Generating Pandigital Numbers

Question

A Base10 Pandigital Number is a number which uses all the digits 0-9 once:

• 1234567890
• 2468013579
• etc...

My naive solution was just to use a bunch of nested loops to do this but it's quite slow. And I'm blanking on a more efficient way to do it? The following takes ~6 seconds.

IEnumerable<long> GeneratePandigital()
{
 var other=Enumerable.Range(0,10);
foreach(var a in other)
foreach(var b in other.Except(new int [] {a}))
foreach(var c in other.Except(new int [] {a,b}))
foreach(var d in other.Except(new int [] {a,b,c}))
foreach(var e in other.Except(new int [] {a,b,c,d}))
foreach(var f in other.Except(new int [] {a,b,c,d,e}))
foreach(var g in other.Except(new int [] {a,b,c,d,e,f}))
foreach(var h in other.Except(new int [] {a,b,c,d,e,f,g}))
foreach(var i in other.Except(new int [] {a,b,c,d,e,f,g,h}))
foreach(var j in other.Except(new int [] {a,b,c,d,e,f,g,h,i}))
{
 yield return a * 1000000000L + 
 b * 100000000L+
 c * 10000000L+
 d * 1000000+
 e * 100000+
 f * 10000+
 g * 1000+
 h * 100+
 i * 10+
 j;
} 
}

• Dave Jarvis: users.telenet.be/vdmoortel/dirk/Maths/permutations.html

Answer 1

The easiest way to do this is to note that each value that is generated will be a permutation of the values 0 - 9. This will generate 10! = 3628800 possibilities. There are a number of algorithms to generate permutations, the easiest in this case would simply be to generate them in lexicographic order.

Wiki has a rundown of some algorithms you can use to generate permutations.

• Eoin Campbell: sigh how did I not spot this... and I actually curate a combinatorics library on Nuget that's perfect for this. nuget.org/packages/Combinatorics - I blame the early hour. Thanks