The New Sudoku Players' Forum

[jovi_al01]

gave myself a challenge to explore a certain deduction today and am incredibly pleased with this result! i hope you enjoy solving, and i look forward to your solutions!
my solution is one step (two steps if you include basics)!

Code: Select all

,-------,-------,-------,
| . 1 . | . 2 . | . 3 . |
| . . 4 | . . 5 | . . . |
| 6 . . | . . . | 7 . . |
+-------+-------+-------+
| 3 . . | 2 . . | . 1 . |
| 1 . . | . 3 . | . . 8 |
| . 8 . | . . 9 | . . 3 |
+-------+-------+-------+
| . . 7 | . . . | . . 5 |
| . . . | 6 . . | 4 . . |
| . 2 . | . 9 . | . 8 . |
'-------'-------'-------'

--jovi

[Leren]

Code: Select all

*--------------------------------------------------------------------------------*
| 5789    1       589      | 4789    2       4678     | 5689    3       469      |
| 2789    379     4        | 13789   1678    5        | 12689   269     1269     |
| 6       359     23589    | 13489   148     1348     | 7       2459    1249     |
|--------------------------+--------------------------+--------------------------|
| 3      T56-479  56-9     | 2       45678   4678     | 569     1       4679     |
| 1       45679   2569     |B5-47    3      B6-47     | 2569    245679  8        |
| 2457    8       256      | 1457    14567   9        | 56-2   T56-247  3        |
|--------------------------+--------------------------+--------------------------|
| 489     3469    7        | 1348    148     12348    | 12369   269     5        |
| 589     359     13589    | 6       1578    12378    | 4       279     1279     |
| 45      2       1356     | 13457   9       1347     | 136     8       167      |
*--------------------------------------------------------------------------------*

J Exocet: r5c4 r5c6 r4c2 r6c8 (4567) : Target cell eliminatons : - 9 r4c2, - 2 r6c8; Mirror node eliminations : - 9 r4c3, - 47 r6c8, - 47 r4c2, - 2 r6c7; Reduce Base cells digits to Target cells digits : - 47 r5c46; btte.

Leren

[jovi_al01]

my solution as well-- thank you!

[Leren]

You can also make 6 S cell eliminations - 5 r3c3, r8c13, - 6 r2c79, r7c7, but it's still btte

Leren

[RSW]

By my count, there are 32 eliminations in total from the JE:

Rule 2: Base candidates 4,7 don't appear in both target and mirror node, and are therefore invalid.
-47r5c46
Thus 5,6 are true base digits.
Rule 3: Non-base candidates 4,7,9 are invalid in target cells.
-479r4c2, -247r6c8.
Rule 7: If one cell of a mirror node has only non-base candidates, then all non-base candidates can be removed from the other cell of the mirror node.
-2r6c7, -9r4c3.
Rule 10a: True base digits 5,6 can be removed from all cells in sight of both base cells.
-5r4c5, -5r5c2, -5r5c3, -5r6c4, -5r5c7, -5r6c5, -5r5c8, -6r4c5, -6r5c2, -6r4c6, -6r5c3, -6r5c7, -6r6c5, -6r5c8.
Rule 10b: True base digits 5,6 can be removed from cells in sight of both target cells.
-5r4c7, -5r6c1, -5r6c3, -6r4c7, -6r4c9, -6r6c3.

But it's still btte.

(Edit: with these plus Leren's S-cell eliminations, I believe it's now stte.)

[Leren]

I just checked by hand and I think it's stte without the S cell eliminations, but still they should be part of the Exocet move.

There must be a Rule somewhere in the J Exocet Compendium that says that when a digit is known to be in the Exocet solution its S cell eliminations can be made.

Leren

[denis_berthier]

Mirror node eliminations : - 9 r4c3, - 47 r6c8, - 47 r4c2, - 2 r6c7; Reduce Base cells digits to Target cells digits : - 47 r5c46; btte.

Hi Leren,
Is there any reference to these additional eliminations in the (very long) J-Exocet thread?

[RSW]

Mirror node eliminations : - 9 r4c3, - 47 r6c8, - 47 r4c2, - 2 r6c7; Reduce Base cells digits to Target cells digits : - 47 r5c46; btte.

Is there any reference to these additional eliminations in the (very long) J-Exocet thread?

These are rules 2 & 3 in David P. Bird's treatise.

[yzfwsf]

I adjusted the calling sequence of the JE elimination rule and got the following results .JE->stte

Hidden Text: Show

Code: Select all

Junior Exocet:Base Cells-r5c4,r5c6;Target Cells-r4c2,r6c2,r4c8,r6c8;Cross Cells-r123789c258   
Target Cells Check: r4c2<>9,r6c8<>2
Mirror Check:r4c2<>47,r6c7<>2,r6c8<>47,r4c3<>9
True Base digits seen by base cells or both targets: r4c5<>5,r4c7<>5,r5c2<>5,r5c3<>5,r5c7<>5,r5c8<>5,r6c1<>5,r6c3<>5,r6c4<>5,r6c5<>5,r9c4<>5r1c6<>6,r4c5<>6,r4c6<>6,r4c7<>6,r4c9<>6,r5c2<>6,r5c3<>6,r5c7<>6,r5c8<>6,r6c3<>6,r6c5<>6
True Base digits in non-'S' cells: r3c3<>5,r8c1<>5,r8c3<>5r2c7<>6,r2c9<>6,r7c7<>6
Compatibility Test:r5c46,r4c2,r6c8<>4,r5c46,r4c2,r6c8<>7
Base disappear from both targets(opposite mirror nodes): r5c4<>47 r5c6<>47 

[Leren]

denis-berthier Wrote : Is there any reference to these additional eliminations in the (very long) J-Exocet thread?

Hi denis. Lot's of cross posting going on here. I haven't looked at the compendium for ages. You should be able to download it at various pages from here.

In the meantime why don't I just cut to the chase and just show it works in this example.

Code: Select all

*--------------------------------------------------------------------------------*
| 5789    1       589      | 4789    2       4678     | 5689    3       469      |
| 2789    379     4        | 13789   1678    5        | 12689   269     1269     |
| 6       359     23589    | 13489   148     1348     | 7       2459    1249     |
|--------------------------+--------------------------+--------------------------|
| 3      T4567    56-9     | 2       45678   4678     | 569     1       4679     |
| 1       45679   2569     |B457     3      B467      | 2569    245679  8        |
| 2457    8       256      | 1457    14567   9        | 256    T56-47   3        |
|--------------------------+--------------------------+--------------------------|
| 489     3469    7        | 1348    148     12348    | 12369   269     5        |
| 589     359     13589    | 6       1578    12378    | 4       279     1279     |
| 45      2       1356     | 13457   9       1347     | 136     8       167      |
*--------------------------------------------------------------------------------*

Exocet 1: r5c4 r5c6 r4c2 r6c8 4567 ; r4c3==r6c8 ; Eliminate non-common candidates.

Here is how it works. We know that in the J Exocet, the values in the base cells must appear in the target cells. Let's look at the mirror node in r4c123, and suppose that the solution for the J Exocet was X and Y in the base cells, X in r4c2 and Y in r5c8, where X and Y can be any two of 4567. Now note that there is a non-base digit 3 in r4c1. Suppose r4c2 is X and r6c8 is Y.

Where can Y go in Box 4 ? It can't go in r4c1, which is 3, it can't go in r4c2, which is X. It can't go in r5c123 because X and Y are in the base cells r5c46. It can't go in r6c123 because r6c8 is Y.

So there is only one place it can go in Box 4 - in r4c3. Cool, huh ? We say r4c3==r6c8 to mean that non-target mirror node cell r4c3 will eventually hold the same base digit that is in the opposite target cell r6c8.

The net result of this is that you can remove all non-common digits from those two cells. There is one extra trick that does not appear here. Suppose r4c1 was not solved as 3, but r4c13 contained the only two 3s in Row 4.

In that case, you know that one of r4c13 will be 3. The net result of this case is that you can remove all non-common digits from r4c13 and r6c8, but not the linking digit 3 from r4c13.

Thus endeth the lesson. Hopefully that will save you a lot of downloading files and reading.

Leren

[denis_berthier]

Mirror node eliminations : - 9 r4c3, - 47 r6c8, - 47 r4c2, - 2 r6c7; Reduce Base cells digits to Target cells digits : - 47 r5c46; btte.

Is there any reference to these additional eliminations in the (very long) J-Exocet thread?

These are rules 2 & 3 in David P. Bird's treatise.

Thanks
I'll have a look at this. I followed the J-Exocet thread until there was a clean definition. I wasn't aware additional eliminations had been found later.

[denis_berthier]

Hi Leren, Thanks also; I'll take time to read this example.

[RSW]

I can confirm that the JE eliminations using David P. Bird's rules 2,3,7,10, are sufficient for a stte finish.
JE2: (4567)r5c46, r4c2 r6c8 => -2r6c7 -247r6c8 -47r5c46 -479r4c2 -5r6c14 -56r5c2378 -56r6c35 -56r4c57 -6r4c69 -9r4c3
Occasionally, for some unknown reason, my solver doesn't always check for singles after a complex move, and instead, skips ahead to basics.

To avoid having to refer back to referenced material, here are the "rules" that I refer to from David P. Bird's JE article:

1. A base candidate that is restricted to only one 'S' cell cover house is invalid and is false in the base mini-line and target cells.
2. Any base candidate that isn't capable of being simultaneously true in at least one target cell and its mirror node is false.
3. Any non-base candidate in a target is false.
4. A base digit in a target that must be true in the other target is false.
5. A base candidate that has a cross-line as an 'S' cell cover house must be false in the target cell in that cross-line
which may make other simple colouring eliminations available.
6. Any base candidate that can't be true in the mirror node for a target cell is false in the target cell.
7. If one mirror node cell can only contain non-base digits, the second one will be restricted to the base digits in
the opposite object cells.
8. If a mirror node contains only one possible non-base digit value, it is true in that node and false in the cells in
sight of it.
9. If a mirror node contains a locked digit, any other digits it contains of the same type (known base or non-base)
are false.
10. A known base digit is false in any cells in full sight of either a) both base cells or b) both target cells.
11. A known base digit, or one that can only occur once in the 'escape' cells in the cross-lines, is false in the non-'S'
cells in its cover houses.
12. For a known base digit, any digit instance that would prevent two of its 'S' cells being true is false.

In addition to these, based on some experiences with various puzzles, I've added a couple of additional rules to my solver. As with the other rules, they follow directly from the fundamental inferences of the exocet.

13. If one cell in a mirror node contains only non base candidates, then any base candidate, absent from the corresponding target, may be removed from the other mirror cell in that mirror node.

14. (Marek's rule) If, in a target box, a base candidate occurs only along the base line (and optionally in the target cell), then it is invalid in the baseline of the other target box, and invalid in the base box base line escape cell.