The New Sudoku Players' Forum

[David P Bird]

Champagne, I'm pleased I spotted it before you did!

But in your reply to Leren

I feel uncomfortable with your proof.
what if '8' is in both bases.
For me, the only thing to do is to prove that this can't be. In your writing, I don't see such a proof.

But I showed earlier that that can't happen with 4 target cells when

The two targets cells for A see base set B, and vice versa, so the two JEs must contain 4 different digits. This is true for all double JExocets that have 4 target cells, so there is no need to dive into examining cases.

I don't think Leren's approach is wrong, but what I'm aiming for is a concise justification that's easy to follow and in line with our traditional approaches. I'm using a short-cut that depends on a) 4 target cells, and b) 9 cells holding the partial fish for all 4 base digits, which I guess won't always be true (I've only looked at 5 of your puzzles), but what we'd like is a simple test that possibly can be registered as a pattern.

Unfortunately my Sudoku time has been limited recently and that's probably going to continue for another 2 months.

David

[champagne]

Champagne, I'm pleased I spotted it before you did!

But in your reply to Leren

I feel uncomfortable with your proof.
what if '8' is in both bases.
For me, the only thing to do is to prove that this can't be. In your writing, I don't see such a proof.

But I showed earlier that that can't happen with 4 target cells when ....


David

Hi David,

I would fully agree with you when JE's are there.

As written in my last post, I don't see JE's here, only partial ones.

In such a case, nothing prevent "a priori" the digit '8' to occupy both bases (as far as I understand the logic of that position).

[David P Bird]

Champagne, I see your problem but lets put it this way:

Whichever of the 2 JEs doesn't contain (8) will be valid and will resolve the base and target cells to (ab)
The other JE will have both these digits eliminated because the base cells will see one target holding (a) and the other target holding (b)
It will then reduce them to a pair (c8), so (8) must exist in the other base pair.

(It looks as if this the basis of the required pattern when there are 4 base digits and 4 target cells, but we need to test it properly)

David

[Leren]

Hi Champagne,

This is an excerpt of what I said in my last post.

Now it's only necessary to prove that the Spoiler digit 8 can't appear in both of r12c5 or r89c6 for the ADE move to work.

Why is that ? Well, here's the key observation:

Suppose 8 can't appear in r12c5. Then a (269) Jexocet exists in r1c5 r2c5 r5c6 r7c4.

On the other hand suppose 8 can't appear in r89c6. Then a (269) Jexocet exists in r8c6 r9c6 r3c4 r5c5.

So one of two (269) Jexocets exists but we don't know which one it is. No problem !

In other words I am agreeing with you that you must prove that the Spoiler digit (to use David's term), in this case 8, can't occupy both base pairs. In your earlier post to David you
appeared to have also provided proofs for digits 2, 6 and 9, which I don't think is necessary, because you have already established the ADE position. For the Spoiler digit proof I use Scenario testing
in a similar fashion to what I do for SK loop scenario testing etc. Hopefully this has cleared up any misunderstandings.

Perhaps you can provide a link to your full list of similar puzzles.

Leren

[champagne]

In your earlier post to David you
appeared to have also provided proofs for digits 2, 6 and 9, which I don't think is necessary, because you have already established the ADE position.

In that process, I make no assumption about other patterns, so I did not check for partial JE's.

I am using the fish analysis as shown in my post to verify if some digits can be validated through that simple process first, then I try a chain net expansion.

This is IMO the most general process I can implement to locate such pattern. JE partial analysis works as well in some cases.

You'll see later in the list some puzzles where 3 of the 4 digits must be checked through a chain expansion (I did not noticed a puzzle where the 4 digits must be checked using the expansion process).

Perhaps you can provide a link to your full list of similar puzzles.

This was included in one of the recent posts


I loaded the entire file in my google drive
with the file name ph_band_comp_AAHS.zip

[champagne]

Champagne, I see your problem but lets put it this way:

Whichever of the 2 JEs doesn't contain (8) will be valid and will resolve the base and target cells to (ab)
The other JE will have both these digits eliminated because the base cells will see one target holding (a) and the other target holding (b)
It will then reduce them to a pair (c8), so (8) must exist in the other base pair.

(It looks as if this the basis of the required pattern when there are 4 base digits and 4 target cells, but we need to test it properly)

David

Hi David,

Sorry David, but if a digit does not verify the JE rule, you can not say that if the digit occupy the base it will occupy one of the target. Nothing prevent such a digit to occupy both bases and none of the target cells. Normally, in such a case, the other digit of the base occupies both targets. We have seen long ago such an example on a single partial JE;

[JC Van Hay]

A condensed presentation of the solutions of puzzles #3 and #5 :

Puzzle #5 : ..............1.23..2.345.....6.......3.27.5..8...37.....9.54....4.7...51....627.;759669;dob;12_12_19;156;1689 ;169;1689

#1. LC(4B3,2B8) :=> -4r1c12,-2r8c12; UP25
#2. Analysis of (abcd)r12c7+(abcd)r79c9 where {a,b,c,d}={1,6,8,9}

2a. r12c7+r79c9=1/6/9 -> Invalid Swordfish in rows 358

Code: Select all

+------------------------+-----------------------+--------------------+
| 356789  135679  156789 | 2578    5689     289  | 1689  1469   14679 |
| 456789  45679   56789  | 578     5689     1    | 689   2      3     |
| 679(8)  1679    2      | 7(8)    3        4    | 5     169    1679  |
+------------------------+-----------------------+--------------------+
| 24579   124579  1579   | 6       (14589)  (89) | 139   13489  1249  |
| 469     1469    3      | 14-8    2        7    | 169   5      1469  |
| 24569   8       1569   | 145     (1459)   3    | 7     1469   12469 |
+------------------------+-----------------------+--------------------+
| 23678   2367    678    | 9       (18)     5    | 4     136    168   |
| 369(8)  369     4      | 123(8)  7        2(8) | 1369  1369   5     |
| 1       359     589    | 34-8    (48)     6    | 2     7      89    |
+------------------------+-----------------------+--------------------+

2b. r12c7+r79c9=8 -> Chain[7] : 8r3c4=8r3c1-8r8c1=8r8c46-(8=14)r79c5-(14=589)r46c5r4c6 :=> LC(8r4c45) :=> 8R5 is empty

Code: Select all

+------------------------+-----------------------+--------------------+
| 356789  135679  156789 | 2578    5689     289  | ab    cd     47-169|
| 456789  45679   56789  | 578     5689     1    | ab    2      3     |
| 679(8)  1679    2      | 7(8)    3        4    | 5     cd     7-169 |
+------------------------+-----------------------+--------------------+
| 24579   124579  1579   | 6       (14589)  (89) | 139   34-189 1249  |
| 469     1469    3      | 14-8    2        7    | 169   5      1469  |
| 24569   8       1569   | 145     (1459)   3    | 7     4-169  12469 |
+------------------------+-----------------------+--------------------+
| 23678   2367    678    | 9       (18)     5    | 4     ab     cd    |
| 369(8)  369     4      | 123(8)  7        2(8) | 3-169 ab     5     |
| 1       359     589    | 34-8    (48)     6    | 2     7      cd    |
+------------------------+-----------------------+--------------------+

Conclusion : r12c7=ab -> r79c9=cd; r13c8=cd and r78c8=ab (because r3c7=5=r8c9 !) :=> -1689[r13c9(=47) and r8c7(=3) and r46c8(=34)]; UP34; NP(69)r35c1; UP81

Puzzle #3 : ..............1..2..3.4..15......6....14...37.8..7...1..5..4.7..3..5.9.42..73...6;246934;dob;12_12_03;158;2689 ;269;2689

#1. Analysis of (abcd)r12c5+(abcd)r89c6 where {a,b,c,d}={2,6,8,9}

1a. r12c5+r89c6=2/6/9 -> Invalid Swordfish in rows 357

Code: Select all

+--------------------------+----------------------+-----------------------+
| 1456789  1245679  246789 | 23569   2689  235679 | 3478    4689     389  |
| 456789   45679    46789  | 3569    689   1      | 3478    4689     2    |
| 679(8)   2679     3      | 269     4     2679   | 7(8)    1        5    |
+--------------------------+----------------------+-----------------------+
| 34579    24579    2479   | 123589  129   2359   | 6       (24589)  (89) |
| 569      2569     1      | 4       269   2569   | 25-8    3        7    |
| 34569    8        2469   | 23569   7     23569  | 245     (2459)   1    |
+--------------------------+----------------------+-----------------------+
| 169(8)   169      5      | 1269    1269  4      | 123(8)  7        3(8) |
| 1678     3        678    | 126     5     268    | 9       (28)     4    |
| 2        149      489    | 7       3     89     | 15-8    (58)     6    |
+--------------------------+----------------------+-----------------------+

1b. r12c5+r89c6=8 -> Chain[7] : 8r3c7=8r3c1-8r7c1=8r7c79-(8=25)r89c8-(25=489r46c8r4c9 :=> -8r5c7 :=> 8R5 is empty

Code: Select all

+--------------------------+------------------------+------------------+
| 1456789  1245679  246789 | cd.     ab     357-2689| 3478  4689   389 |
| 456789   45679    46789  | cd.     ab     1       | 3478  4689   2   |
| 6789     2679     3      | cd.     4      7-2689  | 78    1      5   |
+--------------------------+------------------------+------------------+
| 34579    24579    2479   | 135-289 1289   23589   | 6     24589  89  |
| 569      2569     1      | 4       2689   25689   | 258   3      7   |
| 34569    8        2469   | 35-269  7       23569  | 245   2459   1   |
+--------------------------+------------------------+------------------+
| 1689     169      5      | ab      1-2689 4       | 1238  7      38  |
| 1678     3        678    | ab      5      cd      | 9     28     4   |
| 2        149      489    | 7       3      cd      | 158   58     6   |
+--------------------------+------------------------+------------------+

Conclusion : r12c5=ab -> r89c6=cd; r123c4 contains cd and r78c4=cd (because r3c5=4=r7c6 !) :=> -2689[r12c6(=357) and r46c4(=135) and r7c5(=1)]; UP34; sk-basics and 1 XYWing to the end

[champagne]

I'll never try to fight against JC to show a short notation

To stay within the same group that puzzle

........1..2..3.4..5..6.72....7....4.87......2...8.67...8.2.5..5....6...6..9.....;689575;dob;12_12_19;12;568 ;4;1349

The "hard position" is rated 10.9 by skfr.

This puzzle is in the list because it shows some potential of elimination with digits 568, but the complementary AAHS are with digits 1349

Only one digit satisfies to the property through a fish analysis.
After the property has been established with relatively easy chain nets, we go to the end with easy moves.

[JC Van Hay]

What about :

........1..2..3.4..5..6.72....7....4.87......2...8.67...8.2.5..5....6...6..9.....;689575;dob;12_12_19;12;568 ;4;1349

#1. r5c4=6,r1c4=2, UP25; HP(56)r1c8r2c9, SF(5B3C43) :=> -5r4c8
#2. Chain[4] : (5=6)r1c8-6r1c3=(6-5)r4c3=5r6c4=5r2c4 :=> r1c8=5; UP28
#3. Chain[3] : 5r2c4=5r2c5-5r9c5=(5-8)r9c6=8r8c4 :=> -8r2c4
#4. FXWing(8C49) :=> -8r8c78
.... ?
#5. Double Exocet : r45c1+r89c3=1/3/4/9 -> Invalid Swordfish in rows 367 -> -1349[r7c1+r46c3+r12c2]; UP35; NP(39)r67c9; UP81

PS
Read #2. Chain[4] : (5=6)r1c8-6r1c3=(6-5)r4c3=5r6c3-5r6c4=5r2c4 :=> r1c8=5; UP28

[champagne]

What about :

#5. Double Exocet : r45c1+r89c3=1/3/4/9 -> Invalid Swordfish in rows 367 -> -1349[r7c1+r46c3+r12c2]; UP35; NP(39)r67c9; UP81

I have to decipher the full solution, but I would prefer

#5 Complementary AAHS ....

I could change my mind after a deeper study of your post.

On my side, this is a door open on may-be a significant lot of puzzles in the grey zone. As far as I can see, most of the puzzles in the entire list have ratings on the low side of the potential hardest file. (just referring to the sequence number)

I have still pending to work on the first example you gave.

[champagne]

What about :

........1..2..3.4..5..6.72....7....4.87......2...8.67...8.2.5..5....6...6..9.....;689575;dob;12_12_19;12;568 ;4;1349

#1. r5c4=6,r1c4=2, UP25; HP(56)r1c8r2c9,

SF(5B3C43) :=> -5r4c8

Question:

after HP(56) is my start.

what means SF(5B3C43) :=> -5r4c8

thanks

champagne

[JC Van Hay]

what means SF(5B3C43) :=> -5r4c8

SF(5B3C43) :=> -5r4c8 means Swordfish(5B3C43) : on 5s : r1c8=r2c9-r2c4=r6c4-r6c3=r4c3 :=> -5r4c8

[David P Bird]

Champagne, right, I misinterpreted Leren's proof. The scenario still to be proved false is for one of the digits to be true in both pairs of base cells.

I quickly looked at the restrictions on placing the other "base" candidates in the JE band. These are very severe, but depend on the positions of the givens. The only way I see to analyse the options is to iterate cases which would be very tedious. It might be that if only one of the 4 potential base candidates doesn't comply with the partial fish requirements, it doesn't matter.

So the question is, do you have an example in your collection where an Almost DJE (with 4 base digits, 4 target cells, and just 1 digit that doesn't comply with the pattern requirements) is false?

JC

That’s very clever. It's the first time I've seen a Double JExocet being developed using preliminary eliminations.

David

[champagne]

SF(5B3C43) :=> -5r4c8 means Swordfish(5B3C43) : on 5s : r1c8=r2c9-r2c4=r6c4-r6c3=r4c3 :=> -5r4c8

OK The chain makes sense for me, but I can't see that as a swordfish.

Generally speaking, you show, as noticed David, that an early stop in the solving process can hide an exotic pattern.

This should be true for a significant part of the file and my current solving code does not detect it because the check of the Exocet pattern (in the solving process) is done once for all before looking for XY chains.

The solution specific to that search was the following

start point after your
#1. r5c4=6,r1c4=2, UP25; HP(56)r1c8r2c9

Code: Select all

A     B      C     |D    E     F     |G      H     I     
34789 34679  3469  |2    4579  45789 |389    56    1     
1789  1679   2     |158  1579  3     |89     4     56   
13489 5      1349  |148  6     1489  |7      2     389   
--------------------------------------------------------
139   1369   13569 |7    1359  1259  |12389  13589 4     
1349  8      7     |6    13459 12459 |1239   1359  2359 
2     1349   13459 |1345 8     1459  |6      7     359   
--------------------------------------------------------
13479 13479  8     |134  2     147   |5      1369  3679 
5     123479 1349  |1348 1347  6     |123489 1389  23789
6     12347  134   |9    13457 14578 |12348  138   2378 

digit 4 is derived from the classical JE elimination

4r45c1+4r89c3 => deadly pattern 4r367

for other digits, we need more but this remains easy

Code: Select all

1r45c1+1r89c3
 =>1r2c2 -> 6r2c9 -> 6r7c8
 =>1r2c2 <1>r5c2   |
   1r6c46 <1>r45c5 | 1r89c5 =>1r7c8


Code: Select all

3r45c1+3r89c3
 => 3r3c9 -> 3r6c4 ->3r7c8 ->6r1c8
 => 3r3c9 -> 3r6c4 ->5r1c4 ->5r1c8


Code: Select all

9r45c1+9r89c3
 => HP 69r7c89           |
    XW 9r36c69-> <9>r7c9 | 6r7c9 ;5r2c9 | 5r6c4 5r4c3 
                                        | 6r1c8 6r4c3


once the property established, the game is over

A manual player can freely take one of these 2 ways. I had a private exchange with "abi" on that puzzle and I feel she should like that early exit

[champagne]

So the question is, do you have an example in your collection where an Almost DJE (with 4 base digits, 4 target cells, and just 1 digit that doesn't comply with the pattern requirements) is false?

I am not sure I catch your point.

I extracted puzzles where the 2 AAHS have complementary digits in the solution. In that sense, I would say No

In most cases, 3 of the 4 digits comply to the property through a fish analysis. In that sense, the biggest part should meet your spec, but this is the situation of the short list I posted here, so I doubt that this is the correct understanding.