The New Sudoku Players' Forum

[denis_berthier]

You would have to show, that r1c3=3. But i think, that's almost as hard as solving the puzzle in a conventional way.

Well, as the sticks symmetry is a uniqueness technique, couldn't we say: I add 3r1c3 as extra clue, and if the new puzzle has a solution, it is necessary the same as the unique solution of the original one ?

As soon as you accept uniqueness, this guessing technique is logically valid. Guessing has always been considered as still worse than T&E (although many real players use it). In the present situation, it's not arbitrary guessing, it's educated guessing - so, it may not be worse than T&E, after all.

My question: is it possible to build a non symmetric minimal puzzle (having a symmetric solution) ?

Almost obviously yes, by starting from a symmetric full grid. What might be interesting in the context of symmetry is a notion of symmetrically minimal puzzle (not necessarily minimal in the general sense).

[champagne]

You permute the bands and the columns within a band, to line up with any column Sticks symmetry case.

hmmm, how many different "column Stick symmetry cases" are there? Are they listed anywhere?

If you look at things this way, you can, with some practice, work out a Sticks symmetry by pattern recognition, you don't have to repeat the proof every time, just like with other patterns, like fish or whatever.

Yes, provided that the number of cases is low enough.

we have four classes having a possible symmetry of given

180%
diagonal
rotational (90°)
stick

But in each class, for a given puzzle, all morphs of the puzzle have the symmetry property.

Each class has a classical/canonical form showing better the property.

180% r5c5 is the central point for the symmetry
diagonal here, 2 forms used, main diagonal symmetry and anti-diagonal symmetry
rotational r5c5 is the central point for the rotation
stick usually shown with stick in columns c2, c5, c8 in band 2, band 1 and 3 having the proper order in rows.

In each class, part of the morphs have a similar view of the property. for example, you can exchange rows 2/3 and rows 8/9 in a central symmetry, the symmetry is still there.

As we can see the symmetry property in the main diagonal and anti diagonal form, you can see the stick symmetry with sticks in rows or in columns.

Finding a symmetry in a morph where the property is not in an acceptable form is not easy, so IMO, this should be excluded in puzzles posted and is not considered in my solver. A separate process morph puzzles in the "canonical form".

At the end, I would answer {one form per class}, but this can be discussed one alternative is "thousands" in fact
number of morphs of a puzzle/number of morphs showing easily the property.

[champagne]

Then I have a question to symmetry theory experts. Here we got example of a minimal symmetric puzzle made "less" symmetric by adding clue(s) in non stick-cell(s). My question: is it possible to build a non symmetric minimal puzzle (having a symmetric solution) ?

Easy answer. Don't mix symmetry of given and symmetric solution.
Take any symmetrical solution. You can build billions of minimal puzzles producing this result. Only a tiny part of then will have a symmetry of given.

But I have doubts that you can produce a minimal puzzle done of a symmetry of given plus one clue.

EDIT:
Just assume that you have a symmetry of given with multiple solutions.
adding one clue, you come to a unique solution. Then, all remaining unavoidable sets had a hit to this cell and to the symmetric one. Not so easy to figure out, but I have no proof that this can not be

[Leren]

Cenoman wrote : Just a comment to Leren's description of the pattern. Only the canonicalised version is described. What about puzzles mingled by an isomorphism, as were the examples posted the last weeks ?

Far too much information to address all that was posted while I was asleep but I thought that these two issues are easy to answer.

The short answers are that 1. You can get away with just a canonicalised exemplar, and 2. You can't mingle (scramble) Sticks symmetry if it exists, you can just move it about. Lets show this by an example. The original puzzle:

Code: Select all

...|...|...
...|..1|.23
1..|23.|456
---+---+---
.12|7..|5.4
64.|1..|.72
7.5|.2.|.6.
---+---+---
.31|.8.|64.
5.4|.1.|.38
86.|3..|..5

By inspection of your solution, swap Bands 1 & 2, Columns 1 & 2, 5 & 6, 7 & 8 and you get this, in canonicalised format.

Code: Select all

1.2|7..|.54
46.|1..|7.2
.75|..2|6..
---+---+---
...|...|...
...|.1.|2.3
.1.|2.3|546
---+---+---
3.1|..8|46.
.54|..1|3.8
68.|3..|..5

It's easy to check that a Sticks symmetry exists with isomorphism [1][4][9] [23][56][78], because this is consistently observed in the abcd and X cells of the canonicalised exemplar.

The kool (sic) thing about this way of looking at things is that the isomorphism just works itself out. So eg r1c1 = r7c3 = 1 and these are symmetry cells b1 & b2 of the canonicalised exemplar, so you know that 1 is a Sticks digit.

You then just reverse the band and column swaps to find the true location of the Sticks, and you are done.

The fact that Sticks can't be "scrambled", or moved into more than one band, should be obvious, because no morphing operation can "scramble" a row/column box intersection and obviously you can't move a part of a band or stack.

Leren

[champagne]

reading the comments of cenoman, I had a look to the solution grid seen here

Here is the solution, slightly reshaped to show the sticks columns

Code: Select all

9 23   65 4   1 87
4 56   87 1   9 23
1 87   23 9   4 56

3 12   76 8   5 94
6 48   19 5   3 72
7 95   42 3   8 61

2 31   58 7   6 49
5 74   91 6   2 38
8 69   34 2   7 15


band 1 has the sticks.
the rest is done of the paired digits, each pair appearing in a different row in each box.
This is a property of the symmetry. BTW, band 1 (the band where are the sticks) can not have an unavoidable set of size 4.

in bands 2;3, the three "stick digits" can show an unavoidable set of size 4 as 19 in stack 2 or 49 in stack 3. Then, the minimum is to have one of the digits appearing twice as given (1 in stack 2 and 4 in stack 3)
This is likely the only possible form of unavoidable set of size 4 in the solution

[eleven]

You would have to show, that r1c3=3. But i think, that's almost as hard as solving the puzzle in a conventional way.

Well, as the sticks symmetry is a uniqueness technique, couldn't we say: I add 3r1c3 as extra clue, and if the new puzzle has a solution, it is necessary the same as the unique solution of the original one ?

As soon as you accept uniqueness, this guessing technique is logically valid. Guessing has always been considered as still worse than T&E (although many real players use it). In the present situation, it's not arbitrary guessing, it's educated guessing - so, it may not be worse than T&E, after all.

One could argue, that placing the 3 into r1c3 is a legal uniqueness move, because it leads straightforward to the single solution, proving it's correctness.
Same with a backdoor digit.

Personally i think, that the correctness has to be proven completely at the time, the move is made. In this sense those moves are very complex.

[Leren]

351.8..6...45.21....71..5..9187.6.........8.9.........143.9..7.5..2....16....1..4

Code: Select all

*-----------*
|351|.8.|.6.|
|..4|5.2|1..|
|..7|1..|5..|
|---+---+---|
|918|7.6|...|
|...|...|8.9|
|...|...|...|
|---+---+---|
|143|.9.|.7.|
|5..|2..|..1|
|6..|..1|..4|
*-----------*

This puzzle has a unique solution and careful examination of the clues will show that it looks to have a Sticks symmetry arrangement in canonical form with isomorphism [1][2][3][45][67[89] except for one clue 5 in r2c4 that does not have its partner digit 4 as a clue in its "Symmetry cell" r8c6. Here is the puzzle PM after basics :

Code: Select all

*--------------------------------------------------------------------------------*
| 3       5       1        | 49      8       7        | 49      6       2        |
| 8       69      4        |*5       36      2        | 1       39      7        |
| 2       69      7        | 1       346     349      | 5       3489    38       |
|--------------------------+--------------------------+--------------------------|
| 9       1       8        | 7       2345    6        | 34      2345    35       |
| 7       23      6        | 34      12345   345      | 8       12345   9        |
| 4       23      5        | 89      123     89       | 7       123     6        |
|--------------------------+--------------------------+--------------------------|
| 1       4       3        | 6       9       58       | 2       7       58       |
| 5       78      9        | 2       347    *348      | 6       38      1        |
| 6       78      2        | 38      357     1        | 39      3589    4        |
*--------------------------------------------------------------------------------*

This looks great ! 1,2 & 3 are clues or candidates in the "Sticks cells" r456c258 and 4 is a candidate in r8c6. So why not set r8c6 to be 4 and see what happens?

Leren

[champagne]

351.8..6...45.21....71..5..9187.6.........8.9.........143.9..7.5..2....16....1..4

This puzzle has a unique solution and careful examination of the clues will show that it looks to have a Sticks symmetry arrangement in canonical form with isomorphism [1][2][3][45][67[89] except for one clue 5 in r2c4 that does not have its partner digit 4 as a clue in its "Symmetry cell" r8c6. Here is the puzzle PM after basics :


Leren

Hi Leren,

good example, but not exactly what I described.
without 4r2c5, the puzzle has multiple solutions, adding 4r2c5, the solution is unique, but has not a stick symmetry. This is a classical status.

In fact, I think that this could come easily with a symmetric solution grid if the last clue is "self symmetric"(in a stick), more interesting would be an example with the last clue having a symmetric cell as here.