
}
return CompleteTree;
}
2nd one to sum up the maximum of the lines:
int GetSumOfTree(List<List<int>> tree)
{
int sum = 0;
foreach (List<int> line in tree)
{
line.Sort();
int max = line[line.Count - 1];
sum += max;
}
return sum;
}
I have a query which I get as:
var query = Data.Items
.Where(x => criteria.IsMatch(x))
.ToList<Item>();
This works fine.
However now I want to break up this list into x number of lists, for example 3. Each list will therefore contain 1/3 the amount of elements from query.
Can it be done using LINQ?
You can use PLINQ partitioners to break the results into separate enumerables.
var partitioner = Partitioner.Create<Item>(query);
var partitions = partitioner.GetPartitions(3);
You'll need to reference the System.Collections.Concurrent namespace. partitions will be a list of IEnumerable<Item> where each enumerable returns a portion of the query.
I think something like this could work, splitting the list into IGroupings.
const int numberOfGroups = 3;
var groups = query
.Select((item, i) => new { item, i })
.GroupBy(e => e.i % numberOfGroups);
You can use Skip and Take in a simple for to accomplish what you want
var groupSize = (int)Math.Ceiling(query.Count() / 3d);
var result = new List<List<Item>>();
for (var j = 0; j < 3; j++)
result.Add(query.Skip(j * groupSize).Take(groupSize).ToList());
If the order of the elements doesn't matter using an IGrouping as suggested by Daniel Imms is probably the most elegant way (add .Select(gr => gr.Select(e => e.item)) to get an IEnumerable<IEnumerable<T>>).
If however you want to preserve the order you need to know the total number of elements. Otherwise you wouldn't know when to start the next group. You can do this with LINQ but it requires two enumerations: one for counting and another for returning the data (as suggested by Esteban Elverdin).
If enumerating the query is expensive you can avoid the second enumeration by turning the query into a list and then use the GetRange method:
public static IEnumerable<List<T>> SplitList<T>(List<T> list, int numberOfRanges)
{
int sizeOfRanges = list.Count / numberOfRanges;
int remainder = list.Count % numberOfRanges;
int startIndex = 0;
for (int i = 0; i < numberOfRanges; i++)
{
int size = sizeOfRanges + (remainder > 0 ? 1 : 0);
yield return list.GetRange(startIndex, size);
if (remainder > 0)
{
remainder--;
}
startIndex += size;
}
}
static void Main()
{
List<int> list = Enumerable.Range(0, 10).ToList();
IEnumerable<List<int>> result = SplitList(list, 3);
foreach (List<int> values in result)
{
string s = string.Join(", ", values);
Console.WriteLine("{{ {0} }}", s);
}
}
The output is:
{ 0, 1, 2, 3 }
{ 4, 5, 6 }
{ 7, 8, 9 }
You can create an extension method:
public static IList<List<T>> GetChunks<T>(this IList<T> items, int numOfChunks)
{
if (items.Count < numOfChunks)
throw new ArgumentException("The number of elements is lower than the number of chunks");
int div = items.Count / numOfChunks;
int rem = items.Count % numOfChunks;
var listOfLists = new List<T>[numOfChunks];
for (int i = 0; i < numOfChunks; i++)
listOfLists[i] = new List<T>();
int currentGrp = 0;
int currRemainder = rem;
foreach (var el in items)
{
int currentElementsInGrp = listOfLists[currentGrp].Count;
if (currentElementsInGrp == div && currRemainder > 0)
{
currRemainder--;
}
else if (currentElementsInGrp >= div)
{
currentGrp++;
}
listOfLists[currentGrp].Add(el);
}
return listOfLists;
}
then use it like this :
var chunks = query.GetChunks(3);
N.B.
in case of number of elements not divisible by the number of groups, the first groups will be bigger. e.g. [0,1,2,3,4] --> [0,1] - [2,3] - [4]
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;
}
}
F# has sequences that allows to create sequences:
seq { 0 .. 10 }
Create sequence of numbers from 0 to 10.
Is there something similar in C#?
You can use Enumerable.Range(0, 10);. Example:
var seq = Enumerable.Range(0, 10);
MSDN page here.
Enumerable.Range(0, 11);
Generates a sequence of integral numbers within a specified range.
http://msdn.microsoft.com/en-us/library/system.linq.enumerable.range.aspx
You could create a simple function. This would work for a more complicated sequence. Otherwise the Enumerable.Range should do.
IEnumerable<int> Sequence(int n1, int n2)
{
while (n1 <= n2)
{
yield return n1++;
}
}
Linq projection with the rarely used indexer overload (i):
(new int[11]).Select((o,i) => i)
I prefer this method for its flexibilty.
For example, if I want evens:
(new int[11]).Select((item,i) => i*2)
Or if I want 5 minute increments of an hour:
(new int[12]).Select((item,i) => i*5)
Or strings:
(new int[12]).Select((item,i) => "Minute:" + i*5)
In C# 8.0 you can use Indices and ranges
For example:
var seq = 0..2;
var array = new string[]
{
"First",
"Second",
"Third",
};
foreach(var s in array[seq])
{
System.Console.WriteLine(s);
}
// Output: First, Second
Or if you want create IEnumerable<int> then you can use extension:
public static IEnumerable<int> ToEnumerable(this Range range)
{
for (var i = range.Start.Value; i < range.End.Value; i++)
{
yield return i;
}
}
...
var seq = 0..2;
foreach (var s in seq.ToEnumerable())
{
System.Console.WriteLine(s);
}
// Output: 0, 1
P.S. But be careful with 'indexes from end'. For example, ToEnumerable extension method is not working with var seq = ^2..^0.
My implementation:
private static IEnumerable<int> Sequence(int start, int end)
{
switch (Math.Sign(end - start))
{
case -1:
while (start >= end)
{
yield return start--;
}
break;
case 1:
while (start <= end)
{
yield return start++;
}
break;
default:
yield break;
}
}
I have these functions in my code
private static IEnumerable<int> FromZero(this int count)
{
if (count <= 0)
yield break;
for (var i = 0; i < count; i++)
{
yield return i;
}
}
private static IEnumerable<int> FromOne(this int count)
{
if (count <= 0)
yield break;
for (var i = 1; i <= count; i++)
{
yield return i;
}
}
This helps to reduce some for(i) code.
In case you wish to also save the generated sequence in a variable:
using System.Collections.Generic;
using System.Linq;
IEnumerable<int> numbersToPrint = Enumerable.Range(1, 11);
This is implicit in other solutions shown above, but I am also explicitly including the needed namespaces for this to work as expected.
Originally answered here.
If you want to enumerate a sequence of numbers (IEnumerable<int>) from 0 to a 10, then try
Enumerable.Range(0, ++10);
In explanation, to get a sequence of numbers from 0 to 10, you want the sequence to start at 0 (remembering that there are 11 numbers between 0 and 10, inclusive).
If you want an unlimited linear series, you could write a function like
IEnumerable<int> Series(int k = 0, int n = 1, int c = 1)
{
while (true)
{
yield return k;
k = (c * k) + n;
}
}
which you could use like
var ZeroTo10 = Series().Take(11);
If you want a function you can call repeatedly to generate incrementing numbers, perhaps you want somthing like.
using System.Threading;
private static int orderNumber = 0;
int Seq()
{
return Interlocked.Increment(ref orderNumber);
}
When you call Seq() it will return the next order number and increment the counter.