## Almost Fishes patterns

Advanced methods and approaches for solving Sudoku puzzles
Allan Barker wrote:So, in my country we say "If you want to talk to the nicest girl, maybe you bring your own SLG".

I am slighly diappointed that you could not do with a smaller number of additions.

I hoped that one would be small enough to be attractive.

May be you can find a nicest girl for our friend ronk

I have an hypothesis to explain may be why we have so big SLG in these puzzles.

As long as the logic uses the first property shown, you have to insert all the sets necessary to re establish that property.

This is one reason why I am expecting some goodies when inserting them in AIC's.

champagne
champagne
2017 Supporter

Posts: 7138
Joined: 02 August 2007
Location: France Brittany

champagne wrote:I am slighly diappointed that you could not do with a smaller number of additions.

I was unaware that you expected any additions would be required. No wonder I couldn't find anything useful in your base/cover set diagrams.

Allan, so you and your software were able to split out the cover sets ... and add 10 more that were missing. Very good!
ronk
2012 Supporter

Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA

Allan Barker wrote:As Pasted
39 Sets = {1R6 3R24568 4R7 5R1269 7R24568 1C2 3C1567 4C6 5C124568 7C357 6N56 357B4 357B5}
Elimination --> none Permutations 1,518,512

39 Sets = {1R6 3R24568 4R7 5R1269 7R24568 1C2 3C1567 4C6 5C124568 7C357 6N56 357B4 357B5}
10 Links = {7c6 567n2 28n5 28n6 46n8}
Elimination --> r7c2<>4 Permutations 5408

Still unable to finish the task, but we where not so far from each other.

I added to your final data a vector 'NV' of potential nodes to be added as seen by my solver.
This is in fact the assumed "better situation" at the end of the global analysis (both eliminations).

Yours are in.

I have 5557 permutations compared to 5408 for you.
I guess the difference comes from the linkset you added in column.

Code: Select all
`NV......... ....XX... .X......X ...X...X. .X.X..... .X..XX.X. .X.....X. ....XX... .X....... N:......... ....oo... ......... .......o. .o....... .o..XX.o. .o....... ....oo... .........R:.....X... ......... .X.XXX.X. ......X.. XX...X..X ......... .X.XXX.X. ......... .........C:.X....... ......... X...XXX.. .....X... XX.XXX.X. ......... ..X.XoX.. ......... .........B:......... ......... ...XX.... ......... ...XX.... ......... ...XX.... ......... .........  1         2         3         4         5         6         7         8         9  `

champagne
champagne
2017 Supporter

Posts: 7138
Joined: 02 August 2007
Location: France Brittany

Allan Barker wrote:Final Set-Linkset Group, SLG
24 Sets = {3R24568 5R2 7R24568 1C2 4C6 5C124568 6N56 357B4}
25 Links = {1r6 4r7 5r169 3c1567 7c3567 567n2 28n5 28n6 46n8 357b5}
Elimination--> (4r7*7n2) => r7c2<>4 Permutations 3614

Note: this is a rank 1 elimination by the overlap of 4r7 and 7n2.

Leaving 5r245c1 without a cover set seems bizaare, but it's certainly less bizaare than the following, with no cover sets at all.

Also, my solver does not require linksets, the follow will work as well. I checked, the number of permutations is really 9999!

As a Set Group
49 Sets = {1R6 3R24568 4R7 5R1269 7R24568 1C2 3C1567 4C6 5C124568 7C3567 2N56 4N8 5N2 6N2568 7N2 8N56 357B4 357B5}
Elimination--> [4R7*7N2] => r7c2<>4
Last edited by ronk on Sun Jan 04, 2009 11:43 pm, edited 1 time in total.
ronk
2012 Supporter

Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA

ronk wrote:
Allan Barker wrote:Final Set-Linkset Group, SLG
24 Sets = {3R24568 5R2 7R24568 1C2 4C6 5C124568 6N56 357B4}
25 Links = {1r6 4r7 5r169 3c1567 7c3567 567n2 28n5 28n6 46n8 357b5}
Elimination--> (4r7*7n2) => r7c2<>4 Permutations 3614

Note: this is a rank 1 elimination by the overlap of 4r7 and 7n2.

Leaving 5r245c1 without a cover set seems bizaare, but it's certainly less bizaare than the following, with no cover sets at all.

Also, my solver does require linksets, the follow will work as well. I checked, the number of permutations is really 9999!

As a Set Group
49 Sets = {1R6 3R24568 4R7 5R1269 7R24568 1C2 3C1567 4C6 5C124568 7C3567 2N56 4N8 5N2 6N2568 7N2 8N56 357B4 357B5}
Elimination--> [4R7*7N2] => r7c2<>4

RE: "Leaving 5r245c1 without a cover set". This is un-common but can happen for two reasons I have seen. 1) The embedding of broken wing type logic, i.e., loops of odd numbers of strong sets. In this case row set 5r1 is a "strong" cover set, as well as several others. 2) Two "out of phase" AICs that travel through (partially or all) the same candidates.

RE: "the following, with no cover sets at all". (Edit: my solver does not require cover sets, I have changed my original post). The solver does not use a base/cover set model, rather some of the logic it finds can also be expressed as a base/cover set model. This accounts for about 95% of all the logic it does find.

Champagne wrote:I am slighly diappointed that you could not do with a smaller number of additions.

I hoped that one would be small enough to be attractive.

There may be, or probably are, several other logical arrangements. This is only the one I got by filling in extra linksets.
Allan Barker

Posts: 266
Joined: 20 February 2008

Previous