The New Sudoku Players' Forum

[ronk]

This Sudoku can be solved by some triplets and in the end a bi-directional cycle with a pair as node. It eliminates 4 candidates leaving a BUG Type 1.

I haven't found that bi-directional cycle yet, at least not for 4 eliminations. Are these the correct pencilmarks?

Code: Select all

 9   134 13  | 78  6   2   | 48  35  357
 68  57  57  | 4   1   3   | 9   68  2
 68  34  2   | 78  5   9   | 468 1   37
-------------+-------------+------------
 1   25  58  | 9   4   7   | 268 368 38
 3   27  79  | 1   8   6   | 5   29  4
 4   6   89  | 3   2   5   | 18  7   19
-------------+-------------+------------
 7   8   13  | 5   39  4   | 12  29  6
 5   13  6   | 2   39  8   | 7   4   19
 2   9   4   | 6   7   1   | 3   58  58

[claudiarabia]

This Sudoku can be solved by some triplets and in the end a bi-directional cycle with a pair as node. It eliminates 4 candidates leaving a BUG Type 1.

I haven't found that bi-directional cycle yet, at least not for 4 eliminations. Are these the correct pencilmarks?

Code: Select all

 9  14-3 #13  | 78  6   2   |  48  #35  #57-3
 68  57   57  | 4   1   3   |  9    68   2
 68  34   2   | 78  5   9   |  468  1   #37
--------------+-------------+-------------
 1   25   58  | 9   4   7   | 26-8 36-8 #38
 3   27   79  | 1   8   6   |  5    29   4
 4   6    89  | 3   2   5   | #18   7    19
--------------+-------------+-------------
 7   8   #13  | 5   39  4   | #12   29   6
 5   13   6   | 2   39  8   |  7    4    19
 2   9    4   | 6   7   1   |  3    58   58

Hi ronk,

your grid is totally correct. The clue is that r1c9 functions as a cycle cell while one clue is eliminated right there.

this is the cycle:

Code: Select all

r1c3 r1c8 r1c9 r3c9 r4c9 r6c7 r7c7 r7c3
13   35   57   37   38   18   1    1   

Eliminated are 3 in r1c29 and 8 in r4c78. The 8 in r3c7 is the BUG-solving candidate then.

Claudia

[daj95376]

The theme for this thread is puzzles that are solved with (one or more) Continuous Simple Nice Loops as the highest ranked (most difficult) technique - not quite precise because ranking of techniques is debatable. Some properties of such chains are:
1. they describe a full/continuous cycle
2. they can freely start anywhere within the cycle (no specific starting point)
3. they work equally in both directions.
4. candidates may be eliminated from cells sharing a unit with two linked cycle nodes, or from nodes in the cycle itself (see the theoretical descriptions of nice loops elsewhere; links to be provided later).

Has property #2 been cancelled for Bicycles? Otherwise, claudiarabia can't have [r1c9] in her Bicycle.

[ronk]

claudiarabia can't have [r1c9] in her Bicycle.

[edit: I also had trouble understanding (see here) what Claudia was doing.

Another continuous loop (bidirectional cycle) is ...

Code: Select all

 9    14-3 *13   | 78   6    2    | 48   35    57-3
 68   57    57   | 4    1    3    | 9    68    2
 68  *34    2    | 78   5    9    | 468  1    *37
-----------------+----------------+----------------
 1    25    58   | 9    4    7    | 26-8 36-8 *38
 3    27    79   | 1    8    6    | 5    29    4
 4    6     89   | 3    2    5    |*18   7     19
-----------------+----------------+----------------
 7    8    *13   | 5    39   4    |*12   29    6
 5    13    6    | 2    39   8    | 7    4     19
 2    9     4    | 6    7    1    | 3    58    58

 r1c3 -3- r3c2 =3= r3c9 -3- r4c9 -8- r6c7 -1- r7c7 =1= r7c3 =3= loop

Due to the continuous loop, the weak bilocation links are converted to conjugate links, eliminating other 3s in b1, 3s in c9, 8s in b6, and 1s in c7 (if there were any). Moreover, if r7c3 weren't already bivalued, candidates other than bivalue 13 could be eliminated.

[edit: as noted by daj95376, continuous loop incorrectly used r1c8; replaced with r3c2]

[re'born]

Alternatively, there is the potential deadly pattern:

Code: Select all

  *--------------------------------------------------*
 | 9    134  13   | 78   6    2    | 48  *35  *357  |
 | 68   57   57   | 4    1    3    | 9    68   2    |
 | 68   34   2    | 78   5    9    | 468  1    37-  |
 |----------------+----------------+----------------|
 | 1    25   58   | 9    4    7    | 268 *368 *38   |
 | 3    27   79   | 1    8    6    | 5    29   4    |
 | 4    6    89   | 3    2    5    | 18   7    19   |
 |----------------+----------------+----------------|
 | 7    8    13   | 5    39   4    | 12   29   6    |
 | 5    13   6    | 2    39   8    | 7    4    19   |
 | 2    9    4    | 6    7    1    | 3   *58  *58   |
 *--------------------------------------------------*


We then get
[r3c9]-7-[r1c9]=7|6=[r4c8]-6-[r2c8]=6=[r2c1]-6-[r3c1]-8-[r3c4]-7-[r3c9], =>r3c9<>7, solving the puzzle.
(There is also a basic xy-chain that solves the puzzle -4-[r3c2]-...-[r1c7]-4-, =>r1c2<>4).
Edit: You can also add a few cells (in row 7) to claudia's chain to turn it into an almost xy-ring.

[claudiarabia]

The theme for this thread is puzzles that are solved with (one or more) Continuous Simple Nice Loops as the highest ranked (most difficult) technique - not quite precise because ranking of techniques is debatable. Some properties of such chains are:
1. they describe a full/continuous cycle
2. they can freely start anywhere within the cycle (no specific starting point)
3. they work equally in both directions.
4. candidates may be eliminated from cells sharing a unit with two linked cycle nodes, or from nodes in the cycle itself (see the theoretical descriptions of nice loops elsewhere; links to be provided later).

Has property #2 been cancelled for Bicycles? Otherwise, claudiarabia can't have [r1c9] in her Bicycle.

Hi daj,

property #2 is still valid. I can start from any point of my cycle in both directions. The only pecularity is that when starting anticlockwise from 7 in r3c9 then at first a naked pair of 35 is emerging in r1c98 which is forming one node in this situation. But at least when the cycle has arrived in his anticlockwise turn in 3 in r4c9 it turns out that there is no pair anymore in r1c89 but instead a 3 in r1c9 to be eliminated which is confirmed in the clockwise direction where there is no pair anymore.

Code: Select all

 9  14-3 #13  | 78  6   2   |  48  #35  #57-3
 68  57   57  | 4   1   3   |  9    68   2
 68  34   2   | 78  5   9   |  468  1   #37
--------------+-------------+-------------
 1   25   58  | 9   4   7   | 26-8 36-8 #38
 3   27   79  | 1   8   6   |  5    29   4
 4   6    89  | 3   2   5   | #18   7    19
--------------+-------------+-------------
 7   8   #13  | 5   39  4   | #12   29   6
 5   13   6   | 2   39  8   |  7    4    19
 2   9    4   | 6   7   1   |  3    58   58

But anyway, ronks cycle is much more elegant because it has 7 nodes only and by adding r3c2 to the cycle he uses the bi-location of 3 in some units too while I used only the bi-location of 1. And ronk confines the xy-(bivalue-cell) element in the cycle to the cells r1c3, r4c9 and r6c7.

In many situations there are several possibilities to form cycles which eliminate the same candidates. But set the case that there is really no possibility to form the classical elegant cycle like ronk did here, I may take resort to pairs as nodes of cycles in one direction, which aren't pairs in the other direction. And if I can have a cycle with this - why not? It works.

Claudia

[ronk]

The only pecularity is that when starting anticlockwise from 7 in r3c9 then at first a naked pair of 35 is emerging in r1c98 which is forming one node in this situation.

Using your ALS in the continuous loop is perfectly legit, but it changes the eliminations a bit. That pair in r1c89 for the anti-clockwise direction means r1c9<>3 is an invalid deduction, as shown in this counter-example.

Code: Select all

 .  .  1 | .  .  . | .  5  3
 .  .  . | .  .  . | .  .  .
 .  .  . | .  .  . | .  .  7
---------+---------+---------
 .  .  . | .  .  . | .  .  8
 .  .  . | .  .  . | .  .  .
 .  .  . | .  .  . | 1  .  .
---------+---------+---------
 .  .  3 | .  .  . | 2  .  .
 .  .  . | .  .  . | .  .  .
 .  .  . | .  .  . | .  .  .

[edit: Since digit 1 does not exist in r7 as required by the nice loop below, this counter-example actually illustrates a contradiction. This means r1c9<>3 is indeed a valid deduction for this set of cells. For the why, see the "edit" below.]

Your continuous loop is expressible as:

r1c3 -3- (ALS:r1c89=3|5|7=r1c89) -7- r3c9 -3- r4c9 -8- r6c7 -1- r7c7 =1= r7c3 =3= loop

This loop [edit: also] potentially eliminates other 5's in r1 and b3.

[edit: This is an example of "cannibalism" -- cannibalism of the chain. The continuous loop converts weak link r3c9 -3- r4c9 into a conjugate link. In other words, exactly one of r3c9=3 and r4c9=3 is ultimately true. This yields the elimination r1c9<>3 even though r1c9 is part of the chain.

My apologies to Claudia for doubting her deduction.]

[claudiarabia]

Could it be qualified ?

Code: Select all

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


JPF

Code: Select all

 *-----------------------------------------------------------*
 | 59    56    8     | 13    7     13    | 2     6-4  *49    |
 | 67    1     4     | 89    2     89    | 56    3     57    |
 | 37    29    23    | 4     5     6     | 1     79    8     |
 |-------------------+-------------------+-------------------|
 | 19    24    7     | 1259  6     1249  | 8     25    3     |
 | 24    3     5     | 27    8     247   | 9     1     6     |
 | 189   89    6     | 1259  3     129   | 4     257   27    |
 |-------------------+-------------------+-------------------|
 | 38    48   #23-1  | 6     9     5     | 7    *42   #12    |
 | 46    7     9     | 23    1     23    | 56    8     5-4   |
 | 25    56   #12    | 78    4     78    | 3    269  *#29-1  |
 *-----------------------------------------------------------*

Next to re'borns solution I found one of two small steps. There is an Almost Unique Rectangle (AUR) in r79c39 of 12 (#). Because of the bi-value cells in r7c9 and r9c3 of 12 we can deduct that a placement of 1 either in r7c3 or in r9c9 will lead to the deadly pattern (the notorious deadly rectangle) Thus we can place the 1 into r7c9 as well as in r9c3.

Thus in the second and last step we have an xy-wing (*) now upon the numbers 249 in r1c9, r7c8 and r9c9 by which the 4 in r1c8 and in r8c9 will be eliminated.

Now this nice puzzle is practically solved.

Claudia

[daj95376]

Why is this an Almost Unique Rectangle?

Either [r7c3]=3 or [r9c9]=9

Code: Select all

# [r7c3]=3 => [r7c9]=1,[r9c3]=1 => [r9c9]<>1
# [r9c9]=9 => [r7c9]=1,[r9c3]=1 => [r7c3]<>1
# either way, X-Wing on <1> inside the UR forces [r7c3]<>1,[r9c9}<>1
 *-----------------------------------------------------------*
 | 59    56    8     | 13    7     13    | 2     46    49    |
 | 67    1     4     | 89    2     89    | 56    3     57    |
 | 37    29    23    | 4     5     6     | 1     79    8     |
 |-------------------+-------------------+-------------------|
 | 19    24    7     | 1259  6     1249  | 8     25    3     |
 | 24    3     5     | 27    8     247   | 9     1     6     |
 | 189   89    6     | 1259  3     129   | 4     257   27    |
 |-------------------+-------------------+-------------------|
 | 38    48    123   | 6     9     5     | 7     24    12    |
 | 46    7     9     | 23    1     23    | 56    8     45    |
 | 25    56    12    | 78    4     78    | 3     269   129   |
 *-----------------------------------------------------------*


[re'born]

Why is this an Almost Unique Rectangle?

Either [r7c3]=3 or [r9c9]=9

Code: Select all

# [r7c3]=3 => [r7c9]=1,[r9c3]=1 => [r9c9]<>1
# [r9c9]=9 => [r7c9]=1,[r9c3]=1 => [r7c3]<>1
# either way, X-Wing on <1> inside the UR forces [r7c3]<>1,[r9c9}<>1
 *-----------------------------------------------------------*
 | 59    56    8     | 13    7     13    | 2     46    49    |
 | 67    1     4     | 89    2     89    | 56    3     57    |
 | 37    29    23    | 4     5     6     | 1     79    8     |
 |-------------------+-------------------+-------------------|
 | 19    24    7     | 1259  6     1249  | 8     25    3     |
 | 24    3     5     | 27    8     247   | 9     1     6     |
 | 189   89    6     | 1259  3     129   | 4     257   27    |
 |-------------------+-------------------+-------------------|
 | 38    48    123   | 6     9     5     | 7     24    12    |
 | 46    7     9     | 23    1     23    | 56    8     45    |
 | 25    56    12    | 78    4     78    | 3     269   129   |
 *-----------------------------------------------------------*


Using the notation from Mike Barker's post, I would call it two applications of UR+2D/1SL. Carcul's original defintion of AUR's

Almost Unique Rectangle (AUR): a set of four cells populated in such a way that if one or two of them does not have a specific candidate 'x', then those four cells form a unique rectangle.

doesn't strictly omit this, but I think that his intention was not that the 'x''s were removed by setting one of the cells in the UR to something not x, but rather by setting a cell outside of a pattern to something that excluded x. So, IMO, it is not an almost unqiue rectangle.

[ronk]

Why is this an Almost Unique Rectangle?

Good question, but I don't think it's possible to define Unique Rectangle (UR) and Almost Unique Rectangle (AUR) in a way that would be consistent with all of today's usages -- even just the common usages -- of these terms.

[claudiarabia]

Why is this an Almost Unique Rectangle?

Good question, but I don't think it's possible to define Unique Rectangle (UR) and Almost Unique Rectangle (AUR) in a way that would be consistent with all of today's usages -- even just the common usages -- of these terms.

Today I searched for the source of my definition of

Code: Select all

. 12 . | . 12x.
. 12x. | . 12 .

as an Almost unique Rectangle. I didn't find the page, where I found it at first but I found this posting by Keith defining this constellation as a Unique Rectangle Type 5.
http://forum.enjoysudoku.com/viewtopic.php?p=29105#p29105

Claudia

[daj95376]

The Type 5 UR only allows one extra value to be present. That means 'x' must be the same value in your preceeding pattern.

In your puzzle, there is an X-Wing constraint and two different values for 'x' in diagonal corner cells. This makes it a diagonal variant of a Type 4 UR.

[Edit: Changed Type 3 to Type 4]

[claudiarabia]

Hi,

this sudoku contains for it's solution next to a wxyz-wing a bi-directional cycle, a bi-directional y-cycle and a bi-directional x-cycle.

Code: Select all

. 7 . . 3 . . 8 .
9 . . . . . . . 4
. . 1 6 . 5 9 . .
. . 4 . . . 7 . .
3 . . . 5 . . . 9
. . 6 . . . 5 . .
. . 7 4 . 3 1 . .
5 . . . . . . . 2
. 8 . . 2 . . 6 .


Enjoy

Claudia

[re'born]

Hi,

this sudoku contains for it's solution next to a wxyz-wing a bi-directional cycle, a bi-directional y-cycle and a bi-directional x-cycle.

Code: Select all

. 7 . . 3 . . 8 .
9 . . . . . . . 4
. . 1 6 . 5 9 . .
. . 4 . . . 7 . .
3 . . . 5 . . . 9
. . 6 . . . 5 . .
. . 7 4 . 3 1 . .
5 . . . . . . . 2
. 8 . . 2 . . 6 .


Enjoy

Claudia

At the point the puzzle becomes non-trivial, the following xy-chain solves the puzzle:
- 8 - [r5c3 - 2 - r1c3 - 5 - r1c9 - 1 - r6c9] - 8 -, => r5c7<>8.