Jacko

IJ · Posted: Sat Apr 30, 2005 11:16 am Post subject: Jacko

I've been meaning to post this for ages, but I've been distracting myself with the problem of how to set a puzzle containing a specific pattern, but have failed miserabley with several different approaches.

So, here by popular demand, we have the Jacko:

This is a pattern I spotted in the Very Hard example on the Times Web site.

The original puzzle is like so:

* 4 3 | * 8 * | 2 5 *
6 * * | * * * | * * *
* * * | * * 1 | * 9 4
-----------------------
9 * * | * * 4 | * 7 *
* * * | 6 * 8 | * * *
* 1 * | 2 * * | * * 3
-----------------------
8 2 * | 5 * * | * * *
* * * | * * * | * * 5
* 3 4 | * 9 * | 7 1 *

This is easily resolved this far:

* 4 3 | 9 8 * | 2 5 *
6 * * | 4 2 5 | * * *
2 * * | * * 1 | * 9 4
-----------------------
9 * * | * * 4 | * 7 *
3 * * | 6 * 8 | * * *
4 1 * | 2 * 9 | * * 3
-----------------------
8 2 * | 5 * * | * * *
* * * | * 4 * | * * 5
5 3 4 | 8 9 * | 7 1 *

At which point there is an X-wing of 6's. But, ignoring that, there is also an interesting pattern of 7s.

Consider columns 1 and 6, both have only two candidate cells for 7 - r1c1, r8c1 and r1c6 and r7c6.

Due to the arrangement of r8c1 and r7c6, these can be described as "Tied" for the digit 7. In other words, if one is 7, then they both are, and if one is not 7, then neither are.

Given that this is true, the fact the other two candidate cells in cols 1 and 6 are on the same row, and must therefore be opposites (one, and only one can be a 7), then r8c1 and r7c6 must both be 7 (If they were not, then both r1c1 and r1c6 would have to 7 due to the column contraints).

So, the key to identifying this pattern is finding two columns with only two candidate cells for a digit. Two of the four cells must be on one row, and the other two must be tied.

I identify tied cells with the following logic, though there may be another way:

a) The two cells must be in the same horizontal chute (line of 3 boxes)
b) One cell is on box A, and the other in box B (i.e. different boxes)
c) There must be only one candidate cell in box B that is not on the same row as the candidate cell in box A (and obviously vice versa).

So, there could be either 2 or 3 candidate cells in both boxes A & B. In this example there are 2.

That is the Jacko.

There is a derivative (Jacko's Cousin), where if you have two sets of tied cells, covering two columns (but four rows), and one of the columns has only two candidate cells, then the digit can be eliminated from the other column.

Both these can obviously be flipped to work for rows Vs columns.

As I said, I have failed wholeheartedly in my quest to produce puzzles of a specific difficulty, or containing particular patterns Sad

, so I can't give you another or example, maybe someone else could...

AMcK · Posted: Tue May 10, 2005 10:03 am Post subject: Colouring Jacko

Re IJ's post on Jacko, here's the colouring algorithm for the interesting pattern of 7's.

I still assert that the colouring algorithm is a general mechanical process suggested by Rubylips for discovering all such "opposites" of "tied" values as IJ calls them. Sadly none of these approaches - xwings, swordfish or colouring - seem to be able to solve practical puzzles

If you look at my c/d colouring for 7 below, you will see that the "opposite" sets are even larger than IJ suggests.

r1c1, r1c6, r7c3, r8c4 are all "tied" together as one possible placement of 7 - r3c4, r7c6, r8c1 are all "tied" together for the alternative placement of digit 7. So all other possible 7's can be removed from r7, r8, c1, c4, c6, b7, b8. Sadly in this puzzle, although there are 12 other 7's on the board, none of them are in any of the above exclusion regions and so we have no logic from this rule that enables us to reduce any of them.

Regards
Andrew

Partial transcript

Conjugate row 6: digit 7 cells {6,3} {6,5}
Colour: {6,3} (a) {6,5} (b)
Conjugate row 7: digit 7 cells {7,3} {7,6}
Colour: {7,3} (c) {7,6} (d)
Conjugate row 8: digit 7 cells {8,1} {8,4}
Colour: {8,1} (e) {8,4} (f)
Conjugate column 1: digit 7 cells {1,1} {8,1}
Colour: {1,1} (f) {8,1} (e)
Conjugate column 4: digit 7 cells {3,4} {8,4}
Colour: {3,4} (e) {8,4} (f)
Conjugate column 5: digit 7 cells {5,5} {6,5}
Colour: {5,5} (a) {6,5} (b)
Conjugate column 6: digit 7 cells {1,6} {7,6}
Colour: {1,6} (c) {7,6} (d)
Conjugate column 9: digit 7 cells {1,9} {2,9}
Colour: {1,9} (g) {2,9} (h)
Conjugate box 2: digit 7 cells {1,6} {3,4}
Recolour: {1,1} (f->c)
Recolour: {3,4} (e->d)
Recolour: {8,1} (e->d)
Recolour: {8,4} (f->c)
Colour: {1,6} (c) {3,4} (d)
Conjugate box 3: digit 7 cells {1,9} {2,9}
Colour: {1,9} (g) {2,9} (h)
Conjugate box 5: digit 7 cells {5,5} {6,5}
Colour: {5,5} (a) {6,5} (b)
Conjugate box 7: digit 7 cells {7,3} {8,1}
Colour: {7,3} (c) {8,1} (d)
Conjugate box 8: digit 7 cells {7,6} {8,4}
Colour: {7,6} (d) {8,4} (c)
Exclusion in row 1: colours c g
Exclusion in row 1: colours c g
Exclusion in column 3: colours a c
Colouring for digit: 7

c - - - - c - - g
- * * - - - - - h
- * * d - - - - -
- - - - - - - * -
- * * - a - - - -
- - a - b - - - -
- - c - - d - - -
d - - c - - - - -
- - - - - - * - -

IJ's original post:

* 4 3 | 9 8 * | 2 5 *
6 * * | 4 2 5 | * * *
2 * * | * * 1 | * 9 4
-----------------------
9 * * | * * 4 | * 7 *
3 * * | 6 * 8 | * * *
4 1 * | 2 * 9 | * * 3
-----------------------
8 2 * | 5 * * | * * *
* * * | * 4 * | * * 5
5 3 4 | 8 9 * | 7 1 *

At which point there is an X-wing of 6's. But, ignoring that, there is also an interesting pattern of 7s.

Consider columns 1 and 6, both have only two candidate cells for 7 - r1c1, r8c1 and r1c6 and r7c6.

Due to the arrangement of r8c1 and r7c6, these can be described as "Tied" for the digit 7. In other words, if one is 7, then they both are, and if one is not 7, then neither are.

Given that this is true, the fact the other two candidate cells in cols 1 and 6 are on the same row, and must therefore be opposites (one, and only one can be a 7), then r8c1 and r7c6 must both be 7 (If they were not, then both r1c1 and r1c6 would have to 7 due to the column contraints).

Mark Summerville · Posted: Wed May 11, 2005 3:50 pm Post subject:

All,

As a newcomer to these fora I have been reading through the ones on www.sudoku.com and google groups. Apologies if what I'm about to say duplicates something I've missed.

I have been working with the guys at sudokusolver.co.uk to enhance the solving rules. It is with great interest that I've read about the Jacko, the Xwings and others. They mirror some of the concepts that we've unimaginatively called solve rules A-F!

See http://www.sudokusolver.co.uk/codeit.html

I think Solve Method E relates to the Jacko pattern. I have tried to rationalise it in terms of 'cyclical non-slack groups'. The colour painting method seems to be relatively similar and may be more generic. I don't think I fully understand the impact of the painting method yet.

Solve Method E relates to an X-Wing and we have expanded it in a similar fashion to include all areas and a more generic state where there are n values available.

It would be interesting to hear peoples opinions on solve method E and to see if it is possible to code it efficiently.

Also does anyone have any problems that can be soved by logic only, but the logic is not known yet? (Such as the Times very hard one was for a period until the discovery of x-wings and other methods)

Regards,

Mark