The New Sudoku Players' Forum

[Leren]

Re daj95376's response to my last post, puzzle 3811 isn't an exception to my constraint (2) but, re-reading it, I can see why my wording may have caused confusion. What I meant to say was (2) Each Base candidate must appear in at at least one Target cell (ie a base candidate cannot be absent from both Target cells). Are there exceptions to this (re-worded) constraint?

I also get a base candidate elimination for this Exocet <5> r7c8. For the reasoning behind this, see Champagne's post "what to do after an exocet has been seen" on p1 of the Exotic Patterns thread.

Leren

[ronk]

r9c2=6 implies r9c1=3 which leaves no room for base digits 1, 5, or 7 in r9c12 of this "JExocet" (with a twin target). This faulty common outcome argument by itself seems sufficient reason to a discard the twin "JExocet" POV.

Yes I jumped a bit too quickly with that answer. We only get a twin JExocet if we can eliminate (6)r9c2 first, but the point stands it can't be eliminated because of any JExocet pattern.

Despite daj95376's qExocet term, he, Xsudo and I say that it can. As I've hinted before, the r9c12 twin target cells are analogous to a distributed hub in an HSR (hub, spokes, and rim) pattern.

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Code: Select all

     12 Truths = {157R349 3R9 12N3}
     15 Links = {1c379 5c346 7c379 49n1 9n2 157b1}
     3 Eliminations --> r4c1<>48, r9c2<>6

[Leren]

In puzzle 18371 there is a fourth elimination <6> r8c2. This is because, whichever of r9c12 eventually holds 3 the other eventually holds whatever base digit is not eventually held by r4c1.

So eventually r8c2 = r4c1 (there is nowhere else for the digit held by r4c1 to go in Box 7) and you can eliminate non-base digits there.

Leren

[David P Bird]

ronk Your truth and link set diagram has me bewildered. I've been looking at it for some time, but with my limited experience with this approach haven't got very far. Consequently I tried setting (3)r9c1 and (6)r9c2 and got no immediate contradictions. A (157) multi-fish then eliminates 20 candidates in 13 cells including (48)r4c1. The only wrong elimination in that set is (5)r6c6, but again this produces no immediate contradiction.

It would help me if you could try those wrong assignments and see what contradictions you can find with Xsudo.

[ronk]

ronk Your truth and link set diagram has me bewildered. I've been looking at it for some time, but with my limited experience with this approach haven't got very far. Consequently I tried setting (3)r9c1 and (6)r9c2 and got no immediate contradictions. A (157) multi-fish then eliminates 20 candidates in 13 cells including (48)r4c1. The only wrong elimination in that set is (5)r6c6, but again this produces no immediate contradiction.

Then first use something you're familiar with. Assert an ultimately excluded candidate in 1) a "classical JExocet" with 2) a familiar tool, and find the contradiction within that pattern.

It would help me if you could try those wrong assignments and see what contradictions you can find with Xsudo.

Asserting r9c2=6 in puzzle #18371 causes an impossible pattern, which is often difficult to analyze, but it's no different in a "classical JExocet" AFAICS. In such instances, I find it easier to use a proof by cases. There are only three different pairwise combinations of base digits (ab) = 15, 17, or 57 in r13c3. Each asserted combination results in a remote hidden triple (ab3)[r4c1,r9c12]. In all three cases, r4c1<>48 and r9c2<>6.

[daj95376]

In puzzle 18371 there is a fourth elimination <6> r8c2. This is because, whichever of r9c12 eventually holds 3 the other eventually holds whatever base digit is not eventually held by r4c1.

So eventually r8c2 = r4c1 (there is nowhere else for the digit held by r4c1 to go in Box 7) and you can eliminate non-base digits there.

Good catch. My analyzer doesn't check for that scenario. I'll add it to my ToDo list.

However, your logic is a two-way street. Whatever is in r8c2 must be in r4c1 as well. This leads to r4c1<>5.

Here's my latest grid for this puzzle.

Code: Select all

Puzzle:   ;18371

98.7.....6.....5....4.6..8..9..3..2...3..74.....1......4...13....23....5....8..4.

 +--------------------------------------------------------------------------------+
 |  9       8      *15      |  7       1245    2345    |  126     136     12346   |
 |  6       1237   *17      |  2489    1249    23489   |  5       1379    123479  |
 |  12357   12357   4       |  259     6       2359    |  1279    8       12379   |
 |--------------------------+--------------------------+--------------------------|
 | #17-48-5 9       15678   |  4568    3       4568    |  1678    2       1678    |
 |  1258    1256    3       |  25689   259     7       |  4       1569    1689    |
 |  24578   2567    5678    |  1       2459    245689  |  6789    35679   36789   |
 |--------------------------+--------------------------+--------------------------|
 |  578     4       56789   |  2569    2579    1       |  3       679     26789   |
 |  178     17-6    2       |  3       479     469     |  16789   1679    5       |
 | #157+3  #157+3-6 15679   |  2569    8       2569    |  12679   4       12679   |
 +--------------------------------------------------------------------------------+
 # 182 eliminations remain

 #3# -157- qExocet   Base = r12c3   Target = (3)r9c12,r4c1==r8c2   aligned


[Edit: included missing details that I must add manually.]

[David P Bird]

ronk,Thank you for your response, and I'll follow your suggestions later when I have time.

In the meantime I can see checks to be made before accepting this configuration as a regular pattern which you may like to consider.
1) Does it still apply when there are 4 digits in the base set
2) Can there be further candidates in r9c2
3) Can r4c12 also contain a locked non-member candidate with further non-member candidates in one of the cells

From your diagram, the answer to all these questions would appear to be yes, and the criteria would still be contained enough for the pattern to be "recognisable". In that case rather than deserving our derision, daj deserves our congratulations.

Fundamentally I think these extended JExocets would allow the population of the target and companion cell pairs to be increased if they form, or are part of, an Almost Hidden Set. Eg in the grid if we had (38x)r9c5 then (8) could also exist in r9c12, but not in any other cells in the row.

[ronk]

I can see checks to be made before accepting this configuration as a regular pattern which you may like to consider.
1) Does it still apply when there are 4 digits in the base set
2) Can there be further candidates in r9c2
3) Can r4c12 also contain a locked non-member candidate with further non-member candidates in one of the cells

From your diagram, the answer to all these questions would appear to be yes, and the criteria would still be contained enough for the pattern to be "recognisable".

"Yes" to all three. See daj95376's example here [edit: for answers to 2) and 3). daj95376, do you have, or can you find, a real-world example for the first question?]

In that case rather than deserving our derision, daj deserves our congratulations.

"Our derision?" Do you have a mouse in your pocket?

[daj95376]

daj95376, do you have, or can you find, a real-world example for the first question?

Code: Select all

98.7.....6..85......4..3...7...6.8...5.9.......6..8.3..6......2..1..24.........13;85

 +----------------------------------------------------------------------------------+
 |  9       8       235     |  7       124     146     |  12356   2456     1456     |
 |  6       1237    237     |  8       5       149     |  12379   2479     1479     |
 |  125     127     4       |  126     129     3       |  125679  5679+8-2 5679+8-1 |
 |--------------------------+--------------------------+----------------------------|
 |  7       12349   239     |  12345   6       145     |  8       2459     1459     |
 |  12348   5       238     |  9       12347   147     |  1267    2467     1467     |
 |  124     1249    6       |  1245    1247    8       |  12579   3        579-14   |
 |--------------------------+--------------------------+----------------------------|
 |  3458    6       35789   |  1345    134789  14579   | *579     5789     2        |
 |  358     379     1       |  356     3789    2       |  4       56789    56789    |
 |  2458    2479    25789   |  456     4789    45679   | *5679    1        3        |
 +----------------------------------------------------------------------------------+
 # 177 eliminations remain

 #4# -5679- qExocet   Base = r79c7   Target = r6c9,(SL=8)r3c89


Although I have several examples of qExocet for abc and abcd scenarios, I don't have an abcd scenario with two strong links. I suspect that it's because so many of the puzzles have "similar" patterns ... and there's actually a small sample set from which to select unique qExocet patterns.

[Leren]

ronk, if you look at your grid for 18371 you'll see that there are some elementary eliminations that follow.

There is a locked quin 13578 in Box 7 so r79c3 reduces to a locked pair 69 and also r46c3 <> 6. Perhaps you could update your grid to reflect this.

Your elimination r4c1<>5 is clearly a general property of "non-aligned" exocets ie

If an Exocet Target cell shares a mini-row or mini-column with a cell with no base digits then

1) non-base digits can be eliminated from the other cell in the mini-row or mini-column

2) the other cell must eventually take the same value as the opposing Target cell and;

3) base digits that don't appear in the other cell can be eliminated from the opposing Target cell

Is any of this new ? I new about 1) and I'd forgotten about 2) but I was unaware of 3) until you pointed it out.

What may be new is that the result still applies in this twin exocet case, since one of the half? exocets is non-aligned.

Leren

[ronk]

Although I have several examples of qExocet for abc and abcd scenarios, I don't have an abcd scenario with two strong links.

Thanks, the one strong link scenario was all I expected.

ronk, if you look at your grid for 18371 ...

I think you mean daj95376, not me. Also, keep in mind that this thread is about the exocet technique, but not including an unlimited number of follow-on moves.

[daj95376]

If an Exocet Target cell shares a mini-row or mini-column with a cell with no base digits then

1) non-base digits can be eliminated from the other cell in the mini-row or mini-column
2) the other cell must eventually take the same value as the opposing Target cell and;
3) base digits that don't appear in the other cell can be eliminated from the opposing Target cell

Is any of this new ? I new about 1) and I'd forgotten about 2) but I was unaware of 3) until you pointed it out.

What may be new is that the result still applies in this twin exocet case, since one of the half? exocets is non-aligned.

Subsequent steps are rarely discussed ... unless they're interesting or you intend to provide a full solution for the puzzle.

I can't follow all of your (1-3) details, but you appear to be on the right track.

For a puzzle with Exocet(s), I break the eliminations down in the following order:

1) Direct eliminations associated with the Exocet target cells; r4c1<>48 and r9c2<>6.
2) Secondary eliminations from Exocet forced equivalences; r4c1==r8c2 => ( r4c1<>5 and r8c2<>6 ).
3) Cross-eliminations associated with two (a "double") Exocets for the same values in the same chute.

[Leren]

My apologies daj95376, I should have credited you with the r4c1<>5 elimination in puzzle 18371.

I also have to modify what I said in my previous post. 1) - 3) would not apply if the opposing Target cell

could hold a non-base digit. For example, in puzzle 18371 r6c4 becomes 4 but r6c3 <> r9c2.

Leren

[David P Bird]

Right, I've had a shot at re-working of the JExocet definition

The rest of this post has been withdrawn as being outdated by <this one>

[ronk]

Fundamentally I think these extended JExocets would allow the population of the target and companion cell pairs to be increased if they form, or are part of, an Almost Hidden Set. Eg in the grid if we had (38x)r9c5 then (8) could also exist in r9c12, but not in any other cells in the row.

I somehow missed this earlier, but I totally agree. [edit: IYO can 'x' include base set candidates?]

My apologies daj95376, I should have credited you with the r4c1<>5 elimination in puzzle 18371.

I also have to modify what I said in my previous post. 1) - 3) would not apply if the opposing Target cell

could hold a non-base digit. For example, in puzzle 18371 r6c4 becomes 4 but r6c3 <> r9c2.

Instead of letting the errors stand, I recommend editing to remove errors, and leaving an "edit trail" as appropriate. Then future readers don't have to stumble over the same thing.

[edit: Dropped the notion of "remote hidden sets", which may not be valid for sets larger than pairs.]