The New Sudoku Players' Forum

[champagne]

If the following is redundant or the timing is inappropriate, please send me a private message and I'll delete this posting.

This is likely not redundant.

It's not what I call a double exocet, although we could have an equivalence with a JE + an exocet,

Here, my question will be how that pair of exocets eases the solution.
I have an optical answer, but I want before to check it carefully

[Leren]

daj 95376 wrote : ### -1235- JExocet Base = r1c89 Target = r2c3==r3c4,r3c6

I was wondering why you don't report this as ### -1235- JExocet Base = r1c89 Target = r2c3==r3c4,r3c6==r2c1 although you mention the result of this 2nd secondary equivalence, r3c6 = 5, later on.

Leren

[daj95376]

daj 95376 wrote : ### -1235- JExocet Base = r1c89 Target = r2c3==r3c4,r3c6

I was wondering why you don't report this as ### -1235- JExocet Base = r1c89 Target = r2c3==r3c4,r3c6==r2c1 although you mention the result of this 2nd secondary equivalence, r3c6 = 5, later on.

As of now, my solver doesn't resolve secondary equivalences with solved/given cells. Something I need to correct.

I was so fixed on showing the (SL) property of the JExocet that I forgot to include the fact that each SL is part of a discontinuous loop.

Code: Select all

 r1c89 = 1  =>  (1)r2c3 = (1)r3c6   -   (1)r2c4 = (1)r2c3   (forced)
 r1c89 = 2  =>  (2)r2c3 = (2)r3c6   -   (2)r2c4 = (2)r2c3   (forced)
 r1c89 = 3  =>  (3)r2c3 = (3)r3c6   -   (3)r2c4 = (3)r2c3   (forced)

 r1c89 = 5  -> - (5)r1c56 = (5)r3c6   (forced)


BTW: I just recently included the reporting of values being locked to target cells. Here's what my solver reports on the JExocet.

Code: Select all

 ***   value   = <1> true   locks cells true:   r2c3
 ***   value   = <2> true   locks cells true:   r2c3
 ***   value   = <3> true   locks cells true:   r2c3

 ***   value   = <5> true   locks cells true:   r3c6


This explains my JExocet assignment r3c6=5 ... and constraint (r2c3==r3c4)=<123>.

_

[eleven]

Hm, your notations confuse me more than the exocets.

Code: Select all

     +--------------------------------------------------------------------------------+
     |  9       8       1234    |  7       12345  t1235    |  6      B235    B135     |
     |  5       6      T123     |  123     9       8       | t123     4       7       |
     | b124    b1234    7       |  1234    6      T1235    |  12358   23589   13589   |
     |--------------------------+--------------------------+--------------------------|
     |  3       1479    149     |  1468    145     156     |  158     5678    2       |
     |  1267    127     8       |  1236    1235    9       |  4       3567    1356    |
     |  1246    124     5       |  123468  1234    7       |  9       368     1368    |
     |--------------------------+--------------------------+--------------------------|
     |  128     12359   6       |  1239    123     4       |  7       23589   3589    |
     |  1248    12349   12349   |  5       7       1236    |  238     23689   34689   |
     |  247     234579  2349    |  2369    8       236     |  235     1       34569   |
     +--------------------------------------------------------------------------------+

What i can see is, that digits 123 have the classical JExocet properties in both cases with targets r3c6,r2c3 and r1c6,r2c7 resp.
Moreover a 123 in r3c6 would force the same digit into r2c3 (last digit in r2) in the first case, and in the second case 123 in r1c6 also would appear in r2c7.
So not 2 of 123 can be in the base cells in both cases, which implies r3c6=5 and r1c5=4.

[David P Bird]

This is a substitute for the post I made yesterday and later deleted when I realised the errors I'd made.

The purpose of the 'S' cell JE condition is to ensure that each base digit must be forced into one of the two target cells if it's true. When the JE digits include one that is locked in a mini-line there may be other ways to show that this must be true for that digit which will allow its 'S' cell check to be by-passed.

Code: Select all

     +--------------------------------------------------------------------------------+
     |  9       8       1234    |  7       12345   1235    |  6      B235    B135     |
     |  5       6      T123     |  123     9       8       |  123     4       7       |
     |  124     1234    7       |  1234    6      T1235    |  12358   23589   13589   |
     |--------------------------+--------------------------+--------------------------|
     |  3       1479    149     |  1468    145     156     |  158     5678    2       | 1  5
     |  1267    127     8       |  1236    1235    9       |  4       3567    1356    |
     |  1246    124     5       |  123468  1234    7       |  9       368     1368    |    5
     |--------------------------+--------------------------+--------------------------|
     |  128     12359   6       |  1239    123     4       |  7       23589   3589    |
     |  1248    12349   12349   |  5       7       1236    |  238     23689   34689   | 1235
     |  247     234579  2349    |  2369    8       236     |  235     1       34569   |  235
     +--------------------------------------------------------------------------------+
                        CL1                        CL2        CLB

The extended JE is:
(1235)JE:r1c89,r2c3,r3c6+(5)r2c1 => r3c6 <> 123, r3c4 <> 4

This can't be a regular JE because of the 'S' cells for (5). However because (5)r2c1 is a given and (5) is also missing in r3c45, if it's true in the base cells it must be true in target r3c6. This makes the 'S' cell cover count for (5) unnecessary. Therefore whatever digits are true in the base cells must also be true in the target cells and all the regular JE eliminations will apply.

The eliminations then follow from comparing the digits in the target cells with those in the diagonal mini-row cells that must mirror them.

Yesterday I assumed the pattern would be automatically be completed because the target r2c3 only contained base digits (123), but that alone doesn't guarantee the digit in the target must also be true in the base cells, so these digits must still comply to the 'S' cells requirement. I also overlooked the (4)r3c4 elimination.

As this logic would still apply if (5) was locked in r2c12, I now consider a better name for this would be the 'Locked Base Digit Extension'

[daj95376]

champagne: You may wish to ask JasonLion to split off the discussion of this puzzle into a separate thread.

Code: Select all

 Exocet Single-Target Pattern
 +-----------------------------------------------+
 | abcd abcd .   |   .   .   .   |   .   .   .   |
 |   .   .   .   |   .   .   .   |   .   .   .   |
 |   .   .   .   |   .   .   .   |   .   .   .   |
 |---------------+---------------+---------------|
 |   .   .   .   |   .   .   .   |   .   .   .   |
 |   .   .   .   |   .   T   .   |   .   .   .   |
 |   .   .   .   |   .   .   .   |   .   .   .   |
 |---------------+---------------+---------------|
 |   .   .   .   |   .   .   .   |   .   .   .   |
 |   .   .   .   |   .   .   .   |   .   .   .   |
 |   .   .   .   |   .   .   .   |   .   .   .   |
 +-----------------------------------------------+

 If any of abc is assumed true in the base cells,
 and T is subsequently forced true for that value,
 then d must be true in the base cells
 and T may only contain candidates abc.


In the puzzle posted above, both the JExocet and its "cousin" qualify as an Exocet Single-Target Pattern.

If any of 123 is true in r1c89, then that value is forced true in r2c3 through a discontinuous loop. This forces 5 to be true in the base cells and r2c3=123.

Similarly, if any of 123 is true in r3c12, then that value is forced true in r2c7 through a discontinuous loop. This forces 4 to be true in the base cells and r2c7=123.

Since r2c3 and r2c7 are in [r2], their solutions must be different. Thus, none of the solved values in r1c89 match any of the solved values in r3c12.

_

[champagne]

champagne: You may wish to ask JasonLion to split off the discussion of this puzzle into a separate thread.
_

Hi Danny,

I locked enough posts in the head of the thread to make the summary and let people discuss freely on any subject not too far from the topic and this is "within the scope" .

However, these summary posts are locked. If anybody has some remarks about errors or missing points in these summary posts, I am open to catch proposals and to put them in due place (may be after a discussion through pm).

I have on my side a pending draft of code looking for exocets with 2 locked digits. I stopped that task (doubling something blue has already done) for a moment, but I should start the tests next week.

[eleven]

The eliminations then follow from comparing the digits in the target cells with those in the diagonal mini-row cells that must mirror them.

Please explain, i have no idea, what this means.

[David P Bird]

The eliminations then follow from comparing the digits in the target cells with those in the diagonal mini-row cells that must mirror them.

Please explain, i have no idea, what this means.

I'm trying to keep to my word to provide a summary of all the various Junior Exocet varieties which I hope to post when I've finished but not before. To make the descriptions easier I'm calling the cells in the diagonal mini-lines mirror nodes.

Code: Select all

  *---------*----------*----------*  *---------*----------*----------*
  | B  B  . | .  .  .  | .  .  .  |  | B  B  . | .  .  .  | .  .  .  | T1, T2 Target Cells 
  | .  .  . | T1 m2 m2 | /  .  .  |  | .  .  . | T1 .  .  | T2 .  .  | /      Companion Cells
  | .  .  . | /  .  .  | T2 m1 m1 |  | .  .  . | /  m2 m2 | /  m1 m1 | m1, m2 Mirror Nodes   
  *---------*----------*----------*  *---------*----------*----------*

Mirror nodes have 2 cells and must eventually contain the same true base digit as the diagonal target plus one other digit which I call a non-base digit. A non-base digit is one that is either missing from the base cells or false in them. This wording allows me to describe the inferences they provide.

Code: Select all

     *--------------------------*--------------------------*--------------------------*
     |  9       8       1234    |  7       12345   1235    |  6      B235    B135     |
     |  5       6      T123     |  123     9       8       |  123     4       7       |
     |  124     1234    7       |  1234    6      T1235    |  12358   23589   13589   |
     *--------------------------*--------------------------*--------------------------*

Here the mirror node for the r3c6 target is r2c12. (6)r2c2 is a non-base digit so (5)r2c1 must be the true base digit
=> r3c6 <> 123
The mirror node for the r2c3 target is r3c45. Again (6)r3c4 is a non-base digit so r3c4 must contain the true digit in r2c3
=> r3c4 <> 4

Here are the mirror node inferences:
1. Any base digit candidate that can't be true in both a target cell and its mirror node is false in these three cells.
2. If one mirror node cell can only contain non-base digits, the non-base digits in the other mirror cell are false.
3. 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 it.
4. If a mirror node contains a locked digit, any other digits it contains of the same type (known-base or non-base) are false.

I don't use the term 'equivalence' in my write up because I use it in its traditional sense for different digits that must be true or false together, not cells that must hold the same digit.

[Leren]

David P Bird wrote : This can't be a regular JE because of the 'S' cells for (5). However because (5)r2c1 is a given and (5) is also missing in r3c45, if it's true in the base cells it must be true in target r3c6. This makes the 'S' cell cover count for (5) unnecessary. Therefore whatever digits are true in the base cells must also be true in the target cells and all the regular JE eliminations will apply.

To my mind the wording of this paragraph needs improving. When testing a potential Exocet digit I check for: 1. A valid S cell count. Failing that, I try: 2. Some "other" method to prove that if the digit is True in the base cell it must be True in a target cell. So 1. is never unnecessary but 2. is unnecessary if 1 is passed. The point being that if a digit passes 1. then in addition to target cell and secondary equivalence (mirror?) eliminations, if that digit is subsequently proven to be True in the Base then S cell style eliminations can be made for that digit. On the other hand if the digit fails 1. but passes 2. target cell and secondary equivalence (mirror?) eliminations can be made but if that digit is subsequently proven to be True in the Base, no S cell style eliminations can be made. This is the case for digit 5 in this example. Your wording doesn't seem to reflect this.

Leren

[eleven]

Thanks David,

so the mirror node consists of the 2 cells in the other target's minirow (or column, if the JE is in a stack). If a base digit is true, it also must be there (for targets in different boxes).

Added: Can't see inference 3 now (would mean, that the mirror node must contain a non-base digit ?), but have no time to think about it.

[blue]

Here's a (rare?) instance of Danny's "Single-Target" exocet, that doesn't involve "mirror nodes".
The puzzle is from an old version of champagne's database.

Code: Select all

+----------------------+--------------------+----------------------+
| 9      8      2345   | 7     135    135   | 6      234    1234   | (S)
| 234    124    7      | 128   6      1389  | 12489  5      123489 |
| 2356   1256   2356   | 1258  13589  4     | 1289   2389   7      |
+----------------------+--------------------+----------------------+
| 8      3     B456    | 1456  157    2     | 149    4679   149    |
| 2467   9     B246    | 1468  1378   13678 | 1248   24678  5      |
| 24567  24567  1      | 9     578    5678  | 3      24678  248    | (S)
+----------------------+--------------------+----------------------+
| 1     T2456   245689 | 3     589    5689  | 7      2489   2489   | (S)
| 23567  2567   235689 | 568   4      56789 | 2589   1      2389   |
| 3457   457    34589  | 158   2      15789 | 4589   3489   6      |
+----------------------+--------------------+----------------------+
                               5      5              24     24

Base cells: r45c3
Target cell: r7c2
Normal digits: <245>
Extra digit: <6>

For digits <245>, it's like a normal JExocet, except that it's missing the usual target in r1c12.

Below, is an XSudo analysis.

Code: Select all

+-------------------------------+---------------------+------------------------+
| 9         8         3(245)    | 7     13(5)  13(5)  | 6      3(24)    13(24) | (S)
| 234       124       7         | 128   6      1389   | 12489  5        123489 |
| 2356      1256      235-6     | 1258  13589  4      | 1289   2389     7      |
+-------------------------------+---------------------+------------------------+
| 8         3         (456) B   | 1456  157    2      | 149    4679     149    |
| 247-6     9         (246) B   | 1468  1378   13678  | 1248   24678    5      |
| 7-6(245)  7-6(245)  1         | 9     78(5)  678(5) | 3      678(24)  8(24)  | (S)
+-------------------------------+---------------------+------------------------+
| 1       T -6(245)   89-6(245) | 3     89(5)  689(5) | 7      89(24)   89(24) | (S)
| 23567     2567      23589-6   | 568   4      56789  | 2589   1        2389   |
| 3457      457       34589     | 158   2      15789  | 4589   3489     6      |
+-------------------------------+---------------------+------------------------+
                                        5      5               24       24

11 Truths = {2R167 4R167 5R167 45N3}
15 Links = {2456c3 5c5 5c6 24c8 24c9 7n2 2456b4}
7 Eliminations --> r378c3<>6, r6c12<>6, r5c1<>6, r7c2<>6,

[eleven]

Nice example.

I supposed, that examples with exocet properties would be rare. But of course those are not needed for the conclusions.
E.g. if you have 123,123 in a minirow and you can show with any technique, that both a 1 in these cells forces a 1 in T, and a 2 would force a 2 in T, then you know, that a 3 must be in the 2 cells and T reduces to 12.

I don't think, that this "technique" has been tried already. (Probably with good reasons, i guess it will be rare too)

[David P Bird]

Quickly analysing Blues example in the same way as for JE:

Code: Select all

  *-------*--------*--------*
  | B B H1| H2 . . | H3 . . |  B = Base Cells holding (abcw) 
  | . . H4| T  . . | \  . . |  \ = (abc) absent 
  | . . H5| \  . . | \  . . |  T = Target Cell
  *-------*--------*--------*  H = Host Cells 
  | . . S | S  . . | S  . . |       
  | . . S | S  . . | S  . . |       
  | . . S | S  . . | S  . . |   
  *-------*--- ----*--------*   
  | . . S | S  . . | S  . . | 
  | . . S | S  . . | S  . . |     
  | . . S | S  . . | S  . . |   
  *-------*--------*--------*

The base cell must hold 4 digits (abcw), 3 of which must satisfy the JE 'S' cell requirement in three cross lines (abc) and one which needn't (w).
The cross line through the base box misses both the base cells
The other two cross lines must only have a single Target cell out of sight of the base cells capable of holding any of the digits (abc), and they could both cross the same box.

Each of (a) (b) & (c) can only be true twice at most in their 'S' cells so must be true at least once in the pattern band in the target cell and the five H cells.
As only the target cell is out of sight of the base cells, it must hold the digit out of (abc) that is a true base digit.
The other two digits must each occupy at least one of the other H cells so must be false in the base cells, leaving (w) to be the other true base digit.

There will many variations depending on the availability of the H cells.
Strangely enough if (w) were to pass the 'S' cell test it would be a hindrance rather than a help.
Just how common will this pattern be?

[daj95376]

Originally, I was concerned about contradictions while examining some JExocets. The JExocet path would lead to a strong link between the target cells, and another path would end with the candidate being false in both target cells. I recently converted my QExocet solver's logic over to utilize templates. Additional diagnostics explained the contradictions as discontinuous loops that forced a target cell true for one or more values. When (N-1) of N values exist as discontinuous loops to the same target cell, then the Exocet Single-Target pattern is present in the JExocet ... and additional eliminations result.

Here is a JExocet returned by champagne's solver. As a JExocet, I only found a few eliminations. However, 15 eliminations appeared when I applied the output from my QExocet solver.

Code: Select all

9876..5..4...9........87...6.5...9...3.....7....5....82.6.5.8...1...2........6..4

   c4b8  Locked Candidate 1              <> 8    r45c4
 r5  b4  Locked Candidate 1              <> 8    r5c46

 +--------------------------------------------------------------------------------+
 |  9       8       7       |  6       1234   Q14-3    |  5       1234    123     |
 |  4       26      123     |  123     9       5       |  12367   8       12367   |
 |  135     256     123     |  1234    8       7       |  12346   123469  12369   |
 |--------------------------+--------------------------+--------------------------|
 |  6       24-7    5       | R7-1234  1234-7  8       |  9       1234    123     |
 |  18      3       12489   | Q14-29   1246    149     |  1246    7       5       |
 |  17      2479    1249    |  5       12346-7 1349    |  12346   12346   8       |
 |--------------------------+--------------------------+--------------------------|
 |  2       479     6       |  1349-7  5       1349    |  8       139     1379    |
 |  3578    1       3489    |  3489-7 B47-3    2       |  367     3569    3679    |
 |  3578    579     389     |  1389-7 B17-3    6       |  1237    12359   4       |
 +--------------------------------------------------------------------------------+
 # 153 eliminations remain

 ***   value   = <3> true   locks cells true:   r1c6,r4c4   ( contradiction -- eliminate in )
                                                            ( base/target/secondary cells   )

 ( dropping the JExocet from four values to three )

 ***   value   = <1> true   locks cells true:   r1c6        ( Single-Target cell )
 ***   value   = <4> true   locks cells true:   r1c6        ( Single-Target cell )

 ***   value   = <7> true   locks cells true:   r4c4        ( value forced true in base cells and r4c4 )
                                                            ( exists as target cell due to JExocet )

 ### -1347- QExocet   Base = r89c5   Target = r4c4,r1c6==r5c4


_