This post has remained unanswered for quite a few months now so I wonder if an answer is still wanted. Regardless of my hesitation over the past few days, I’ll try to answer.
I believe the reason why you needed to ask how to generate a wheel is primarily due to not fully understanding a few questions.
Question 1). What is a wheel?
An abbreviated wheel is a list of sets structured such that any k numbers from a range of v numbers shares at least t number with one or more sets listed.
Using real numbers, 5 numbers are randomly chosen ranging from 1 to 12. When a list of sets (a wheel) are individually analyzed, at least one of them will contain 4 numbers from the randomly chosen 5. This is represented as C(v, k, t)=n where n is the number of sets in the list required to achieve a complete cover.
Eg: C(12, 5, 4)=15
01, 02, 03, 04, 05, 06
01, 02, 07, 08, 09, 10
03, 04, 07, 08, 11, 12
05, 06, 09, 10, 11, 12
01, 02, 03, 09, 11, 12
01, 04, 05, 07, 10, 11
02, 04, 06, 08, 10, 12
03, 05, 06, 07, 08, 09
01, 03, 05, 08, 10, 12
01, 04, 06, 07, 09, 12
02, 03, 06, 07, 10, 11
02, 04, 05, 08, 09, 11
01, 02, 03, 04, 09, 10
01, 03, 06, 08, 09, 11
02, 03, 05, 07, 09, 12
Choose any 5 numbers from 1 to 12 inclusive and at least one of the above sets will have 4 of those numbers.
Question 2). How to check a wheel?
While there are many methods, using the above example, a simple overview of the procedure is to generate all possible combinations of 5 numbers from a range of 12.
12C5 = 792. There are 792 combinations of 5 numbers chosen from 12.
Once you have your list of combinations, compare each one to every set in the generated wheel and count that combination as being covered if it shares 4 numbers with any set in the wheel list.
A pseudo method would be:
Start with 1, 2, 3, 4, 5
Compare with every set in the wheel list
If there were one or more wheel sets which shared 4 numbers
Count this combination
Increment to the next combination: 1, 2, 3, 4, 6
Loop and do it again until the combination set reaches 8, 9, 10, 11, 12
If there are 792 combinations of 5 from 12 which share 4 numbers from one or more sets in the wheel list, then your wheel is complete.
Question 3). How to generate a wheel.
Well, I just showed you; sort of.
There are 12C4=495 combinations of 4 numbers from a range of 12.
Using a pick 6 number set:
Start with 1, 2, 3, 4, 5, 6
How many unique sets of 4 numbers are there? (6C4=15)
If this less than 495
Increment to the next set of 1, 2, 3, 4, 5, 7 and add that to your list
1, 2, 3, 4, 5, 6
1, 2, 3, 4, 5, 7
How many unique sets of 4 numbers are there in this list of sets? 6C4:15+5C4:5=20
5 better than one set alone.
Try the sets:
1, 2, 3, 4, 5, 6
1, 2, 3, 4, 5, 8
Same result, only 5 better than one set alone.
Keep going until you get to:
1, 2, 3, 4, 5, 6
7, 8, 9, 10, 11, 12
Now there are a total of 30 unique sets of 4 numbers within both of these sets.
Add a third unique set of 6 numbers:
1, 2, 3, 4, 5, 6
7, 8, 9, 10, 11, 12
1, 2, 3, 4, 5, 7
This is an improvement of 5 sets of 4 when compared to 2 sets alone.
Try all possible combinations of 6 from 12 and keep the one which adds more value to the total cover.
Seems pretty straight forward? You’ve asked for a 4 if 5 from 35. Well there are 324,632 combinations of 5 from 35 and 52,360 combinations of 4 from 35. That’s a hell of a lot of sets to try.
From the above C(12, 5, 4)=15 example, consider these facts:
There are 924 combinations of 6 numbers from a range of 12. ( 12C6=924 )
The wheel presented as a solution for C(12, 5, 4)=15 uses only 15 of the 924 possible sets.
There are 2.08x10^32 different sets of 15 which can be produced from the 924 total.
The largest 4 if 5 cover I know of is for 24 numbers with a condition of pick 6 and contains 267 wheeled sets: C(24, 5, 4)=267.
There are 24C6=134,596 combinations of 6 numbers from 24.
267 sets from a total of 134,596 can be produced a quintibazillion ways plus 1.
So finding 267 sets of 6 numbers from which at least one set contains 4 numbers from any random set of 5 chosen from a range of 24 is a masterful job.
If you read JavaScript, try analyze this:
http://www.iinet.net.au/~htjs/javascript/jswheel.htm
And Visual Basic:
http://www.iinet.net.au/~htjs/lottery/oscover.zip
Draco Merest.
