The New Sudoku Players' Forum

[daj95376]

Please save me from searching through previous discussions on double Exocets. The following has two Exocets in the same band ... and three base cell values in common. Is it a double Exocet ... and why? TIA!

Code: Select all

98.7..6..5...9......7..6...4..5...3..3......2..5.2.7....9.7.8.....1...2......4..1

;21117;GP;KZ1C;1;r3c1 r3c2 r1c5 r2c7 1234;;

 +--------------------------------------------------------------------------------+
 |  9       8       1234    |  7       1345    1235    |  6       145     345     |
 |  5       1246    12346   |  2348    9       1238    |  1234    1478    3478    |
 |  123     124     7       |  2348    13458   6       |  123459  14589   34589   |
 |--------------------------+--------------------------+--------------------------|
 |  4       12679   1268    |  5       168     1789    |  19      3       689     |
 |  1678    3       168     |  4689    1468    1789    |  1459    145689  2       |
 |  168     169     5       |  34689   2       1389    |  7       14689   4689    |
 |--------------------------+--------------------------+--------------------------|
 |  1236    12456   9       |  236     7       235     |  8       456     3456    |
 |  3678    4567    3468    |  1       3568    3589    |  3459    2       345679  |
 |  23678   2567    2368    |  23689   3568    4       |  359     5679    1       |
 +--------------------------------------------------------------------------------+
 # 175 eliminations remain

 ### -1234- QExocet   Base = r3c12   Target = r2c7==r1c6,r1c5

 ### -1345- QExocet   Base = r1c89   Target = r2c3==r3c4,r3c5


_

[champagne]

Hi Danny,

Relying on your findings, I see r3c5=5 forcing '5' in the base of your second exocet.

This is enough to be sure that the other digits can not be shared in both exocets.

Surely enough to do more eliminations, but I would not call it a double exocet, despite the similarities.

Nice puzzle, quickly killed if I am correct

[Leren]

Hi Danny,

The problem I see is that one of the three common base cell values (3) doesn't end up in the base cells solution. If you could prove up front that r1c9 and r3c1 <> 3 then you could then make Double Exocet style eliminations for the consequently known base cell values 1245.

Leren

[FLOCKMAN]

Here is another example solvable via a double Exocet

1.....7.9.5...91...98....5....3......4..2.3..8....1.9...26.........4...69....5.7.

[JC Van Hay]

Here is another example solvable via a double Exocet

1.....7.9.5...91...98....5....3......4..2.3..8....1.9...26.........4...69....5.7.

! !

[daj95376]

It seems a waste to have a double JExocet and not make an attempt at finding its eliminations.

My solver finds the patterns and flags secondary equivalences, but it doesn't try to find the eliminations. Those I found manually ... and so there may be errors. The hardest part was finding common eliminations in what DPB calls the non-S cells. Since both JExocets share [c6] for one of their target cells, I knew that this fell under one of his special patterns for non-S cell eliminations.

Code: Select all

 +--------------------------------------------------------------------------------+
 |  1      B236    B346     |  245-8   356-8  q346-28  |  7       8-2346  9       |
 | r2346-7  5      q346-7   |  78-24   78-36   9       |  1      Q2346-8 R234-8   |
 |  7-2346  9       8       |  124-7   136-7  R234-67  | b246     5      b234     |
 |--------------------------+--------------------------+--------------------------|
 |  2567    17-26   1579-6  |  3       5789-6  4678    |  58-246  12468   1578-24 |
 |  567     4       1579-6  |  5789    2       678     |  3       168     1578    |
 |  8       2367    3567    |  457     567     1       |  2456    9       2457    |
 |--------------------------+--------------------------+--------------------------|
 |  3457    178-3   2       |  6       1789-3  378     |  589-4   1348    158-34  |
 |  357     178-3   157-3   |  1789-2  4       2378    |  589-2   1238    6       |
 |  9       1368    1346    |  128     138     5       |  248     7       12348   |
 +--------------------------------------------------------------------------------+
 # 177 eliminations remain


 ### -2346- QExocet   Base = r1c23 <23>   Target = r2c8,r3c6==r2c9

 -8 r2c89 & -67 r3c6 & -7 r3c45


 ### -2346- QExocet   Base = r3c79 <46>   Target = r1c6==r2c3,r2c1

 -28 r1c6 & -7 r2c13 & -8 r1c45


 *** double QExocet

 -2346 r1c8,r2c45,r3c1

 -2    r4c279,r8c47
 - 3   r7c259,r8c23
 -  4  r47c79
 -   6 r4c2357,r5c3


I've noticed, from past experience, that it makes more sense to perform double JExocet eliminations first in cells where all four values would be eliminated. In the case of this puzzle, that's r1c8, r2c45, and r3c1. Now, solve for any Singles in these cells. All that's left to derive in [band 1] is -2r1c6 and -6r3c6.

_

[JC Van Hay]

... All that's left to derive in [band 1] is -2r1c6 and -6r3c6.

They are included in the basic steps after the exclusions by the double exocet : HT(246)r4c168, NT(178)r478c2, NT(589)r478c7, Swordfish(2R48C2)-2r1c6, XWing(6C57)-6r3c6, Swordfish(8C267).