The New Sudoku Players' Forum

[civiliza]

Still bolting odds and ends on my generator.

The last two were 4/5 Cell XY Chains and 6 Cell Type 1 Unique Loops.

Now that the UL code gobbles up the stray UL's that were being detected by 5 Cell XY Chains, I am finding that all the remaining XY Chains either produce bidirectional Turbot Fish (I now know some of these are Remote Pairs), or coexist with / "mirror" a Turbot Fish using a different route.

Is this just a coincidence based on a small number of runs, or a natural consquence of a generation method that starts with an answer and then successively takes away cells one at a time that can each be "solved" by a mixture of:

Hidden Singles,
Direct Techniques,
Pointing / Claiming
XY Wings
XYZ Wings
Naked & Hidden Sets up to Quads
Fish up to Jellyfish
UR's Type 1, 2 & 5
6 Cell UL Type 1's
4 & 5 Cell XY Chains
Bug+1's

Anyway, having just cured a bug in the new UL detector, here's my first puzzle with a (deliberate, as opposed to accidental) Unique Loop:

Code: Select all

3..|.17|...
8.1|5..|4.7
65.|49.|...
-----------
5..|...|3..
..4|...|..8
..8|..5|.1.
-----------
...|6..|...
..6|2.9|...
7..|1..|.86


Code: Select all

3...17...8.15..4.765.49....5.....3....4.....8..8..5.1....6.......62.9...7..1...86

--

Using GSF's backtracker to ensure uniqueness, would there be any advantage in taking away cells two at a time instead of one at a time? eg Check whether the puzzle would still be unique if the two cells were removed, and only then check if they can both be "detected" - I have a hunch that I would hit some of my higher rated tecniques more often if I skipped over a low value single cell technique and went straight to a higher valued two cell technique.

(I have noted of late that by making my low level techniques more bulletproof, I am removing some of the loopholes that allowed escape routes to higher techniques).

[champagne]

Hi,

If I catch your point I suggest the following

pick up in the results of the pattern game any puzzle rating 7.1 to 7.4

you should find a puzzle with an XY chain as hardest step and no turbot

a turbot in serate is rated in the range 6.6 to 7.0, 7.1 at most

[daj95376]

I couldn't follow what you need answered. However, I do have an XY-Chain example provided by Mike Barker. He rates it at "11-nodes", but all my solver found was 10-cells. It's possible that he included the elimination cell as well. For sure, there's no way it can be confused with a Turbot. (Basic steps are not included.)

Code: Select all

 +-----------------------+
 | . . 6 | 5 . . | . . . |
 | 1 8 . | . 4 . | 9 . 5 |
 | . . . | . . . | 3 . . |
 |-------+-------+-------|
 | . . 8 | 4 6 . | 5 . . |
 | . . 2 | . . . | . . 4 |
 | . 5 9 | . 8 . | . 6 . |
 |-------+-------+-------|
 | . . . | . . 6 | 1 4 . |
 | . 3 . | . . . | 6 . . |
 | . . . | . . 7 | . 8 3 |
 +-----------------------+

 +-----------------------------------------------------+
 |  279  79-2 6    |  5    3   b19   |  4   a12   8    |
 |  1    8    3    |  6    4    2    |  9    7    5    |
 |  259  249  45   |  179  179  8    |  3    12   6    |
 |-----------------+-----------------+-----------------|
 | e37  d17   8    |  4    6   c19   |  5    39   2    |
 | f36   16   2    |  179  179  5    |  8    39   4    |
 |  4    5    9    |  2    8    3    |  7    6    1    |
 |-----------------+-----------------+-----------------|
 |  8   j29  h57   |  3    25   6    |  1    4   i79   |
 |  29   3    17   |  8    12   4    |  6    5    79   |
 | g56   46   145  |  19   159  7    |  2    8    3    |
 +-----------------------------------------------------+
 # 42 eliminations remain

 (2=1)r1c8 - (1=9)r1c6 - (9=1)r4c6 - (1=7)r4c2 - (7=3)r4c1 - ...
 (3=6)r5c1 - (6=5)r9c1 - (5=7)r7c3 - (7=9)r7c9 - (9=2)r7c2        =>  r1c2<>2

 -- followed by --

 <79+1>  XY-Wing  r1c2/r1c6+r4c2         <> 1    r4c6   complementary
 <17+9>  XY-Wing  r4c2/r1c2+r4c6         <> 9    r1c6         "

Solution:
976531428183642975524798316718469532362175894459283761895326147237814659641957283


As for removing pairs of cells to reduce a solved grid to a puzzle state; if done properly, that's one way to get symmetry in a resulting puzzle. However, it doesn't guarantee that the resulting puzzle will be "minimal" -- a term that I consider to be obliquitous and irrelevant, but not by others.

[civiliza]

Thanks Champagne and DAJ, part of the reason I have shyed away from chains in the past was their potential length. The examples you mention do seem to use longer chains than the 4/5 Cell limit I have set myself. (One joy of having my own generator is being able to dumb it down to my own comfort levels).

That said, they did trigger a suppressed memory of a period when my early (and buggy) XY Chain code created XY Chains that Sudoku Explainer could only solve using Forcing Chains. While the code was buggy, I remember following the XY Chains, and they were valid.

Which, I guess, answers my own question, even with 4 / 5 cell chains, there are some XY Chains that do not result in / reduce to one or more Turbot Fish.

Over the last few days, I had formed the half baked idea that all (4 & 5) cell XY Chains either equated to or came with a companion Turbot Fish in the same way that a Hidden Set comes with a companion Naked Set. (Not helped by all my recent XY Chain puzzles getting a Sudoku Explainer rating of 6.6).

==

Thanks for bringing up symmetry DAJ, I had not thought of that, but removing cells in pairs does seem like an ideal way of producing puzzles with single plane or rotational symettry.

I was wondering whether there was a potential situation when either of two cells could not be deduced if removed on it's own, but could be deduced if both were removed at the same time. (By creating a recognisable elimination pattern somewhere else in the puzzle). - A long shot I know, but if there's one thing early coding errors have taught me, it's that a low level technique missed / deferred at a crucial stage can result in a biased puzzle that uses much higher techniques.

[ArkieTech]

There are two Remote Pairs on 29 that solve the puzzle. Is a remote pair an XY Turbot?

[daj95376]

There are two Remote Pairs on 29 that solve the puzzle. Is a remote pair an XY Turbot?

Code: Select all

 +-----------------------------------------------------------------------+
 |  3      4     d29     |  8      1      7      |  2569   2569   259    |
 |  8     e29     1      |  5      26     236    |  4      239    7      |
 |  6      5      7      |  4      9      23     |  8      23     1      |
 |-----------------------+-----------------------+-----------------------|
 |  5      67    c29     |  79     8      1      |  3      24679  249    |
 |  1      37     4      |  379    26     26     |  579    579    8      |
 | b29     367    8      |  37     4      5      |  67     1     a29     |
 |-----------------------+-----------------------+-----------------------|
 |  29     1      35     |  6      357    8      |  2579   24579  23459  |
 |  4      8      6      |  2      357    9      |  1      57     35     |
 |  7     f29     35     |  1      35     4      | g29     8      6      |
 +-----------------------------------------------------------------------+
 # 64 eliminations remain


A classic Remote Pair is simply two, 2-value XY-Chains using the same cells.

Code: Select all

 RP (2=9)r1c3 - (9=2)r2c2 - (2=9)r9c2 - (9=2)r9c7  =>  r1c7<>2
 RP (9=2)r1c3 - (2=9)r2c2 - (9=2)r9c2 - (2=9)r9c7  =>  r1c7<>9

 RP (2=9)r1c3 - (9=2)r4c3 - (2=9)r6c1 - (9=2)r6c9  =>  r1c9<>2
 RP (9=2)r1c3 - (2=9)r4c3 - (9=2)r6c1 - (2=9)r6c9  =>  r1c9<>9


When the chain consists of four cells, then the elimination cell forms the "oddagon" pattern of a Turbot.

Code: Select all

    (2)r1c3   = (2)r2c2   - (2)r9c2   = (2)r9c7    =>  r1c7<>2
    (2)r1c3   = (2)r4c3   - (2)r6c1   = (2)r6c9    =>  r1c9<>2

    (9)r1c3   = (9)r2c2   - (9)r9c2   = (9)r9c7    =>  r1c7<>9
    (9)r1c3   = (9)r4c3   - (9)r6c1   = (9)r6c9    =>  r1c9<>9


What should be notes is that each Turbot/Remote Pair is part of a 7-cell XY-Chain (a..g -- with no endpoint eliminations).