Adjacent Numbers +/-2 and +/-3

[stan]

03, 06, 21, 25, 38, 16 => 2+1+0+1+2
02, 11, 15, 26, 40, 16

all six numbers are adjacent to the previous draw in this case. and your filter settings specify that at most two adjacent numbers are allowed in a matching ticket.

[Gatsby]

so there would be a series of transformations changing each pool number into one or more other numbers and the filter would accept tickets that have 0 to 6 transformed numbers (in a 6/xx lottery), right?

Yes Stan. And the number of accepted/rejected maybe could be defined for each permutation, though generally no more than 3 numbers occur.

[Gatsby]

so there would be a series of transformations changing each pool number into one or more other numbers and the filter would accept tickets that have 0 to 6 transformed numbers (in a 6/xx lottery), right?

Yes Stan. Adajacents, knight move, or any other move as a matter of fact, that´s all permutations.

[teoman]

03, 06, 21, 25, 38, 16 => 2+1+0+1+2
02, 11, 15, 26, 40, 16

all six numbers are adjacent to the previous draw in this case. and your filter settings specify that at most two adjacent numbers are allowed in a matching ticket.

Stan, the bonus is not taken into account in statistics, but to no longer be any doubt, give another example from 6/49, where is not bonus:

07, 08, 15, 26, 39, 40 => 2+2+2+2+2
03, 05, 06, 09, 34, 36

[stan]

07, 08, 15, 26, 39, 40 => 2+2+2+2+2
03, 05, 06, 09, 34, 36

there are four adjacent numbers in this case (7,8,39,40) so the the ticket is failing the condition that at most 2 adjacent numbers are allowed
(btw, don't know what your 2+2+2+2+2 means)

[stan]

, or any other move as a matter of fact, that´s all permutations.

so what would be those other permutations. do you have any examples?

[Gatsby]

Sure,

Same numbers from previous draws: 1 "permutes" with 1; 2 permutes with 2, ... 50 permutes with 50

Adjacents A (+1): 1 permutes with 2, ..., 49 permutes with 50, 50 permutes with 1
Adjacents b (-1):2 permutes with 1, ...., 50 permutes with 49
Permutations by 2: 1 permutes with 3, 2 permutes with 4,...., 47 permutes with 50, 48 permutes with 1, ...
Permutations by 3: 1 permutes with 4, 2 permutes with 5, ...., 46 permutes with 50, ...
Permutations by 4:
... etc...

Permution with odds: 1 permutes with 3, 3 permutes with 5, 5 permutes with 7, etc, (and all the evens are maintained)
Permutation with evens: 2 permutes with 4, 4 permutes with 6, etc (all odds are mantained)

etc (i think you see the logic).

I know it may seem strange but generally in most cases the combinations originated by this transformations (from the last draw combination) do not carry more than 3 numbers to the next draw. In a 50 set number there are many different ways to do permutations, adjacents, plus 2,3,4, ... till 25 (1 permutes with 25, 2 permutes 26, ...), are the most obvious permutations, others can be envisaged, and originate a set of combinations from wich we can saffely reduce the package, permuting ones and keeping others equal

As you understand these are all complete circular patterns (in the sense that you permute all or some numbers from 1 to 50, if you choose to let some stay equal), but it´s also possible that you can permute some
sub-groups, like 1 to 10, 11 to 20, and permute only within this subsets. Once you start considering it, it really makes sense because lottery isn´t predictable, and this way we are aiming at some repetitive comportment, which is not expectable. I can´t say if it will remove many tickets from package, but i can assure that for each permutation normally no more than 3 (usually 2) numbers come in the next draw. So we want to keep withdraw from the package combinations that have 5, 4, 3 or 2 numbers equal from each of this transformed combinations of 5 numbers.
Best Regards,

F.

[teoman]

07, 08, 15, 26, 39, 40 => 2+2+2+2+2
03, 05, 06, 09, 34, 36

there are four adjacent numbers in this case (7,8,39,40) so the the ticket is failing the condition that at most 2 adjacent numbers are allowed
(btw, don't know what your 2+2+2+2+2 means)

Give another example, without consecutive double:

04, 15, 18, 24, 31, 35 => 2+1+2+2+2
01, 12, 19, 23, 29, 35

2+1+2+2+2 means:

+/-1 => 2 => (18, 24)
+/-2 => 1 => (31)
+/-3 => 2 => (04, 15)
+/-4 => 2 => (15, 31)
+/-5 => 2 => (18,24)

or

[+/-1 => 2 => (18, 24)]+[+/-2 => 1 => (31)]+[+/-3 => 2 => (04, 15)]+[+/-4 => 2 => (15, 31)]+[+/-5 => 2 => (18,24)].

[stan]

07, 08, 15, 26, 39, 40 => 2+2+2+2+2
03, 05, 06, 09, 34, 36

there are four adjacent numbers in this case (7,8,39,40) so the the ticket is failing the condition that at most 2 adjacent numbers are allowed
(btw, don't know what your 2+2+2+2+2 means)

Give another example, without consecutive double:

04, 15, 18, 24, 31, 35 => 2+1+2+2+2
01, 12, 19, 23, 29, 35

2+1+2+2+2 means:

+/-1 => 2 => (18, 24)
+/-2 => 1 => (31)
+/-3 => 2 => (04, 15)
+/-4 => 2 => (15, 31)
+/-5 => 2 => (18,24)

or

[+/-1 => 2 => (18, 24)]+[+/-2 => 1 => (31)]+[+/-3 => 2 => (04, 15)]+[+/-4 => 2 => (15, 31)]+[+/-5 => 2 => (18,24)].

you probably misunderstood how min/max option works. it specifies the 'total' minimum and maximum count of numbers with any adjacent distance. not the min/max of a single selected distance.

[stan]

Sure,

Same numbers from previous draws: 1 "permutes" with 1; 2 permutes with 2, ... 50 permutes with 50

Adjacents A (+1): 1 permutes with 2, ..., 49 permutes with 50, 50 permutes with 1
Adjacents b (-1):2 permutes with 1, ...., 50 permutes with 49
Permutations by 2: 1 permutes with 3, 2 permutes with 4,...., 47 permutes with 50, 48 permutes with 1, ...
Permutations by 3: 1 permutes with 4, 2 permutes with 5, ...., 46 permutes with 50, ...
Permutations by 4:
... etc...

Permution with odds: 1 permutes with 3, 3 permutes with 5, 5 permutes with 7, etc, (and all the evens are maintained)
Permutation with evens: 2 permutes with 4, 4 permutes with 6, etc (all odds are mantained)

etc (i think you see the logic).

I know it may seem strange but generally in most cases the combinations originated by this transformations (from the last draw combination) do not carry more than 3 numbers to the next draw. In a 50 set number there are many different ways to do permutations, adjacents, plus 2,3,4, ... till 25 (1 permutes with 25, 2 permutes 26, ...), are the most obvious permutations, others can be envisaged, and originate a set of combinations from wich we can saffely reduce the package, permuting ones and keeping others equal

As you understand these are all complete circular patterns (in the sense that you permute all or some numbers from 1 to 50, if you choose to let some stay equal), but it´s also possible that you can permute some
sub-groups, like 1 to 10, 11 to 20, and permute only within this subsets. Once you start considering it, it really makes sense because lottery isn´t predictable, and this way we are aiming at some repetitive comportment, which is not expectable. I can´t say if it will remove many tickets from package, but i can assure that for each permutation normally no more than 3 (usually 2) numbers come in the next draw. So we want to keep withdraw from the package combinations that have 5, 4, 3 or 2 numbers equal from each of this transformed combinations of 5 numbers.
Best Regards,

F.

thanks for the examples, i'll revisit that issue in some of the future releases.

[Gatsby]

Ok, Stan, please do.

Even if you do it as a plugin, that can be used to create the file with the transformations. If then we put the name of this file in a complex filter, then i suppose as the plugin is applied and the results are saved under the same name, the filter will be autommaticlay updated and ready to use.

Best Regards,

F.

[teoman]

you probably misunderstood how min/max option works. it specifies the 'total' minimum and maximum count of numbers with any adjacent distance. not the min/max of a single selected distance.

Finally all you're absolutely right, but, it is possible to implement in this filter an option, as sum to the tickets accepted between 2 of 'Adjacent Numbers' any, 3 of 'Adjacent Numbers' any, 4 of 'Adjacent Numbers' any or all 5 'Adjacent Numbers'?
Only so, this filter should become on fully operational, comprising a larger number of operations.
For details see here: http://expertlotto.com/issues/view.php?id=148