The New Sudoku Players' Forum

[Cenoman]

EDIT2. Warning to readers. In the first version of this post I raised questions whether my treatment of incompatibility of the pair 25 was acceptable. The target was missed due to a logic fault I made in the use of the JE pattern for the pair 23 (thanks to champagne for spotting) Each occurrence of the fault is mentionned with strikes and blue types.
Nevertheless, a repair for the incompatibility of the pair 23 has been proposed (Thanks to eleven) This third version takes such repair solution into account.

My attention has been drawn a few weeks ago on the "Weekly Extreme Unsolvable" #281 on Andrew Stuart's site
(Sudokuwiki) To access #281, select it in the archive zone & click "show puzzle"

.2.34....4...5......6..17..5..2......3.....4...1..68....9....7....1..9.......9.68

This puzzle is rated S.E. 9.2
It can be solved several ways. I have re-opened David P Bird's Junior Exocet Compendium (here)
The notation and vocabulary follow this document.

The puzzle after basics (2 placements, locked sets)

Code: Select all

 +--------------------------+-------------------------+---------------------------+
 |  19      2      (5)78    |  3      4       78      | (5)16    1589    1569     |
 |  4       19     (3)78    |  6      5      (2)78    | (23)1    12389*  1239*    |
 |  38      58      6       |  89     289     1       |  7       2358    4        |
 +--------------------------+-------------------------+---------------------------+
 |  5       46789   48      |  2      13789  (3)48    | (3)16    139     13679    |
 |  26789   3      (2)8     |  5789   1789   (5)8     | (25)16   4       125679   |
 |  279     479     1       |  4579   379     6       |  8       2359*#  23579*#  |
 +--------------------------+-------------------------+---------------------------+
 |  12368   14568   9       |  458    2368  t<235>-48 |  12345   7       1235     |
 |  23678   45678   23458   |  1      23678   23458   |  9     B<235     235>     |
 |  1237    1457  t<235>-4  |  457    237     9       |  12345   6       8        |
 +--------------------------+-------------------------+---------------------------+

Exocet (235)JE2: r8c89, r9c3, r7c6
Direct eliminations: -48 r7c6, -4 r9c3 (non base digits in target cells)
Three base digits, hence three possible pairs for base cells r8c89: 23, 25, 35.

EDIT
For the pair (23)r8c89, two UR threats (*) exist in r2c89 and r6c89.
Their S-cells in target cross-lines (c3, c6) occupy four boxes
=>pair 23 is incompatible.
For the pair (23)r8c89, only one UR threat (*) exist in r6c89.
If target cells are assigned the values r7c6=2 and r9c3=3, then r6c89=23 (and UR(23)r68c89).
If target cells are assigned the values r7c6=3 and r9c3=2, then r3c8=23 (contradiction, r3c8 only cell in box 3 for digits 2 & 3) [eleven's repair, see his post below]
=>pair 23 is incompatible

For the pair (25)r8c89, only one UR threat (#) exists in r6c89.
If target cells are assigned the values r7c6=2 and r9c3=5, then r6c89=25 (and UR(25)r68c89).
If target cells are assigned the values r7c6=5 and r9c3=2, then r5c3=8 and r5c6=8 (contradiction)
=>pair 25 is incompatible
EDIT: this could be presented using JE pattern digits only.
If target cells are assigned the values r7c6=5 and r9c3=2, then r5c7=25 (contradiction, r5c7 only cell in box 6 for digits 2 & 5)

Only the pair (35)r8c89 is compatible => -2 r8c89, puzzle now solved with basics.

My questions to JE users are:
- is my interpretation of §6 in JE Compendium correct ("Incompatible base pairs") ? I had to expanse a bit the
threat for pair (25)r8c89...
- is not such an exploration of the diverse pairs and target cells assignment, assimilated to some T&E ?
- in the end, if the method is acceptable, how about the result presentation ?

Thanks in advance.

[David P Bird]

Hi Cenoman,

I have posted a response in the JExocet Compendium thread. The conversation in that thread has since been deleted.

David PB

[Leren]

Withdrawn

[David P Bird]

Leren,
the aim of JExocet Compendium was to produce squeaky clean patterns that could be recognised without making any assumptions.

The cost of this approach is that rarely occuring special cases which involve further analysis to to prove won't be picked up. To be able to extend the definitions to include them would require:
a) Definable pattern elements to exist that make the extension recognisable without following logical streams.
b) A high enough recurrence frequency to make the extra search effort worthwhile - if not any extra wrinkle will soon be lost in obscurity.

Although this further point doesn't apply to Cenoman's discovery, it's worth mentioning: How does the the extension affect the the range of inferences the pattern provides?
For example, I have come to think that a single inference arising from proving two target cells in different bands from the base cells must hold the same digits is unlikely to be very significant. It just relies on too many other pieces of good fortune to occur to be useful, and in that case, there will probably be simpler routes.

I recognise that nowadays there are far more members who don't feel obliged to keep to non-assumptive methods, but it is up to them to decide how to define the new limits they set themselves and how to notate their less constrained patterns and solutions.

David
.

[Cenoman]

Withdrawn

[Leren]

Withdrawn

[Cenoman]

EDIT
Deleted David's response stated that the solution proposed in my first post breaks normal solving rules.
The above solution based on the JE pattern uses assumptive steps.



Here is a very nice solution without Exocet, found by player rwm on sudoku.com.au:
The following PM is tagged for step #2. Note the doubly linked ALS in box4 and AALS in box 5

Code: Select all

 +--------------------------+-------------------------+---------------------------+
 |  19      2      d578     |  3      4      a78      |  156     1589    1569     |
 |  4       19     d378     |  6      5      a278     |  123     12389   1239     |
 | c38     c58      6       | b89    b289     1       |  7       2358    4        |
 +--------------------------+-------------------------+---------------------------+
 |  5       46789  e48      |  2      13789  f348     |  136     139     13679    |
 |  26789   3      e28      |  5789   1789    58      |  1256    4       125679   |
 | e279    e479     1       | f479   f379     6       |  8       2359    23579    |
 +--------------------------+-------------------------+---------------------------+
 |  12368   14568   9       |  458    2368    23458   |  12345   7       1235     |
 |  23678   45678   23458   |  1      23678   23458   |  9       235     235      |
 |  1237    1457    2345    |  457    237     9       |  12345   6       8        |
 +--------------------------+-------------------------+---------------------------+

1. (4)r6c4 = r4c6 - (4=8)r4c3 - r12c3 = r3c12 - r3c45 = r12c6 - (8=5)r5c6 => -5 r6c4
2. Loop (8)r12c6 = r3c45 - r3c12 = r12c3 - (8=2479)b4p3678 - (734|934=348)b5p378@ =>
-8 r578c6,r8c3,r3c8, -2 r5c1, -4 r4c2, -7 r6c9, -9 r6c89, -3 r4c5; 1 placement + basics
3. Kite (5)r1c7 = r79c7 - r8c89 = r8c3 => -5 r1c3; 2 placements + basics
4. Kraken column (2)r2579c7
(2)r2c7 - (2=3)r3c8 - r3c1 = (3)r2c3
(2)r5c7 - (2=8)r5c3 - (87=3)r12c3
(2)r79c7 - (2=53*)r8c89 - (5)r8c3 = r7c3
=>-3 r9c3,-3 r8c3*; 8 placements + basics
5. Skyscraper (3)r6c5 = r6c9 - r8c9 = r8c6 => -3 r4c6, r79c5; stte

On sudokuwiki site, I have also seen a solution using MSLS.

[Leren]

Withdrawn

[champagne]

Hi Cenoman,
For the pair (23)r8c89, two UR threats (*) exist in r2c89 and r6c89.
Their S-cells in target cross-lines (c3, c6) occupy four boxes
=>pair 23 is incompatible. Hence +5r8c89 => -5 r8c236, r9c7, r7c79

Code: Select all

 +-------------------------+------------------------+--------------------------+
 |  19      2       578    |  3      4       78     |  156    1589    1569     |
 |  4       19      378    |  6      5       278    |  123    12389   1239     |
 |  38      58      6      |  89     289     1      |  7      2358    4        |
 +-------------------------+------------------------+--------------------------+
 |  5       46789   48     |  2      13789   348    |  136    139     13679    |
 |  26789   3       28     |  5789   1789    58     |  1256   4       125679   |
 |  279     479     1      |  4579   379     6      |  8      2359    23579    |
 +-------------------------+------------------------+--------------------------+
 |  12368   14568   9      |  458    2368    235    |  1234   7       123      |
 |  23678   4678    2348   |  1      23678   2348   |  9      235     235      |
 |  1237    1457    235    |  457    237     9      |  1234   6       8        |
 +-------------------------+------------------------+--------------------------+

Hi cenoman

I enter late in the discussion, but I have a problem with your elimination of the pair 23.

23 r2c89 is not in my understanding a UR threat. the cell r2c7 contains digits 23.

PS; remembering

A UR threat tells that one of the digits of the UR (here 23) must be outside of the UR threat in the unit(s) containing the 2 cells
in row 6 we have a typical UR threat seen in exocets we must have 2r6c1 | 3r6c5 a strong inference forced by the UR threat
This is not true in row 2.

[champagne]

after the exocet clearings, you can have this rank 0 logic

SLG rank 0
14 Truths = {2R36 2C36 3R36 3C36 5R36 5C36 8N89 }
14 Links = {2r8 2b24 3r8 3b15 5r8 5b15 3n8 6n89 7n6 9n3 }
12 elims 8r3c8 9r6c8 7r6c9 9r6c9 2r5c1 3r4c5 5r5c4 2r8c1 2r8c5 3r8c1 3r8c5 5r8c2
8r3c8 9r6c8 7r6c9 9r6c9 2r5c1 3r4c5 5r5c4 2r8c1 2r8c5 3r8c1 3r8c5 5r8c2

Code: Select all

19    2     578   |3    4     78    |156   1589  1569   
4     19    378   |6    5     278   |123   12389 1239   
38    58    6     |89   289   1     |7     2358  4     
-------------------------------------------------------
5     6789  48    |2    13789 348   |136   139   13679 
26789 3     28    |5789 1789  58    |1256  4     125679
279   479   1     |4579 379   6     |8     2359  23579 
-------------------------------------------------------
12368 14568 9     |458  2368  235   |12345 7     1235   
23678 5678  23458 |1    23678 23458 |9     235   235   
1237  1457  235   |457  237   9     |12345 6     8     


floors PM 235

Code: Select all

X    X    5+   |X    X    X    |5+   5+   5+   
X    X    3+   |X    X    2+   |23+  23+  23+ 
3+   5+   X    |X    2+   X    |X    235+ X   

X    X    X    |X    3+   3+   |3+   3+   3+   
2+   X    2+   |5+   X    5+   |25+  X    25+ 
2+   X    X    |5+   3+   X    |X    235+ 235+

23+  5+   X    |5+   23+  235  |235+ X    235+
23+  5+   235+ |X    23+  235+ |X    235  235 
23+  5+   235  |5+   23+  X    |235+ X    X   

A common 2 rows x 2 columns logic with 2 set_cells added as sets in the linksets

[Cenoman]

Withdrawn

[Cenoman]

Hi cenoman

I enter late in the discussion, but I have a problem with your elimination of the pair 23.

23 r2c89 is not in my understanding a UR threat. the cell r2c7 contains digits 23.

PS; remembering

A UR threat tells that one of the digits of the UR (here 23) must be outside of the UR threat in the unit(s) containing the 2 cells
in row 6 we have a typical UR threat seen in exocets we must have 2r6c1 | 3r6c5 a strong inference forced by the UR threat
This is not true in row 2.

Hi champagne,

I already wrote in a response to Leren that "The devil is in the detail".
Here, I clearly missed a detail: the UR pair must not be aligned with the cover houses of the UR digits.

OK, I will edit my post accordingly.

In your further post, you propose a solution. Thanks.

[ghfick]

I have returned to U281. I posted a [partial] solution path back then.
I noted back then that the path appeared to require at least one assumptive step [after the MSLS and the JE].
I now have a path but I think I need a single finned Kraken X Wing and a POM step.
I believe that the single finned Kraken X Wing is non-assumptive since one can present it as a bi-directional chain. I am far from secure about a POM step.
May I ask the contributors to this thread for their opinion on Krakens and POMs?

[eleven]

Code: Select all

 +--------------------------+-------------------------+---------------------------+
 |  19      2      (5)78    |  3      4       78      | (5)16    1589    1569     |
 |  4       19     (3)78    |  6      5      (2)78    |X(23)1    12389*  1239*    |
 |  38      58      6       |  89     289     1       |  7       2358    4        |
 +--------------------------+-------------------------+---------------------------+
 |  5       46789   48      |  2      13789  (3)48    |Y(3)16    139     13679    |
 |  26789   3     X(2)8     |  5789   1789   (5)8     | (25)16   4       125679   |
 |  279     479     1       |  4579   379     6       |  8       2359*#  23579*#  |
 +--------------------------+-------------------------+---------------------------+
 |  12368   14568   9       |  458    2368  t<235>-48 |  12345   7       1235     |
 |  23678   45678   23458   |  1      23678   23458   |  9     B<235     235>     |
 |  1237    1457  t<235>-4  |  457    237     9       |  12345   6       8        |
 +--------------------------+-------------------------+---------------------------+

However, the 23 elimination in r8c89 can be repaired.
We know from the Exocet, that one would have to go to r9c3 and the other to r7c6, and the other digit must be in c36 of bands 1,2.
But 2r9c3 & 3r7c6 imply 32r2c36-r2c789=23r3c8 (already a contradiction) -(2|3)=5r8c8.
And 3r9c3 & 2r7c6 imply 2r5c3 & 3r4c6 => 23r6c89, 5r8c89 from the UR23r68c89.

[champagne]

But 2r9c3 & 3r7c6 imply 32r2c36-r2c789=23r3c8 (already a contradiction) -(2|3)=5r8c8.

Well done Eleven

Compared to the classical abi loop the UR threat is replaced by the fact that only one cell out of r2 contains candidates 23