The New Sudoku Players' Forum

[StrmCkr]

deleted: rehashed in an up coming post.

[blue]

Hi David,

Just as I found the first of these, JCVH posted his full set of solutions, so I watched the Tour de France instead! My solution was identical to JC's and was an adaptation of the 4 box cover pattern used for SK loops.

Now I notice that each of his and your solutions includes a box cover set. This might be why Champagne & Leren couldn't identify them. The question therefore is – is it possible to get these results using only row and column cover sets?

I don't think it's possible without introducing box-type base sets too.
These were some odd cases, and the patterns that I posted were each "strange" in some way.
The first one in each pair, had lots of added cell-type base sets, with box-type cover sets added to cover "extra" candidates in those cells. The 2nd one in each pair, was 5x5, with a missing cell-type base set.

Below, are some equally odd patterns with box truths -- complementary pairs again (basically).

The first in each pair is odd, since two box-type base sets and one cell-type cover set have been added (to a 4x4 cell-type base set). The second in each pair is odd, since a cell-type base set has been replaced by two box-type base sets (in a 5x5 cell-type base set). The replaced cell-type base set here, is for the same cell as the cell-type cover set in the first pattern. The cover sets, though, are all row or column-type.

Puzzle #2 is an odd case for the 2nd pair, since one of the box-type base sets from the 1st pair, can be taken as a column-type base set, and like that, it cancels with a column type cover set that would normally be there.

Hidden Text: Show

98.7..6..5.7....4..3.......2...9.1....86...3......5..2.4.8...7.....1...9.....65..

18 Truths = {4689N2 4689N3 4689N4 4689N8 46B1}
18 Links = {5r4 19r6 25r8 129r9 3c34 4c34 6c238 7c2 8c8 3n1}
15 Eliminations --> r369c1<>1, r8c67<>2, r7c3<>36, r2c4<>3, r3c4<>4, r3c8<>8, r4c9<>5,
r5c2<>7, r6c7<>9, r9c9<>1, r9c5<>2,

23 Truths = {57N1 12357N5 12357N6 2357N7 12357N9 46B1}
23 Links = {34r1 368r2 4678r3 47r5 36r7 1c169 2c567 5c59 9c67}
15 Eliminations --> r369c1<>1, r8c67<>2, r7c3<>36, r2c4<>3, r3c4<>4, r3c8<>8, r4c9<>5,
r5c2<>7, r6c7<>9, r9c9<>1, r9c5<>2,


98.7..6......6.........5.4.8..3..7...2...1.6...3......7...1..5..9..4...6..82..9..

18 Truths = {1469N5 1469N6 1469N8 1469N9 79B4}
18 Links = {23r1 29r4 2789r6 37r9 5c5 46c6 1c8 145c9 5n3}
20 Eliminations --> r5c359<>5, r2c89<>1, r5c39<>4, r6c47<>8, r9c12<>3, r1c3<>2, r2c6<>4,
r2c9<>5, r3c9<>1, r6c7<>2, r6c4<>9, r7c9<>4, r7c6<>6, r8c8<>1,

22 Truths = {2358N1 237N2 2378N3 23578N4 23578N7 7B4}
22 Links = {1r238 4r257 5r258 6r37 2c137 3c127 7c23 8c47 9c4}
20 Eliminations --> r5c359<>5, r2c89<>1, r5c39<>4, r6c47<>8, r9c12<>3, r1c3<>2, r2c6<>4,
r2c9<>5, r3c9<>1, r6c7<>2, r6c4<>9, r7c9<>4, r7c6<>6, r8c8<>1,


98.7..6......6.........5.4.8..3..2...2...1..6..3......7...1...5.3..4..6...82..9..

18 Truths = {1469N5 1469N6 1469N8 1469N9 79B4}
18 Links = {23r1 79r4 2789r6 37r9 1c89 4c69 5c58 6c6 5n3}
16 Eliminations --> r5c358<>5, r238c9<>1, r6c47<>8, r2c8<>15, r1c3<>2, r2c6<>4, r5c3<>4,
r6c7<>7, r6c4<>9, r7c6<>6,

23 Truths = {2358N1 237N2 2378N3 23578N4 23578N7 79B4}
23 Links = {1r238 4r257 5r258 6r37 2c13 3c17 7c237 8c47 9c234}
16 Eliminations --> r5c358<>5, r238c9<>1, r6c47<>8, r2c8<>15, r1c3<>2, r2c6<>4, r5c3<>4,
r6c7<>7, r6c4<>9, r7c6<>6,

Regards,
Blue.

[champagne]

I don't think it's possible without introducing box-type base sets too.
These were some odd cases, and the patterns that I posted were each "strange" in some way.

congratulations first to JC and blue for these multiple cells base solutions.

I had a look to the first one and checked whether these solutions fitted with David's split "Home" "Away" which is the base of leren's lay-out.

In JC's solution, Digit 6 is both "'Home" and "Away". Enough to be out of the scope.

I just considered blue's 4 rows solution, but I guess the 5 rows would be complementary. Here, the digits 12 are both "Home" and "Away" The fact is hidden in the box cover 12b1.

No chance in both cases to catch these cells bases using the "home" "away" logic.

[Leren]

I also add my congratulations JC and blue for their MSLS solutions. Also to Champagne who encouraged me to post those puzzles for the experts to find them.

I've been working on a modification to my code to include solutions along the lines of blue's post, since that lines up most easily with my code.

JC's solutions are equally good but will take me more time to implement since they are based on a pattern that I have not coded for.

Leren

[David P Bird]

I had a look to the first one and checked whether these solutions fitted with David's split "Home" "Away" which is the base of leren's lay-out.

In JC's solution, Digit 6 is both "'Home" and "Away". Enough to be out of the scope.

I just considered blue's 4 rows solution, but I guess the 5 rows would be complementary. Here, the digits 12 are both "Home" and "Away" The fact is hidden in the box cover 12b1.

No chance in both cases to catch these cells bases using the "home" "away" logic.

Champagne, that's not so! You seem to have misidentified the sets being used.

98.7..6..5.7....4..3.......2...9.1....86...3......5..2.4.8...7.....1...9.....65..;4;4;10.2;10.2;9.2;fl; 1259;tr; 16;
MS-NS r57c15679,r4690c58 (18 Cells)
(12)b5,(59)b6,(259)b8,(12)b9,(1)c1 (47)r5,(36)r7, (34)c4,(68)c8, (18 Digit Covers)
=> Elims:3r2c4, 1r3c1, 4r3c4, 8r3c8, 5r4c9, 7r5c2, 1r6c1, 9r6c7, 36r7c3, 2r8c6, 2r8c7, 1r9c1, 2r9c5, 1r9c9,
(15 Digits in 14 cells)

Notating JC's solution as above, the Home Set is (1259) used for b5689 & c1, and the Away set is (34678) used for r57 & c48

In all three of his solutions the home set is used to cover 4 boxes and the away set is used to cover 2 rows and two columns (which is the basis for an SK loop solution) and a home row or column is added as an adjustment. Well, at least that's the way I approached them. For his second and third puzzles the division of the house covers are identical presumably because these puzzles originate from the patterns game.

All his solutions keep to the spirit of my approach in that the base set is confined to cells and the cover set to houses. I believe that this should reduce the search effort needed to find if a MSLS exists. As I understand it, Xsudo is geared up to automatically add link sets as truth sets are specified, but for these patterns it's easier to do the opposite.

David

[champagne]

Champagne, that's not so! You seem to have misidentified the sets being used.

I don't, but that does not mean that I am fully in line with you.

AFAIK, leren, as me, is working with "Home" in one direction and "Away" in the other.

With that specification, the process is already 10 times slower than my search on rows columns or "X" base.
Opening the specification would be a disaster in terms of run time.

[David P Bird]

Champagne, that's not so! You seem to have misidentified the sets being used.

I don't, but that does not mean that I am fully in line with you.

AFAIK, leren, as me, is working with "Home" in one direction and "Away" in the other.

With that specification, the process is already 10 times slower than my search on rows columns or "X" base.
Opening the specification would be a disaster in terms of run time.

As I remember it, your X base is equivalent to the 4 box cover I'm using for SK loops, but I'm not quite sure how you implement it. You may get similar results to me though. After assigning the home set to cover the 4 boxes and the away set to cover 2 rows and 2 columns: if a rank1 pattern results, quickly check if adding an extra Home set anywhere produces a rank 0 pattern.

[David P Bird]

Blue, Thanks for your response and I get the overall approach

Using puzzle 3, here's how I see it being turned into a practical procedure:
98.7..6......6.........5.4.8..3..2...2...1..6..3......7...1...5.3..4..6...82..9..

Code: Select all

     *----------------------------*----------------------------*----------------------------*
23   | <9>      <8>      145-2    | <7>      #23      #234     | <6>      #1235    #123     |
     | 12345    1457     12457    | 1489     <6>      2389-4   | 13578    23789-15 23789-1  |
     | 1236     167      1267     | 189      2389     <5>      | 1378     <4>      23789-1  |
     *----------------------------*----------------------------*----------------------------*
79   | <8>      1456-79  1456-79  | <3>      #579     #4679    | <2>      #1579    #1479    |
     | 45       <2>      4579     | 4589     789-5    <1>      | 34578    3789-5   <6>      |
2789 | 1456     1456-79  <3>      | 456-89   #25789   #246789  | 145-78   #15789   #14789   |
     *----------------------------*----------------------------*----------------------------*
     | <7>      469      2469     | 689      <1>      389-6    | 348      238      <5>      |
     | 125      <3>      1259     | 589      <4>      789      | 178      <6>      278-1    |
37   | 1456     1456     <8>      | <2>      #357     #367     | <9>      #137     #1347    |
     *----------------------------*----------------------------*----------------------------*
                                               5        46                 15        14

MS-Almost Naked Set: r1469c5689 (16 Cells)
Home (5)c5,(46)c6,(15)c8,(14)c9 Away (23)r1,(79)r4,(2789)r6,(37)r9 (17 Digit Covers)
=> PEs: 2r1c3, 4r2c6, 15r2c8, 1r2c9, 1r3c9, 79r4c2, 79r4c3, 5r5c5, 5r5c8, 79r6c2, 89r6c4, 78r6c7, 6r7c6, 1r8c9 (20 Candidates in 14 Cells)

This is the row/column cover set pattern you give which turns out to be rank 1. Checking the boxes it becomes clear that in b4 (7) & (9) would both have to occupy r5c3 if all the PEs were made and so one of the eliminations in that box must be invalid. This would allow the PEs in the other boxes to be made but adding a home set cover for b4 adds 4digit covers and 5 cell covers (r5c3 will still have 79 uncovered) to correct the count. The (79) PEs are still lost but (45)r5c3 are now gained.

Your alternative adjustment to give the same effect is to add 79b4 to the truth sets and 5n3 to the link sets which I can understand but obviously wish to avoid as it complicates the model. However our programmers may be far less fussy.

Am I the only one to spot a (1456)JExocet in this puzzle that makes further eliminations?

Regards
David

[pjb]

Code: Select all

*9      8      *124    | 7     *1236   1346   |*1235  *125    345   
 6      1234    1247   | 2348   5      134    | 9      1278   3478   
 12347  1234    5      | 23489  1238   1349   | 12378  6      3478   
 ----------------------+----------------------+---------------------
 5      129     1289   | 689    4      1679   | 2678   3      6789   
*348    7      *489    | 35689 *368    2      |*568   *589    1     
 1238   1239    6      | 3589   1378   13579  | 4      25789  5789   
 ----------------------+----------------------+---------------------
 1278   1256    3      | 256    9      567    | 15678  4      5678   
*478    4569   *4789   | 1     *367    34567  |*35678 *5789   2     
*1247   124569 *12479  | 23456 *2367   8      |*13567 *1579   35679 



98.7.....6...5.9....5....6.5...4..3..7...2..1..6...4....3.9..4....1....2.....8...;327;GP;H51;2;1278;R;C; ; ;; MSLS 1 : Base 1278; r1589 c13578: 19 Links; 12r1 8r5 78r8 127r9; 34c1 49c3 36c5 356c7 59c8;

I apologize if I'm missing something obvious, but I've run into a few of these grids where there is one given, and thus one less cell truth, in this case 19. The link set as described by Leren above has in the column with the given (at r1c1) the inclusion of a number that appears only once. In the example above we have 34c1, but only one 3 in r2c1. My problem is don't you need a number occurring in two cell truths to form a link? I assume there is an answer since the eliminations in this situation are always OK.

Phil

[David P Bird]

Hi Phil

I'm not sure that I fully get your point. Leren's reporting a cell set of 20 cells where the Home and Away sets intersect which should be taken to represent 20 cell truths. However the digit covers in each house (the links) total 19. But (9)r1c1 is a given and so shouldn't be included in the set of cell truths under accepted practice. This reduces them to 19 and we get a balance. This could be overcome either by listing the cells properly or adding "(19 available cells)". The alternative is to add (9) to the digit covers shown for c1 to add another link (which is already resolved by the given). Mathematically it makes no difference but many pundits would object to that.

Switching the Home sets to cover columns and the Away sets to cover rows the problem vanishes!
MS-NS r23467c2469 (20 cells)
(34)r2,(349)r3,(69)r4,(359)r6,(56)r7,(12)c2,(28)c4,(17)c6,(78)c9 (20 digit covers)
=> Elims:1r1c6, 4r2c3, 34r3c1, 3r3c5, 3r3c7, 9r4c3, 6r4c7, 8r5c4, 3r6c1, 3r6c5, 59r6c8, 56r7c7, 7r8c6, 12r9c2, 2r9c4, 7r9c9 (20 candidates in 16 cells).

I'm beyond arguing about notation issues, but in my view if a further explanation is needed about what it means, the notation used for a step isn't fit for purpose.

David

[champagne]

I apologize if I'm missing something obvious, but I've run into a few of these grids where there is one given, and thus one less cell truth, in this case 19. The link set as described by Leren above has in the column with the given (at r1c1) the inclusion of a number that appears only once. In the example above we have 34c1, but only one 3 in r2c1. My problem is don't you need a number occurring in two cell truths to form a link? I assume there is an answer since the eliminations in this situation are always OK.

Phil

Hi Phil

A link touching only one set candidate is something very common in all chain eliminations.
These are usually some rank 1 logic and only candidates seen by both end are eliminated.

Here we are in a rank 0 logic where one candidate is assigned and somehow isolated. The raw situation is a rank -1 logic with an uncovered candidate. You can use the appropriate(s) row/column or box link to come to the rank 0 logic clearing the corresponding region. Usually, any of the 3 possible links works.