Question

Possible Duplicate:
Generating permutations of a set (most efficiently)

I was looking at an old programming challenge, and I was trying to come up with a solution.
The challenge is expired, and years old, and I’m doing it just to build skill at this point.

I need to generate numbers in the following pattern:

• 123456789
• 123456798
• 123456879
• 123456897
• 123456978
• 123456987

Continuing onward, always using the same 9 numbers, never duplicating them, and always grabbing the next one in line.

I’ve been wracking my brain for the last 2 hours, and can’t figure out a good programming pattern to tackle this.

Any suggestions?

Answer 1

How about this as a solution:

var numerals = Enumerable.Range(1, 9).ToArray();

var query =
    from n1 in numerals
    from n2 in numerals.Except(new [] { n1, })
    from n3 in numerals.Except(new [] { n1, n2, })
    from n4 in numerals.Except(new [] { n1, n2, n3, })
    from n5 in numerals.Except(new [] { n1, n2, n3, n4, })
    from n6 in numerals.Except(new [] { n1, n2, n3, n4, n5, })
    from n7 in numerals.Except(new [] { n1, n2, n3, n4, n5, n6, })
    from n8 in numerals.Except(new [] { n1, n2, n3, n4, n5, n6, n7, })
    from n9 in numerals.Except(new [] { n1, n2, n3, n4, n5, n6, n7, n8, })
    select n1 * 100000000
        + n2 * 10000000
        + n3 * 1000000
        + n4 * 100000
        + n5 * 10000
        + n6 * 1000
        + n7 * 100
        + n8 * 10
        + n9;

This turns out to be quite fast producing all of the results in 864 milliseconds on my computer.

Here are the first 10 results:

123456789 
123456798 
123456879 
123456897 
123456978 
123456987 
123457689 
123457698 
123457869 
123457896