**Last attempt**
**...some method only displacing a number from one line and put it on another will affect the garantee...**

It may be my poor english, but I'm not talking about changing numbers in the wheel at all.

Maybe I should use a different example using a wheel with letters instead of numbers.

The two combinations in an arbitrary wheel are:

A B C D E F

G H I J K L

I have 12 numbers I would like to play with:

03 08 09 12 15 23 28 31 37 39 48 49

First, I assign my numbers to the letters. I'm young, free and single, so I can arrange my numbers the way I see fit or maybe even let two dice do the trick. Here's one way of assigning them:

A=03

B=09

C=15

D=28

E=37

F=48

G=08

H=12

I=23

J=31

K=39

L=49

A little puzzling with wheel and numbers will give me the following two combinations:

03 09 15 28 37 48

08 12 23 31 39 49

In my previous message, the puzzling stage was called "populating the wheel". I've called the resulting set of combinations a "population".

Now let's assume (using filters, common sense or the input of other people) that I don't like these combinations at all. I'll just go back to my dice, and arrange my numbers differently. For example:

A=03

B=08

C=09

D=28

E=31

F=37

G=12

H=15

I=23

J=39

K=48

L=49

After puzzling, my combinations suddenly look like this:

03 08 09 28 31 37

12 15 23 39 48 49

As you can see, these are two totally different combinations. And guess what? If I don't like 'em, I'll go back to the drawing board and throw the dice again. I can repeat this until I'm satisfied.

The wheel never changes, I just use a different assignment. There's absolutely no way that assigning numbers in a different sequence violates the guarantee of the wheel. In fact, the beauty of a wheel is that it doesn't depend on numbers at all.

I know that there are wheels specifically designed so that the first number is the most important one. Is the designer going to take me to court if I swap my numbers the way I want?

**Filtering**
I never said that all filters should be fully satisfied. Nick is absolutely right in stating that it might be impossible. In fact, the more filters you use, the more difficult (if not impossible) it will become to find combinations that satisfy them ALL.

For instance: I would like to play 20 numbers.

I want to use two very silly filters:

Low numbers only (1..24)

Odd numbers only (...)

As you can plainly see, I've shot myself in the foot.

The combination of these filters only allow me 12 numbers to play.

I can "wheel myself silly", but I can never get to a C(20,...).

The solution is not to difficult.

A filter can give a combination a "score".

In the simplest way the score can be either 0 or 1, indicating "bad" or "good", but that seems to be a little too simple.

In a little more complex situation the score could be a number from 0 to 9, where 0 is "please throw this one out" and 9 is "you're a fool if you don't use this one".

Or maybe a percentage indicating the level of satisfaction.

In fact, the value itself is not important at all, as long as you can compare them. This rules out the values "apple" and "pear" because these two can not be compared.

If a "score" can be given to a combination, a "score" can be given to a set of combinations too.

Remember the first set? It was:

03 09 15 28 37 48

08 12 23 31 39 49

Using a combination of all kinds of arbitrary filters and averaging the scores, the "overall score" for this set is "7".

The second set:

03 08 09 28 31 37

12 15 23 39 48 49

The overall score results to "5".

If the value "7" is considered better than the value "5", I would have to say that the first set is the one we should play.

**The strategy revisited**
When we want to play with 12 numbers, there are 12 numbers that can be assigned to "A", leaving 11 numbers to be assigned to "B", 10 to "C", etc.

The total number of permutations is: 479001600.

In other words, there are 479001600 different "populations".

**It's all a matter of finding the "best" population.**
The difference between this strategy and the one Nick has described, is that "my" strategy can be applied to an existing wheel whereas Nick builds a new wheel from the found combinations.

Take a look at the following scenario:

I look in my wallet and see enough money to play 48 combinations.

I can spend a lot time thinking if I want a lot of different numbers or if I want a certain guarantee.

The most important thing is: I'm going to play 48 combinations, no more and no less.

If I use "my" strategy, I pick a wheel that satisfies my needs and I'll find the best possible "population" in accordance to my filters.

Nick, I have two questions for you.

1) Using your method, ALL filters may like combinations consisting of a total of 24 different numbers. How can you get this in a C(18,...) wheel?

2) The only other program I know that handles the "L" parameter is the cover32 program by Kari Nurmela and Patric Ostergard. This program can optimize wheels to the bare minimum. Can your program build wheels as dense as cover32?