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wheel algorithms?

Topic closed. 42 replies. Last post 2 years ago by SergeM.

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bgonçalves
Brasil
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Posted: February 22, 2013, 1:56 pm - IP Logged
XXXOOOO
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Ok, sergem, fanstastico, the rows and columns can monter in 4 pairs and three odd and vice versa too, so as the perfect matrix sergem this ai, everything depends on distributing in the rows and columns of the array, the array is 100% IE hitting within the thread, small temosprêmios almost all sweepstakes, few cards
Thanks sergem
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    bgonçalves
    Brasil
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    Posted: February 22, 2013, 2:13 pm - IP Logged
    Hello then sergem crossing segments of rows and columns and mirror
    We Have =
    (0000) lines 35 times 35 columns (xxx) = 1225
    (xxx) lines 35 times 35 columns (0000) = 1225
    Total = 2,450 filtering can reduce lines at 80% of this
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      bgonçalves
      Brasil
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      Posted: February 22, 2013, 2:43 pm - IP Logged
      Ok sergem, but is a combination of a line segment with column, because we're playing
      With all the numbers of the 49/6, i.e. combining per array, with line segment with columns, what will depend on and how do we assemble the rows and columns, because it covers the numbers 49 49/6, can you explain your list?
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        bgonçalves
        Brasil
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        Posted: February 22, 2013, 2:56 pm - IP Logged
        Hello sergem has much repeated positional training
        Has to do
        row and column xxx 0000
        Or xxxxlinha and 000coluna
        The lines have become so
        0000xxx = column and row
        Xxxx000 = row and column
        The training must be of 7 positional numbers without repetition in the array
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          bgonçalves
          Brasil
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          Posted: February 22, 2013, 4:41 pm - IP Logged
          Hello, sergem training has to be 7 digits and have to take care to give positions hit.
          Example of segment xxx took the 3rd position in the column segment 0000 cannot repeat the positions, is not so easy to match by position, to give hit within the 7 x 7 array cannot be linear combinations, must be positional, won't be easy
          on your list has a lot of repeated formations, tends to take advantage of the positional distrbuidoem coodenadas factor in matrix 7 x 7

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            Posted: February 22, 2013, 5:38 pm - IP Logged

            I want to play tic tac toe too!!!!

            XOXOXOXOXOXOXO

            OXOXOXOXOXOXOX

            XOXOXOXOXOXOXO

            Jester Laugh

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              bgonçalves
              Brasil
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              Posted: February 22, 2013, 7:19 pm - IP Logged

              I want to play tic tac toe too!!!!

              XOXOXOXOXOXOXO

              OXOXOXOXOXOXOX

              XOXOXOXOXOXOXO

              Jester Laugh

              Hello, lottoboner, the lines are ready for 7 digits
              So, xxx of the row with the column 1 = 0000 mirror xxx0000
              1 line long column two .....
              Vice versa online with 0000 xxx column 1 = mirror 0000xxx
              And so with each column crossing the segments with the lines,
              After and see in matrix 7 x 7 numbers and change, is this
              example 1, 2, 7 or this on xxx or this in the mirror 0000
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                bgonçalves
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                Posted: February 22, 2013, 9:30 pm - IP Logged
                1ªpos2ª pos3ª pos4ª pos5ª pos6ª pos7ª poslinesXXXOOOO
                1ª pos1234567XXOXOOO
                2ª pos891011121314XXOOXOO
                3ª pos15161718192021XXOOOXO
                4ª pos22232425262728XXOOOOX
                5ª pos29303132333435XOXXOOO
                6ª pos36373839404142XOXOXOO
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                OOXXXOO
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                OOXXOOX
                OOXOXXO
                OOXOXOX
                OOXOOXX
                OOOXXXO
                OOOXXOX
                OOOXOXX
                OOOOXXX
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                  Posted: February 27, 2013, 11:20 pm - IP Logged

                  The only known to me available source for wheeling code is the old Cover32 program (not sure if it is still somewhere available). What source code do you refer to, because such things are really hard to find mostly because abbreviated wheel development is based on unique heuristics and nobody really wants to share his own work (I also have developed mine from scratch because I really couldn't find any source to work on - the same problem you face).

                  Also Iliya Bluskov's books do not contain any info on wheel development. Most of his wheels are based on theories and are optimal or near-optimal constructions. I also have a few wheels of mine in his lotto 6 book made by my program simply because no optimal theory-based solution has been found for those few instances.

                  For your final question on 4if4if8, the general case is that we cannot approach the theoretical minimum. Given a covering v,k,t,m=b, the computed general theoretical minimum assumes the covering can be made as "pack design" which means every combination that must be covered to offer 100% guarantee, it is covered by one and only one block of the covering. Such a property is possible in some t=m constructions and maybe very few t < m constructions (if any). Therefore the general theoretical minimum is just a lowest boundary which can't be reached in most cases. Very advanced mathematics can compute a better lowest bound (higher than the general theoretical) for some classes of v,k,t,m coverings but even these may not be possible to reach. So your 4if8, given it is a t < m construction it will not approach the theoretical minimum simply because there is an unavoidable overlap among the blocks of the covering.

                  By the way, 4if4of8 means nothing, you need 4 parameters to define a wheel.

                  lottoarchitect wrote:  "By the way, 4if4of8 means nothing, you need 4 parameters to define a wheel."

                  Which is why I wrote:  "a 7-ticket wheel for 4-if-4-of-8 for a 6-draw lottery".

                  The "4-if-4-of-8" notation is derived from the terminology described in this forum under the title "What's a Lottery Wheel" at the "Lottery Wheel" link.  So I thought it might be better understood than the algebraic notation (8,6,4,4)=7.

                   

                  lottoarchitect wrote:  "Given a covering v,k,t,m=b,  [.... for] your 4if8, given it is a t < m construction [...]".

                  Actually, I wrote 4-if-4-of-8.  And as I understand the (v,k,t,m)=b notation [1], that is a t=m construction.

                  That is a small detail.  I mention it only to clarify my next question.

                  -----

                  In fact, I do not understand the t < m condition per se.  Can someone explain it?

                  Consider the very excellent step-by-step explanation of the JADELottery algorithm starting at www.lotterypost.com/thread/228056/2017518, referenced by a Ramijami in an earlier response in this discussion.

                  The JADELottery example is for (9,5,3,4)=b.  That is, a wheel that "guarantees" a 3-win if 4 of 9 match 5 drawn.  It appears that b=5.

                  I don't exactly gronk the implications of the requirement "3 if 4 ...".

                  With "4 if 4 of 8", that is (8,6,4,4)=b, I believe that means the wheel of pick-6 tickets must cover all of the 70 4-tuples from the pool of 8 numbers.

                  But with "3 if 4 of 9", that is (9,5,3,4)=b, it seems that must mean the wheel of pick-5 tickets covers only some of the 126 4-tuples from the pool of 9 numbers.

                  Is that right?!

                  I'd be surprised if it is because that does not seem to "guarantee" as 3-win if (any of) 4 of the 9 numbers match part of the pick-5 drawing.

                  But if the requirement were different -- if (9,5,3,4) means that all of the 4-tuples must be covered by the wheel -- that would "guarantee" a 4-win.  So why bother with (also) "guaranteeing" a 3-win?

                  If I'm wrong, I would appreciate it someone would post some counter-examples, namely:  a (9,5,4,4) wheel, a (9,5,3,4) wheel, and a (9,5,3,3) wheel.

                  Unfortunately, the JADELottery explanation is not sufficient because the resulting 5-ticket wheel does not cover all 3-tuples or all 4-tuples.

                  So it does not seem to "guarantee" a 3-win at all, as I (mis?)understand the wheel requirements.

                  (Note:  I have not yet worked through all of the JADELottery details to see if perhaps there is a minor typo that misrepresents the final wheel.)

                   

                  -----

                  [1] I am familiar with the notation (n,k,t,m)=b used by an apparently different "Lotto Architect".  This forum's silly rules do not permit me, a "new" member, to post a URL.  Google for "lottery wheel terminology", and look for "anastasios tampakis" in the URL.

                  For that notation, n = #numbers to wheel (aka "pool"); k = #numbers drawn (aka "ticket size"); t = minimum "guaranteed" win (aka "prize category"); m = #numbers that must match (are among) the drawn numbers; and b = #tickets to purchase (aka "wheel").

                  In other words, a b-ticket wheel that guarantees a t-win if m of n match k drawn numbers.

                  A Google search also finds the notation (v,k,t,m)=b, with the same interpretation of the common variables, and v = n.

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                    Posted: February 27, 2013, 11:54 pm - IP Logged

                    lottoarchitect wrote:  "By the way, 4if4of8 means nothing, you need 4 parameters to define a wheel."

                    Which is why I wrote:  "a 7-ticket wheel for 4-if-4-of-8 for a 6-draw lottery".

                    The "4-if-4-of-8" notation is derived from the terminology described in this forum under the title "What's a Lottery Wheel" at the "Lottery Wheel" link.  So I thought it might be better understood than the algebraic notation (8,6,4,4)=7.

                     

                    lottoarchitect wrote:  "Given a covering v,k,t,m=b,  [.... for] your 4if8, given it is a t < m construction [...]".

                    Actually, I wrote 4-if-4-of-8.  And as I understand the (v,k,t,m)=b notation [1], that is a t=m construction.

                    That is a small detail.  I mention it only to clarify my next question.

                    -----

                    In fact, I do not understand the t < m condition per se.  Can someone explain it?

                    Consider the very excellent step-by-step explanation of the JADELottery algorithm starting at www.lotterypost.com/thread/228056/2017518, referenced by a Ramijami in an earlier response in this discussion.

                    The JADELottery example is for (9,5,3,4)=b.  That is, a wheel that "guarantees" a 3-win if 4 of 9 match 5 drawn.  It appears that b=5.

                    I don't exactly gronk the implications of the requirement "3 if 4 ...".

                    With "4 if 4 of 8", that is (8,6,4,4)=b, I believe that means the wheel of pick-6 tickets must cover all of the 70 4-tuples from the pool of 8 numbers.

                    But with "3 if 4 of 9", that is (9,5,3,4)=b, it seems that must mean the wheel of pick-5 tickets covers only some of the 126 4-tuples from the pool of 9 numbers.

                    Is that right?!

                    I'd be surprised if it is because that does not seem to "guarantee" as 3-win if (any of) 4 of the 9 numbers match part of the pick-5 drawing.

                    But if the requirement were different -- if (9,5,3,4) means that all of the 4-tuples must be covered by the wheel -- that would "guarantee" a 4-win.  So why bother with (also) "guaranteeing" a 3-win?

                    If I'm wrong, I would appreciate it someone would post some counter-examples, namely:  a (9,5,4,4) wheel, a (9,5,3,4) wheel, and a (9,5,3,3) wheel.

                    Unfortunately, the JADELottery explanation is not sufficient because the resulting 5-ticket wheel does not cover all 3-tuples or all 4-tuples.

                    So it does not seem to "guarantee" a 3-win at all, as I (mis?)understand the wheel requirements.

                    (Note:  I have not yet worked through all of the JADELottery details to see if perhaps there is a minor typo that misrepresents the final wheel.)

                     

                    -----

                    [1] I am familiar with the notation (n,k,t,m)=b used by an apparently different "Lotto Architect".  This forum's silly rules do not permit me, a "new" member, to post a URL.  Google for "lottery wheel terminology", and look for "anastasios tampakis" in the URL.

                    For that notation, n = #numbers to wheel (aka "pool"); k = #numbers drawn (aka "ticket size"); t = minimum "guaranteed" win (aka "prize category"); m = #numbers that must match (are among) the drawn numbers; and b = #tickets to purchase (aka "wheel").

                    In other words, a b-ticket wheel that guarantees a t-win if m of n match k drawn numbers.

                    A Google search also finds the notation (v,k,t,m)=b, with the same interpretation of the common variables, and v = n.

                    I wrote:  ``Unfortunately, the JADELottery explanation is not sufficient because the resulting 5-ticket wheel does not cover all 3-tuples or all 4-tuples.  So it does not seem to "guarantee" a 3-win at all, as I (mis?)understand the wheel requirements.``

                    I meant to add the following footnote; just missed the 20-minute edit window.

                    JADELottery wrote that the 5-wheel solution is:

                    05 06 07 08 09
                    01 03 04 06 09
                    02 03 04 08 09
                    01 02 06 07 08
                    01 02 03 04 05

                    That does not cover the 4-tuple 01 02 03 06, for example.  And it does not cover the 3-tuple 01 02 09, for example.

                    (Since JADELottery's step-by-step explanation shows that all 4-tuples are covered, I wonder if there is a typo in JADELottery's final wheel.  I have not checked yet.  And I have not run JADELottery's program; I don't want to install SilverLight.  So I don't know if the mistake, if any, is simply in the explanation.)

                      lakerben's avatar - Lottery-020.jpg
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                      Posted: February 28, 2013, 12:22 am - IP Logged

                      I wrote:  ``Unfortunately, the JADELottery explanation is not sufficient because the resulting 5-ticket wheel does not cover all 3-tuples or all 4-tuples.  So it does not seem to "guarantee" a 3-win at all, as I (mis?)understand the wheel requirements.``

                      I meant to add the following footnote; just missed the 20-minute edit window.

                      JADELottery wrote that the 5-wheel solution is:

                      05 06 07 08 09
                      01 03 04 06 09
                      02 03 04 08 09
                      01 02 06 07 08
                      01 02 03 04 05

                      That does not cover the 4-tuple 01 02 03 06, for example.  And it does not cover the 3-tuple 01 02 09, for example.

                      (Since JADELottery's step-by-step explanation shows that all 4-tuples are covered, I wonder if there is a typo in JADELottery's final wheel.  I have not checked yet.  And I have not run JADELottery's program; I don't want to install SilverLight.  So I don't know if the mistake, if any, is simply in the explanation.)

                      Put all your numbers on a piece of paper. Next, sprinkle some chicken feed on top of the paper.  Turn the chicken loose and wait while the paper is being poked.  Take the 5 or six biggest holes and use the numbers adjacent to them.  Alas , you have your wheel.  Don't laugh it worked for a farmer in Texas, he won big.

                       

                      Hiding Behind Computer

                      Use what works for your state and good luck.

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                        Put all your numbers on a piece of paper. Next, sprinkle some chicken feed on top of the paper.  Turn the chicken loose and wait while the paper is being poked.  Take the 5 or six biggest holes and use the numbers adjacent to them.  Alas , you have your wheel.  Don't laugh it worked for a farmer in Texas, he won big.

                         

                        Hiding Behind Computer

                        That's an ironic comment, considering the "systems" that you suggest in other discussions (e.g, click here).

                        Humor aside, I want to reiterate that I am not interested in using wheels.  I am interested in understanding the algorithms for generating wheels.  It is an academic exercise for my own edification.

                        So comments like yours are neither helpful nor wanted in this discussion.  In fact, they are never constructive, IMHO.

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                          Posted: February 28, 2013, 6:49 am - IP Logged

                          I wrote:  ``Unfortunately, the JADELottery explanation is not sufficient because the resulting 5-ticket wheel does not cover all 3-tuples or all 4-tuples.  So it does not seem to "guarantee" a 3-win at all, as I (mis?)understand the wheel requirements.``

                          I meant to add the following footnote; just missed the 20-minute edit window.

                          JADELottery wrote that the 5-wheel solution is:

                          05 06 07 08 09
                          01 03 04 06 09
                          02 03 04 08 09
                          01 02 06 07 08
                          01 02 03 04 05

                          That does not cover the 4-tuple 01 02 03 06, for example.  And it does not cover the 3-tuple 01 02 09, for example.

                          (Since JADELottery's step-by-step explanation shows that all 4-tuples are covered, I wonder if there is a typo in JADELottery's final wheel.  I have not checked yet.  And I have not run JADELottery's program; I don't want to install SilverLight.  So I don't know if the mistake, if any, is simply in the explanation.)

                          I wrote earlier:  ``I don't exactly gronk the implications of the requirement "3 if 4 ...".``

                          And I wrote of the JADELottery example:  ``That does not cover the 4-tuple 01 02 03 06, for example.  And it does not cover the 3-tuple 01 02 09, for example.``

                          Never mind!  I get it now.

                          If the drawing is 01 02 09 xx yy, xx or yy (or both) must be one of 03 through 08 in order to meet the match-4 condition.

                          So it is sufficient that 01 02 03, 01 02 04, 01 02 05, 01 02 06, 01 02 07 and 01 02 08 are covered by the wheel to meet the win-3 "guarantee".

                          The JADELottery algorithm is similar to the one that I developed independently for a (8,6,4,4) wheel.  Now I need to adapt my algorithm to handle the t<m scenario.

                            lottoarchitect's avatar - waveform

                            Greece
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                            Posted: February 28, 2013, 8:11 am - IP Logged

                            The principle and the maths are straightforward to this. We have a wheel C(v,k,t,m)=b where

                            v=the total balls we want to wheel
                            k=the ticket size (e.g. a 6 ball game has k=6)
                            t=the prize division we want to guarantee a win
                            m=the condition that has to be met, in order to guarantee the t prize division win; m defines the least number of balls from our v set that must be correct.
                            b=the total tickets required to play.

                            We are interested to find the coverage achieved in a certain category e.g. t if m.
                            Then the total combinations that need to be covered are nCk(v,m) = A (combination formula).
                            Thus, you have to test A combinations, each one containing m numbers against the tickets b of your wheel (where each ticket contains k numbers).
                            A combination of those A is covered if there is at least one ticket in your wheel, that contains at least t numbers in common.
                            All you have to do is to go through all A combinations and test each of them if it contains at least t numbers in common with at least one ticket of your wheel. If it does, then it is covered.
                            Now, you have to convert this in code in such a way that minimizes b. In order to do that you have to find a way to make the least overlap possible among the wheel tickets, which means a combination from the set A is covered by only one wheel ticket. The quick approach is brute force (scan all tickets) but this works only for very small wheels. Another good option is the use of an annealing process (Cover32 implements that). A 3rd option is to find construction methodologies derived by theory and implement that in code but this works optimally for few coverings only. Annealing is a all-good overall approach which can give good results most of the time but not optimal in most cases. You can of course explore other ideas like mutation/genetic algorithms, aggressive optimization etc.

                            By the way, I'm the author of Lotto Architect and Wheel Generator so I'm the same person. I also used back then the (n,k,t,m)=b and before that I used (n,k,p,m)=b but the official expression is c(v,k,t,m,L)=b and when L=1 it is omitted.

                              JADELottery's avatar - EyePhnoeyReality6
                              The Quantum Master
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                              Posted: February 28, 2013, 10:01 am - IP Logged

                              I wrote:  ``Unfortunately, the JADELottery explanation is not sufficient because the resulting 5-ticket wheel does not cover all 3-tuples or all 4-tuples.  So it does not seem to "guarantee" a 3-win at all, as I (mis?)understand the wheel requirements.``

                              I meant to add the following footnote; just missed the 20-minute edit window.

                              JADELottery wrote that the 5-wheel solution is:

                              05 06 07 08 09
                              01 03 04 06 09
                              02 03 04 08 09
                              01 02 06 07 08
                              01 02 03 04 05

                              That does not cover the 4-tuple 01 02 03 06, for example.  And it does not cover the 3-tuple 01 02 09, for example.

                              (Since JADELottery's step-by-step explanation shows that all 4-tuples are covered, I wonder if there is a typo in JADELottery's final wheel.  I have not checked yet.  And I have not run JADELottery's program; I don't want to install SilverLight.  So I don't know if the mistake, if any, is simply in the explanation.)

                              The wheel you've generated should have the additional information as follows:

                              Wheel
                              - Pick 5
                              - 9 Numbers
                              - 5 Combinations
                              - 3.968254% Coverage of 126 Combinations

                              Condition
                              - If 4 Drawn Numbers are in the Set of 9 Numbers,
                              - Then at least 1 Combination has 3 Winning Numbers.
                              ____________________________________________
                              05 06 07 08 09
                              01 03 04 06 09
                              02 03 04 08 09
                              01 02 06 07 08
                              01 02 03 04 05
                              ____________________________________________

                              When reading the condition you should understand the second part '- Then at least 1 Combination has 3 Winning Numbers.' to mean that a line will have At Least that many or more winning numbers.

                              This means if the condition is '3 Winning Numbers', then there could be a winning line with 3 or 4 winning numbers, because the other part of the condition is '4 Drawn Numbers are in the Set of 9 Numbers,".

                              Also, these mean the following:

                              Set - The wheel pool of numbers you are playing; typically less than the lottery pool of numbers being drawn.
                              Pick - The number of balls being drawn for that lottery pool of numbers.
                              Match - The quantity of drawn numbers that match your wheel pool numbers.
                              Win - Guaranteed to have at least 1 line with that many matching numbers in the wheeled combination or more.

                              If there is a line with 4 winning numbers, it definitely has any combination of 3 winning numbers, and thus qualifies as having 3 Winning numbers; even if there are no lines with just 3 winning numbers.

                              We'd prefer the 4 winning numbers line than just 3.

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