Pencil marks, gangsters and sockets

For most readers,  this is trivial and can be skipped. A summary of the definitions is in the previous page.

 

Split of the Pencil Marks

 

We restart from our example

 

123 456 789

457 189 623

986 327 415

 

235 874 196

614 293 857

879 615 342

 

392 741 568

561 938 274

748 562 931

 

The brute force starts with a Pencil Marks (PM) status very simple. Each cell can contain the 9 digits

 

If we now assume that the 2 first bands are fully given, The third band has 6 digits known per column and the residual PM  (A) for the band is

 

357 469 128 | 579 346 128 | 259 367 148

357 469 128 | 579 346 128 | 259 367 148

357 469 128 | 579 346 128 | 259 367 148

 

And reversely, if the band 3 is known, the residual PM  (B) for bands 1+2 is

 

124689 123578 345679 | 123468 125789 345679 |134678 124589 235679

124689 123578 345679 | 123468 125789 345679 |134678 124589 235679

124689 123578 345679 | 123468 125789 345679 |134678 124589 235679

 

124689 123578 345679 | 123468 125789 345679 |134678 124589 235679

124689 123578 345679 | 123468 125789 345679 |134678 124589 235679

124689 123578 345679 | 123468 125789 345679 |134678 124589 235679

 

And if the given define a Sudoku;

Given in band 3 will solve the PM (A)

Given in bands 1+2 will solve the PM (B)

 

Note : if a set of given solve PM(A) and PM(B), it does not mean that the global PM will be solved. Many UAs having cells in both subsets can still exist.

 

 In our example, we can easily understand this using a diagonal morph of our example

 

  149 268 357

 258 317 964

 376 549 218

 

 413 826 795

 582 791 436

 697 435 182

 

 762 183 529

 821 954 673

 935 672 841

 

Even with all these given

 

 ..9 268 357

 258 317 964

 376 549 218

 

 ..3 826 795

 582 791 436

 697 435 182

 762 183 529

 821 954 673

 935 672 841

 

The  global puzzle still has 2 solutions but each PM(A) and PM(B) is solved

 

 

Gangster

 

The reduction of a solved band to the column digits has been named gangster

We have 416 lexically minimal bands, we have 44 lexically minimal gangster.

 

In the band PM, the gangster appears in each cell at the start

 

There is no agreed name for the equivalent situation in 2 bands AFAIK, I’ll use the expression  “gangster 2 bands” or in short gangster2 to qualify it.

 

Socket

 

The next key concept to understand is the socket view. From experience, UAs limited to one or 2 mini rows in a band have a great interest.

 

We will use this property in 2 different cases.

 

Band3 toward bands 1+2

This is by far the most important case.

The elementary step in the 17 clues search is to consider in once all solution grids having the same  bands 1+2 and the relevant group of bands 3.

All these bands 3 have the same gangster and are seen equally by the bands 1+2.

This is a true socket view.

 

Band1 <->band2

In the primary UA collection for bands 1+2, UAs limited in one of the 2 bands to one or 2 mini rows have (again through experience) a pretty good filtering effect. They are searched .

This is a very similar concept, although we have here only one band 1 and one band 2.

 

Socket 2 and socket 3

 

If a UA is located in a mini row in a specific band, we can have 2 or 3 cells belonging to the UA In our process, we have the same situation in bands 1+2 if different bands3 have a similar mini row in the same stack.

 

At the start, analyzing the gangster in band 3 we can define

 

81 ways to have 2 cells with a given pair of digits

  9 pairs of column x 3^2 pairs of digits

81 ways to have a mini row with a given triplet of digits

 9 mini rows x 3^3 triplets

 

We call socket 2 the {pair of column x pair of digits}

And socket 3 the {mini row plus triplet of digits

 

Socket and UA search

 

Just consider the gangster band 3 and the gangster 2 bands of our example

 

 357      469        128      | 579       346       128       | 259      367       148

 

124689 123578 345679 | 123468 125789 345679 |134678 124589 235679

 

Socket 2

 

Consider the pair 56 in columns 1 2 (band3)

To find UAs of the socket, we can exchange 56 in the corresponding columns of the gangster bands 1+2.

 

All valid solutions collected with the PM

 

124589 123678 345679 | 123468 125789 345679 |134678 124589 235679

124589 123678 345679 | 123468 125789 345679 |134678 124589 235679

124589 123678 345679 | 123468 125789 345679 |134678 124589 235679

 

124589 123678 345679 | 123468 125789 345679 |134678 124589 235679

124589 123678 345679 | 123468 125789 345679 |134678 124589 235679

124589 123678 345679 | 123468 125789 345679 |134678 124589 235679

 

are UAs of the socket for all cells having a digit not in the solution grid.

 

Socket 3

 

 357      469        128      | 579       346       128       | 259      367       148

 

124689 123578 345679 | 123468 125789 345679 |134678 124589 235679

 

 

For each triplet eg 562 in band 3 stack1, we can have 2 ways to have UAs not pair

 652and 256  (each digit must move to another column)

In the same way, we can define the new PM and work out UAs of the socket 3

 

GUAs

 

GUA2 GUA3

 

Once a band 1+2 has been seen as “valid” against the ganster 2, all “socket UAs” not hit are reduced to the band3 cells. This piece (it is the same piece for all UAs not hit of a given socket), must be hit by the given in band 3 to have a valid Sudoku. This is 100% equivalent here to an UA of size 2/3.

 

In each band, the pair/triplet  of the socket can be seen in one mini-row. I use the term GUA2 GUA3 to identify such “UAs”.

 

GUA4 GUA6

 

We have another interesting situation with sockets 2.

If in a  band 3 the 2 digits are in the right columns, but not in the right mini-rows, we can still have a piece of socket UA  of interest in band 3

 

Here are the 3 corresponding blueprints

 

  a.. ... b..

 .b. ... a..

 ... ... ...

 

  a.. ... b..

 .b. a.. ...

 ... b.. a..

 

  a.. b.. c..

 .b. c.. a..

 ... ... ...

 

 ab. interface to bands 1+2

 

We have GUAs with 4 or 6 cells. 90% of the job is done if the socket has been searched.

These GUAs gua4 gua6 depending on the number of cells  can be added to the list of UAs of the band for the specific band 3.

 

 

Band 3 sockets

 

For a given band, not all of the 81 sockets2 81 sockets3 are of interest

Only 9 of the sockets 3 are active.

For sockets 2, the situation is slightly more complex.

Only 27 pure sockets 2 are active, but sockets leading to GUA4s GUA6s must be considered as well.

 

A quick check of all bands 3 is done to see which socket2 and sockets 3 are needed before the generation of sockets UAs