Compute a set of concatenated sets of n sets

Question

Good - I’m not even sure that this term is right, and I’m sure that there is a certain term for this, but I will do my best to explain it. This is not a cross-product, and the order of the results is absolutely important.

Given:

IEnumerable<IEnumerable<string>> sets = 
      new[] { 
              /* a */ new[] { "a", "b", "c" },
              /* b */ new[] { "1", "2", "3" },
              /* c */ new[] { "x", "y", "z" }
            };

If each internal enumeration is a command to create a set of concatenations as follows (meaning is important here):

set a* = new string[] { "abc", "ab", "a" };
set b* = new string[] { "123", "12", "1" };
set c* = new string[] { "xyz", "xy", "x" };

I want to create the specified ordered concatenations as follows:

set final = new string { a*[0] + b*[0] + c*[0], /* abc123xyz */
                         a*[0] + b*[0] + c*[1], /* abc123xy  */
                         a*[0] + b*[0] + c*[2], /* abc123x   */
                         a*[0] + b*[0],         /* abc123    */
                         a*[0] + b*[1] + c*[0], /* abc12xyz  */
                         a*[0] + b*[1] + c*[1], /* abc12xy   */
                         a*[0] + b*[1] + c*[2], /* abc12x    */
                         a*[0] + b*[1],         /* abc12     */
                         a*[0] + b*[2] + c*[0], /* abc1xyz   */
                         a*[0] + b*[2] + c*[1], /* abc1xy    */
                         a*[0] + b*[2] + c*[2], /* abc1x     */
                         a*[0] + b*[2],         /* abc1      */
                         a*[0],                 /* abc       */
                         a*[1] + b*[0] + c*[0], /* ab123xyz  */

                         /* and so on for a*[1] */
                         /* ... */

                         a*[2] + b*[0] + c*[0], /* a123xyz   */

                         /* and so on for a*[2] */
                         /* ... */

                         /* now lop off a[*] and start with b + c */

                         b*[0] + c*[0],         /* 123xyz    */

                         /* rest of the combinations of b + c
                            with b on its own as well */

                         /* then finally */
                         c[0],
                         c[1],
                         c[2]};

It’s so clear that there will be many combinations!

I see similarities with numerical fundamentals (since order is also important), and I'm sure there are permutations / combinations here.

The question is how to write such an algorithm that can cope with any number of rowsets. Linq, not Linq; I did not fuss.

Why am I doing this?

In fact, why !?

Asp.Net MVC - , . View View-en-GB, View-en, View-GB View (, , / , - a Distinct() ).

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Answer 1

:

void Main()
{
    IEnumerable<IEnumerable<string>> sets = 
          new[] { 
                  /* a */ new[] { "a", "b", "c" },
                  /* b */ new[] { "1", "2", "3" },
                  /* c */ new[] { "x", "y", "z" }
                };

    var setCombinations = from set in sets
                          select (from itemLength in Enumerable.Range(1, set.Count()).Reverse()
                                  select string.Concat(set.Take(itemLength).ToArray()));

    IEnumerable<string> result = new[] { string.Empty };

    foreach (var list in setCombinations) {
        result = GetCombinations(result, list);
    }
    // do something with the result
}

IEnumerable<string> GetCombinations(IEnumerable<string> root, IEnumerable<string> append) {
    return from baseString in root
           from combination in ((from str in append select baseString + str).Concat(new [] { baseString }))
           select combination;
}

Answer 2

, :

using System;
using System.Linq;
using System.Collections.Generic;

namespace SO3014119
{
    class Program
    {
        private static IEnumerable<string> GetStringCombinations(
            string prefix, 
            IEnumerable<string>[] collections, int startWithIndex)
        {
            foreach (var element in collections[startWithIndex])
            {
                if (startWithIndex < collections.Length - 1)
                {
                    foreach (var restCombination in
                        GetStringCombinations(prefix + element, collections,
                            startWithIndex + 1))
                    {
                        yield return restCombination;
                    }
                }

                yield return prefix + element;
            }
        }

        public static IEnumerable<string> GetStringCombinations(
            params IEnumerable<string>[] collections)
        {
            while (collections.Length > 0)
            {
                foreach (var comb in GetStringCombinations("", collections, 0))
                    yield return comb;

                // "lop off" head and iterate
                collections = collections.Skip(1).ToArray();
            }
        }

        static void Main(string[] args)
        {
            var a = new string[] { "a1", "a2", "a3" };
            var b = new string[] { "b1", "b2", "b3" };
            var c = new string[] { "c1", "c2", "c3" };

            foreach (string combination in GetStringCombinations(a, b, c))
            {
                Console.Out.WriteLine(combination);
            }
        }
    }
}

( , , , ):

a1b1c1
a1b1c2
a1b1c3
a1b1
a1b2c1
a1b2c2
a1b2c3
a1b2
a1b3c1
a1b3c2
a1b3c3
a1b3
a1
a2b1c1
a2b1c2
a2b1c3
a2b1
a2b2c1
a2b2c2
a2b2c3
a2b2
a2b3c1
a2b3c2
a2b3c3
a2b3
a2
a3b1c1
a3b1c2
a3b1c3
a3b1
a3b2c1
a3b2c2
a3b2c3
a3b2
a3b3c1
a3b3c2
a3b3c3
a3b3
a3
b1c1
b1c2
b1c3
b1
b2c1
b2c2
b2c3
b2
b3c1
b3c2
b3c3
b3
c1
c2
c3

Answer 3

()

Add an empty string a *, b *, c * to the end of each array

string[] a* = { "abc","ab","a","" };
string[] b* = { "123","12","1","" };
string[] c* = { "xyz","xy","x","" };

List<string> final = new List<string>();

Now create a new version for three arrays

foreach(string aMember in a*)
foreach(string bMember in b*)
foreach(string cMember in c*)
final.add(aMember+bMember+cMember);

An additional empty line at the end of *, b * and c * will generate special lines, such as [0] (= a [0] + b [3] + c [3]) in the required order,

EDIT: this code will create extra lines. See Comment below.