The New Sudoku Players' Forum

[sultan vinegar]

Danny, I disagree about there being only one ADJE. Suppose that (4)r7c3 and (12)r8c5 are all false. Then we would have a DJE with box singles extension on both candidates 3 and 4 in boxes 2 and 1 respectively, regardless of whether or not (4) occupies cell 8N5. (We could eliminate (4) from r3c5 as a secondary inference if this DJE were true). But wait, there's more! Suppose that (4)r67c7 and (12)r8c5 are all false. Then there's another DJE. But wait, there's more! Suppose (1)r46c37 and (24)r8c5 are all false. Then there's another DJE. There's heaps more too! Do we have to consider them all???

That’s why I think we need to be very careful when we use DJE in chains. Just writing:

Code: Select all

DJE – (4)r7c468

Isn’t good enough. Which DJE does DJE refer to? The one with box singles extension on candidate 3, or the one with box singles extension on candidates 3 and 4, or one of the other ones? Maybe even one that hasn't been invented yet???

Reading from left to right you would need to use memory to know that you meant the former.

Reading from right to left you have no idea what DJE by itself means. If it meant all of the possible DJE, then the inference would be invalid since you could have (4)r7c468 true and a true DJE with box singles extension on candidates 3 and 4, i.e. both true, so not a weak inference.

Provided that you are careful with which DJE you mean, then I agree with what Danny says:

*) if the DJE is true then all of its eliminations occur. Even subsets like -(4)r7c468 and -(124=56)r67c6.
*) If any eliminations associated with the DJE are assumed true, then it follows that the DJE must be false.

DPB, this is the bit that I don’t understand at all:

By setting the parameters you have for the DJE, you are only considering the case when r8c357 doesn't hold any (1) (2) or (4). This means this chain therefore proves nothing new.

If (124)r8c5 [r8c37 are givens], then read right to left and discover that (9)r7c4, so r7c4 <> 4. No consideration of exocets is required.

[David P Bird]

SV You're right, and I'm wrong. (Thanks for keeping me on my toes.)

My objection was that when you specified which partial fish cells belong to the DJE you considered, that by itself was insufficient to show that the elimination would also hold for all the other possible valid partial fish configurations.

Now you've brought me to realise what I was doing before, when I first noted that chain!
(9)r7c4 = (9 - 7)r7c5 = (7 - 124)r8c5
Proves that when (9)r7c4 is false so must the potential spoilers be false too. This would force the DJE with the specified partial fish true (with all other configurations being impossible). So (4)r7c4 will either be eliminated by (9)r7c4 or by the specified DJE.

The same consideration holds for my 4th chain and also yours that eliminates (9)r2c4,r3c9

DPB

[daj95376]

As for the weak link discussions:

*) if the DJE is true then all of its eliminations occur. Even subsets like -(4)r7c468 and -(124=56)r67c6.

*) If any eliminations associated with the DJE are assumed true, then it follows that the DJE must be false.

If anything, I was expecting discussions on the r-to-l reading of the strong link (7-124)r8c5 = DJE in DPB's loop. _ _

Your second bullet point doesn't hold with respect to the eliminations in the non JE bands because it's possible for (124)r8c5 and the DJE to be true together. For example if (4)r8c5 and (4)r7c7 were both true the DJE would be forced true as well. If (1)r8c5 and (1)r6c7 were both true it would depend on (1)r4c3 whether the DJE was true or false.

Hmmm!!! I must be missing something about the definition of a DJE. I though it meant that both JEs had to valid.

When the JE in base cells r3c12, w/target cells r1c5 and r2c7, are tested for r3c12={1,4}, each of your your scenarios prevent the JE from being valid for the value in the target cells. This leads to the DJE not being valid from my perspective.

[daj95376]

[Withdrawn and replaced with a later post.]

[daj95376]

[Withdrawn and replaced with a later post.]

[David P Bird]

As for the weak link discussions:

*) if the DJE is true then all of its eliminations occur. Even subsets like -(4)r7c468 and -(124=56)r67c6.

*) If any eliminations associated with the DJE are assumed true, then it follows that the DJE must be false.

If anything, I was expecting discussions on the r-to-l reading of the strong link (7-124)r8c5 = DJE in DPB's loop. _ _

Your second bullet point doesn't hold with respect to the eliminations in the non JE bands because it's possible for (124)r8c5 and the DJE to be true together. For example if (4)r8c5 and (4)r7c7 were both true the DJE would be forced true as well. If (1)r8c5 and (1)r6c7 were both true it would depend on (1)r4c3 whether the DJE was true or false.

Hmmm!!! I must be missing something about the definition of a DJE. I though it meant that both JEs had to valid.

When the JE in base cells r3c12, w/target cells r1c5 and r2c7, are tested for r3c12={1,4}, each of your your scenarios prevent the JE from being valid for the value in the target cells. This leads to the DJE not being valid from my perspective.

There are multiple possible ways to select 2 houses to hold the truths for the different base digits in the S cells. OBVIOUSLY as this is a valid puzzle, only one of them will be true and the others will lead to contradictions as you have found. What I was pointing out was that until the correct one has been isolated, it's unsafe to assume what eliminations would be available in the non-JE bands if the DJE is true.

< This Link > lists different JExeocet varieties.

[daj95376]

Sultan,

I've only been considering DJEs composed of JE3+JE3 or JE4+JE4. In light of your examples, I've discovered that I was missing JE3+JE4 as being a valid DJE.

I withdraw my original assertion that there's only one ADJE and one DJE associated with it. However, actually attaining the initial eliminations in your examples may prove to be difficult outside of a network (using memory).


David, Thanks for the link. I'll copy it into my notes.


Regards, Danny

[sultan vinegar]

However, actually attaining the initial eliminations in your examples may prove to be difficult outside of a network (using memory).

Indeed. I would like to get views as to the acceptability/unacceptability of the following unconventional rank 2 multi-fish. First though, let's consider a different example of the same structure:

Code: Select all

+---------------------+------------------------+----------------------+
| 9       8     234   | 7       6(3)    1234   | 1236   125    12356  |
| 127(3)  127   6     | 18(3)   5       128(3) | 4      12789  129(3) |
| 12347   5     2347  | 13468   68(3)   9      | 12367  1278   1236   |
+---------------------+------------------------+----------------------+
| 8       1279  2479  | 14569   679     147    | 1269   3      12569  |
| 1246    3     249   | 145689  689     18     | 1269   12459  7      |
| 1467    179   5     | 13469   2       137    | 8      149    169    |
+---------------------+------------------------+----------------------+
| 5       6     23789 | 389     1       378    | 2379   279    4      |
| 237     4     23789 | 389     789(3)  6      | 5      1279   129-3  |
| 7(3)    79    1     | 2       4       5      | 79(3)  6      8      |
+---------------------+------------------------+----------------------+

3 Truths = {3R29 3C5}
5 Links = {3r8 3c19 3b29}
1 Elimination --> r8c9<>3

The above is a rank 2 fish with no triplets, so r8c9 <> 3 as it a member of 0 truth-sets and 3 link-sets. I think this is an acceptable technique. Now, I would like to stretch the envelope and change the technique from a fish to a multi-fish and to add unconventional elements to the set mix such as DJE and NP:

Leren's ADJE puzzle:

Code: Select all

+----------------------+---------------------+-----------------------+
| 9      8       12345 | 7      1234   1245  | 6      1234    234    |
| 7      1345    12345 | 12459  6      12458 | 1248   123489  23489  |
| 123    134     6     | 1249   12349  1248  | 12478  5       234789 |
+----------------------+---------------------+-----------------------+
| 4      1567    125   | 3      12     9     | 1278   12678   25678  |
| 126    169     8     | 1246   5      7     | 3      12469   2469   |
| 12356  135679  1235  | 8      124    1246  | 1247   124679  245679 |
+----------------------+---------------------+-----------------------+
| 3568   2       345   | 4569   479    456   | 478    34678   1      |
| 1368   1346    9     | 1246   1247   1246  | 5      234678  234678 |
| 156    1456    7     | 12456  8      3     | 9      246     246    |
+----------------------+---------------------+-----------------------+


My proposed multi-fish has 2 truths (SIS), 4 links (WIS), 0 triplets, so requires 3 overlap link-sets to cause the elimination of (4)r6c689, so is the same structure as the rank 2 fish example. Note that SIS = strong inference set (at least one truth), WIS = weak inference set (at most one truth).

1st truth SIS = {DJE*, (12)r8c5, (4)r6c7}
2nd truth SIS = {(12)NPr46c5, (4)r6c5}
1st link WIS = {DJE*, (4)r6c689}
2nd link WIS = {(4)r6c5, (4)r6c689}
3rd link WIS = {(4)r6c7, (4)r6c689}
4th link WIS = {(12)r8c5, (12)NPr46c5}

1 Elimination --> r6c689<>4

The DJE* is DJE[B:r1c89,r3c12;T:r1c5,r2c37,r3c5;C:12r46,4r7c5], i.e. using a row and a column as the covering lines for digit (4).

One might try to draw it as a multi-fish to see the three WIS converging on (4)r6c689:

Code: Select all

DJE*==(12)r8c5==(4)r6c7
|          |        |
|   (12)NPr46c5     |
|          ||       |
|      (4)r6c5      |
|           |       |
\-----(4)r6c689-----/     

So, acceptable/unacceptable?

[ronk]

Now, I would like to stretch the envelope and change the technique from a fish to a multi-fish and to add unconventional elements to the set mix such as DJE and NP:

Leren's ADJE puzzle:My proposed multi-fish has 2 truths (SIS), 4 links (WIS), 0 triplets, so requires 3 overlap link-sets to cause the elimination of (4)r6c689, so is the same structure as the rank 2 fish example. Note that SIS = strong inference set (at least one truth), WIS = weak inference set (at most one truth).

1st truth SIS = {DJE*, (12)r8c5, (4)r6c7}
2nd truth SIS = {(12)NPr46c5, (4)r6c5}
1st link WIS = {DJE*, (4)r6c689}
2nd link WIS = {(4)r6c5, (4)r6c689}
3rd link WIS = {(4)r6c7, (4)r6c689}
4th link WIS = {(12)r8c5, (12)NPr46c5}

1 Elimination --> r6c689<>4

The DJE* is DJE[B:r1c89,r3c12;T:r1c5,r2c37,r3c5;C:12r46,4r7c5], i.e. using a row and a column as the covering lines for digit (4).[/code]

So, acceptable/unacceptable?

I can only guess as to how to read this notation. For example, why are base digits <1234> not specified somewhere in the DJE*? Why are truths 379c5, necessary to construct the ADJE, omitted?

More importantly, (12)r8c5 makes no sense to me either. AFAIK one can't select a few candidates from a cell and call it a SIS or a WIS. The valid sets are either all the candidates in the cell, or all like candidates in its row, column or box. [edit: Oh, these are "JExocet covers", which helps my understanding just a little.]

[David P Bird]

SV, Your (3) elimination in the first example can be taken as a 3x5 mutant fish.
Personally I have acceptability problems with this pattern class as I can't find a sure fire way to explore them other than to enumerate a vast number of possibilities which smacks of too much T&E - but that's just me.

Your work on Leren's tough puzzle looks promising but I have problems with it how valid it is!

You specify the DJE to use partial fish lines (12)r46 and (4)r7,c5, and then declare a truth set for that DJE and (12)r8c5,(4)r6c7, but I think that (3)r2c3 needs to be added to that set (it can be eliminated from r1c3). If (3)target:r2c3 is true then it will also be true in one of the targets r13c5 making the DJE false. r67c3 must therefore hold a (3) and so can only hold one (1), (2), or (4). I don't know what that does to the ranking etc.

I can now use these constraints to make the following assignments for the base digits in tiers 2 & 3 for c357 for the specified DJE to be true.

Code: Select all

+----------------------+---------------------+-----------------------+
| 9      8       12345 | 7      1234   1245  | 6      1234    234    |
| 7      1345    12345 | 12459  6      12458 | 1248   123489  23489  |
| 123    134     6     | 1249   12349  1248  | 12478  5       234789 |
+----------------------+---------------------+-----------------------+
| 4      1567    1-25  | 3      2-1    9     | 1278   12678   25678  |
| 126    169     8     | 1246   5      7     | 3      12469   2469   |
| 12356  135679  3-125 | 8      1-24   1246  | 2-147  124679  245679 |
+----------------------+---------------------+-----------------------+
| 3568   2       4-35  | 4569   479    456   | 478    34678   1      |
| 1368   1346    9     | 1246   4-127  1246  | 5      234678  234678 |
| 156    1456    7     | 12456  8      3     | 9      246     246    |
+----------------------+---------------------+-----------------------+

1) This doesn't eliminate (4)r6c689! What have I missed?

2) When the DJE is true each base digit will be true exactly twice in the 'S' cells, but when it's false, one base digit must be true 3 times in the 'S' cells which allows another base digit to be true just once in them. So, assuming (1) or (2) is true three times (which will include r8c5) then r67c357 need only hold one instance of (4) which provides further cases to consider.

3) As r7c5 <> 4 has already been proved, the DJE you specify considers 4 possible cases when the 'S' cells hold two (4)s : (r6c5,r7c3), (r6c5,r7c7), (r7c3,r8c5), & (r7c7,r8c5). However there are two more (r6c7,r7c3) & (r6c7,r8c5). If r6c7 is true, your specified DJE is false but then there is the case when another, as yet unspecified, DJE could be true.

Sorry to be so brutal. I appear to have wielded the lead pipe in the library, the candlestick in the lounge, and the dagger in the study, just to be sure, but this is dangerous territory!

DPB

[sultan vinegar]

DPB (and others), keep being brutal. I want any technique I use to stand up to the toughest scrutiny. We’ve got some of the best Sudoku minds in the world on this forum; that’s why I joined up.

Whilst you are entitled to have acceptability problems with the 3 x 5 fish, I would argue that anyone who accepts the conventional 3 x 3 swordfish pattern must also accept the 3 x 5 fish pattern. For the 3 x 3 pattern, I give two approaches: one that uses T&E and one that doesn’t.

Hidden Text: Show

T&E approach: One can consider a “Kraken” approach to the 3 x 3 (basic) swordfish pattern by understanding that at least one truth (exactly one in this particular case) exists in one particular base set. This base set contains 3 elements, so there are 3 cases to enumerate. Set one of those elements true, eliminate everything sharing a unit, discover an X-Wing left over and do those eliminations too. Repeat for the other 2 base elements, intersect the three cases for common eliminations and get your result.

Non T&E approach: Count the truths (3), calculate the rank (0), observe that there are no triplets, apply the rank 0 rule to get your result. There is no branching, no considering cases nor T&E anywhere in this approach.
The non T&E approach can also be used for the 3 x 5 fish. The only alteration is that instead of applying the rank 0 rule, you apply the rank 2 rule. Again, no branching, no considering cases nor T&E anywhere in this approach.

You specify the DJE to use partial fish lines (12)r46 and (4)r7,c5, and then declare a truth set for that DJE and (12)r8c5,(4)r6c7, but I think that (3)r2c3 needs to be added to that set (it can be eliminated from r1c3). If (3)target:r2c3 is true then it will also be true in one of the targets r13c5 making the DJE false. r67c3 must therefore hold a (3) and so can only hold one (1), (2), or (4). I don't know what that does to the ranking etc.

Hidden Text: Show

Any Sudoku grid contains lots of information. I understand that if you use Xsudo in “automatic link” mode, and specify every single remaining native truth set in the grid then it can solve any puzzle , memory permitting. That’s too hard for a human, so we need to consider little bits of the grid at a time. The trick then is to understand that a particular little bit of the grid has everything you need to justify an elimination, even though there are more nonetheless true facts out there that you do not consider in your little bit of the grid. I believe that your consideration of (3)r2c3 is one of those nonetheless true facts, but it is not in fact necessary to just my eliminations.

I believe that my SIS and WIS stack up, as in, satisfy the required truth counts. I think that with the exception of the 1st truth SIS, that they are straightforward to verify. In this less straightforward case, note that there are three elements, as (12)r8c5 is a grouped single element. Consider:

1. (12)r8c5 and (4)r6c7 both false. Then all potential spoilers for the DJE* are false, so clearly the DJE* must be true.
2. DJE* false. Then at least one of the DJE* potential spoilers must be true, i.e. at least one of (12)r8c5 and (4)r6c7 must be true.

This covers all cases. Arguing as in the above, I guess that I could have grouped (12)r8c5 and (4)r6c7 into a single element, and just had two elements in that first SIS, but never mind. One can see that in general there is an SIS between a DJE* and the group of all it’s potential spoilers. The constructive argument between several members on p44/45 of this thread came to the conclusion that provided you specify exactly which DJE you mean (i.e. the covering lines) that this relationship may be used.
Once you have all the SIS/WIS, use the non T&E approach as for the 3 x 5 fish: Count the truths (2), calculate the rank (2), observe that there are no triplets, apply the rank 2 rule to get your result. There is no branching, no considering cases nor T&E anywhere in this approach.

I can only guess as to how to read this notation. For example, why are base digits <1234> not specified somewhere in the DJE*? Why are truths 379c5, necessary to construct the ADJE, omitted?

Hidden Text: Show

Finally, apologies to ronk for my shoddy notation. AFAIK, the nomenclature for notating the use of ADJE in chains has not been settled upon. As there is a lot of information to specify for an ADJE, we need to work out how to do this concisely. I think that as you have to specify the base cells, that also specifying the base digits <1234> is an unnecessary duplication. You can tell what the base digits are by looking at the cells!

On your point regarding the truths 379c5, I don’t think I need them? I have specified that the covering lines for my ADJE are r46 for candidates (12), r7c5 for candidate (4), and candidate (3) gets a free ride via the box singles extension. Under this specification, the potential spoilers are (12)r8c5 and (4)r6c7 as none of them are members of the partial fish covering lines (they would be exo-fins in UFG notation if we had full fish and not partial fish, but I’m not sure we should overload the definition of exo-fins as it might get confusing).

Anyone, feel free to fire back with brutal constructive criticism and alternate views, and eventually we can settle on something that is acceptable.

[David P Bird]

SV, I don't want to take up too much space here, but your eliminations appear to depend on (4)r6c5 being true when the specified partial fish is true. I believe the assignments in the grid I posted show that this doesn't always hold.

I still have problems regarding omitting (3)r2c3 from the first SIS. If (3) is true in two target cells, the pattern is blown in the DJE band. This would allow (4) to be true twice in c357 in that band and once in the in the partial fish cells. Therefore it needn't be true in either r6c5 or r6c7.

DPB

[sultan vinegar]

DPB, here is the situation from my POV. We don’t quite have a DJE. What extra conditions do we need to make it one? We have some theorems on DJE. One about a condition on the occurrences in the S cells and one about giving a digit a free ride due to the box singles extension. The theorem says that if these conditions are met, then we have a DJE, that is, whatever is true in the base is true in the target for each JE. So should the DJE be false, such as if (3)r2c3 or (124)r1c3 or (124)r3c7 or (234)r3c9 were true, then at least one of the theorem’s conditions must be false. This is how we can find all of the spoilers.

In any case, there is a huge error in my work. I got mixed up between the DJE* using r7c5 as the covering lines for digit (4), and the DJE** which uses r67 as the covering lines for digit (4). I wrongly assumed that a true DJE* would eliminate (4)r6c689 by the:

When both true base digits are confined to two instances in the S cells, within the 3 cross lines, each digit must occur once in the JE band and twice in the other bands, in the S cells.

rule. Of course, r6 is not a covering line for the DJE*, thus the inference between DJE* and (4)r6c689 is not necessarily a weak inference, so the logic doesn’t work.

[ronk]

Of course, r6 is not a covering line for the DJE*, thus the inference between DJE* and (4)r6c689 is not necessarily a weak inference, so the logic doesn’t work.

Xsudo shows 4r6 to be 0-rank for an ADJE, the only useful ADJE I see BTW. I'm surprised the alternate method you're using doesn't highlight this weak link.

#975059
980700600700060000006000050400309000008057300000800000020000001009000500007083900

The ADJE
16 Truths = {124C357 379C5 1N89 3N12}
28 Links = {12r1346 3r13 4r1367 2n3 13678n5 26n7 124b13}
11 Eliminations --> r6c689<>4, r4c28<>1, r4c89<>2, r12c3<>3, r2c3<>5, r7c5<>4

Unfortunately Xsudo identifies 0-rank WIS (links) only in a graphic and not in a listing. My manual listing is 13678n5, 2n3, 3r1, 12r4 and 4r6. The graphic would be more impressive.

The ADJE graphic: Show

Adding links 12b5 yields r6c6<>12. Adding SIS (truth) 9R7 yields r7c4<>4. Adding SIS 67N6 yields r6c6=6 and r7c4<>5.

[sultan vinegar]

980700600700060000006000050400309000008057300000800000020000001009000500007083900

I can explain the 0-rank link-set (4)r6 and the elimination of (12)r6c6 but I need branching. 10 points to the first person who can do it without branching!

I need to refer to two different exocets here, so let’s be careful and lay them out.

Hidden Text: Show

The first one is the “obvious” ADJE for base candidates <1234>. Call this (A)DJE.
Base B: r1c89, Target T: r2c3, r3c5.
Base b: r3c12, Target t: r1c5, r2c7.
The covering lines are r46 for <12>, r67 for <4>, and <3> gets a free ride via the box singles extension. The spoilers are (124)r8c5.

The second exocet is a single JE with AHS extension for base candidates <1234>. Note that there is no almost in this exocet, it is true for sure. Call this JE+.
Base: r1c89, Target 1:r2c3, Target 2: r3c5/r8c5.
We use the same covering lines as for the DJE above. The AHS is (79)r378c5. The way it works is that r2c3 is always one target cell, but only one of r3c5/r8c5 is the other target cell. One of (79) must be true in r7c5 [if (4) were here then there wouldn’t be two target cells and thus we would have an illegal multi-fish, or if you don’t buy that, use DPBs chain to eliminate the (4) in a separate step], and the candidate that isn’t true in r7c5 must be true in either r3c5 if it is digit (9), or r8c5 if it is digit (7).

OK, so with the JE+ only, we can eliminate (3)r1c3 and (35)r2c3.

Hidden Text: Show

r1c3 <> 3:
If (3) is in the base, then r1c3 <> 3.
If (3) is not in the base, then (3) is not in the target cell r3c5, so (3) is true in cell r1c5 by the bilocation, then r1c3 <> 3.

r2c3 <> 3:
If (3) is in the base, then (3) is not in r1c5, so (3) is true in r3c5 which is a target cell so (3) can’t be true in the other target cell at r2c3.
If (3) is not in the base, then (3) can’t be in the target cell at r2c3.

r2c3 <> 5:
As (5) is not a member of the base, it can’t be the member of the target cell at r2c3.

Now, we need to use both the DJE and JE+ to eliminate (4)r6c689:

Hidden Text: Show

I use a kraken argument from kraken cell r8c5:
If r8c5 = 7, then r8c5 <> (124), so the DJE is true, and r6c689 <> 4.

If r8c5 = 1 or 2, then we have a naked triple (124)r468c5, meaning r6c5 = 4, so r6c689 <> 4.

If r8c5 = 4, then as r8c5 is a target cell for the JE+, we must have (4) true in the base r1c89. This eliminates (4) from r1c3, r23c7 by sight, and also from the other target cell at r2c3. This leaves an X-Wing for (4) in c37/r67, so r6c689 <> 4.

A similar kraken argument can show r6c6 <> (12).

Hidden Text: Show

If r8c5 = 7, then r8c5 <> (124), so the DJE is true, and r6c6 <> (12).

If r8c5 = 4, then we get a naked pair (12) in r46c5, so r6c6 <> (12).

If r8c5 = 1, then we get a naked triple (124) in r468c5, so r4c5 = 2 and so r6c6 <> 2.
Also as r8c5 is a target cell for the JE+, we must have (1) true in the base r1c8. This eliminates (1) from r1c3, r23c7 by sight, and also from the other target cell at r2c3. This leaves an X-Wing for (1) in c37/r46, so r6c6 <> 1.

If r8c5 = 2, then we get a naked triple (124) in r468c5, so r4c5 = 1 and so r6c6 <> 1.
Also as r8c5 is a target cell for the JE+, we must have (2) true in the base r1c89. This eliminates (2) from r1c3, r23c7 by sight, and also from the other target cell at r2c3. This leaves an X-Wing for (2) in c37/r46, so r6c6 <> 2.

For anyone interested, if we add (3)c3 as a truth set, with cell link sets in r67c3, Xsudo eliminates (4) in r2c2, r1c56, r3c5, and eliminates (5) from r7c3.