The New Sudoku Players' Forum

[David P Bird]

In the world of cover set logic if, for any exocet pattern digit, there exists 1) a column void of candidates (as in the 2nd diagram above), or 2) less than two rows with less than two candidates (in the three columns as in the 3rd diagram above), those candidates becomes irrelevent to the logic set. Either 1) or 2) occurs in 41 of the 42 puzzles listed here.

A much simpler approach is possible:
For a candidate (a) that's true in the base cells, the JExocet pattern limits it to just two target cells in intersections of the 3 cross-lines with the JExcoet band.
Condition 3 limits the 18 external cross line cells to holding a maximum of two instances of (a), so ensuring it must occupy at least one of these target cells.
With two true base digits subject to this limitation, each one must occupy a different target cell, so eliminating all non-members from both of them.
There is therefore no need to resort to different SLG diagrams each time to prove this.

I know what you're trying to say, but I find those diagrams confusing. If you're going to show only six in the left, why show more than four in the others?

I was showing the most cells that could be occupied in the cross lines in each case.

Does that mean you don't consider base r9c13 and target r8c89 to be an "Exocet" either?

That doesn't comply with the JExocet pattern requirements where the base cell pair and the two target cells must all be in different boxes in the same band.

By champagne's rules, if combining the template options for a set of 3 or 4 digits succeeds in showing that two cells must hold the same digits as a selected pair of cells, that arrangement is an Exocet pattern, otherwise it's not. For an Exocet these cells are identified as targets where non-member digits can be eliminated. I therefore feel unqualified to answer.

[ronk]

In the world of cover set logic if, for any exocet pattern digit, there exists 1) a column void of candidates (as in the 2nd diagram above), or 2) less than two rows with less than two candidates (in the three columns as in the 3rd diagram above), those candidates becomes irrelevent to the logic set. Either 1) or 2) occurs in 41 of the 42 puzzles listed here.

A much simpler approach is possible:
For a candidate (a) that's true in the base cells, the JExocet pattern limits it to just two target cells in intersections of the 3 cross-lines with the JExcoet band.
Condition 3 limits the 18 external cross line cells to holding a maximum of two instances of (a), so ensuring it must occupy at least one of these target cells.
With two true base digits subject to this limitation, each one must occupy a different target cell, so eliminating all non-members from both of them.
There is therefore no need to resort to different SLG diagrams each time to prove this.

You and I apparently live on two different planets, so I give up trying to explain my POV to you.

[David P Bird]

You and I apparently live on two different planets, so I give up trying to explain my POV to you.

My objective was to describe condition 3 in such a way a player could check if it was satisfied as simply as possible, without needing a computer solver.

This can be done by jotting down the member digits that occupy the critical lines at the side of the grid and counting. (You can see this in the last puzzle grid I posted.)

Your posts have given me the impression that the only way you can navigate the inferences for a new JExocet pattern is to produce an SLG diagram for it.

If you consider that your approach is simpler, then please re-write condition 3 as you would like to see it and post it here.

If you do, I promise before I say anything else in response I will at least say thank you for your efforts.

This would then result in us having a clear winner or two alternative versions which would allow players to select the one that suits them best.

[daj95376]

While testing puzzles that I gathered from the Hardest Sudokus Thread, I ran across this Double-J Exocet puzzle with secondary placements -- r2c8==r3c4 and r3c1==r1c7. There's also r3c6==r2c9, but it doesn't appear to be part of the Exocet logic.

Code: Select all

HardestSudokusThread-02092

1..4....9.56..9.......1..6..6....8..5....4.9.9....5.1..7....2..6....1.5....3.....

 +---------------------------------------------------------------------------+
 |  1      238    2378    |  4      235678   23678  |  357    2378   9       |
 |  23478  5      6       |  278    2378     9      |  1347   23478  123478  |
 |  23478  23489  234789  |  2578   1        2378   |  3457   6      23478   |
 |------------------------+-------------------------+------------------------|
 |  2347   6      12347   |  1279   2379     237    |  8      2347   5       |
 |  5      1238   12378   |  12678  23678    4      |  367    9      2367    |
 |  9      2348   23478   |  2678   23678    5      |  3467   1      23467   |
 |------------------------+-------------------------+------------------------|
 |  348    7      134589  |  5689   45689    68     |  2      348    13468   |
 |  6      23489  23489   |  2789   24789    1      |  3479   5      3478    |
 |  248    12489  124589  |  3      2456789  2678   |  14679  478    14678   |
 +---------------------------------------------------------------------------+
 # 197 eliminations remain

 ### <2378> JExocet:   Base = r1c23   Target = r2c8==r3c4,r3c6
 ### <2378> JExocet:   Base = r2c45   Target = r1c8,r3c1==r1c7


Code: Select all

Eliminations:

r2c8<>34, r3c4<>5, r1c7<>5, r3c1<>248, r1c56,r2c1<>2378; STE


[David P Bird]

There's also r3c6==r2c9, but it doesn't appear to be part of the Exocet logic.

Interesting, that puzzle is the first double Exocet I've seen with the base cells on different lines.

For a pair of targets on different lines, there will always be an equivalence between target (a) and one of the two cells in the same mini-line as the target (b). When one of these cells can be eliminated because it can't hold a member digit it must be the other. At the start that will usually be because there's a given involved.

When that's not the case, as soon as one of the two possibilities is eliminated, the equivalence will be revealed and could be used. But generally by that stage I'd guess that much simpler methods should be available to finish the solution.

[JC Van Hay]

While testing puzzles that I gathered from the Hardest Sudokus Thread, I ran across this Double-J Exocet puzzle with secondary placements -- r2c8==r3c4 and r3c1==r1c7. There's also r3c6==r2c9, but it doesn't appear to be part of the Exocet logic.

Code: Select all

HardestSudokusThread-02092

1..4....9.56..9.......1..6..6....8..5....4.9.9....5.1..7....2..6....1.5....3.....

 +---------------------------------------------------------------------------+
 |  1      238    2378    |  4      235678   23678  |  357    2378   9       |
 |  23478  5      6       |  278    2378     9      |  1347   23478  123478  |
 |  23478  23489  234789  |  2578   1        2378   |  3457   6      23478   |
 |------------------------+-------------------------+------------------------|
 |  2347   6      12347   |  1279   2379     237    |  8      2347   5       |
 |  5      1238   12378   |  12678  23678    4      |  367    9      2367    |
 |  9      2348   23478   |  2678   23678    5      |  3467   1      23467   |
 |------------------------+-------------------------+------------------------|
 |  348    7      134589  |  5689   45689    68     |  2      348    13468   |
 |  6      23489  23489   |  2789   24789    1      |  3479   5      3478    |
 |  248    12489  124589  |  3      2456789  2678   |  14679  478    14678   |
 +---------------------------------------------------------------------------+
 # 197 eliminations remain

 ### <2378> JExocet:   Base = r1c23   Target = r2c8==r3c4,r3c6
 ### <2378> JExocet:   Base = r2c45   Target = r1c8,r3c1==r1c7


Code: Select all

Eliminations:

r2c8<>34, r3c4<>5, r1c7<>5, r3c1<>248, r1c56,r2c1<>2378; STE


I am may be missing something, but what is the logic behind the JExocet that justifies -3r2c8, -28r3c1 ?

The simplest I can see in this puzzle :

#1. Double exocet :

AALS(2378)r1c23 [SSF(C168) => Target : r2c8=r3c6]
AALS(2378)r2c45 [SSF(C168) => Target : r1c8=r3c1]

=> the 2 AALS cannot contain the same digits

:=> -2378r1c56.r2c1; 28 Singles

#2. FXW(8R19) :=> -8r3c1; ste

Best Regards, JC.

[ronk]

I ran across this Double-J Exocet puzzle with secondary placements -- r2c8==r3c4 and r3c1==r1c7
....
### <2378> JExocet: Base = r1c23 Target = r2c8==r3c4,r3c6
### <2378> JExocet: Base = r2c45 Target = r1c8,r3c1==r1c7

A description David P Bird might write, but with (hopefully) common wording: With the two base sets r1c23 and r2c45 ultimately holding all four exocet digits <2378>, minirow r1b2 cannot hold any of them. Then, since two of the remaining six cells in minirows r1b13 are void of exocet digits, there is a hidden quad in r1c2378. Similarly, since two of the remaining six cells in minirows r23b2 are also void of exocet digits, there is a second hidden quad in [r2c45,r3c46].

Two hidden sets in r1c2378 and r23b2 is reminiscent of the Law of Leftovers.

[edits: "hidden quad" was "hidden set"; add the following]

I am may be missing something, but what is the logic behind the JExocet that justifies -3r2c8, -28r3c1 ?

Addressing only the first instance, primary [r2c8,r3c6] and secondary r3c46 target cells for base r1c23 must ultimately hold the same digits. After applying the hidden quad in b2 per the above, r3c4=278. Then, since the primary and secondary targets share r3c6, r2c8=278 as well.

Centerfold: Show

____

Truth/Link Solutions: Show

Code: Select all

100400009056009000000010060060000805500004090900005010070000200600001050000300000

     24 Truths = {2378R1 2378C1 2378C6 2378C8 1N23 2N45 2378B2}
     30 Links = {2r2349 3r2347 7r2349 8r2379 1n78 2n8 3n146 2378b1 2378b3}
     49 Eliminations --> r7c3459<>8, r9c2359<>8, r1c5<>2378, r1c6<>2378, r2c1<>2378, r3c9<>2378,
     r4c345<>2, r4c345<>7, r9c235<>2, r9c579<>7, r3c1<>248, r4c35<>3, r7c39<>3,
     r2c8<>34, r3c7<>37, r1c7<>5, r3c4<>5

Letting XSUDO go crazy with these same 24 Truths:
     24 Truths = {2378R1 2378C1 2378C6 2378C8 1N23 2N45 2378B2}
     43 Links = {2r249 3r2347 7r2349 8r279 7c7 3479n1 3n4 3479n6 1n7 1249n8 2b1348 3b167 7b159 8b1378}
     111 Eliminations, 11 Assignments --> r7c134589<>8, r9c235689<>8, r235689c7<>7, r3c12379<>3, r4c34568<>2,
     r4c13458<>7, r1c5678<>3, r1c3568<>7, r3c3469<>7, r4c1356<>3, r8c2345<>8,
     r9c1235<>2, r3479c1<>4, r2c1<>2378, r3c169<>2, r3c169<>8, r7c389<>3,
     r9c569<>7, r249c8<>4, r2c9<>278, r1c56<>2, r1c56<>8, r2c58<>3, r5c23<>2,
     r5c79<>3, r5c45<>7, r6c23<>2, r6c79<>3, r6c45<>7, r8c45<>2, r8c23<>3,
     r79c6<>6, r1c7<>5, r2c8<>7, r3c4<>5, r8c9<>7,
     r1c7=7, r3c6=3, r3c1=7, r4c1=2, r4c8=3, r4c6=7, r7c1=3, r7c6=8, r9c6=2,
     r9c8=7, r9c1=8

[daj95376]

I may be missing something, but what is the logic behind the JExocet that justifies -3r2c8, -28r3c1 ?

It's not the individual JExocet's that force the secondary placements; but, rather, an effect of the Double Exocet and the mini-units r1b3 and r3b2 only being able to hold two assignments.

Code: Select all

 Double Exocet: r1c23=ab, r2c8=a, r3c6=b => r2c45=cd, r1c8=c, r3c1=d

 r3c4 only place remaining for "a" in [b2]
 r1c7 only place remaining for "d" in [b3]

 the only base digits common to r2c8 and r3c4 are <278>
 the only base digits common to r3c1 and r1c7 are <37>
 +-----------------------------------------------+
 |   /  ab  ab   |   .   .   .   |   d   c   /   |
 |   .   .   .   |  cd  cd   /   |   .   a   .   |
 |   d   .   .   |   a   /   b   |   .   /   .   |
 |---------------+---------------+---------------|
 |   .   .   .   |   .   .   .   |   .   .   .   |
 |   .   .   .   |   .   .   .   |   .   .   .   |
 |   .   .   .   |   .   .   .   |   .   .   .   |
 |---------------+---------------+---------------|
 |   .   .   .   |   .   .   .   |   .   .   .   |
 |   .   .   .   |   .   .   .   |   .   .   .   |
 |   .   .   .   |   .   .   .   |   .   .   .   |
 +-----------------------------------------------+


Regards, Danny

[JC Van Hay]

Ronk, Danny,


Many thanks for your thorough and crystal clear explanations.

I hope you will forgive me for asking as I was only confused by the term JExocet in place of Exocet, so expecting explanations different from what you gave.


Best Regards, JC.

[daj95376]

I'd like to include my template solver's results.

Templates w/UR reduction included:

Hidden Text: Show

Code: Select all

 Templates: 5 212 196 105 3 16 258 308 11   *** 2-template (reduction) completed
 Templates: 5 200 183  91 3 16 236 253  9   *** 3-template (reduction) completed
 Templates: 5 148 154  86 3 14 181 189  9   *** 3-template (reduction) completed
 Templates: 5 141 152  86 3 14 176 188  9

 <2378>   accepted = 16 combinations

 <2378>   <>1  r2c9
 <2378>   <>2  r1c56,r2c149,r3c169,r4c34568,r56c23,r8c45,r9c1235
 <2378>   <>3  r1c5678,r2c1578,r3c12379,r4c1356,r56c79,r7c389,r8c239
 <2378>   <>4  r3479c1,r8c57,r249c8,r268c9
 <2378>   <>5  r1c7,r3c4
 <2378>   <>6  r56c59,r79c6
 <2378>   <>7  r1c3568,r2c15789,r3c34679,r4c13458,r56c457,r8c479,r9c5679
 <2378>   <>8  r1c56,r2c149,r3c169,r7c134589,r8c2345,r9c235689
 <2378>   <>9  r8c57


Note: I created the Naked Singles results manually after performing the above eliminations.

Of interest, only 16 combinations of templates for <2378> were accepted out of ( 141*152*176*188 ) possible combinations.

[Edit: removed superfluous information about Naked Singles.]

[ronk]

I'd like to include my template solver's results.
...
Resulting Naked Singles: 23

r4c1,r9c6 = 2
r2c9,r3c6,r4c8,r7c1,r8c7 = 3
r1c7,r2c4,r3c1,r4c6,r8c5,r9c8 = 7
r7c6,r8c9,r9c1 = 8

r2c1,r3c9,r7c8 = 4
r1c6,r5c7 = 6
r4c5,r8c4 = 9
[/code]
Note: I created the Naked Singles results manually after performing the above eliminations.

The digit <2378> placements are provided by inspection of the template combinations, so I'm assuming "manually" applies to the other digits. Naked singles for these other digits are indirect results, and IMO including them overstates the results for templating. Of course, that might be the reason for your 'Note.'

[daj95376]

The digit <2378> placements are provided by inspection of the template combinations, so I'm assuming "manually" applies to the other digits. Naked singles for these other digits are indirect results, and IMO including them overstates the results for templating. Of course, that might be the reason for your 'Note.'

You are correct. That's what I get for working while tired and not "thinking" about what caused the results that I had gathered. Sorry!!!

[champagne]

back home, I can read many discussions.

I don't see that much to comment, so I continue with the data base analysis.

After I released the constraint to have all digits in the target, I have many more exocets, but still one or two bases, except for 3 new oddities

Code: Select all

..3..6...4.67...3..7....5...6...597..3...7.4.9............2......8.....1.4...3.5.;6519;elev;3594;
  r2c7 r2c9 r1c1 r1c2;
  r2c7 r2c9 r1c2 r3c6;
  r3c1 r3c3 r1c8 r2c6;
  r3c1 r3c3 r2c5 r2c6;
  r1c1 r3c1 r5c3 r6c2;
  r7c2 r8c2 r3c3 r5c3
..3..6...4.67...3..7....5...6...5.7..3.9.7.4.9............2......8.....1.4...3.5.;6527;elev;3595
  r2c7 r2c9 r1c1 r1c2;
  r2c7 r2c9 r1c2 r3c6;
  r3c1 r3c3 r1c8 r2c6;
  r3c1 r3c3 r2c5 r2c6;
  r1c1 r3c1 r5c3 r6c2;
  r7c2 r8c2 r3c3 r5c3
..3.5.......7...2.7...31..5....4..9..6.........51....2.....8.....73....18.25....7;6530;elev;3613
  r2c5 r2c6 r1c9 r3c3;
  r2c5 r2c6 r3c2 r3c3;
  r3c7 r3c8 r1c4 r1c6;
  r3c7 r3c8 r1c4 r2c3;
  r4c4 r5c4 r2c5 r8c5;
  r1c6 r2c6 r7c4 r8c5

these 3 puzzles seem to have (unchecked) interleaved exocets from different bases.

I intend by to morrow, to publish an updated version of the data base (31804 puzzles) with a full analysis of the exocets complying with the revised definition

champagne

PS: next week, I'll have heavy other constraints

[ronk]

I intend by to morrow, to publish an updated version of the data base (31804 puzzles) with a full analysis of the exocets complying with the revised definition
...
PS: next week, I'll have heavy other constraints

Sometime soon, maybe even before doing all of the above, would you please look at "G1, G2, ... multi-fish" detection. I've been picking puzzles out of the "03 NN nothing special" file, hoping to find new interesting 0-rank logic patterns. So far, every time I thought I'd found one, out pops this row-only or column-only multi-digit fish, usually while preparing a post.

If you need more examples, I have several.

[champagne]

Sometime soon, maybe even before doing all of the above, would you please look at "G1, G2, ... multi-fish" detection. I've been picking puzzles out of the "03 NN nothing special" file, hoping to find new interesting 0-rank logic patterns. So far, every time I thought I'd found one, out pops this row-only or column-only multi-digit fish, usually while preparing a post.

If you need more examples, I have several.

first of all, the next step in my "to do" list is to work on the multi fish detection.

but I have difficulties to understand what is your point.

If a puzzle is in the list "03 NN nothing special" (the reduced one or the true one built out of the entire list less the puzzles having something special)
it means that my rank_0_logic detection did not find a known pattern in it.

My goal, in the revision is
a) to include existing process in the new code, that should perform better
b) to extend the search to some patterns you have found


I have small chances in that run to achieve more than to extract patterns similar to those you already used

but as you write, sometimes, examples help to understand.

champagne

PS The search for an exocet with a 5 digits base did not gave any positive answer
I run the test on the file of puzzle having no exocet 3/4 digits