The New Sudoku Players' Forum

[Leren]

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..............1.23...23.456.127..5.464.1...727.5.2..6..31.8.64.5.4.1..3886.3....5
.2............1.23...23.456.127..5.464.1...727.5.2..6..31.8.64.5.4.1..3886.3....5

[denis_berthier]

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..............1.23...23.456.127..5.464.1...727.5.2..6..31.8.64.5.4.1..3886.3....5
.2............1.23...23.456.127..5.464.1...727.5.2..6..31.8.64.5.4.1..3886.3....5

I'm sure you know "harder" doesn't mean anything if you don't specify the criteria of comparison. So, there must be some trick.

The two puzzles are identical but for the additional clue n2r1c2 in the second. A priori, a puzzle with additional clues is easier - unless some rules that disrupt confluence are used, e.g. uniqueness rules. This is confirmed by the following ratings:
First puzzle: SER = 9.0, W = 7, gW = 7, B = 6, gB = 6
Second puzzle: SER = 8.4, W = 5, gW = 5, B = 5, gB = 5
So, the second puzzle is easier by all these ratings.

The question suggests some inversion of this hierarchy when other rules are applied.

I tried to add a few uniqueness rules (BUG, UR 1 to 5), but nothing changed.

My conclusion is, there must be some other uniqueness rule (not coded in CSP-Rules) OR some symmetry revealed by the additional clue.

[Leren]

An easy hint. What are Phasmatodeans ? Leren

[urhegyi]

An easy hint. What are Phasmatodeans ? Leren

walking stick

[denis_berthier]

An easy hint. What are Phasmatodeans ? Leren

https://en.wikipedia.org/wiki/Phasmatodea
I guess this is an hint to stick symmetry.

[Cenoman]

Using a table presentation of the puzzles:
Puzzle 1

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

Map of givens per box

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| 0 | 3 | 5 |
| 6 | 3 | 5 |
| 6 | 3 | 5 |

=> suggests a column sticks symmetry, with sticks in band 1
Map of givens per column

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| 4 | 4 | 4 | 4 | 4 | 1 | 3 | 6 | 6 |

Using a map of givens + singles does not alter the suggestion (single: +1r3c1, in band 1)

Puzzle 2

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

Map of givens per box

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| 1 | 3 | 5 |
| 6 | 3 | 5 |
| 6 | 3 | 5 |

Map of givens givens + singles per box

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| 3 | 3 | 5 |
| 6 | 3 | 5 |
| 7 | 3 | 5 |

This second map loses the similarity of bands 2, 3

Have these puzzles actually a sticks symmetry ?
Map of givens per column, puzzle 1

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| 4 | 4 | 4 | 4 | 4 | 1 | 3 | 6 | 6 |

; suggests sticks in r123c6 and r123c7 (with 1, 4 as invariant digits). 14 are forbidden in cols 23 => third stick in r123c1.
At that point, sticks symmetry and sticks identification are assumptions.
Start puzzles (givens and singles)

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    PUZ 1                      PUZ 2
 ...|...|...                .2.|...|...
 ...|..1|.23                .5.|..1|.23
 1..|23.|456                1..|23.|456
 ---+---+---                ---+---+---
 .12|7..|5.4                .12|7..|5.4
 64.|1..|.72                64.|1..|.72
 7.5|.2.|.6.                7.5|.2.|.6.
 ---+---+---                ---+---+---
 .31|.8.|64.                231|.8.|64.
 5.4|.1.|.38                5.4|.1.|.38
 86.|3..|..5                86.|3..|..5

Swap bands [23]

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    PUZ 1                      PUZ 2
 ...|...|...                .2.|...|...
 ...|..1|.23                .5.|..1|.23
 1..|23.|456                1..|23.|456
 ---+---+---                ---+---+---
 .31|.8.|64.                231|.8.|64.
 5.4|.1.|.38                5.4|.1.|.38
 86.|3..|..5                86.|3..|..5
 ---+---+---                ---+---+---
 .12|7..|5.4                .12|7..|5.4
 64.|1..|.72                64.|1..|.72
 7.5|.2.|.6.                7.5|.2.|.6.

then swap cols [23][45][89]

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    PUZ 1                      PUZ 2
 ...|...|...                ..2|...|...
 ...|..1|.32                ..5|..1|.32
 1..|32.|465                1..|32.|465
 ---+---+---                ---+---+---
 .13|8..|6.4                213|8..|6.4
 54.|1..|.83                54.|1..|.83
 8.6|.3.|.5.                8.6|.3.|.5.
 ---+---+---                ---+---+---
 .21|.7.|54.                .21|.7.|54.
 6.4|.1.|.27                6.4|.1.|.27
 75.|2..|..6                75.|2..|..6

then digit permutation [1][23][4][56][78][9]

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    PUZ 1                      PUZ 2
 ...|...|...                ..3|...|...
 ...|..1|.23                ..6|..1|.23
 1..|23.|456                1..|23.|456
 ---+---+---                ---+---+---
 .12|7..|5.4                312|7..|5.4
 64.|1..|.72                64.|1..|.72
 7.5|.2.|.6.                7.5|.2.|.6.
 ---+---+---                ---+---+---
 .31|.8.|64.                .31|.8.|64.
 5.4|.1.|.38                5.4|.1.|.38
 86.|3..|..5                86.|3..|..5
 equals start               differs start
 puzzle                     puzzle @r1c23, r2c23, r47c1


The sticks symmetry is clearly demonstrated for puzzle 1.
Puzzle 2 has the same symmetry. But I have not found the trick, how to prove it manually.

Solution for puzzle 1:

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 +-------------------------+---------------------------+---------------------+
 | *49-23  25789   36789   |  45689   45679  *49-5678  | *19-78  189   179   |
 | *49     5789    6789    |  45689   45679  *1        | *9-78   2     3     |
 | *1      789     789     |  2       3      *9-78     | *4      5     6     |
 +-------------------------+---------------------------+---------------------+
 |  39     1       2       |  7       69      3689     |  5      89    4     |
 |  6      4       389     |  1       59      3589     |  389    7     2     |
 |  7      89      5       |  489     2       3489     |  1389   6     19    |
 +-------------------------+---------------------------+---------------------+
 |  29     3       1       |  59      8       2579     |  6      4     79    |
 |  5      279     4       |  69      1       2679     |  279    3     8     |
 |  8      6       79      |  3       479     2479     |  1279   19    5     |
 +-------------------------+---------------------------+---------------------+

Column sticks symmetry: -23 r1c1, -56 r1c6, -78 r13c6, -78 r12c7; ste

So, for the moment, to me puzzle 2 is harder.

[eleven]

Puzzle 2 has the same symmetry. But I have not found the trick, how to prove it manually.

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

[yzfwsf]

Puzzle 2 has the same symmetry. But I have not found the trick, how to prove it manually.

I think you can try to remove 2r1c2 and see if the candidates of r1c2 and r1c2 meet the one-to-one correspondence relationship (the same as that of the clue).

Well, there is a bug in this way, because removing a clue cannot guarantee that the puzzle have only one solution.

[Leren]

Code: Select all

.2............1.231..23.456.127..5.464.1...727.5.2..6..31.8.64.5.4.1..3886.3....5

But wait, it gets worse. This puzzle has a second extra clue (1 in r3c1) and it's still a brute (according to Hodoku).

Leren

[denis_berthier]

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.2............1.231..23.456.127..5.464.1...727.5.2..6..31.8.64.5.4.1..3886.3....5

But wait, it gets worse. This puzzle has a second extra clue (1 in r3c1) and it's still a brute (according to Hodoku).

SER = 8.4, W = 5, same as your second puzzle.
Indeed, this is not surprising, as the resolution of your second puzzle starts that way:
hidden-single-in-a-column ==> r7c1 = 2
hidden-single-in-a-row ==> r3c1 = 1 <=== the clue you added in the third puzzle
hidden-single-in-a-column ==> r2c2 = 5
You could still add two more clues without changing the SER or W ratings (but I didn't check anything about re-introducing some symmetry).

[Leren]

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..............1.23...23.456.127..5.464.1...727.5.2..6..31.8.64.5.4.1..3886.3....5
.2........5...1.231..23.456.127..5.464.1...727.5.2..6.231.8.64.5.4.1..3886.3....5
 ^        ^       ^                                   ^

The original puzzle and the one with 4 added clues (3 perhaps trivial) and there is no "Symmetry", so it is still arguably harder than the first puzzle.

The real point I was trying to make was that, on this revisit (for me at least ) of GSP puzzles, was that they assume that the "symmetrically clued" puzzle has a unique solution.

Eleven makes this point for Sticks Symmetry here and I'd completely missed it way back when. I assume the other GSP Symmetries rely on uniqueness also.

So a GSP puzzle, apart from being "symmetrically clued" is arguably a giant UR, so to speak. Knowing from experience that adding a clue in the right place to block a critical UR can make a puzzle harder to solve (if you use UR's) I thought it might work here and it appears to do so (if you assume that the finding of the Sticks Symmetry isn't too "hard").

Leren

[denis_berthier]

Hi Leren,
I understand what you mean and your two original puzzles are a good example of it, but any rating such as SER or W that doesn't take symmetries into account is, almost by definition, a bad rating for puzzles whose solution relies on symmetry.

I haven't thought of how to define the complexity of using symmetry and to modify the ratings accordingly. If Cenoman's solution can be considered as a general start for it, it shows that taking it into account is not free lunch (besides the fact you are right to recall, that it assumes uniqueness).

[Leren]

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' Sticks symmetry - example by Mauricio

'9 . .|. . 3|. . 6    b a c|. . .|. . .
'. 6 .|. . 1|. 5 .    b a c|. . .|. . .
'. . 4|6 . .|2 . .    b a c|. . .|. . .
'-----+-----+-----    -----+-----+-----
'. 7 .|. 1 .|9 . 8    d X d|. X .|. X .
'. . .|. . .|. . .    d X d|. X .|. X .
'3 . 2|. . .|. 7 .    d X d|. X .|. X .
'-----+-----+-----    -----+-----+-----
'. . 8|2 . .|5 . .    c a b|. . .|. . .
'. 5 .|1 . .|. 6 .    c a b|. . .|. . .
'4 . .|. . 5|. . 3    c a b|. . .|. . .
'
'BxCx (exchange bands 13 and Columns 13,46,79) + Isomorphism '(1)(4)(7)(23)(56)(89)

Hi denis,

I todally agree that the detection of Sticks Symmetry is no free lunch but most solvers would agree that it is a shortcut of some sort. Also, there is another way of thinking about it that might make it a bit easier to use after you've had some practice. The above diagram is an example by eleven based on a puzzle by Mauricio that enabled me to cut through the jargon and finally figure out to at least code it up. I've modified the exemplar part and added some extra stuff to make it easier to follow.

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b1 a1 c1|. . .|. . .
b3 a3 c3|. . .|. . .
b5 a5 c5|. . .|. . .
--------+-----+-----
d1 X  d2|. X .|. X .
d3 X  d4|. X .|. X .
d5 X  d6|. X .|. X .
--------+-----+-----
c2 a2 b2|. . .|. . .
c4 a4 b4|. . .|. . .
c6 a6 b6|. . .|. . .

In this canonicalised version of a Sticks symmetry puzzle the X's are the Sticks, and the cells labelled abcd are the "symmetry" cells. By adding the numbering I can say that the Symmetry cell for abcd(n) is abcd(n+1 or n-1) as appropriate.

Also there are abcd's in Stacks 2 and 3 but eleven and I have left them out to reduce the clutter of the diagram. You permute the bands and the columns within a band, to line up with any column Sticks symmetry case. For row Sticks symmetry just rotate or flip the puzzle. In this way of looking at it you can identify a sticks symmetry if there is an automorhism, that is faithfully represented by the clues.

You can see how it works in eleven's example. eg 9 is in cell b1 and its partner digit 8 is in cell b2. The other instance of 9 & 8 are in "symmetry" cells d1 and d2 of Stack 3, so we are OK.

In my first puzzle the Sticks were in band 1 and columns 1 and 2 were swapped, so my extra clue was in d1 = r1c2 = 2 and eleven said that you could restore the Symmetry by putting a 3 in d2 which was r1c3. True.

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.

Maybe this way of looking at things will reduce the "hardness value" you have to place on detecting a Sticks symmetry

Leren

[denis_berthier]

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.

[Cenoman]

Hi eleven, yzfwsf, Leren, Denis (in the order of your messages...)
I don't know which hour each of you has written (don't know your location), but for me it was just sleeping time !
First of all, a clarification:

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.

Agreed. I posted a detailed proof in recent posts (here, here, and above), for three reasons: the first one to consolidate my own learning, the second one to show that the sticks symmetry can be handled manually, the third one to help newcomers' understanding of the pattern (and jco's comment showed it was useful).

A solution presented with the digit permutation, and a PM grid with the sticks and eliminations (such as Leren's) is enough.

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 ? I need to locate the sticks before looking at box pattern similarity and digit permutation. My maps of box given counts and then of line given count takes no more than two minutes. It's a hint, not a demonstration.

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 ? Adding a clue can't yields multi-solutions, but possibly a no-solution puzzle.

I think you can try to remove 2r1c2 and see if the candidates of r1c2 and r1c3 meet the one-to-one correspondence relationship (the same as that of the clue).

It is the same as to assume that the puzzle is not minimal. As you say, the new puzzle has to be checked for unique solution.

BTW, is this what your solver does ? I have fed it with the "non-symmetric" puzzle, and the solver (version 6.18) finds the GSP.

Code: Select all

.2............1.231..23.456.127..5.464.1...727.5.2..6..31.8.64.5.4.1..3886.3....5

But wait, it gets worse. This puzzle has a second extra clue (1 in r3c1) and it's still a brute (according to Hodoku).

Not with this second extra clue. Look at my puzzle 2: 1r3c1 is included as solved cell (hidden single), but yields no extra issue, as r3c1 is in a Stick. (Note: my puzzle 2 is exactly your second example with 4 extra clues)

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) ?