D01

 

In V2, the start is quite new compared to V1.

We start with a “bi bi” pattern

 

25678 14568 124567 |268   2467  4678  |1247   3     9    

26789 4689  2467   |23689 23467 1     |247    2467  5    

2679  1469  3      |269   5     4679  |8      12467 1247 

---------------------------------------------------------

235   345   8      |135   9     357   |123457 1247  6    

3569  7     456    |13568 136   2     |13459  1489  1348 

1     3569  256    |4     367   35678 |23579  2789  2378 

---------------------------------------------------------

367   136   9      |1236  8     346   |12347  5     12347 

3578  2     157    |1359  134   3459  |6      14789 13478

4     13568 156    |7     1236  3569  |1239   1289  1238

 

R12c7 is the first “bi” (red). 2 cells in which 2 digits among four can be the solution.

Each pair of digits is a “super candidate”.

 

Note: the most general situation is “n” cells with “n-2” compulsory digits. Up to now, I have seen two useful patterns, always of 2 cells: 123 123  and 1234 1234

 

R4c8 r7c9 is the second “bi”. 2 single cells having the following property;

 

We will show that whatever if the super candidate in the red cells, the blue cells are filled with the same 2 digits

 

 Digit 1          digit 2       digit 4         digit 7

.XX ... X..      X.X XX. X.. .XX .XX X..      X.X .XX X..

... ..o ...      X.X XX. XX. .XX .X. XX.      X.X .X. XX.

.X. ... .XX      X.. X.. .XX .X. ..X .XX      X.. ..X .XX

 

... X.. XX.      X.. ... XX. .X. ... XX.      ... ..X XX.

... XX. XXX      ... ..o ... ..X ... XXX      .o. ... ...

o.. ... ...      ..X ... XXX ... o.. ...      ... .XX XXX

 

.X. X.. X.X      ... X.. X.X ... ..X X.X      X.. ... X.X

..X XX. .XX      .o. ... ... ... .XX .XX      X.X ... .XX

.XX .X. XXX      ... .X. XXX o.. ... ...      ... o.. ...

 

 

In each sub map, assumption is red  “True” and Blue “False” and we show this can’t be.

 

 Digit 1          digit 2       digit 4         digit 7

... ... X..      X.X XX. X.. .XX .XX X..      X.X .XX X..

... ..o ...      X.X XX. X.. .XX .X. X..      X.X .X. X..

.X. ... ...      X.. X.. ... .X. ..X ...      X.. ..X ...

 

... X.. ...      X.. ... ... .X. ... ...      ... ..X ...

... XX. .XX      ... ..o ... ..X ... .XX      .o. ... ...

o.. ... ...      ..X ... .XX ... o.. ...      ... .XX .XX

 

.X. X.. ...      ... X.. ... ... ..X ...      X.. ... ...

..X XX. .XX      .o. ... ... ... .XX .XX      X.X ... .XX

.XX .X. .XX      ... .X. .XX o.. ... ...      ... o.. ...

 

we do what is possible at that point

 

 Digit 1          digit 2       digit 4         digit 7

... ... X..      X.X XX. X.. .XX .XX X..      X.X .XX X..

... ..o ...      X.X XX. X.. .XX .X. X..      X.X .X. X..

.X. ... ...      X.. X.. ... .X. ..X ...      X.. ..X ...

 

... X.. ...      X.. ... ... .X. ... ...      ... ..X ...

... XX. .XX      ... ..o ... ..X ... .XX      .o. ... ...

o.. ... ...      ..X ... .XX ... o.. ...      ... .XX .XX

 

.X. X.. ...      ... X.. ... ... ..X ...      X.. ... ...

..X XX. .XX      .o. ... ... ... .XX .XX      X.X ... .XX

.XX .X. .XX      ... .X. .XX o.. ... ...      ... o.. ...

row 7 dead       row 3 dead   row 3 dead       row 3 dead

 

No possibility to have “Blue False” . At least one blue cell is occupied.

For each super candidate, the 2 blue cells are occupied by the same digits.

 

 <3> r7c9

 

Next step is to investigate these different super candidates to check whether they are all valid and what can be the associated pattern, using the property found for the blue cells.

 

Here are the sub maps in case red digit is occupied.

One and only one of the blue cells is valid

For any digit except ’1’, only 2 patterns are possible in boxes 6;9.

 

 Digit 1          digit 2       digit 4         digit 7

... ... X..      X.X XX. X.. .XX .XX X..      X.X .XX X..

... ..o ...      X.X XX. X.. .XX .X. X..      X.X .X. X..

.X. ... ...      X.. X.. ... .X. ..X ...      X.. ..X ...

 

... X.. .X.      X.. ... .X. .X. ... .X.      ... ..X .X.

... XX. .X.      ... ..o ... ..X ... .X.      .o. ... ...

o.. ... ...      ..X ... .X. ... o.. ...      ... .XX .X.

 

.X. X.. ..X      ... X.. ..X ... ..X ..X      X.. ... ..X

..X XX. ..X      .o. ... ... ... .XX ..X      X.X ... ..X

.XX .X. ..X      ... .X. ..X o.. ... ...      ... o.. ...

 

 

Lets now combine digits 47 and digits 27

 

 digit 2         digit 7       digit 2         digit 7

X.X XX. X..    X.X .XX X..  ||  X.X XX. X..    X.X .XX X..

X.X XX. X..    X.X .X. X..  ||  X.X XX. X..    X.X .X. X..

X.. X.. ...    X.. ..X ...  ||  X.. X.. ...    X.. ..X ...

 

... ... .X.    ... ..X ...  ||  X.. ... ...    ... ... .X.

... ..o ...    .o. ... ...  ||  ... ..o ...    .o. ... ...

..X ... ...    ... ... .X.  ||  ... ... .X.    ... .XX ...

 

... X.. ...    ... ... ..X  || ... ... ..X    X.. ... ...

.o. ... ...    X.X ... ...  ||  .o. ... ...    ... ... ..X

... ... ..X    ... o.. ...  ||  ... .X. ...    ... o.. ...

Conflicting in r3c1           double XWing r1c37 r2c37

 

  digit 4         digit 7

.XX .XX X..      X.X .XX X..  ||  .XX .XX X..      X.X .XX X..

.XX .X. X..      X.X .X. X..  ||  .XX .X. X..      X.X .X. X..

.X. ..X ...      X.. ..X ...  ||  .X. ..X ...      X.. ..X ...

 

.X. ... ...      ... ... .X.  ||  .X. ... .X.      ... ..X ...

... ... .X.      .o. ... ...  ||  ..X ... ...      .o. ... ...

... o.. ...      ... .XX ...  ||  ... o.. ...      ... ... .X.

 

... ... ..X      X.. ... ..||  ... ..X ...      X.. ... ..X

... .XX ...      ... ... ..X  ||  ... ... ..X      X.X ... ...

o.. ... ...      ... o.. ...  ||  o.. ... ...      ... o.. ...

Conflicting in r3c6              double XWing r1c57 r2c57

 

For each of these 2 super candidates, one of the 2 possibilities on r4c8 r7c9 creates a direct conflict in a cell, the other one creates a double XWing, a deadly pattern for UR’s. (see comments at the end of that page)

None of these super candidates is valid. Consequently, the last possibility for digit ‘7’ in r12c7 is 1&7 r12c7 => 1r1c7 7r2c7.

 

<7> r1c7is established.

 

Let’s now check the four remaining super candidates

 

 

 

Super candidate 1&7 two scenarios for r4c8 r7c9

 Digit 1        digit 7       Digit 1        digit 7

 

... ... X..    ... .X. ... || ... ... X..    ..X ... ...

... ..o ...    ... ... X.. || ... ..o ...    ... ... X..  

.X. ... ...    X.. ... ... || .X. ... ...    ... ..X ...

 

... ... .X.    ... ..X ... || ... X.. ...    ... ... .X.

... .X. ...    .o. ... ... || ... ... .X.    .o. ... ...

o.. ... ...    ... ... .X. || o.. ... ...    ... .X. ...

 

... X.. ...    ... ... ..X || ... ... ..X    X.. ... ...

..X ... ..X    ..X ... ... || ..X .X. ...    ... ... ..X

..X ... ..X   ... o.. ... || ..X .X. ...   ... o.. ...

1 r9c3 forced by 7 r8c3      we are left with an Xwing for 1  

 

Super candidate 1&2

 

 Digit 1         digit 2           Digit 1         digit 2

... ... X..   X.X XX. ...     || ... ... X..   X.X XX. ...

... ..o ...   ... ... X..     || ... ..o ...   ... ... X..

.X. ... ...   X.. X.. ...    || .X. ... ...   X.. X.. ...

 

... ... .X.   X.. ... ...     || ... X.. ...   ... ... .X.

... XX. ...   ... ..o ...     || ... ... .X.   ... ..o ...

o.. ... ...   ... ... .X.    || o.. ... ...   ..X ... ...

 

... X.. ...   ... ... ..X     || ... ... ..X   ... X.. ...

..X XX. ..X   .o. ... ...     || ..X XX. ...   .o. ... ...

..X .X. ..X   ... .X. ...     || ..X .X. ...   ... ... ..X

In both scenarios, all ‘2’ are known, still an XWing pattern for ‘1’

Means four possibilities

 

 

Super Candidate 1&4

 Digit 1        digit 4     

... ... X..   .XX .XX ...   || ... ... X..   .XX .XX ...

... ..o ...   ... ... X..   || ... ..o ...   ... ... X..

.X. ... ...   .X. ..X ...    || .X. ... ...   .X. ..X ...

 

... ... .X  .X. ... ...   || ... X.. ..  ... ... .X.

... XX. ...   ... ... .X.   || ... ... .X.   ..X ... ...

o.. ... ...   ... o.. ...    || o.. ... ...   ... o.. ...

 

... X.. ...   ... ... ..X    || ... ... ..X   ... ..X ...

..X XX. ..X   ... .XX ...    || ..X XX. ...   ... .XX ..X

..X .X. ..X   o.. ... ...    || ..X .X. ...   o.. ... ...

Same situation as above.

 

Last super candidate 2&4

  digit 2       digit 4     

X.X XX. X.. .XX .XX X..   || X.X XX. X.. .XX .XX X..

X.X XX. X.. .XX .X. X..   || X.X XX. X.. .XX .X. X..

X.. X.. ... .X. ..X ...    || X.. X.. ... .X. ..X ...

 

... ... .X. .X. ... ...   || X.. ... ... ... ... .X.

... ..o ... ... ... .X.   || ... ..o ... ..X ... ...

..X ... ... ... o.. ...    || ... ... .X. ... o.. ...

 

... X.. ... ... ... ..X    || ... ... ..X ... ..X ...

.o. ... ... ... .XX ...    || .o. ... ... ... ... ..X

... ... ..X o.. ... ...    || ... .X. ... o.. ... ...

2imbricated XWings             same.

2 possibilities

 

At the end of that vicinity analysis done without any use of other constraints existing in the puzzle,

· We have eliminated 2 candidates

· We have reduced the scope to less than 20 scenarios. In each of these scenario, two digits are nearly fully assigned.

 

This gives a completely different approach of Golden Nugget (compared to V1 as published on that site).

 

The solver can be aware of that situation thru a small number of weak links and equivalence links (same tag).

 

We have eliminated 3r7c9. That was done at step 105 in V1.

We have eliminated 7r1c7, still there close to the end in V1

 

We have eliminated super candidates 2&7;4&7 in r12c, a key AC2 in the former version.

In V1, we found at step 24 (1&4)r12c7

This is expected to come sooner, reducing the possibilities to 1&2 1&7 2&4 in r12c7.

In that situation are missing 1&4 2&7

  giving strong links 1r12c7 = 4r12c7 and  2r2c7 = 7r12c7

 

AT the time this is written, the solver is not yet fully aware of that scope, nevertheless, the path is about halved.

 

 

This is the most important outlet of the “bi bi” analysis.

 

Many partial patterns have been found as for example

 

25678 14568 124567 |268   2467  4678  |1247   3     9    

26789 4689  2467   |23689 23467 1     |247    2467  5    

2679  1469  3      |269   5     4679  |8      12467 1247 

---------------------------------------------------------

235   345   8      |135   9     357   |123457 1247  6    

3569  7     456    |13568 136   2     |13459  1489  1348 

1     3569  256    |4     367   35678 |23579  2789  2378 

---------------------------------------------------------

367   136   9      |1236  8     346   |12347  5     12347 

3578  2     157    |1359  134   3459  |6      14789 13478

4     13568 156    |7     1236  3569  |1239   1289  1238 

 

2r2c78   => r3c1  r9c5 => r3c4  r6c3 => r6c3  r9c5

This is an example of elementary “bi bi”.

The only thing we can get out of it is that “if 2r2c78 is valid”

candidate ‘2’  in cells seeing a couple can be eliminated.

Here nothing useful.

 

Next one is the full situation starting from r12c7

1r1c7    => r4c8  r7c9 => r5c5  r7c9

2r12c7   => r3c1  r7c9 => r3c1  r9c5 => r3c4  r4c8 => r3c4  r6c3

         => r4c8  r7c9 => r4c8  r9c5 => r6c3  r7c9 => r6c3  r9c5

4r12c7   => r3c2  r7c9 => r3c2  r8c5 => r3c6  r4c8 => r3c6  r5c3

         => r4c8  r7c9 => r4c8  r8c5 => r5c3  r7c9 => r5c3  r8c5

7r12c7   => r3c1  r4c8 => r3c1  r6c5 => r3c6  r7c9 => r3c6  r8c3

         => r4c8  r7c9 => r4c8  r8c3 => r6c5  r7c9 => r6c5  r8c3

 

The four digits share =>r4c8 r7c9. This is why we have a final “full bi bi”.

We have in Golden Nugget no example of a partial match, intermediate stage with part of the total consequences.

 

Examples of partial useful pattern will come later. (I have examples in other puzzles).

 

The intermediate level is obtained when two digits share the same couple. Then the second part of the “bi bi” is also occupied by these 2 digits, but the first clearing done here <3>r7c9 is only possible if all the super candidates are sharing the same couple.

 

 

Double XWING pattern

 

We have seen twice what I have called a “double XWing pattern.

 

  digit 4         digit 7

..X .X. X..      ... .X. X..  O given

... .X. X..      ... .X. X..  X(black) assigned when XX true

... ... ...      X.. ... ...  X unassigned giving with XX

                                   an XWing pattern

.X. ... .X.      ... ..X ...  X one of the 2 cells in the ‘bi bi’ pattern

..X ... ...      .o. ... ...

... o.. ...      ... ... .X.

 

|  ... ..X ...      ... ... ..X

... ... ..X      ..X ... ...

|  o.. ... ...      ... o.. ...

double XWing r1c57 r2c57

 

We have here the same four cells of a UR pattern in a XWing configuration on each side.

 

This is a deadly UR pattern.

 

This seem to come easily in a “bi bi” pattern. The same can be seen in Fata Morgana and leads to the elimination of two super candidates out of three.