The New Sudoku Players' Forum

[Leren]

As far as I can remember all three 6 eliminations are valid whether or not 6 is true in the base. They are purely a consequence of the orthogonal lines cover for 6. The moves were in my solver and as far as I can remember it was you and I who were working on this quite a while ago, before there was a long period of no activity on Exocets. It took me a while to re-work out why the eliminations are valid.

If 6 is in the base r9c3 = 6. If 6 is not in the base, target cell r2c4 is not 6, so r9c4 = 6. Thus, 6 can be removed from all other cells in Row 9.

If r2c1 = 6, r3c23 <> 6 => r9c3 = 6. But also target cell r2c4 is not 6, so r9c4 = 6 which is a contradiction, so r2c1 <>6. In general 6 could be removed from any cells in the base box but not in the base or Column Clb.

Leren

[David P Bird]

Leren yes, I had forgotten our exchange. Those additional eliminations will always exist when the two cover houses are orthogonal.

In the example puzzle
(6)JE:r1c12,r2c4 = (6)r9c4 - (6)r9c3 = (6)r23c3 - Loop => r2c1 <> 6, r9c6 <> 6

I've drafted this to be inserted into the definition file:

"Normally when the cover houses for digits are parallel to the JE band, non-'S' cell eliminations can only be made when the digit is known to be true in the base cells. However when a cover house for a digit is also a cross-line some immediate eliminations may be available.

a) it can't be true in the target cell in that line because that would make it true in the base cells which in turn would then make it false when that target is considered as a non-'S' cell.
b) if one cover house is a cross-line and the other is parallel to the JE band, a colouring loop can be constructed to eliminate it in the non-'S' cells in the parallel cover house and also in the four cells in the base cell box that aren't either in the base cells or the base cross-line.

Example 3 in the JE2 examples file illustrates these points."

I'll modify example 3 accordingly.

DPB

[Leren]

David P Bird wrote :
Example 4 Having a Digit With One Cover House

..7..5..89.....3..63.7.........4....8....1..2..35..7....96..5...4...2..1.......2.

Code: Select all

 *-----------------------*-----------------------*-----------------------*
 | 124 b  12 b   <7>     | 12349  12369  <5>     | 12469  1469   <8>     |
 | <9>    1258   12458   | 1248 t 1268   468     | <3>    14567  4567    |
 | <6>    <3>    12458   | <7>    1289   489     | 1249 t 1459   459     |
 *-----------------------*-----------------------*-----------------------*
 | 1257   125679 1256    | 2389   <4>    36789   | 1689   135689 3569    | 12
 | <8>    5679   456     | 39     3679   <1>     | 469    34569  <2>     |   4
 | 124    1269   <3>     | <5>    2689   689     | <7>    14689  469     |
 *-----------------------*-----------------------*-----------------------*
 | 1237   1278   <9>     | <6>    1378   3478    | <5>    3478   347     |
 | 357    <4>    568     | 389    35789  <2>     | 689    36789  <1>     |
 | 1357   15678  1568    | 13489  135789 34789   | 4689   <2>    34679   | 1 4 
 *-----------------------*-----------------------*-----------------------*
                  CLb      CL1                      CL2 

(124)JE2:r1c12,r2c4,r3c7
=> r1c12, <> 2, r2c4 <> 2, r3c7 <> 2 (base digit with a single 'S' cell cover house)
=> r2c4 <> 8, r3c7 <> 9 (non-base digits in the target cells)
=> r1c4578,r2c238,r3c358 <> 1, r1c478,r2c389,r3c36 <> 4, (known base digits seen by both base cells or both target cells)
=> r4c128,r9c125 <> 1, r5c8,r9c69 <> 4 (known base digits in non-'S' cells in their cover houses)

I agree with all these eliminations except r3c8 <> 1 - possible just a typo.

Leren

[daj95376]

I'm curious why -4 r5c7,r6c1 were not included.

_

[Leren]

David P Bird wrote;

Example 2 More Advanced I

..7.59.6.5........84.6.......85..4...2.....7......1..8..67..98...49..5......3...7

Code: Select all

 *-----------------------------*-----------------------------*-----------------------------*
 | 123 b    13 b     <7>       | 12348    <5>      <9>       | 1238     <6>      1234      |
 | <5>      6        1239      | 12348 t  12478    23478     | 12378 q  12349    12349     |
 | <8>      <4>      1239      | <6>      127      237       | 1237  q  12359    12359     |
 *-----------------------------*-----------------------------*-----------------------------*
 | 13679    1379     <8>       | <5>      2679     2367      | <4>      1239     12369     |
 | 13469    <2>      135       | 348      4689     3468      | 136      <7>      13569     | 1 3
 | 34679    3579     35        | 234      24679    <1>       | 236      2359     <8>       |  23
 *-----------------------------*-----------------------------*-----------------------------*
 | 123      135      <6>       | <7>      124      245       | <9>      <8>      1234      |
 | 1237     1378     <4>       | <9>      1268     268       | <5>      123      1236      |
 | 129      1589     125       | 1248     <3>      24568     | 126      124      <7>       | 12
 *-----------------------------*-----------------------------*-----------------------------*
                     CLb         CL1                           CL2

(123)JE2+:r1c12,r2c4,(7)r23c7

As (7) is locked in r23c7 only one of these cells can be a target cell holding base digit, but which one it will be is uncertain.
=> r2c4 <> 48 (non-base digits in a target cell)
=> r2c7 <> 8 (object cells restricted to a base digit + locked(7))
=> r2c89 <> 9 (mirror node restricted to a base digit + locked(5) )

Another typo ; r2c89 <> 9 should be r3c89 <> 9

Leren

[Leren]

daj95376 wrote : I'm curious why -4 r5c7,r6c1 were not include

I think what you are saying is that because there is only one 4 in the S cells in the Base cells stack it must be True once 4 is proven True in the Base, so in this puzzle r5c3 = 4.

It looks to me like this will always follow trivially so maybe David thinks this is covered by basic follow on moves.

I should stop being a butinsky and let David answer the question

Leren

[Leren]

David p Bird wrote:

JE3 Examples Example 1

..79.8...54.......6...7.5.....3...7......2.1...4.9.8.....1...2...8.4.9...6......3

Code: Select all

 *-----------------------*-----------------------*-----------------------*
 | 123 b  123 b  <7>     | <9>    5      <8>     | 12346  346    1246    |
 | <5>    <4>    1239 T  | 26     1236 t 136     | 1237 t 389    1278    |
 | <6>    8      1239    | 24     <7>    134     | <5>    39 B   12 B    |
 *-----------------------*-----------------------*-----------------------*
 | 1289   1259   1256    | <3>    168    1456    | 246    <7>    24569   | 12
 | 3789   3579   356     | 45678  68     <2>     | 346    <1>    4569    |   3
 | 1237   12357  <4>     | 567    <9>    1567    | <8>    356    256     |    9
 *-----------------------*-----------------------*-----------------------*
 | 3479   3579   35      | <1>    368    35679   | 467    <2>    45678   |   3
 | 1237   12357  <8>     | 2567   <4>    3567    | <9>    56     1567    |    9
 | 12479  <6>    125     | 2578   28     579     | 147    458    <3>     | 12
 *-----------------------*-----------------------*-----------------------*
                 CLb              CL1              CLB 

(1239)JE3:r1c12,r2c7,r2c5; r3c89,r2c3,r2c5 (r2c5 is the common target cell that will holding the common digit)
=> r2c5 <> 6, r2c7 <> 7 (non-base digits in target cells)
Singles (7)r2c9, (8)r2c8,r7c9, (9)r3c8,r2c3
=> [JE] r2c5 <> 3 (non-base digit in target cell - as (3) now absent from r3c89)
Singles (3)r7c5, (5)r7c3
(6)Line-Box:c5b5 => r56c4,r46c6 <> 6
(6)Line-Box:r6b6 => r45c79 <> 6
The grid is now substantially reduced.

Your solution seems unnecessarily complex (in my opinion). Only the first Exocet r1c1 r1c2 r2c5 r2c7 123 is required to completely solve the puzzle.

1. r2c5 <> 6 r2c7 <> 7 (non-base candidates in Target cells)

2. r2c5 <> 3 (mirror cell inference with r3c9), followed by some follow on basics

3. r1c12, r2c57 <> 1. Incompatible Base candidate 1 (according to the method detailed by you in your Junior Exocet definition file)

4. stte

Leren

[Leren]

David P Bird wrote :

JE3 Examples Example 2

..75...4.6.....7..98..7....3..2.......4.6.8.......8.1..2.3...7...9.5.4.......1..8

Code: Select all

 *-----------------------*-----------------------*-----------------------*
 | 12 t   13 t   <7>     | <5>    8      239     | 12369  <4>    12369   |
 | <6>    4      235 T   | 19     129 tT 239     | <7>    8      2359    |
 | <9>    <8>    1235    | 6      <7>    4       | 1235 t 235 B  1235 B  |
 *-----------------------*-----------------------*-----------------------*
 | <3>    1679   8       | <2>    19     57      | 569    569    4       |     
 | 1257   179    <4>     | 19     <6>    57      | <8>    2359   23579   | 1
 | 257    679    26      | 4      3      <8>     | 2569   <1>    25679   |  23
 *-----------------------*-----------------------*-----------------------*
 | 8      <2>    16      | <3>    4      69      | 1569   <7>    1569    | 1
 | 17     1367   <9>     | 8      <5>    26      | <4>    236    1236    |
 | 4      5      36      | 7      29     <1>     | 2369   2369   <8>     |  23 
 *-----------------------*-----------------------*-----------------------*
                                   5                5
                 CLb              CL1               CLB

JE3:(123)r1c12,r2c5,r2c7; (1235)r3c89,r2c5,r2c4 (common target r2c5)
=> r2c5 <> 9, r3c7 <> 5 (non-base digits in targets)
The common target r2c5 now can only hold two of the base digits and becomes a source of AICs using JE inferences.
(1)r7c3 = (1)r3c3 - (1)JE:r1c12,r3c9,r2c5 = (2)r2c5 - (2=9)r9c5 - (9=6)r7c6 => r7c3 <> 6

*The JE group node in this chain is has three locations (for the two base pairs and the common target) so will be true when each of them contains (1). This excludes the case when (1)r1c12 (base cells) and (1)r3c7 (target) are true when (1) would only be true in the first JE2 pattern.

Singles (1)r7c3,r8c9,r2c7
=> r2c5 <> 1 (non-base digit in target (as (1) has been eliminated from r3c9))
=> [JE] r6c1,r9c8 <> 2 (base digit in non-'S' cells (as (2) is known to be the common digit))
Which solves the puzzle

There are typos in this line (corrections in red) : JE3:(123)r1c12,r2c5,r3c7; (1235)r3c89,r2c5,r2c3 (common target r2c5)

Also again your solution seems unnecessarily complex to me. My solution was :

1. Eliminate the non base candidates in the first JE3

2. A finned XWing => - 1 r3c9 and reduces the second Exocet to a second JE3 (123) r3c89 r2c5 r2c3

3. Eliminate the non base candidates in the second JE3. This solves r2c5 = 1.

4. A few other finned XWings and Skyscrapers are all that is required to complete the puzzle.

Is it possible to find examples where the combined JE inferences are "necessary" ie the puzzle is not readily solvable by less complex methods ?

Leren

[Leren]

David P Bird Wrote :

08. JE4 Examples with 3 Cross-lines Example 1

..2....61....2....57....2..1....8.....7.6.1...9...5.....5.7.6....1.5.7.4...3.4...

Code: Select all

 *--------------------------*--------------------------*--------------------------*
 | 3489 b  348 b   <2>      | 4589    3489 T  7        | 34589   <6>     <1>      |
 | 34689   1       34689 T  | 45689   <2>     369      | 34589 t 345789  35789    |
 | <5>     <7>     34689    | 14689   13489 t 1369     | <2>     3489 B  389 B    |
 *--------------------------*--------------------------*--------------------------*
 | <1>     23456   346      | 2479    349     <8>      | 3459    234579  235679   |34 9
 | 2348    23458   <7>      | 249     <6>     239      | <1>     234589  23589    |
 | 23468   <9>     3468     | 1247    134     <5>      | 348     23478   23678    |348
 *--------------------------*--------------------------*--------------------------*
 | 23489   2348    <5>      | 1289    <7>     129      | <6>     12389   2389     |
 | 2389    238     <1>      | 2689    <5>     269      | <7>     2389    <4>      |
 | 7       268     689      | <3>     189     <4>      | 589     12589   2589     |  89
 *--------------------------*--------------------------*--------------------------*
                    CLb               CL1                CLB         
Double JE: (3489)r1c12, r2c7, r3c5   &    (3489)r3c89, r1c5, r2c3

As there are four base candidates and four targets each of which will contain a different base digit, all the digits will be true in one or other of the base pairs.

=> r3c5 <> 1, r2c7 <> 7, r2c3 <> 6 (non-base digits in target cells)
=> r1c7,r2c36,r2c3 <> 3489 (known base digits that see all base cells or all target cells)
=> r4c2489 <> 349, r6c13489 <> 348, r9c2389 <> 89 (known base digits in non-'S' cells in their cover houses)

Typos in this line (corrections shown in red) => r1c7,r2c46,r3c3 <> 3489 (known base digits that see all base cells or all target cells)

Also this is yet another puzzle where I can achieve all the above eliminations using mirror node inferences from the first Exocet, plus follow on basics.

Apart from eliminating the non=base digits in the target cells the mirror node inferences => r3c46 <> 6 and r2c89 <> 9.

In your words, after the follow on basics the puzzle is effectively demolished. Similar comments apply for the second Exocet.

Similar comments also apply to Example 2 in this file. You'll be thrilled to know that for Examples 3 and 4 the double Exocet eliminations were actually necessary !

Leren

[David P Bird]

Leren, I'm delighted that you're scrutinising my work, so please keep going.

I'm very busy today with DIY and tomorrow I have a family do (its father's day here) so I can't say when I'll be able to respond.

For now I'd just like to point out that I tried to find examples that demonstrated the inferences that are available from different JE varieties and have tried to list them all in each case, particularly when the pattern is first identified. So, unlike the curveball puzzle I posted, I wasn't looking for the most efficient solutions.

David

[Leren]

David P Bird wrote :

08. JE4 Examples with 3 Cross-lines Example 4

..5....36....5....89....5..4..........9.8.6.....4.1.2..7.....1....2....7..3.6.9.. Coly 150 (Morphed)

Code: Select all

 *--------------------------*--------------------------*--------------------------*
 | 127 b   124 b   <5>      | 1789    12479 T 24789    | 12478   <3>     <6>      |
 | 12367   12346   12467    | 1678    <5>     24678    | 12478 t 4789    12489    |
 | <8>     <9>     12467    | 1367    12347 t 23467    | <5>     47 B    124 B    |
 *--------------------------*--------------------------*--------------------------*
 | <4>     12358   1278     | 35679   2379    235679   | 1378    5789    13589    | 12 7   
 | 12357   1235    <9>      | 357     <8>     2357     | <6>     457     1345     |
 | 3567    3568    678      | <4>     379     <1>      | 378     <2>     3589     |    7
 *--------------------------*--------------------------*--------------------------*
 | 2569    <7>     2468     | 3589    349     34589    | 2348    <1>     23458    |  24
 | 159     1458    148      | <2>     1349    34589    | 348     6       <7>      | 1 4
 | 125     12458   <3>      | 1578    <6>     4578     | <9>     458     2458     |
 *--------------------------*--------------------------*--------------------------*
                   CLb                CL1                CLB

JE4 (1247)r1c12,r3c5,r2c7, (1247)r3c78,r2c3,r1c5
=> r1c5 <> 9, r2c3 <> 6, r2c7 <> 8, r3c5 <> 3 (non-base digits in target)
=> r1c7,r3c3 <> 1247, r2c4 <> 17, r2c6 <> 247 (base digits seen by all base or target cells)
=> r4c29,r8c12 <> 1, r4c26,r7c19 <> 2, r7c69,r8c26 <> 4, r4c368,r6c1 <> 7 (base digits in non-'S' cells)
Singles (6)r3c3,r7c1,r6c2,r8c8, (8)r1c7, (9)r8c1
(127)Hidden Triple:r4c357 => r4c3 <> 8, r4c5 <> 39, r4c7 <> 3

Compatibility Check (12) incompatible in base cells r1c12 and an impossible pair in r3c89 therefore each base pair must contain one digit from (12) and one from (47) => r3c9 <> 4

(2)r2c7 = (2-34)r78c7 = (4)r2c7 => JE:r2c7,r3c46 <> 17

1. Simple typo - there should be a T next to r2c3

This is the PM I get just prior to your last line :

Code: Select all

*--------------------------------------------------------------------------------*
|b127    b124     5        | 179    T1247    2479     | 8       3       6        |
| 1237    1234   T1247     | 68      5       68       |t1247    479     1249     |
| 8       9       6        | 137    t1247    2347     | 5      B47     B12       |
|--------------------------+--------------------------+--------------------------|
| 4       358     127      | 3569    27      3569     | 17      589     3589     |
| 12357   1235    9        | 357     8       2357     | 6       457     1345     |
| 35      6       78       | 4       379     1        | 37      2       3589     |
|--------------------------+--------------------------+--------------------------|
| 6       7       248      | 3589    349     3589     | 234     1       358      |
| 9       58      148      | 2       134     358      | 34      6       7        |
| 125     12458   3        | 1578    6       4578     | 9       458     2458     |
*--------------------------------------------------------------------------------*
          CLb                        CL1                       CLB

2. I just don't understand this in the last line (2)r2c7 = (2-34)r78c7 = (4)r2c7. Perhaps you can explain this a bit more.

Leren

[David P Bird]

Leren, I've been able to grab some time today and have started to catch you up.

On re-reading my draft about the extra eliminations from a digit with orthogonal cover houses, I realised its style didn't match very well with the explanations of other pattern eliminations so I've re-worked it.

In the list of all the eliminations available:
5. A base digit candidate that has a cross-line as an 'S' cell cover house must be false in a target cell in that cross line and further simple colouring eliminations may be available.

In the list of those that are explained in more detail:
5. If the digit is true in a cover house target it would also be true in the base cells which would force it in turn to also be true in the other target cell which would invalidate the pattern. This can be proved by simple colouring the cells in the JE band, the target cross lines, and the cover houses. When the second cover house is parallel to the JE band, a simple colouring loop will also eliminate the digit in the non-'S' cells in that cover house and in the four cells in the base cell box that aren't either in the base cells or the base cross-line. Example 3 in the JE2 examples file demonstrates these eliminations.

In the JE2 Examples #3 these points are repeated for convenience and the colouring loop notated to provide the cell addresses involved.
The balance of the notated solution has also been modified to remove duplicates of the eliminations made by the loop.

Reflecting on this I believe this loop should be notated fully in any solution. Although the eliminations arise from a theorem, considering how rarely they occur, listing them out of the blue would lose too many readers.

JE2 Examples No 4 - agreed, a typo
JE+ Examples No 2 - agreed, another one

I'll continue tomorrow.

DPB

[Leren]

Hi David, I won't post any further until you have caught up on all issues - to avoid confusion by me about which of my posts you are addressing.

From your last post, however, you might re-consider this wording : When the second cover house is parallel to the JE band, a simple colouring loop will also eliminate the digit in the non-'S' cells in that cover house

In the example puzzle, had there been a 6 in r9c7, it would have been eliminated, and it's an S cell in the cover house Row 9.

Leren

[David P Bird]

Leren, I've now caught up with you.

JE3 Examples No 2: Typos agreed and edited.

JE4 Examples No 1: Typos you spotted edited along with quite few others.

I missed the mirror node inferences you spotted and 2 more. For completeness I've added them even though none of them would survive the cascade of follow-on eliminations from the others
The amended part solution now reads:
=> r3c5 <> 1, r2c7 <> 5, r2c3 <> 6 (non-base digits in target cells)
=> r1c7,r2c46,r3c3 <> 3489 (known base digits that see all base cells or all target cells)
=> r1c4 <> 5, r2c1 <> 6, r2c89 <> 9, r3c46 <> 6 (either (1) or (7) are locked as the non-base digit in these mirror nodes)
=> r4c2489 <> 349, r6c1489 <> 348, r9c289 <> 89 (known base digits in non-'S' cells in their cover houses)

Regarding your comments on complexity my feeling is that generally solvers should avoid using complex methods when they aren't needed unless they are forewarned about a puzzles difficulty. However once a complex method has been employed why not make use of all the power it provides?

Although your program looks for the component JE2 patterns separately, I believe that a manual solver should notice when there is a double JE in the puzzle. This will obviously influence the order of the steps taken following the initial JE eliminations. Manual solvers would tend to milk the pattern for what its worth before spreading their nets wider which is what I tried to do. However it's all a matter of taste.

JE4 Examples No 4
Missing T corrected in the grid
Your grid has (167)r1c1 which should be (127), and (35)r6c1 which should be (357) according to me.

I agree the chain you find problems with is unsound.
(2)r2c7 = (2-34)r78c7 = (4)r2c7 => JE:r2c7,r3c46 <> 17
I'm clearly treating (234)r78c7 as an Almost Hidden Pair with (3) locked in the cells but that’s wrong as (3)r6c7 exists.
It's such a long time back that I can't recall the circumstances, perhaps it was an oversight or perhaps my grid was wrong at the time. I do vaguely remember being pleased to find the (false) opening in a tough puzzle.

I've spent some time looking for an alternative linear continuation but haven't found one yet. I'll look again tomorrow but if I fail I'll have to admit that from that point on it seems net methods become necessary.

DPB

[edit typo pointed out by Ronk corrected]

[Leren]

David P Bird Wrote : Your grid has (167)r1c1 which should be (127), and (35)r6c1 which should be (357) according to me.

Corrected r1c1 typo. r6c1 <> 7 is correct. You'll find it further up in your own notes : r4c368,r6c1 <> 7 (base digits in non-'S' cells).

Leren