The New Sudoku Players' Forum

[ronk]

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

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.

Good show. It's a bit ironic that your kraken cell is the "spoiler" cell.

[David P Bird]

SV, I've been pressed for time so haven't been able to analyse your extended JE properly but it looks good up to the point when you have to resort to using a kraken node.

I'm trying to simplify the way this JE+ model should be viewed which I would notate as

(1234)JE+:r1c89,r2c3,(79)r378c5 = (digits)PatternID:base_cells,target1,(locked digits)target2

As r8c5 is a member of a target node, it's no longer an 'S' cell which nicely makes the pattern good with respect to condition 3.

We now know that when (ab) are the true base digits, the 4 target cells will hold (ab79) and can make inferences based on that

In a leap of faith we can now apply the same thinking to (1234)JE+r3c12,r2c7,(379)r1378c5

So if (cd) are the true base digits, the five target cells will hold (cd379) which is 5 digits and so requires that (3) is not a member of (cd)

This gives us a proven DJE+ pattern to conjure with that has one target cell, r3c5, common to both which logic dictates must hold (3). Now making all the regular DJE eliminations the puzzle drops out with no need for a kraken node.

So gentlemen can we prove that leap of faith is valid and the elimination of (3)r3c12 is sound?

[added] Actually, all that is needed to solve this puzzle is to show that r1c89 and r3c12 must not have a common digit. This eliminates (1234) from r1c123 and r3c789 which seems to be sufficient, but that might not always be the case.

[sultan vinegar]

Good show. It's a bit ironic that your kraken cell is the "spoiler" cell.

DPB, fantastic idea to add extra target cells that are filled by the AHS candidates. That way, we can derive virtual locked sets, and do stuff with them.

For the exocet with base r1c89, we get four candidates <ab79> in four cells (r2c3, r378c5). It is clear here that <ab79> are four distinct digits, so a virtual locked set ensues.

For the exocet with base r3c12, we get five candidates <cd379> in five cells. Here, it is possible at first glance for one of <cd> to be (3), which would not result in a virtual locked set. But the five cells all share a column, except the one at r2c7. So if one of <cd> was (3), then r2c7 must be (3). But (3) is not an option for r2c7. So this doesn't occur. Hence, we do get a virtual locked set of five distinct candidates in five cells. This proves r2c7 <> 8 and r3c12 <> 3. Then r2c2 = 3 and there's some more easy eliminations after that.

Incidentally , Xsudo gives most of the possible DJE eliminations after eliminating r3c12 <> 3.

[ronk]

For the exocet with base r3c12, we get five candidates <cd379> in five cells. Here, it is possible at first glance for one of <cd> to be (3), which would not result in a virtual locked set. But the five cells all share a column, except the one at r2c7. So if one of <cd> was (3), then r2c7 must be (3). But (3) is not an option for r2c7. So this doesn't occur. Hence, we do get a virtual locked set of five distinct candidates in five cells. This proves r2c7 <> 8 and r3c12 <> 3.

What do you see as an Xsudo logic set to obtain r3c12<>3?

[sultan vinegar]

I haven't found one yet. I'm not sure I will, it only lets you specify one virtual set.

[ronk]

I haven't found one yet. I'm not sure I will, it only lets you specify one virtual set.

Since virtual sets are ultimately based on native sets, the lack of a virtual set, any virtual set at all, should not be a hindrance.

[David P Bird]

Let's give credit where it's due. It was DAJ who spotted that when a potential target cell was a member of an AHS, the target could be extended to the full AHS call set. In the cases considered at that time then the possible locations of the base digit in the AHS were always considered to be inside the JE band, not outside it. (The most common case being when a non-base digit is locked in a target/companion cell pair, so only one base digit can be held between them).

It was at the same time it was suggested the notation of the target node should be extended to "(locked digits)AHS_cells".

It was SV who, using different language, spotted this AHS concept could also be applied when the location of the base digit in the AHS could possibly be outside the JE band.

All I did (when I eventually made the connection) was to unify the two approaches and to wonder how making one of the locked digits a potential member of one of the DJE+ base sets would work out.

I said I was making a leap of faith because I hadn't evaluated all the possibilities. The areas to check would be:
1. for a JE+, can both targets ever hold the same base digit, one in the JE band and the other in a parallel band.
2. for a ADJE+ can the situation ever arise where a digit is common to both base sets?
I'm assuming the answer to both these questions is no (as it is for this puzzle) but this remains to be proved. It may be that some extra requirement needs to be added to the specification to screen out potential false positive results.

[ronk]

It was SV who, using different language, spotted this AHS concept could also be applied when the location of the base digit in the AHS could possibly be outside the JE band.

All I did (when I eventually made the connection) was to unify the two approaches and to wonder how making one of the locked digits a potential member of one of the DJE+ base sets would work out.

I posted an exemplar doing this with two cross-lines of a JE back in July 2012 (see hidden block). After an objection by champagne (IIRC), a tactical decision was made to restrict the AHS [edit2: cells with base candidates] to the JE band.

I said I was making a leap of faith because I hadn't evaluated all the possibilities. The areas to check would be:
1. for a JE+, can both targets ever hold the same base digit, one in the JE band and the other in a parallel band.
2. for a ADJE+ can the situation ever arise where a digit is common to both base sets?
I'm assuming the answer to both these questions is no (as it is for this puzzle) but this remains to be proved. It may be that some extra requirement needs to be added to the specification to screen out potential false positive results.

It is often easier to find a counter-example than to prove a conjecture. For the DJE under discussion ...

Code: Select all

+---------+---------+---------+
| .  .  4 | .  3  . | .  1  2 |
| .  .  1 | .  .  . | 4  .  . |
| 2  3  . | .  9  . | 7  .  . |
+---------+---------+---------+
| .  .  5 | .  1  . | 2  .  . |
| .  .  . | .  .  . | .  .  . |
| .  .  2 | .  4  . | 1  .  . |
+---------+---------+---------+
| .  .  3 | .  7  . | .  .  . |
| .  .  . | .  2  . | .  .  . |
| .  .  . | .  .  . | .  .  . |
+---------+---------+---------+

... note digit <2> in both r1c89 and r3c12 and [deleted]. Does this violate any constraints of the jexocet patterns? I don't see one [edit: after fixing r23c7].

[David P Bird]

Ronk, apologies for overlooking your contribution. The partial grid you posted (thanks) reveals that there is a flaw either in SV's analysis or in my interpretation of it.

In general, when a target cell is a member of a AHS, if the base digit in the AHS is true outside the JE band there is nothing to stop the other target holding the same digit. I don't remember making this line of analysis personally before, but I think that why such cases weren't treated as being JE patterns.

In the case of this particular pattern, if (3)r1c89 is true we get a DJE because it automatically eliminates (9)r3c5 and therefore the spoilers in r8c5. If it's false then the r1c89 JE (and hence the DJE) depends on (9)r3c5 being false too.

The stumbling block I have in SV's logic is how the eliminations r2c3 <> 35 are achieved in light of these points.

DPB

[sultan vinegar]

SV has made a rookie mistake in the second JE+ with the virtual locked set. There are 5 candidates <cd379>, 5 cells and no member can occur more than once, but the error is that not every member has to occur at least once, so it's not a locked set. Hence, ronk can place an outside candidate (4) in one of the cells (r2c7) and get a contradiction.

I think the elimination of (35) r2c3 is still sound though; I can find an Xsudo truth/link set for that one! In this case, the virtual quantum locked set <ab79> is definitely locked, which makes all the difference.

[daj95376]

I deleted the original posting of this message. Upon reflection, it might be relevant.

Code: Select all

 r8c5=7 ==>> "obvious" DJE
 eliminations associated with assuming DJE true in ADJE
 ### -1234- QExocet   Base = r1c89   Target = r2c3,r3c5==r2c2
 ### -1234- QExocet   Base = r3c12   Target = r1c5,r2c7==r1c6
 *** double QExocet
 +--------------------------------------------------------------------------------+
 |  9       8       5-1234  |  7       1234    124-5   |  6       1234    234     |
 |  7       134-5   124-35  |  59-124  6       58-124  |  124-8   123489  23489   |
 |  123     134     6       |  1249    134-29  1248    |  78-124  5       789-234 |
 |--------------------------+--------------------------+--------------------------|
 |  4       567-1   125     |  3       12      9       |  1278    678-12  5678-2  |
 |  126     169     8       |  1246    5       7       |  3       12469   2469    |
 |  356-12  35679-1 1235    |  8       124     6-124   |  1247    679-124 5679-24 |
 |--------------------------+--------------------------+--------------------------|
 |  3568    2       345     |  569-4   479     56-4    |  478     3678-4  1       |
 |  1368    1346    9       |  1246    7-124   1246    |  5       234678  234678  |
 |  156     1456    7       |  12456   8       3       |  9       246     246     |
 +--------------------------------------------------------------------------------+
 # 182 eliminations remain


I'm lost for most of the current discussions. However, I'd like to summarize what I've been able to piece together.

Code: Select all

1_) (3)r1c89 - (3)r1c5 = (3-9)r3c5 = (9-7)r7c5 = (7)r8c5; "obvious" DJE eliminations

2a) <124>r1c89, r2c3=Q, r3c5=R: (R-9)r3c5 = (9-7)r7c5 = (7)r8c5; "obvious" DJE eliminations

2b) <124>r1c89, r2c3=Q, r8c5=R: (R-7)r8c5 = (7-9)r7c5 = (9-3)r3c5 = (3)r1c5; assorted eliminations


So, two-of-three scenarios for r1c89 lead to the "obvious" DJE eliminations. When the third scenario is also considered, then only its eliminations common to the "obvious" DJE apply. Common eliminations associated with the chain segments also apply.

Code: Select all

r1c3<>3, r2c3<>35, r3c5<>2, r7c5<>4


I see how SV derived r6c689<>4 and r6c6<>12 from (2b) scenario <124>r1c89, r2c3=Q, r8c5=R; but it requires a separate network of logic and begs the question ... where do you draw the line?


I tried combining (2b) with r3c12, but my only result was a headache.

[David P Bird]

Coming back to this after a break:

Theorem: for an ADJE with a single spoiler cell, each base digit must occur at least once in the mini-line diagonal containing the two pairs of base cells.

Short Proof:
1. When both JEs are true, each base digit must occur in one base pair.
2. When a JE is false a rogue digit will be true 3 times in the partial fish cells so either
a) pairs of base digits properly repeat together in the mini-lines, but the rogue digit isn't forced to occupy a target cell.
b) the rogue digit occurs in both pairs of base cells and is accompanied by a different base digit in each mini-line in the diagonal.

[Edit]SV's counter-example below shows the clause in red doesn't automatically follow and therefore the 'theorem' is a dud.

For this puzzle:
When (7)r8c5 is true and the DJE is true (3) is locked in one of two target cells and so can't be true in either of the others
When (7)r8c5 is false, (3)r1c5 is forced true eliminating it from mini-rows r1b3 & r2b2. So, from the theorem, (3) must be true in r3b1
Therefore target r2c3 <> 3.

I suppose we can carry on declaring theorems for sub-patterns as long as we want, but as DAJ says, lines must be drawn somewhere!

[sultan vinegar]

Code: Select all

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


This counter-example has no (3) in b1r3, b2r2, b3r1.

[David P Bird]

SV, thanks for spotting my error. I had taken it that with the rogue digit in both sets of base cells, the other base digits would all repeat in the 3 mini-lines in the opposite diagonal direction, but your counter example shows that this doesn't necessarily follow. I've consequently retracted my false theorem.

DPB

[RichardGoodrich]

In SK loops does SK stand for Steve Kurzhal or something else?