How to get all the possible 3 letter permutations?

Question

Possible Duplicate:
Listing all permutations of a string/integer

For example,

aaa .. aaz .. aba .. abz .. aca .. acz .. azz .. baa .. baz .. bba .. bbz .. zzz

Basically, imagine counting binary but instead of going from 0 to 1, it goes from a to z.

I have been trying to get this working to no avail and the formula is getting quite complex. I'm not sure if there's a simpler way to do it.

Edit

I have something like this at the moment but it's not quite there and I'm not sure if there is a better way:

private IEnumerable<string> GetWordsOfLength(int length)
{
    char letterA = 'a', letterZ = 'z';

    StringBuilder currentLetters = new StringBuilder(new string(letterA, length));
    StringBuilder endingLetters = new StringBuilder(new string(letterZ, length));

    int currentIndex = length - 1;

    while (currentLetters.ToString() != endingLetters.ToString())
    {
        yield return currentLetters.ToString();

        for (int i = length - 1; i > 0; i--)
        {
            if (currentLetters[i] == letterZ)
            {
                for (int j = i; j < length; j++)
                {
                    currentLetters[j] = letterA;
                }

                if (currentLetters[i - 1] != letterZ)
                {
                    currentLetters[i - 1]++;
                }
            }
            else
            {
                currentLetters[i]++;

                break;
            }
        }
    }
}

Take a look at [this question](http://stackoverflow.com/questions/756055/listing-all-permutations-of-a-string-integer) — Zbigniew, Dec 15 '12 at 09:56
Also take a look at this question [Generate list of all possible permutations of a string](http://stackoverflow.com/questions/361/generate-list-of-all-possible-permutations-of-a-string) — Ortwin Angermeier, Dec 15 '12 at 10:03

score 33 · Accepted Answer · answered Dec 15 '12 at 10:22

33

For a variable amount of letter combinations, you can do the following:

var alphabet = "abcdefghijklmnopqrstuvwxyz";
var q = alphabet.Select(x => x.ToString());
int size = 4;
for (int i = 0; i < size - 1; i++)
    q = q.SelectMany(x => alphabet, (x, y) => x + y);

foreach (var item in q)
    Console.WriteLine(item);

answered Dec 15 '12 at 10:22

Magnus

45,362
8
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I know this is pretty old but you could have used LINQ to print out the array. `q.ForEach(x => Console.WriteLine(x));` – asdfasdfadsf Oct 20 '16 at 23:02
1

@JasonHeddle There is no ForEach extension for IEnumerable. – Magnus Oct 21 '16 at 07:35
then all you would have to do is add `.ToArray()` before the `.Foreach`, like so: `q.ToArray().ForEach(Console.WriteLine);` – asdfasdfadsf Oct 21 '16 at 11:33
2

@JasonHeddle - `ForEach` method is specific for List. What is the problem with this _foreach_? – Rahul Singh Apr 26 '17 at 07:31

L.B · Answer 2 · 2012-12-15T13:57:53.473

var alphabet = "abcdefghijklmnopqrstuvwxyz";
//or var alphabet = Enumerable.Range('a', 'z' - 'a' + 1).Select(i => (char)i);

var query = from a in alphabet
            from b in alphabet
            from c in alphabet
            select "" + a + b + c;

foreach (var item in query)
{
    Console.WriteLine(item);
}

__EDIT__

For a general solution, you can use the CartesianProduct here

int N = 4;
var result = Enumerable.Range(0, N).Select(_ => alphabet).CartesianProduct();
foreach (var item in result)
{
    Console.WriteLine(String.Join("",item));
}

// Eric Lippert’s Blog
// Computing a Cartesian Product with LINQ
// http://blogs.msdn.com/b/ericlippert/archive/2010/06/28/computing-a-cartesian-product-with-linq.aspx
public static IEnumerable<IEnumerable<T>> CartesianProduct<T>(this IEnumerable<IEnumerable<T>> sequences)
{
    // base case: 
    IEnumerable<IEnumerable<T>> result = new[] { Enumerable.Empty<T>() };
    foreach (var sequence in sequences)
    {
        var s = sequence; // don't close over the loop variable 
        // recursive case: use SelectMany to build the new product out of the old one 
        result =
            from seq in result
            from item in s
            select seq.Concat(new[] { item });
    }
    return result;
}

I thought of something like this but what if I have a variable amount of possible letter combinations? I just used 3 as an example. Might need to be 4 or 5 — Ryan Peschel, Dec 15 '12 at 10:02
@RyanPeschel than use [CartesianProduct here](http://blogs.msdn.com/b/ericlippert/archive/2010/06/28/computing-a-cartesian-product-with-linq.aspx) giving it n alphabet. — L.B, Dec 15 '12 at 10:09

score 4 · Answer 3 · answered Dec 15 '12 at 10:00

4

You have 26^3 counts for 3 "digits". Just iterate from 'a' to 'z' in three loops.

answered Dec 15 '12 at 10:00

ashley

1,177
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score 4 · Answer 4 · answered Dec 15 '12 at 10:08

Here's a very simple solution:

for(char first = 'a'; first <= (int)'z'; first++)
    for(char second = 'a'; second <= (int)'z'; second++)
        for(char third = 'a'; third <= (int)'z'; third++)
            Console.WriteLine(first.ToString() + second + third);

How to get all the possible 3 letter permutations?

Edit

4 Answers4

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