The New Sudoku Players' Forum

[David P Bird]

98.7..6..7..6..5......4..3.6.85..9...2.........1......1...........1...96....571..

Code: Select all

 *-----------------------*-----------------------*-----------------------*
 | <9>    <8>    2345    | <7>    123    1235    | <6>    124    124     |
 | <7>    134    234     | <6>    12389  12389   | <5>    1248   12489   |
 | 25     156    256     | 289    <4>    12589   | 278    <3>    12789   |
 *-----------------------*-----------------------*-----------------------*
 | <6>    347    <8>     | <5>    1237   1234    | <9>    1247   12347   |
 | 345    <2>    34579   | 3489   136789 134689  | 3478   145678 134578  |
 | 345    34579  <1>     | 23489  236789 234689  | 23478  245678 234578  |
 *-----------------------*-----------------------*-----------------------*
 | <1>    3479   23479   | 23489  23689  234689  | 23478  24578  234578  |
 | 23458  3457   23457   | <1>    238    2348    | 23478  <9>    <6>     |
 | 2348   3469   23469   | 23489  <5>    <7>     | <1>    248    2348    |
 *-----------------------*-----------------------*-----------------------*

[Leren]

Code: Select all

*--------------------------------------------------------------------------------*
| 9       8       2345     | 7       123     1235     | 6       124     124      |
| 7       134     234      | 6       12389   12389    | 5       1248    12489    |
| 25      156     256      | 289     4       12589    | 278     3       12789    |
|--------------------------+--------------------------+--------------------------|
| 6       347     8        | 5       1237    1234     | 9       1247    12347    |
| 345     2       34579    | 3489    136789  134689   | 3478    145678  134578   |
| 345     34579   1        | 23489   236789  234689   | 23478   245678  234578   |
|--------------------------+--------------------------+--------------------------|
| 1       3479    23479    | 23489   23689   234689   |T234-78  2458-7  23458-7  |
| 23458   3457    23457    | 1      B238    B2348     | 23478   9       6        |
|T2348    346-9   2346-9   | 23489   5       7        | 1       248     2348     |
*--------------------------------------------------------------------------------*

Seems simple enough. Exocet 1: r8c5 r8c6 r9c1 r7c7 2348

1. r7c7 <> 7 (Non-base candidate in target cell);

2. r9c12 <> 9, r7c7 <> 8 (Mirror node inferences, or dual secondary equivalence - remove non-common digits from r7c7 and r9c12 except for linking digit 6);

3. r7c89 <> 7 (Mirror node inferences, or dual secondary equivalence - remove non-common digits from r9c1 and r7c89 except for linking digit 5);

4. lclste.

For convenience moves 1-3 are shown on the one PM.

Danny should love this one

Leren

[blue]

There are some curious features, though:

There's a 2nd JExocet: <2348>r9c89 with r8c1 and r7c4 as targets.
Together with Leren's "Exocet 1", it makes a JE4 (double exocet).
In the JE4, there's an 8 locked in r89c1,... that allows eliminating 8 from the other two targets.
[ It's an exocet digit, locked in two target cells ... one from each exocet. ]
(See additions below)

Theres a 5-digit exocet, <23479>r7c23 with r9c4 and r8c7 as targets.
It isn't a JExocet, because of the digits 7 and 9.
However, each of those would be (immediately) forced to a target cell as a box or row single, if it was present in the base.
[ It's a QExocet ? ]
The mirror node inferences for that exocet, force r8c7=7 and r9c4=9, and eliminate 2,3 and 4 from the base cells.
After that it's 'lclste'.

Added notes:

The eliminations for 8r7c4 and 8r7c7, from the above, are maybe not so interesting.
They're eliminated individually, anyway ... one by each of the JExocets.
Leren mentioned the one of them .. -8r7c7 .

I should mention though, that mirror node considerations aren't necessary, for those eliminations.
For Leren's "Exocet 1", for example: If an 8 was true in a base cell, then it would be forced to target cell r9c1 as a column single, and the other base digit would be forced into r7c7. On the other hand, 8 wasn't true in a base cell, then (still) one of the other digits would be forced to r7r7. In either case, r7c7 couldn't be an 8.

I should mention too, that there are two "single exocet" eliminations for 8's, that Leren didn't catch: -8r8c7 and -8r9c4.
I don't know, but probably David catches these somewhere in his deep analysis. [ David ? ]
I understand that there are may ways to skin a cat ... but the way I see them (again for Leren's "Exocet 1"):

8r8c7 is eliminated, since it would eliminate 8 from the base cells, and also force it (as a column single) into a target cell (r9c1).
Then 8r9c4 is eliminated, since with 8r7c7 already eliminated, it it were true, then:
In r8, it would be forced into r7c89, which is the mirror node for the r9c1 target, and/but also:
If it was true, it would eliminate 8 from r9c1 (which the base digit in r7c89 is supposed to "mirroring").

[Leren]

Code: Select all

*--------------------------------------------------------------------------------*
| 9       8       2345     | 7       123     1235     | 6       124     124      |
| 7       134     234      | 6       1238    1238     | 5       1248    12489    |
| 25      156     256      | 289     4       12589    | 28      3       12789        |
|--------------------------+--------------------------+--------------------------|
| 6       347     8        | 5       1237    1234     | 9       1247    1234     |
| 345     2       34579    | 3489    136789  134689   | 348     145678  13458    |
| 345     34579   1        | 23489   236789  234689   | 2348    245678  23458    |
|--------------------------+--------------------------+--------------------------|
| 1       3479    23479    |T234-89  2368-9  23468-9  | 2348    2458    23458    |
|T2348-5  345-7   2345-7   | 1       238     2348     | 23478   9       6        |
| 2348    3469    23469    | 23489   5       7        | 1      B248    B2348     |
*--------------------------------------------------------------------------------*

Here's the PM for the second Exocet r9c8 r9c9 r8c1 r7c4 2348.

This also has two dual secondary equivalences r7c4==r8c12 with linking digit 5 and r8c1==r7c56 with linking digit 6.

Presumably the two Exocets are a double, but either one is sufficient to solve the puzzle.

This supports an empirical observation of mine, that dual secondary equivalences are very common in double Exocets. I don't know why this is, there is possibly some deep reasoning why this is so.

Leren

[daj95376]

Aaaaagh !!!

My QExocet solver/analyzer locates patterns with 3/4 values and (currently) only flags an SL involving a target cell and another cell in the same box. Everything else -- including Leren's dual secondary equivalences -- I find manually and break down in steps. Whew !!!

Code: Select all

 +--------------------------------------------------------------------------------------+
 |  9        8        2345    |  7       123      1235     |  6       124      124      |
 |  7        134      234     |  6       12389    12389    |  5       1248     12489    |
 |  25       156      256     |  289     4        12589    |  278     3        12789    |
 |----------------------------+----------------------------+----------------------------|
 |  6        347      8       |  5       1237     1234     |  9       1247     12347    |
 |  345      2        34579   |  3489    136789   134689   |  3478    145678   134578   |
 |  345      34579    1       |  23489   236789   234689   |  23478   245678   234578   |
 |----------------------------+----------------------------+----------------------------|
 |  1        3479     23479   | q234-89  238+6-9  2348+6-9 | Q234-78  248+5-7  2348+5-7 |
 | r2348-5   34+5-7   234+5-7 |  1      B238     B2348     |  7-2348  9        6        |
 | R2348     34+6-9   234+6-9 |  9-2348  5        7        |  1      b248     b2348     |
 +--------------------------------------------------------------------------------------+
 # 201 eliminations remain


Code: Select all

 Step #1:

 ### -2348- QExocet   Base = r8c56 <24>   Target = r9c1,r7c7

 -7 r7c7   --   basic elims in target cells
 -8 r7c7   --   r7c7 combined with <8> not in r9c23


Code: Select all

 Step #2:

 ### -2348- QExocet   Base = r9c89 <38>   Target = r8c1,r7c4

 -9 r7c4; -5 r8c1   --   basic elims in target cells
 -8 r7c4            --   r7c4 combined with <8> not in r8c23


Code: Select all

 Step #3:

 *** double QExocet

 -2348=7 r8c7     --   cell sees both base sets
 -2348=9 r9c4     --   cell sees both base sets

 -7 r7c89,r8c23   --   r8c7=7
 -9 r7c56,r9c23   --   r9c4=9


_

[Edit: completely revamped explanations for many eliminations.]

[Leren]

blue wrote : I should mention too, that there are two "single exocet" eliminations for 8's, that Leren didn't catch: -8r8c7 and -8r9c4.

Now that's a real curve ball - I don't understand those eliminations at all (as single Exocet eliminations). Could you explain the reasoning ?

Leren

[David P Bird]

Code: Select all

*------------------------*------------------------*-----------------------*
 | <9>     <8>    2345    | <7>     123    1235    | <6>    124    124     |
 | <7>     134    234     | <6>     12389  12389   | <5>    1248   12489   |
 | 25      156    256     | 289     <4>    12589   | 278    <3>    12789   | 2..
 *------------------------*------------------------*-----------------------*
 | <6>     347    <8>     | <5>     1237   1234    | <9>    1247   12347   |
 | 345     <2>    34579   | 3489    136789 134689  | 3478   145678 134578  | .34 
 | 345     34579  <1>     | 23489   236789 234689  | 23478  245678 234578  | 234
 *------------------------*------------------------*-----------------------*
 | <1>     79-34  79-234  | 23489 b 23689  234689  | 23478a 24578  234578  |
 | 23458 b 3457   23457   | <1>     238 A  2348 A  | 23478  <9>    <6>     |
 | 2348  a 3469   23469   | 23489   <5>    <7>     | <1>    248 B  2348 B  |
 *------------------------*------------------------*-----------------------*
                             8                        8

(2348)JE4:r8c56,r9c1,r7c7; r9c89,r8c1,r7c4 (cover houses for (8) are c4 & c7)
=> r7c23 <> 2348 (seen by all 4 targets) sste

This is a curious double JExocet as, despite the abundance of deductions available, only one of them is needed, which is why I chose the curveball title.
I don't know if finding a pattern with multiple deductions counts as a single-stepper in this puzzle section or not.

Some interesting observations have been made though.

Actually I was looking for a JE4 where one of the base digits needed three cover houses. That would mean that at least one of the JE2 patterns would have to be true and there would be a strong link between them.

Leren/DAJ if don't you call the pattern a JExocet it means that you found it by logical testing, not by recognising that the JE pattern requirements were satisfied. Being pernickety this would mean you couldn't automatically assume that the JE theorem to eliminate true base digits in non-'S' cells will hold which would include the eliminations of (8)r7c47.

However if your code finds the double Exocet you could say that (8) must be true in r8c1 or r9c1 as it is locked in these targets. Therefore it can't be true in the other two at r7c4 & r7c7.

[daj95376]

I missed blue's single exocet eliminations as well. This is the same pattern that I've also missed previously.

Code: Select all

8r8c7 - 8r8c56,r8c1 = 8r9c1,r8c56 - 8r8c7   (discontinuous loop)

8r9c4 - 8r9c89,r9c1 = 8r8c1,r9c89 - 8r9c4   (discontinuous loop)


_

Leren/DAJ if don't you call the pattern a JExocet it means that you found it by logical testing, not by recognising that the JE pattern requirements were satisfied. Being pernickety this would mean you couldn't automatically assume that the JE theorem to eliminate true base digits in non-'S' cells will hold which would include the eliminations of (8)r7c47.

I got tired of banging heads with you over what constitutes a JExocet pattern. Your Compendium did little to clarify it for me. So, I simply don't mark any pattern as JExocet anymore.

BTW, if you check my eliminations above, you'll see that I performed -8r7c47.

_

[David P Bird]

Leren/DAJ if don't you call the pattern a JExocet it means that you found it by logical testing, not by recognising that the JE pattern requirements were satisfied. Being pernickety this would mean you couldn't automatically assume that the JE theorem to eliminate true base digits in non-'S' cells will hold which would include the eliminations of (8)r7c47.

I got tired of banging heads with you over what constitutes a JExocet pattern. Your Compendium did little to clarify it for me.

DAJ I'm truly sorry about that, I did try.

[Leren]

Thanks Danny - nice explanation of those 8 eliminations - definitely a move I'll add to my solver.

David, I do keep track of which digits satisfy JE pattern requirements and those which require some "other method" to prove that they are in the targets if they are in the base, precisely because you can't make S cell eliminations for the "other method'' digits.

My understanding is that if all digits conform to the JE pattern rules it's a JExocet but if one or more digits requires some ""other method" it isn't. I've just been too lazy to add a J in my printout if all digits conform. That's also something I'll put into my "gunna" list.

Leren