Calculate the Number of Subsets

P

Paul Black

Hi Everyone,

This is Lotto Based.
There is a System in Lotto called "Wheels".
A Lotto Wheeling System is a Special Pattern to Arrange Numbers into
Combinations. This System can be Used Over and Over again with Different
Numbers. A Lotto Wheel is Constructed in such a way that if the Winning
Numbers Fall in the Group of Numbers you have Selected, you will Always
have a Winning Combination Somewhere. If for Example, you have a Group
of 12 Numbers then the Lotto Wheel you Select should have the Numbers
1-12 Arranged in Sets of Numbers ( Each Set is 6 Numbers ). You then
take the Wheel and Substitute your Numbers in the Wheels Pattern and
Simply Replace all the 1's in the Pattern with your First Number, all
the 2's with your Second Number, all the 3's with your Third Number,
etc. All Wheels give a Guarantee. For Example, the Wheel 24,6,3,6,50
Means, there are 24 Different Numbers Used in the Wheel, there are 6
Numbers Drawn, the Guarantee of having 3 Numbers in at Least 1
Combination if ALL the 6 Numbers Drawn are in the Selected 24 Numbers.
I have a Set of 6 Number Combinations in Cells "G13:L27" ( the Number of
Combinations could be More or Less ).
In this Example I am Using a Wheel with 24 Numbers :-

1,3,7,12,15,16
1,4,5,17,20,21
1,8,9,10,19,22
1,13,14,18,23,24
2,3,6,9,21,23
2,10,12,14,16,20
2,11,15,19,20,24
3,4,7,10,18,24
3,5,7,14,17,19
4,6,8,14,15,22
4,9,11,13,16,19
5,10,13,15,17,23
5,11,12,18,21,22
6,8,12,16,17,24
7,8,13,20,22,23

Here is the Code that someone has Kindly Provided which Cycles through
ALL Combinations and Compares ALL the Combinations with ALL the
Combinations in the Above Wheel. The Below Code Finds the Coverage (
Total Combinations ) of 5 Numbers if 5 Numbers are Matched …

Code:
Sub test_5()
Dim a, dic As Object
Set dic = CreateObject("Scripting.Dictionary")
a = Range("g13").CurrentRegion.Value
For i = 1 To UBound(a, 1)
For ii = 1 To 2
For iii = ii + 1 To 3
For iv = iii + 1 To 4
For v = iv + 1 To 5
For vi = v + 1 To 6
z = a(i, ii) & "," & a(i, iii) & a(i, iv) & a(i,
v) & a(i, vi)
If Not dic.exists(z) Then
dic.Add z, Nothing
n = n + 1
End If
Next vi, v, iv, iii, ii, i
Set dic = Nothing
Range("O16") = n
End Sub

… and Produces the Correct Result of 90.
How can the Code be Modified to Also Produce the Combinations Covered
for the Categories …

Matched = Covered Combinations
2 if 5 = 42,504
3 if 5 = 35,720
4 if 5 = 4,140
5 if 5 = 90 ( the Code Already Provides this Result )

… Please.
I was Told for the Interpretation of the 3 if 5 Category that you Need
to Cycle through ALL 5 Number Combinations that can be Constructed from
the Total Numbers Used in the Wheel ( 24 in this Case ). So if the Wheel
Contains "x" Unique Numbers, you Need to Cycle through ALL 5 Number
Combinations from those "x" Numbers. Then you Need to Scan the Wheel for
Each 5 Number Combination Produced and Compare it with Each Line in the
Wheel to see if that Line Matches the 5 Number Combination in *EXACTLY*
3 Numbers. If it does, then that Combination of 3 if 5 is Covered and
Added to the Total and there is NO Need to Continue to Check for that
Particular Combination Any Further. You then go onto the Next
Combination to Check and so on Until ALL Combinations have been Cycled
through and Checked with the Wheel.

I Hope I have Explained this Clear Enough.
Many Thanks in Advance.
All the Best.
Paul
 
P

Paul Black

Hi Everyone,

I think I have probably used the word subset in error. I am basically
just trying to cycle through the combinations already in the Wheel and
find out the coverage ( total combinations for each category ).
Any help will be greatly appreciated.

Thanks in advance.
All the Best.
Paul
 
P

Paul Black

Thanks for the reply Tushar Mehta,

I will try to Explain it a bit Clearer.
Lets take our First 6 Number Combination in the Wheel of 1,3,7,12,15,16.
Now if we take the First 5 Numbers 1,3,7,12,15.
There is 1 Combination of 5 Numbers from 5, it is as Follows :-
Combination 1 = 1,3,7,12,15

There are 5 Combinations of 4 Numbers from 5, they are as Follows :-
Combination 1 = 1,3,7,12
Combination 2 = 1,3,7,15
Combination 3 = 1,3,12,15
Combination 4 = 1,7,12,15
Combination 5 = 3,7,12,15

There are 10 Combinations of 3 Numbers from 5, they are as Follows :-
Combination 1 = 1,3,7
Combination 2 = 1,3,12
Combination 3 = 1,3,15
Combination 4 = 1,7,12
Combination 5 = 1,7,15
Combination 6 = 1,12,15
Combination 7 = 3,7,12
Combination 8 = 3,7,15
Combination 9 = 3,12,15
Combination 10 = 7,12,15

There are 10 Combinations of 2 Numbers from 5, they are as Follows :-
Combination 1 = 1,3
Combination 2 = 1,7
Combination 3 = 1,12
Combination 4 = 1,15
Combination 5 = 3,7
Combination 6 = 3,12
Combination 7 = 3,15
Combination 8 = 7,12
Combination 9 = 7,15
Combination 10 = 12,15

Now this Needs to be Repeated with the Remaining 5 Combinations of 5
Numbers from 6 in Our First Combination in the Wheel :-
Combination 1 = 1,3,7,12,15 ( Already Produced Above )
Combination 2 = 1,3,7,12,16
Combination 3 = 1,3,7,15,16
Combination 4 = 1,3,12,15,16
Combination 5 = 1,7,12,15,16
Combination 6 = 3,7,12,15,16

Then the Whole of the Above Needs to be Repeated for the Remaining 14
Combinations in this Particular Wheel ( Other Wheels Might be Smaller or
Larger ). Obviously if a Combination for a Certain Category has Already
been Counted ( from ANY Other Combination in the Wheel ) then the
Repeated Combination is NOT to be Added to the Total Combinations for
that Particular Category.

I Hope this Make it a Bit Easier to Understand.
All the Best.
Paul
 
T

Tushar Mehta

Take a look at the code for generating the power set. You will have to
adapt it to either not generate sets containing only 1 element or
generate all sets and then throw away those sets.

--
Regards,

Tushar Mehta
www.tushar-mehta.com
Multi-disciplinary business expertise
+ Technology skills
= Optimal solution to your business problem
Recipient Microsoft MVP award 2000-2005
 
P

Paul Black

Thanks for the Reply,

I will Certainly have a Look at Generating the Power Set, although
Adapting it will be Beyond my Capabilities. Do you think that this will
do as I Require.

Thanks VERY Much again.
All the Best.
Paul
 
P

Paul Black

Hi Tushar Mehta,

I have had a Look at the Code for Generating the Power Set and I think
it Might Accomodate my Needs. The Only Problem is that I do NOT have
Enough VBA Skills to be Able to Adapt it to Produce my Required Results.

Any Help will be GREATLY Appreciated.
Thanks in Advance.
All the Best.
Paul
 
T

Tushar Mehta

Why not generate the powerset and then delete all the oneses by hand? It
will take all of a few seconds!

--
Regards,

Tushar Mehta
www.tushar-mehta.com
Excel, PowerPoint, and VBA add-ins, tutorials
Custom MS Office productivity solutions
 
P

Paul Black

Hi Tushar Mehta,

I am still at a loss on how I can use the PowerSet code to find the
information I am looking for if my Data is in Cells "G13:L27".
Can you please advise me on how I can achieve the required results.

Thanks in advance.
All the Best.
Paul
 
P

Paul Black

Hi Tushar Mehta,

Any Help will be greatly appreciated.

Thanks in Advance.
All the Best.
Paul
 
P

Paul Black

Hi Tushar Mehta,

I am still at a loss on how I can use the PowerSet code to find the
information I am looking for if my Data is in Cells "G13:L27".
Can you please advise me on how I can achieve the required results.

Thanks in advance.
All the Best.
Paul
 

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