The New Sudoku Players' Forum

[mith]

In Sudoku Explorer, are you turning on all the solving techniques? It defaults to certain things being turned off, and resets when you exit SE. So you need to turn on the techniques again when you start up SE. I only turn on the optional techniques if SE say's it can't analyze a puzzle.

I use 1.2.1 mode to rate these puzzles initially, but I would be very surprised if this one rated much lower with all of SE's techniques. SE doesn't implement everything (in particular the various Exocet patterns being discussed).

But if you want to try turning everything on in SE and stepping through it, be my guest.

[SpAce]

Hi Cenoman,

Tried to find something funny, but failed !

Don't worry about it. You're doing a great job trying to keep the patterns as close to the definitions as possible. I definitely see value in that because otherwise pattern names stop meaning anything. In this particular case, I can't do the same myself because I don't really look at [J/S]Exocets through David's definitions (I find them too complicated), so I can't claim to even know them accurately. For me the important part is that the Exocet property is there and easily verified, but it means I can't say with confidence if a pattern is a [J/S]Exocet or a generic one.

In the light of these recent extensions it might be a good idea to review those definitions anyway. Personally I'd keep anything that can be verified with single-digit techniques in the [J/S] family, possibly qualified with fishy terms when applicable. Then the truly generic ones would be those that require multi-digit logic to prove (I might call them Alien Exocets).

Of course the standard JExocet should still have a special place, because it has many more inferences available that are missing from the extended patterns. However, as we just saw, even the basic Exocet inferences can be enough to crack a tough puzzle, so they're definitely not useless (unlike David suspected).

I confused blue's post and David's Compendium.

Thanks! I found it. One of David's definite strengths is the ability to find such tell-tale signs that help a manual solver to narrow the search. He's provided such good tips not only for [J/S]Exocets but also SK-Loops and (other) MSLS. That's valuable information.

[yzfwsf]

Code: Select all

.------------------------.-------------------------.--------------------.
| 34568-9 4678-9 345678-9| 567-9  2567-9  T9-25678 | 278-9  2678   1    |
| 568-9   678-9  2       | 567-9  567-9   1        | 9-78   3      4    |
| B69-8   1      B679-8  | 3      4       2678-9   | 5      267-8  26-89|
:------------------------+-------------------------+--------------------:
| 134589  489    34589   | 2      1579    579      | 6      1458   389  |
| 145689  2      45689   | 14569  1569    3        | 1489   1458   7    |
| 7       469    34569   | 14569  8       569      | 12349  1245   239  |
:------------------------+-------------------------+--------------------:
| 468     3      4678    | 1567   12567   2567     | 12478  9      268  |
| 469     5      4679    | 8      123679  2679     | 12347  12467  236  |
| 2       T679-8 1       | 679    3679    4        | 378    678    5    |CL3(69)
'------------------------'-------------------------'--------------------'
          CLB                             CL1                      CL2

CH for 6:r6b89;7 is locked member, so no need check cover house;CH for 8:r47;CH for 9:r46b8

[SpAce]

Hi yzfwsf,

grid: Show

Code: Select all

.------------------------.-------------------------.--------------------.
| 34568-9 4678-9 345678-9| 567-9  2567-9  T9-25678 | 278-9  2678   1    |
| 568-9   678-9  2       | 567-9  567-9   1        | 9-78   3      4    |
| B69-8   1      B679-8  | 3      4       2678-9   | 5      267-8  26-89|
:------------------------+-------------------------+--------------------:
| 134589  489    34589   | 2      1579    579      | 6      1458   389  |
| 145689  2      45689   | 14569  1569    3        | 1489   1458   7    |
| 7       469    34569   | 14569  8       569      | 12349  1245   239  |
:------------------------+-------------------------+--------------------:
| 468     3      4678    | 1567   12567   2567     | 12478  9      268  |
| 469     5      4679    | 8      123679  2679     | 12347  12467  236  |
| 2       T679-8 1       | 679    3679    4        | 378    678    5    |CL3(69)
'------------------------'-------------------------'--------------------'
          CLB                             CL1                      CL2

CH for 6:r6b89;7 is locked member, so no need check cover house;CH for 8:r47;CH for 9:r46b8

Very cool! Thanks for sharing this.

However, I think that's way too little information to understand what's going on. Most critically, it doesn't tell why 8 is an invalid base candidate or why 9 must be true in r1c6. I can see why both happen, but I don't think it's fully depicted here. In fact, I can't see how your choice of CLs and CHs alone prove it.

The way I see it:

grid: Show

Code: Select all

  \b1:6789
           *6789                              *689               \6      *89
.---------------------------.--------------------------.------------------------.
|  345689  *46789   3456789 |  5679   25679  \t9-25678 |  2789   *2678    1     | \6
|  5689    *6789    2       |  5679   5679     1       |  789     3       4     |      *B3:6
| b69-8     1      b679-8   |  3      4       *2678-9  |  5      *2678   *268-9 | \689
:---------------------------+--------------------------+------------------------:
|  134589  *489     34589   |  2      1579    *579     |  6       1458   *389   | \89
|  145689   2       45689   |  14569  1569     3       |  1489    1458    7     |
|  7       *469     34569   |  14569  8       *569     |  12349   1245   *239   | \69
:---------------------------+--------------------------+------------------------:
|  468      3       4678    |  1567   12567   *2567    |  12478   9      *268   |
|  469      5       4679    |  8      123679  *2679    |  12347   12467   236   |
|  2      \t67-89   1       | *679   *3679     4       | *378    *678     5     | *689
'---------------------------'--------------------------'------------------------'
                                  \b8:69                    \b9:8

15x18 {3N13 689R9 6789C2 689C6 89C9 6B3 \ 6r1 689r3 89r4 69r6 6c8 6789b1 69b8 8b9 1n6 9n2}

6 can only be true in target r9c2 (proved non-trivially).
7 can only be true in target r9c2 (proved trivially).
8 must be true in both targets; hence it's an invalid base candidate.
9 is the only valid base candidate capable of being true in either target; hence it must be true in r1c6 and false in the other.

=> -8 r3c13,r1c6,r9c2; +9 r1c6,r3c13; -9 r9c2; btte

digit 6: Show

Code: Select all

        6r3          6r1
  9n2   6b1    6r6   6c8   6b8
==================================
       6r3c13                     | *3N13
 6r9c2 6r12c2 6r6c2               |  6C2
       6r3c89       6r1c8         |  6B3
       6r3c6  6r6c6 6r1c6 6r78c6  |  6C6
 6r9c2              6r9c8 6r9c45  |  6R9
==================================
 6r9c2

digit 7: Show

Code: Select all

  9n2   7b1
==================================
       7r3c13                     | *3N13
 7r9c2 7r12c2                     |  7C2
==================================
 7r9c2

digit 8: Show

Code: Select all

 1n6    8r3
 9n2    8b1    8r4   8b9
==================================
       8r3c13                     | *3N13
 8r1c6 8r3c6                      |  8C6
 8r9c2 8r12c2 8r4c2               |  8C2
       8r3c9  8r4c9 8r7c9         |  8C9
 8r9c2              8r9c78        |  8R9
==================================
&8r1c6
&8r9c2

Note: 8b9, 8R9 could be 8r7, 8B7.

digit 9: Show

Code: Select all

  1n6   9r3
  9n2   9b1    9r4   9r6   9b8
==================================
       9r3c13                     | *3N13
 9r9c2 9r12c2 9r4c2 9r6c2         |  9C2
       9r3c9  9r4c9 9r6c9         |  9C9
 9r1c6 9r3c6  9r4c6 9r6c6 9r8c6   |  9C6
 9r9c2                    9r9c45  |  9R9
==================================
|9r1c6
|9r9c2

Is there another way to see the logic?

If not, I can't see how your choice of bases (CLs) and covers (CHs) locks 6 in r9c2, which is necessary to force 9 into r1c6. I needed box 6 as an extra base and r1 and c8 as extra covers to do that. If (only) C9 and b9 are used for 6 as your model suggests, then both targets remain options for digit 6 just like for digit 9:

Code: Select all

  1n6   6b2
  9n2   6r3    6r6   6b9   6b8
==================================
       6r3c13                     | *3N13
 6r9c2 6r12c2 6r6c2               |  6C2
       6r3c9        6r78c9        |  6C9
 6r1c6 6r3c6  6r6c6       6r78c6  |  6C6
 6r9c2              6r9c8 6r9c45  |  6R9
==================================
|6r1c6
|6r9c2

Am I blind to something?

[yzfwsf]

Is there another way to see the logic?
If not, I can't see how your choice of bases (CLs) and covers (CHs) locks 6 in r9c2, which is necessary to force 9 into r1c6. I needed box 6 as an extra base and r1 and c8 as extra covers to do that. If (only) C9 and b9 are used for 6 as your model suggests, then both targets remain options for digit 6 just like for digit 9:

6 can't true in r1c6 due to the mirror node r2c789 have no 6

[Cenoman]

Hi yzfwsf and SpAce,

Thanks for sharing this.

I add my thanks to SpAce's

A minor comment about the exocet pattern identification:
+8r3c13 -> -8 r3c6 -> +8r1c6, so why not write for digit 8, the same as for digit 7: '8 is a locked member, so no need to check for cover houses' ?

Once the exocet pattern is proved, simple rationales (including contradiction inferences) are suitable to me, for a valid puzzle solution:
1) -25 r1c6 (non base digits in target cell)
2) 8 is a false base digit:
+8r3c13 -> -8 r3c689 -> +8r1c6; therefore -8r9c2 -> (+8r4c2 -> -8r4c9) And (+8r9c78 -> -8r7c9); contradiction (c9 void of 8) => -8 r3c13, -r1c6, r9c2
3) 7 can be true only in target r9c2:
+7r3c3 -> -7r12c2 -> +7r9c2; => -7 r1c6
4) 6 can't be true in target r1c6
+6r3c13 -> -6r3c89 -> +6r1c8; => -6 r1c6 then +9r1c6 (True base digit), +9r3c13, -9r9c2; ste

Otherwise, above moves 1. & 2. and one AIC solve the puzzle:
e.g. after -8 r3c13, -r1c6, r9c2, the following AIC proves that 6 is a true base digit, false in target r1c6.

Hidden Text: Show

Code: Select all

 +----------------------------+-------------------------+------------------------+
 |  3459-6  g4789-6  34579-6  |  5679    2       679    |  789     678     1     |
 |  59-6    g789-6   2        |  5679    5679    1      |  789     3       4     |
 |ha69       1      h679      |  3       4       8      |  5       27-6   b29-6  |
 +----------------------------+-------------------------+------------------------+
 |  13459    49      3459     |  2       1579    579    |  6       145     8     |
 |  145689   2       45689    |  14569   1569    3      |  149     145     7     |
 |  7        469     4569     |  14569   8       569    | c12349   1245   c239   |
 +----------------------------+-------------------------+------------------------+
 |  468      3       4678     |  1567    1567    2567   | d1247    9       26    |
 |  469      5       4679     |  8       13679   2679   | d12347  d12467   236   |
 |  2       f679     1        |  679     3679    4      | e378    e678     5     |
 +----------------------------+-------------------------+------------------------+

(6=9)r3c1 - r3c9 = (9-32)r6c79 = (2-147)b9p145 = r9c78 - r9c2 = r12c2 - (7=96)r3c13 => -6 r3c89, r1c123, r2c12; => +6 r1c8; -6 r1c6, +6r9c2 (True base digit); lclste

[SpAce]

6 can't true in r1c6 due to the mirror node r2c789 have no 6

Ah, yes, of course. Thanks! I never really doubted that you'd have a valid explanation My mistake was that I didn't even consider such derived inferences in a non-Junior context, but you're right that 'mirror node' is still a valid concept for one of the targets. I'll try to remember next time. Nevertheless, it would be helpful to mention that, or at least mark it in the grid:

Code: Select all

.------------------------.-------------------------.--------------------.
| 34568-9 4678-9 345678-9| 567-9  2567-9  T9-25678 | 278-9  2678   1    |
| 568-9   678-9  2       | 567-9  567-9   1        | M9-78  M3     M4   |
| B69-8   1      B679-8  | 3      4       2678-9   | 5      267-8  26-89|
:------------------------+-------------------------+--------------------:


As I recently said elsewhere, I think all but the direct Exocet inferences should be accompanied with an explanation. Mirror inferences are probably the simplest of the derived kind, but even those are easy to forget (as we just saw). The above grid marking should be enough of a reminder for someone well-versed with the JExocet compendium, but for someone who only knows or remembers the basic Exocet property, it's not.

Nothing is too little for anyone. It simply can't be expected that even those familiar with the derived JE concepts can remember them well enough to make the necessary connections automatically, unless they use JEs all the time (who does?). I'm pretty sure most of the readers would rather have more than less information when it comes to Exocet inferences. (That said, personally I probably learned the most by having to work it out myself and failing to see the simpler explanation. I don't really mind as long as I eventually get the right answer!)

Most importantly, thanks for another awesome and educational example! Keep them coming. I think Exocets, especially these non-trivial kinds, are among the most interesting patterns. I'm pretty sure very few people have too much experience with them.

[yzfwsf]

'8 is a locked member, so no need to check for cover houses' ?

Because check regular junior rule(<=2 cover houses) first.

Another one for fun

Code: Select all

98.76.5..75...8.....6......4......7...8..79..........3.27..96....932.7.....1...2.

Single and Locked Candidates to current PM

Code: Select all

.------------------.----------------------.---------------------.
| 9      8    1234 | 7       6      1234  | 5      134    124   |
| 7      5    1234 | 249     1349   8     | 1234   13469  1246  |
| 123    134  6    | 2459    13459  12345 | 12348  13489  7     |
:------------------+----------------------+---------------------:
| 4      9    1235 | 2568    1358   1235  | 128    7      12568 |
| 12356  136  8    | 2456    1345   7     | 9      1456   12456 |
| 1256   7    125  | 245689  14589  1245  | 1248   14568  3     |
:------------------+----------------------+---------------------:
| 135    2    7    | 458     458    9     | 6      1345   145   |
| 1568   146  9    | 3       2      456   | 7      1458   1458  |
| 3568   346  345  | 1       7      456   | 348    2      9     |
'------------------'----------------------'---------------------'

[totuan]

Provide another puzzle:

Code: Select all

........1..2..1.34.1.34.5.....2..6...2...3..77...8.....3.....9..5.8.....2.1..4..5

Thanks for the puzzle.
Based on your hint and Xsudo (v.99) I found as below - don’t know how Xsudo works

Code: Select all

yzfw 60 Candidates, Raw Rank = 11 (linksets - sets)
     18 Sets = {689R2 689R9 6789C2 69C6 689C9 3N13 8B2}
     29 Links = {6r36 8r34 9r346 249n2 13n6 2n7 46n9 6b1289 7b1 8b169 9b123468}
     62 Eliminations, 7 Assignments -->
     r1c123457<>9, r4c13569<>9, r6c23467<>9, r1c6<>25678, r2c1245<>9, r3c1389<>8,
     r4c1238<>8, r3c6<>2679, r1c123<>8, r5c137<>9, r257c7<>8, r2c27<>7,
     r8c56<>9, r46c9<>3, r9c2<>89, r1c3<>7, r2c2<>6, r2c1<>8, r3c9<>9,
     r4c2<>4, r5c8<>8, r6c9<>2, r7c9<>8,
     r1c6=9, r2c2=8, r2c7=9, r3c6=8, r4c9=8, r4c2=9, r6c9=9   


P/s: I don’t know how to post images

totuan

[totuan]

Another one for fun

Code: Select all

98.76.5..75...8.....6......4......7...8..79..........3.27..96....932.7.....1...2.

Again, based on Xsudo:

Hidden Text: Show

Code: Select all

yzf-2  58 Candidates, Raw Rank = 10 (linksets - sets)
     18 Sets = {1234C3 1234C6 1234C7 1N89 1234B1}
     28 Links = {1r1346 2r1346 3r1349 4r13689 3n2 29n3 8n6 6n7 1234b3 4b6 4b8}
     34 Eliminations, 4 Assignments -->
     r3c45678<>4, r2c3789<>4, r6c4568<>4, r8c289<>4, r9c267<>4, r6c7<>128,
     r1c36<>4, r5c89<>4, r7c45<>4, r3c2<>13, r8c6<>56, r9c3<>35,
     r3c2=4, r6c7=4, r8c6=4, r9c3=4 

More detail:

yzf-2  58 Candidates, Raw Rank = 10 (linksets - sets)
     18 Sets = {1234C3 1234C6 1234C7 1N89 1234B1}
     28 Links = {1r1346 2r1346 3r1349 4r13689 3n2 29n3 8n6 6n7 1234b3 4b6 4b8}
     34 Eliminations, 4 Assignments --> [4B1*4r3*3n2] =>
     r3c2=4, [4C7*4r6*6n7*4b6] => r6c7=4, [4C6*4r8*8n6*4b8] => r8c6=4, [4C3*4r9*9n3] => r9c3=4, (4C3*4B1*4r1) =>
     r1c3<>4, (4C6*4r1) => r1c6<>4, (4C3*4B1*2n3) => r2c3<>4, (4C7*4b3) => r2c7<>4, (4b3) => r2c8<>4, (4b3) => r2c9<>4, (1B1*1r3*3n2) =>
     r3c2<>1, (3B1*3r3*3n2) => r3c2<>3, (4r3) => r3c4<>4, (4r3) => r3c5<>4, (4C6*4r3) => r3c6<>4, (4C7*4r3*4b3) => r3c7<>4, (4r3*4b3) =>
     r3c8<>4, (4b6) => r5c8<>4, (4b6) => r5c9<>4, (4r6) => r6c4<>4, (4r6) => r6c5<>4, (4C6*4r6) => r6c6<>4, (1C7*1r6*6n7) =>
     r6c7<>1, (2C7*2r6*6n7) => r6c7<>2, (6n7) => r6c7<>8, (4r6*4b6) => r6c8<>4, (4b8) => r7c4<>4, (4b8) => r7c5<>4, (4r8) =>
     r8c2<>4, (8n6) => r8c6<>5, (8n6) => r8c6<>6, (4r8) => r8c8<>4, (4r8) => r8c9<>4, (4r9) => r9c2<>4, (3C3*3r9*9n3) =>
     r9c3<>3, (9n3) => r9c3<>5, (4C6*4r9*4b8) => r9c6<>4, (4C7*4r9) => r9c7<>4

totuan

[Cenoman]

Another one for fun
98.76.5..75...8.....6......4......7...8..79..........3.27..96....932.7.....1...2.

There is a great temptation to find a nice Double JE (1234)r3c12,r1c6,r2c7; r1c89,r2c3,r3c6 but... it fails !
Surprising, isn't it ? there would be no fun !

Hidden Text: Show

Code: Select all

 +-----------------------+---------------------------+--------------------------+
 |  9       8     1234   |  7        6      T1234    |  5      b134    b124     |
 |  7       5    t1234   |  249      1349    8       | T1234    13469   1246    |
 | B123    B134   6      |  2459     13459  t12345   |  12348   13489   7       |
 +-----------------------+---------------------------+--------------------------+
 |  4       9     1235   |  2568     1358    1235    |  128     7       12568   |123
 |  12356   136   8      |  2456     1345    7       |  9       1456    12456   |
 |  1256    7     125    |  245689   14589   1245    |  1248    14568   3       |12 4
 +-----------------------+---------------------------+--------------------------+
 |  135     2     7      |  458      458     9       |  6       1345    145     |
 |  1568    146   9      |  3        2       456     |  7       1458    1458    |   4
 |  3568    346   345    |  1        7       456     |  348     2       9       |  34
 +-----------------------+---------------------------+--------------------------+
                  CLB                        CL1        CL2
                  CL1                        CL2        CLb
The 4s are present in three lines. No way to an extra cross-line.

So, starting from that point, all you need is to find an alternate target cell in cross lines.
Two possibilities, one with each of both near miss JE2:
With base (1234)r1c89, keep target r2c3, try target pair r89c6 (locked 6 r89c6); (BTW, yzfwsf solver's choice)

Code: Select all

                                     b2: CH 1234
 +-----------------------+---------------------------+--------------------------+
 |  9       8     1234   |  7        6       1234    |  5      b134    b124     |
 |  7       5    t1234   | S249     S1349    8       |  1234    13469   1246    | CL3
 |  123     134   6      |  2459     13459  S12345   |  12348   13489   7       | 
 +-----------------------+---------------------------+--------------------------+
 |  4       9    S1235   |  2568     1358   S1235    | S128     7       12568   | 123
 |  12356   136   8      |  2456     1345    7       |  9       1456    12456   |
 |  1256    7    S125    |  245689   14589  S1245    | S1248    14568   3       | 12 4
 +-----------------------+---------------------------+--------------------------+
 |  135     2     7      |  458      458     9       |  6       1345    145     |
 |  1568    146   9      |  3        2      t456     |  7       1458    1458    |     
 |  3568    346  S345    |  1        7      t456     | S348     2       9       |   34
 +-----------------------+---------------------------+--------------------------+
                  CL1                        CL2        CLb
CH for 1 & 2: r46b2; CH for 3: r49b2; CH for 4: r69b2

With base (1234)r3c12, keep target r1c6, try target pair r89c6 (locked 6 r89c6)

Code: Select all

                                                               b3: CH 1234
 +-----------------------+---------------------------+--------------------------+
 |  9       8     1234   |  7        6      T1234    |  5      S134    S124     | CL3
 |  7       5     1234   |  249      1349    8       | S1234    13469   1246    |
 | B123    B134   6      |  2459     13459   12345   |  12348   13489   7       |
 +-----------------------+---------------------------+--------------------------+
 |  4       9    S1235   |  2568     1358   S1235    | S128     7       12568   | 123
 |  12356   136   8      |  2456     1345    7       |  9       1456    12456   |
 |  1256    7    S125    |  245689   14589  S1245    | S1248    14568   3       | 12 4
 +-----------------------+---------------------------+--------------------------+
 |  135     2     7      |  458      458     9       |  6       1345    145     |
 |  1568    146   9      |  3        2      T456     |  7       1458    1458    |
 |  3568    346  S345    |  1        7      T456     | S348     2       9       |   34
 +-----------------------+---------------------------+--------------------------+
                  CLB                        CL1        CL2
CH for 1 & 2: r46b3; CH for 3: r49b3; CH for 4: r69b3
Note the cross-aligned targets !

Whichever choice is made, the result is the same.
-5 r89c6 (non base digit in target cells, 6 is the only possible NB digit)
4 is the only possible base digit in target cells r89c6 => 4 is a true base digit, false in the other target; -4 r12c3 & other eliminations; 20 placements and basics.

Code: Select all

 +-------------------+----------------------+-------------------+
 |  9     8    123   |  7     6      123    |  5     4     12   |
 |  7     5    123#  |  24*   134    8      |  123*  9     6    |
 |  3-2   4    6     |  259#  1359   1235   |  123#  8     7    |
 +-------------------+----------------------+-------------------+
 |  4     9    235   |  6     135    1235   |  12*   7     8    |
 |  236#  1    8     |  245*  345    7      |  9     56    25*  |
 |  26    7    25    |  89    89     125    |  4     156   3    |
 +-------------------+----------------------+-------------------+
 |  1     2    7     |  58    58     9      |  6     3     4    |
 |  8     6    9     |  3     2      4      |  7     15    15   |
 |  5     3    4     |  1     7      6      |  8     2     9    |
 +-------------------+----------------------+-------------------+

5-link oddagon (2)r25,c47,b6 having four guardians (#): (2)r2c3,r3c47,r5c1 => -2 r3c1; ste
@yzfwsf: thank you ! that was real fun !

[SpAce]

I'm probably doing something wrong in the first step, so feel free to correct me...

Step 1.

Code: Select all

                *                      *        *
.---------------------.-----------------------.----------------------.
|  9      8      1234 | 7       6      m123-4 |  5      134    124   |
|  7      5      1234 | 249     1349    8     | t123-4  13469  1246  |
| b123   b4-13   6    | 2459    13459   12345 |  12348  13489  7     |
:---------------------+-----------------------+----------------------:
|  4      9     *1235 | 2568    1358   *1235  | *128    7      12568 | \123.
|  12356  136    8    | 2456    1345    7     |  9      1456   12456 |
|  1256   7     *125  | 245689  14589  *1245  | *1248   14568  3     | \12.4
:---------------------+-----------------------+----------------------:
|  135    2      7    | 458     458     9     |  6      1345   145   |
|  1568   146    9    | 3       2      t4-56  |  7      1458   1458  |
|  3568   346   *345  | 1       7      *456   | *348    2      9     | \..34
'---------------------'-----------------------'----------------------'

SExocet (1234)r3c12,r2c7,r8c6 => +4 r3c2,r8c6; -4 r1c6,r2c7


4 is the only base digit possible in target r8c6, so it must be true there and in the base cell, and false in the other target and its mirror.


I'm not confident about that at all. My initial logic was the same as with the first puzzle, ie. using impossible JExocets (missing mirror) as DPs:

Code: Select all

![JE2:(123|1234)r3c12,r1c6,r2c7] = (4,4)r3c2,r8c6 => +4 r3c2,r8c6

Step 2 (unnecessary but I used it, so I list it): Show

Code: Select all

.-------------.---------------------.--------------.
| 9    8  123 | 7     6       123   | 5    4    12 |
| 7    5  123 | 24    134     8     | 123  9    6  |
| 23   4  6   | 59+2  59+13  #5-123 | 123  8    7  |
:-------------+---------------------+--------------:
| 4    9  235 | 6     135     1235  | 12   7    8  |
| 236  1  8   | 245   345     7     | 9    56   25 |
| 26   7  25  | 89+   89+     125   | 4    156  3  |
:-------------+---------------------+--------------:
| 1    2  7   | 58+   58+     9     | 6    3    4  |
| 8    6  9   | 3     2       4     | 7    15   15 |
| 5    3  4   | 1     7       6     | 8    2    9  |
'-------------'---------------------'--------------'

BUG-Lite+3 (589)r367c45 => +5 r3c6 (external)


Step 3 (or 2): Show

Code: Select all

.---------------.-----------------.---------------.
| 9     8  13+2 |  7    6    *123 | 5    4    *12 |
| 7     5  123  | *24   134   8   | 123  9     6  |
| 3-2   4  6    |  9+2  139   5   | 123  8     7  |
:---------------+-----------------+---------------:
| 4     9  235  |  6    135   123 | 12   7     8  |
| 36+2  1  8    | *245  345   7   | 9    56   *25 |
| 26    7  25   |  89   89    12  | 4    156   3  |
:---------------+-----------------+---------------:
| 1     2  7    |  58   58    9   | 6    3     4  |
| 8     6  9    |  3    2     4   | 7    15    15 |
| 5     3  4    |  1    7     6   | 8    2     9  |
'---------------'-----------------'---------------'

Oddagon+3 2r15b2 : (2)r1c3|r3c4|r5c1 => -2 r3c1; stte


--

@yzfwsf: thank you ! that was real fun !

Hear hear!