The Bicycle Collection.

Everything about Sudoku that doesn't fit in one of the other sections

Postby re'born » Mon May 14, 2007 7:58 pm

ArkieTech wrote:Just when I think I am beginning to get a grip or understanding.
Code: Select all
 *-----------*
 |...|...|..1|
 |...|2.3|...|
 |.45|...|6..|
 |---+---+---|
 |..7|..6|.3.|
 |.8.|...|.9.|
 |.1.|8..|7..|
 |---+---+---|
 |..3|...|95.|
 |...|4.1|...|
 |2..|...|...|
 *-----------*


 *-----------*
 |.32|...|.71|
 |671|2.3|.89|
 |.45|...|623|
 |---+---+---|
 |.27|..6|138|
 |.86|...|295|
 |.19|8..|746|
 |---+---+---|
 |163|7..|954|
 |7.8|4.1|362|
 |2.4|...|817|
 *-----------*

 
 *-----------------------------------------------------------*
 | 89    3     2     | 569   68    4589  | 45    7     1     |
 | 6     7     1     | 2     45    3     | 45    8     9     |
 | 89    4     5     | 19    17    789   | 6     2     3     |
 |-------------------+-------------------+-------------------|
 | 45    2     7     | 59    459   6     | 1     3     8     |
 | 34    8     6     | 13    17    47    | 2     9     5     |
 | 35    1     9     | 8     23    25    | 7     4     6     |
 |-------------------+-------------------+-------------------|
 | 1     6     3     | 7     28    28    | 9     5     4     |
 | 7     59    8     | 4     59    1     | 3     6     2     |
 | 2     59    4     | 36    36    59    | 8     1     7     |
 *-----------------------------------------------------------*

r1c4-9-r4c4=9=r4c5-5-r8c5=9=r9c6-9-r1c6=5=r1c4

Help! since weak side r1c4, r4c4 is part of the ring I thought one or the other point had to be a 9 and therefore the 9 at r3c4 could be eliminated. I am wrong of course. Can anyone help me?

dan


You've made an error in your chain. After r4c4=9=r4c5, it doesn't make sense to have r4c5-5-r8c5. You've effectively written "if r4c4<>9 then r4c5=9. But if r4c5=5, then r8c5<>5..."

I do see a y-cycle, but it doesn't seem to help matters much:

[r4c4]-9-[r3c4]-1-[r5c4]-3-[r6c5]-2-[r6c6]-5-[r4c4], implying r1c4<>9, r4c5<>5.

However, a related xy-chain does solve the puzzle immediately:

9-[r4c4]-1-[r5c4]-3-[r6c5]-2-[r6c6]-5-[r9c6]-9, implying r13c6<>9.
Last edited by re'born on Tue May 15, 2007 9:45 am, edited 1 time in total.
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Postby ArkieTech » Tue May 15, 2007 12:07 pm

Rep'nA said
You've made an error in your chain. After r4c4=9=[r4c5],

Thanks, I must retreat for contemplation. Maybe I will learn something.:)

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Postby Ocean » Tue May 15, 2007 3:21 pm

claudiarabia wrote:At first, Ocean, let me express my joy over your strong comeback in the forum after a long time of being seldom here. I feel very honored that my incentive to discuss bi-directional cycles fell on such a fertile ground.
Thanks. I also find the bi-directional cycles quite attractive, and worth a closer study.

Had a look at your puzzle, with what I call 'Turbot Fish', accompanied by a 4-node xy-ring (Type I, from the configurations in the figure below). It also has a pleasant symmetrical design. (Thanks rep'nA for showing alternative solution paths.)



If we exclude the naked quadruples, four-rings can have six different configurations (maybe suitable for classification/taxonomi?):
Code: Select all
A..|...|..D    A..|...|..D    A..|...|D..    A..|...|D..    A.C|...|..D    A.C|...|..D
...|...|...    ...|...|...    ...|...|...    ...|...|...    ...|...|...    ...|...|...
...|...|...    B..|...|..C    B..|...|..C    ..B|...|..C    B..|...|...    .B.|...|...
---+---+---    ---+---+---    ---+---+---    ---+---+---    ---+---+---    ---+---+---
B..|...|..C    ...|...|...    ...|...|...    ...|...|...    ...|...|...    ...|...|...
...|...|...    ...|...|...    ...|...|...    ...|...|...    ...|...|...    ...|...|...
...|...|...    ...|...|...    ...|...|...    ...|...|...    ...|...|...    ...|...|...
---+---+---    ---+---+---    ---+---+---    ---+---+---    ---+---+---    ---+---+---
...|...|...    ...|...|...    ...|...|...    ...|...|...    ...|...|...    ...|...|...
...|...|...    ...|...|...    ...|...|...    ...|...|...    ...|...|...    ...|...|...
...|...|...    ...|...|...    ...|...|...    ...|...|...    ...|...|...    ...|...|...

Type I         Type II        Type III       Type IV        Type V         Type VI
#
Last edited by Ocean on Tue May 15, 2007 5:33 pm, edited 3 times in total.
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Postby Ocean » Tue May 15, 2007 3:31 pm

rep'nA wrote:
ArkieTech wrote:Just when I think I am beginning to get a grip or understanding.
Code: Select all
 *-----------------------------------------------------------*
 | 89    3     2     | 569   68    4589  | 45    7     1     |
 | 6     7     1     | 2     45    3     | 45    8     9     |
 | 89    4     5     | 19    17    789   | 6     2     3     |
 |-------------------+-------------------+-------------------|
 | 45    2     7     | 59    459   6     | 1     3     8     |
 | 34    8     6     | 13    17    47    | 2     9     5     |
 | 35    1     9     | 8     23    25    | 7     4     6     |
 |-------------------+-------------------+-------------------|
 | 1     6     3     | 7     28    28    | 9     5     4     |
 | 7     59    8     | 4     59    1     | 3     6     2     |
 | 2     59    4     | 36    36    59    | 8     1     7     |
 *-----------------------------------------------------------*

I do see a y-cycle, but it doesn't seem to help matters much:

[r4c4]-9-[r3c4]-1-[r5c4]-3-[r6c5]-2-[r6c6]-5-[r4c4], implying r1c4<>9, r4c5<>5.

However, a related xy-chain does solve the puzzle immediately:

9-[r4c4]-1-[r5c4]-3-[r6c5]-2-[r6c6]-5-[r9c6]-9, implying r13c6<>9.

Thanks for showing the alternative xy-chain.
After the xy-ring (y-cycle), the puzzle is solved with an xy-wing, which makes two 'advanced' steps, while the xy-chain is one step.

Here is an even shorter chain that solves the puzzle directly:
-4-[r2c5]-5-[r1c4]=5=[r4c4]=9=[r4c5]=4=[r2c5] ->r2c5=4.

I should say, I'm not too satisfied with this example. The best pedagogical examples should have a bidirectional cycle as the clearly preferred technique. In this case that is not so clear. My logical analyzer detects xy-rings, but not every other alternative. When checking the puzzle in Sudoku Explainer, the 'Bidirectional Y-Cycle' was listed as one of the (preferred) techniques. But by manual inspection we see that it can be a matter of taste which technique is the simplest at the 'crucial point'.
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BUG type

Postby claudiarabia » Fri May 18, 2007 2:10 pm

rep'nA wrote:
claudiarabia wrote:
Code: Select all
. 4 . . . . . 1 .
. . . 5 . 7 . . .
8 . . . 9 . . . 7
. 2 . . 1 . . 9 .
. . 5 4 . 6 8 . .
. 3 . . 8 . . 7 .
4 . . . 5 . . . 6
. . . 2 . 9 . . .
. 7 . . . . . 3 .


This one needs one Sky-craper which eliminates two 1s, one bi-directionyl-y-cycle in r5c59 and r8c59 with the numbers 1347. It ends up in a BUG1 Situation.

Claudia


As long as you are ending up in a BUG+1 situation, you might as well solve it before the y-cycle, when you are in a BUG+2 situation.

Code: Select all
 *--------------------------------------------------*
 | 2    4    7    | 68   36   38   | 5    1    9    |
 | 19   6    19   | 5    2    7    | 3    4    8    |
 | 8    5    3    | 1    9    4    | 2    6    7    |
 |----------------+----------------+----------------|
 | 67   2    8    | 37   1    5    | 46   9    34   |
 | 79+1 19   5    | 4    37   6    | 8    2    13   |
 | 16   3    4    | 9    8    2    | 16   7    5    |
 |----------------+----------------+----------------|
 | 4    19   2    | 37   5    13   | 79   8    6    |
 | 3    8    6    | 2    47   9    | 17+4 5    14   |
 | 5    7    19   | 68   46   18   | 49   3    2    |
 *--------------------------------------------------*


This is not a BUG2-Situation. In a BUG type2-Situation you have two times the same candidate three times in one row, column and box, which disturbes the BUG-pattern, where every "ordinary" candidate shows up twice only in every line and box. So you can't have 1 and 4 but only 1 and 1 or 7 and 7 and so on. In a BUG-2 Situation these two eliminate a similar number which sees both of the two "disturbing" candidates. In BUG type 1 the solution starts in the cell with the three candidates while in the BUG type 2-situation the solution starts outside the two or even three Three-candidate-cells.The 1 in r5c1 in this puzzle is the only BUG-solving candidate. The 1 in r5c1 is definitely to remain in its cell while the 1 in r8c7 shows up only twice in its region and the 4 in the same cell can be eliminated and can't be a part of a BUG-solving pattern too. Only in BUG type 4 you have two different candidates disturbing the BUG-pattern, but they are in the same region. See here:
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BUG type

Postby claudiarabia » Fri May 18, 2007 2:14 pm

rep'nA wrote:
claudiarabia wrote:
Code: Select all
. 4 . . . . . 1 .
. . . 5 . 7 . . .
8 . . . 9 . . . 7
. 2 . . 1 . . 9 .
. . 5 4 . 6 8 . .
. 3 . . 8 . . 7 .
4 . . . 5 . . . 6
. . . 2 . 9 . . .
. 7 . . . . . 3 .


This one needs one Sky-craper which eliminates two 1s, one bi-directionyl-y-cycle in r5c59 and r8c59 with the numbers 1347. It ends up in a BUG1 Situation.

Claudia


As long as you are ending up in a BUG+1 situation, you might as well solve it before the y-cycle, when you are in a BUG+2 situation.

Code: Select all
 *--------------------------------------------------*
 | 2    4    7    | 68   36   38   | 5    1    9    |
 | 19   6    19   | 5    2    7    | 3    4    8    |
 | 8    5    3    | 1    9    4    | 2    6    7    |
 |----------------+----------------+----------------|
 | 67   2    8    | 37   1    5    | 46   9    34   |
 | 79+1 19   5    | 4    37   6    | 8    2    13   |
 | 16   3    4    | 9    8    2    | 16   7    5    |
 |----------------+----------------+----------------|
 | 4    19   2    | 37   5    13   | 79   8    6    |
 | 3    8    6    | 2    47   9    | 17+4 5    14   |
 | 5    7    19   | 68   46   18   | 49   3    2    |
 *--------------------------------------------------*


This is not a BUG2-Situation. In a BUG type2-Situation you have two times the same candidate three times in one row, column and box, which disturbes the BUG-pattern, where every "ordinary" candidate shows up twice only in every line and box. So you can't have 1 and 4 but only 1 and 1 or 7 and 7 and so on. In a BUG-2 Situation these two eliminate a similar number which sees both of the two "disturbing" candidates. In BUG type 1 the solution starts in the cell with the three candidates while in the BUG type 2-situation the solution starts outside the two or even three Three-candidate-cells.The 1 in r5c1 in this puzzle is the only BUG-solving candidate. The 1 in r5c1 is definitely to remain in its cell while the 1 in r8c7 shows up only twice in its region and the 4 in the same cell can be eliminated and can't be a part of a BUG-solving pattern too. Only in BUG type 4 you have two different candidates disturbing the BUG-pattern, but they are in the same region. See here:http://forum.enjoysudoku.com/viewtopic.php?t=2352&start=135
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Re: BUG type

Postby ronk » Fri May 18, 2007 2:29 pm

claudiarabia wrote:
rep'nA wrote:As long as you are ending up in a BUG+1 situation, you might as well solve it before the y-cycle, when you are in a BUG+2 situation.
Code: Select all
 *--------------------------------------------------*
 | 2    4    7    | 68   36   38   | 5    1    9    |
 | 19   6    19   | 5    2    7    | 3    4    8    |
 | 8    5    3    | 1    9    4    | 2    6    7    |
 |----------------+----------------+----------------|
 | 67   2    8    | 37   1    5    | 46   9    34   |
 | 79+1 19   5    | 4    37   6    | 8    2    13   |
 | 16   3    4    | 9    8    2    | 16   7    5    |
 |----------------+----------------+----------------|
 | 4    19   2    | 37   5    13   | 79   8    6    |
 | 3    8    6    | 2    47   9    | 17+4 5    14   |
 | 5    7    19   | 68   46   18   | 49   3    2    |
 *--------------------------------------------------*

This is not a BUG2-Situation. In a BUG type2-Situation you have two times the same candidate three times in one row, column and box
[...]
See here:http://forum.enjoysudoku.com/viewtopic.php?t=2352&start=135

According to the topic you linked, rep'nA's illustration is indeed a BUG+2.

I think you're talking about a more narrowly defined "BUG+2 Type 2". AFAIK you are the only one using that terminology on this forum.
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Re: BUG type

Postby claudiarabia » Wed May 23, 2007 10:30 am

ronk wrote:
claudiarabia wrote:
rep'nA wrote:As long as you are ending up in a BUG+1 situation, you might as well solve it before the y-cycle, when you are in a BUG+2 situation.
Code: Select all
 *--------------------------------------------------*
 | 2    4    7    | 68   36   38   | 5    1    9    |
 | 19   6    19   | 5    2    7    | 3    4    8    |
 | 8    5    3    | 1    9    4    | 2    6    7    |
 |----------------+----------------+----------------|
 | 67   2    8    | 37   1    5    | 46   9    34   |
 | 79+1 19   5    | 4    37   6    | 8    2    13   |
 | 16   3    4    | 9    8    2    | 16   7    5    |
 |----------------+----------------+----------------|
 | 4    19   2    | 37   5    13   | 79   8    6    |
 | 3    8    6    | 2    47   9    | 17+4 5    14   |
 | 5    7    19   | 68   46   18   | 49   3    2    |
 *--------------------------------------------------*

This is not a BUG2-Situation. In a BUG type2-Situation you have two times the same candidate three times in one row, column and box
[...]
See here:http://forum.enjoysudoku.com/viewtopic.php?t=2352&start=135

According to the topic you linked, rep'nA's illustration is indeed a BUG+2.

I think you're talking about a more narrowly defined "BUG+2 Type 2". AFAIK you are the only one using that terminology on this forum.


There is a striking difference between the 1 in r5c1 and the 4 in r8c7. Both are the only candidates, who appear three times in its row, column and box. But the 1 in r5c1 is to remain in its cell thus solving the BUG finally, whereas the 4 in r8c7 can be removed by a forcing chain, a cycle, whatever you want. How can such a candidate contribute to the final solution of the bug, if you don't eliminate it, leaving the classical BUG Type 1 Situation after it disappeared? If you want to spare the effort to eliminate it, how can you be sure, which of the two numbers finally will solve the bug and how and where?

I did not invent this pattern of myself. I took it from the Sudoku-Explainer which explains in a logical way the BUG-types. So BUG+n, as it is used in this forum quite often, to me says nothing about the type of the BUG, it merely counts how many cells have more then 2 candidates in the BUG, as says rep'nA. It is up to you, when you like this definition. But in the end it is not so much important how many additional poly-value-cells are in the BUG, but how they are arranged, which of the three values in one cell have to go finally, which will survive and if they bring the solution inside the multi-value-cells or in an outside bi-value cell. Juillerat used roughly the same definitions for the BUG as he did for the Unique Rectangles, which he put into 4 categories too.

Sorry when I was off the cycle-topic that much.

Claudia
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Postby ronk » Wed May 23, 2007 2:21 pm

claudiarabia wrote:But the 1 in r5c1 is to remain in its cell thus solving the BUG finally, whereas the 4 in r8c7 can be removed by a forcing chain, a cycle, whatever you want. How can such a candidate contribute to the final solution of the bug, if you don't eliminate it, leaving the classical BUG Type 1 Situation after it disappeared? If you want to spare the effort to eliminate it, how can you be sure, which of the two numbers finally will solve the bug and how and where?

repn'A wrote:
Code: Select all
 2    4    7    | 68   36   38   | 5    1    9
 19   6    19   | 5    2    7    | 3    4    8
 8    5    3    | 1    9    4    | 2    6    7
----------------+----------------+--------------
 67   2    8    | 37   1    5    | 46   9    34
 79+1 19   5    | 4    37   6    | 8    2    13
 16   3    4    | 9    8    2    | 16   7    5
----------------+----------------+--------------
 4    19   2    | 37   5    13   | 79   8    6
 3    8    6    | 2    47   9    | 17+4 5    14
 5    7    19   | 68   46   18   | 49   3    2

To avoid the deadly BUG, at least one of r8c7=4 and r5c1=1 must be true. This creates the strong inference r8c7=4|1=r5c1 which may be used in a forcing chain, as rep'nA did in ...
repn'A wrote:[r8c9]-4-[r8c7]=4|1=[r5c1]-1-[r5c9]=1=[r8c9], implying r8c9<>4

... which solves the puzzle with cascading singles. IOW the reduction from BUG+2 need not lead to a BUG+1.

However ... r8c7=4|1=r5c1-1-r5c9=1=r8c9-1-r8c7, implying r8c7<>1 ... does lead directly to a BUG+1. Note that this is done without eliminating the extra candidate at r8c7. For all we know at this point, both extra candidates might be true.
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an extraordinary sudoku

Postby claudiarabia » Thu May 24, 2007 2:44 pm

Code: Select all
4 . . . . . . . 2
. 8 . . 9 . . 7 .
. . . 3 . 6 . . .
. . 1 . . . 6 . .
. 6 . . 4 . . 9 .
. . 5 . . . 3 . .
. . . 5 . 2 . . .
. 7 . . 6 . . 8 .
9 . . . . . . . 4


This sudoku has a bi-directional X-Cycle next to two bi-directional Y-cycles next to 2 X-Wings and many other fancy techniques.

Happy solving.

Claudia
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Re: an extraordinary sudoku

Postby Ocean » Thu May 24, 2007 10:18 pm

claudiarabia wrote:
Code: Select all
4 . . . . . . . 2
. 8 . . 9 . . 7 .
. . . 3 . 6 . . .
. . 1 . . . 6 . .
. 6 . . 4 . . 9 .
. . 5 . . . 3 . .
. . . 5 . 2 . . .
. 7 . . 6 . . 8 .
9 . . . . . . . 4


This sudoku has a bi-directional X-Cycle next to two bi-directional Y-cycles next to 2 X-Wings and many other fancy techniques.

Nice puzzle!

The X-wings are also bidirectional x-cycles, the shortest possible (4 nodes, or 2 elements). Then we have this other x-cycle with eight nodes (or 4 elements): [r1c7]-[r3c9]=[r3c5]-[r7c5]=[r7c1]-[r9c3]=[r5c3]-[r5c7]=[r1c7] (8 candidates eliminated). And two xy-rings (eight and nine nodes, respectively). Alltogether five "bidirectional cycles", a fine collection in one single puzzle.

The other techniques (Turbot Fish, XYZ-wing, XY-wing, and naked/hidden pairs) add to the complexity of solving the puzzle. A good challenge, possible to solve manually without any chains, except for the "bicycles".
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Postby Ocean » Thu May 31, 2007 9:25 pm

Observing that posts by rep'nA mysteriously disappeared from the forum; this is an attempt to recreate his six posts in this thread.


rep'nA Thu May 10, 2007 6:29 pm wrote:Ocean,

Thanks for starting this thread. I think it should be a lot of fun and will help many of us increase our Sudoku Kung Fu.

I have a question about the taxonomy of your last puzzle.

Code: Select all
 *-----------*
 |...|...|12.|
 |3..|...|4..|
 |15.|6..|...|
 |---+---+---|
 |...|7.2|3..|
 |...|...|...|
 |..2|4.8|...|
 |---+---+---|
 |...|..1|.69|
 |..8|...|..5|
 |.47|...|...|
 *-----------*


Can you also describe it as a 4-ring XY-cycle example?

Code: Select all
 *--------------------------------------------------*
 | 78   78   69   | 359* 59*  4    | 1    2    36   |
 | 3    2    69   | 1    8    79   | 4    5    67   |
 | 1    5    4    | 6    2    37   | 9    37   8    |
 |----------------+----------------+----------------|
 | 58   69   1    | 7    56   2    | 3    89   4    |
 | 4    78   3    | 59*  1    569  | 56   78   2    |
 | 57   69   2    | 4    3    8    | 56   179  17   |
 |----------------+----------------+----------------|
 | 2    3    5    | 8    4    1    | 7    6    9    |
 | 69   1    8    | 39-  7    369  | 2    4    5    |
 | 69   4    7    | 2    569* 569  | 8    13   13   |
 *--------------------------------------------------*



[r9c5]=9=[r1c5]=5=[r1c4]-5-[r5c4]-9-[r8c4]

and so r8c4<>9, solving the puzzle.

Given that it is extremely likely that I don't know what I'm talking about, perhaps we could add precise definitions of the three cycle types to the beginning of the post.



rep'nA Fri May 11, 2007 4:15 pm wrote:
Ocean wrote:
rep'nA wrote:I have a question about the taxonomy of your last puzzle.
...
Can you also describe it as a 4-ring XY-cycle example?

Code: Select all
 
 *--------------------------------------------------*
 | 78   78   69   | 359* 59*  4    | 1    2    36   |
 | 3    2    69   | 1    8    79   | 4    5    67   |
 | 1    5    4    | 6    2    37   | 9    37   8    |
 |----------------+----------------+----------------|
 | 58   69   1    | 7    56   2    | 3    89   4    |
 | 4    78   3    | 59*  1    569  | 56   78   2    |
 | 57   69   2    | 4    3    8    | 56   179  17   |
 |----------------+----------------+----------------|
 | 2    3    5    | 8    4    1    | 7    6    9    |
 | 69   1    8    | 39-  7    369  | 2    4    5    |
 | 69   4    7    | 2    569* 569  | 8    13   13   |
 *--------------------------------------------------*



[r9c5]=9=[r1c5]=5=[r1c4]-5-[r5c4]-9-[r8c4]

and so r8c4<>9, solving the puzzle.


Thanks for analyzing the puzzle, and showing this alternative solution! The chain you list is shorter than the 8-node xy-ring (which I called 8-ring). But it's a discontinuous "cycle", so I would prefer not to call it a 4-ring as you suggest.




Oh, I wasn't suggesting. More like...guessing.

Ocean wrote:It qualifies as a "5-node discontinuous nice loop", as far as I recall Jeff's descriptions.




Ahh, yes. I always screw that up. I keep forgetting what ronk told me once about these things: the only time strong links should be used is when moving within a bivalue cell.


rep'nA Fri May 11, 2007 4:43 pm wrote:
Ocean wrote:
Effortless Extremes wrote:Locked Candidates plus one Bidirectional Cycle (5-ring) is enough to solve this puzzle:
Code: Select all
   
001002000030040050500600700008000006010000090600000800006007004090010030000500200
#
 *--------------------------------------------------------------------*
 |#479    6      1      | 3789   5      2      |-349   #48     389    |
 | 279    3      279    | 1789   4      189    | 6      5      1289   |
 | 5      8      249    | 6      39     139    | 7      124    1239   |
 |----------------------+----------------------+----------------------|
 | 39     24     8      | 12349  2379   5      | 134    1247   6      |
 | 37     1      357    | 2348   6      348    | 34     9      235    |
 | 6      24     359    | 12349  2379   1349   | 8      1247   1235   |
 |----------------------+----------------------+----------------------|
 | 123    5      6      | 239    2389   7      |-19    #18     4      |
 | 8      9      24     | 24     1      6      | 5      3      7      |
 |#134    7      34     | 5      389    349    | 2      6     #189    |
 *--------------------------------------------------------------------*
#
# [r1c1]=4=[r9c1]=1=[r9c9]-1-[r7c8]=8=[r1c8]-4-[r1c1] => r1c7<>4, r7c7<>1, r9c1<>3.
#



I'm guessing that the chain should be written

[r1c1]=4=[r9c1]=1=[r9c9]-1-[r7c8]-8-[r1c8]-4-[r1c1]


rep'nA Fri May 11, 2007 5:26 pm wrote:
Ocean wrote:...but a strong link can always replace a weak link in a chain.

Of course, any time you have a strong link you get a weak link (with the the definition of strong link being XOR and weak link being NAND), but I've always been told by the notation guru's that in the nice loop notation, you cannot replace a weak link with a strong link.

Hey ref, can I get a ruling on this?


rep'nA Mon May 14, 2007 9:50 pm wrote:
claudiarabia wrote:At first, Ocean, let me express my joy over your strong comeback in the forum after a long time of being seldom here.
I feel very honored that my incentive to discuss bi-directional cycles fell on such a fertile ground.

Code: Select all
 
. 4 . . . . . 1 .
. . . 5 . 7 . . .
8 . . . 9 . . . 7
. 2 . . 1 . . 9 .
. . 5 4 . 6 8 . .
. 3 . . 8 . . 7 .
4 . . . 5 . . . 6
. . . 2 . 9 . . .
. 7 . . . . . 3 .



This one needs one Sky-craper which eliminates two 1s, one bi-directionyl-y-cycle in r5c59 and r8c59 with the numbers 1347. It ends up in a BUG1 Situation.

Claudia



As long as you are ending up in a BUG+1 situation, you might as well solve it before the y-cycle, when you are in a BUG+2 situation.

Code: Select all
 *--------------------------------------------------*
 | 2    4    7    | 68   36   38   | 5    1    9    |
 | 19   6    19   | 5    2    7    | 3    4    8    |
 | 8    5    3    | 1    9    4    | 2    6    7    |
 |----------------+----------------+----------------|
 | 67   2    8    | 37   1    5    | 46   9    34   |
 | 79+1 19   5    | 4    37   6    | 8    2    13   |
 | 16   3    4    | 9    8    2    | 16   7    5    |
 |----------------+----------------+----------------|
 | 4    19   2    | 37   5    13   | 79   8    6    |
 | 3    8    6    | 2    47   9    | 17+4 5    14   |
 | 5    7    19   | 68   46   18   | 49   3    2    |
 *--------------------------------------------------*



[r8c9]-4-[r8c7]=4|1=[r5c1]-1-[r5c9]=1=[r8c9], implying r8c9<>4, solving the puzzle.

Of course, if you're not into the whole uniqueness thing, you could always try the xy-chain

4-[r9c7]-9-[r9c3]-1-[r7c2]-9-[r5c2]-1-[r5c9]-3-[r4c9]-4

implying r4c7, r8c9<>4, solving the puzzle.

Or finally, if you're really lazy, you might find this move before the Skyscraper:
Code: Select all
 
 *--------------------------------------------------*
 | 2    4    7    | 68   36   38   | 5    1    9    |
 | 19   6    19   | 5    2    7    | 3    4    8    |
 | 8    5    3    | 1    9    4    | 2    6    7    |
 |----------------+----------------+----------------|
 | 67A  2    8    | 37-  1    5    | 46   9    34   |
 | 179  19A  5    | 4    37   6    | 8    2    13   |
 | 16A  3    4    | 9    8    2    | 16   7    5    |
 |----------------+----------------+----------------|
 | 4    19B  2    | 37B  5    13B  | 79   8    6    |
 | 3    8    6    | 2    47   9    | 147  5    14   |
 | 5    7    19   | 68   46   18   | 49   3    2    |
 *--------------------------------------------------*


A={1,6,7,9} on r46c1, r5c2
B={1,3,7,9} on r7c246
x=9
z=7

ALS xz-rule gives r4c4<>7, solving the puzzle.


rep'nA Mon May 14, 2007 9:58 pm wrote:
ArkieTech wrote:Just when I think I am beginning to get a grip or understanding.
Code: Select all
 *-----------*
 |...|...|..1|
 |...|2.3|...|
 |.45|...|6..|
 |---+---+---|
 |..7|..6|.3.|
 |.8.|...|.9.|
 |.1.|8..|7..|
 |---+---+---|
 |..3|...|95.|
 |...|4.1|...|
 |2..|...|...|
 *-----------*


 *-----------*
 |.32|...|.71|
 |671|2.3|.89|
 |.45|...|623|
 |---+---+---|
 |.27|..6|138|
 |.86|...|295|
 |.19|8..|746|
 |---+---+---|
 |163|7..|954|
 |7.8|4.1|362|
 |2.4|...|817|
 *-----------*

 
 *-----------------------------------------------------------*
 | 89    3     2     | 569   68    4589  | 45    7     1     |
 | 6     7     1     | 2     45    3     | 45    8     9     |
 | 89    4     5     | 19    17    789   | 6     2     3     |
 |-------------------+-------------------+-------------------|
 | 45    2     7     | 59    459   6     | 1     3     8     |
 | 34    8     6     | 13    17    47    | 2     9     5     |
 | 35    1     9     | 8     23    25    | 7     4     6     |
 |-------------------+-------------------+-------------------|
 | 1     6     3     | 7     28    28    | 9     5     4     |
 | 7     59    8     | 4     59    1     | 3     6     2     |
 | 2     59    4     | 36    36    59    | 8     1     7     |
 *-----------------------------------------------------------*


r1c4-9-r4c4=9=r4c5-5-r8c5=9=r9c6-9-r1c6=5=r1c4

Help! since weak side r1c4, r4c4 is part of the ring I thought one or the other point had to be a 9 and therefore the 9 at r3c4 could be eliminated. I am wrong of course. Can anyone help me?

dan



You've made an error in your chain. After r4c4=9=r4c5, it doesn't make sense to have r4c5-5-r8c5. You've effectively written "if r4c4<>9 then r4c5=9. But if r4c5=5, then r8c5<>5..."

I do see a y-cycle, but it doesn't seem to help matters much:

[r4c4]-9-[r3c4]-1-[r5c4]-3-[r6c5]-2-[r6c6]-5-[r4c4], implying r1c4<>9, r4c5<>5.

However, a related xy-chain does solve the puzzle immediately:

9-[r4c4]-1-[r5c4]-3-[r6c5]-2-[r6c6]-5-[r9c6]-9, implying r13c6<>9.

Last edited by rep'nA on Tue May 15, 2007 3:45 pm; edited 1 time in total
Ocean
 
Posts: 442
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Two bi-directional cycles

Postby claudiarabia » Mon Jun 04, 2007 9:36 am

Code: Select all
. . 9 . . 7 . . .
. 4 . 5 . . . 8 .
. . 3 . . . 1 . 5
1 . . . 3 . . 4 .
. . . 6 . 5 . . .
. 8 . . 9 . . . 2
4 . 5 . . . 6 . .
. 6 . . . 2 . 3 .
. . . 8 . . 9 . .
It solves with a Naked Triplet, an xy-wing and two bi-directional cycles.

Happy solving!

Claudia
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Posts: 288
Joined: 14 May 2006

Re: Two bi-directional cycles

Postby re'born » Mon Jun 04, 2007 10:39 am

claudiarabia wrote:
Code: Select all
. . 9 . . 7 . . .
. 4 . 5 . . . 8 .
. . 3 . . . 1 . 5
1 . . . 3 . . 4 .
. . . 6 . 5 . . .
. 8 . . 9 . . . 2
4 . 5 . . . 6 . .
. 6 . . . 2 . 3 .
. . . 8 . . 9 . .
It solves with a Naked Triplet, an xy-wing and two bi-directional cycles.

Happy solving!

Claudia


Or after the naked triple, you could use the xy-chain:

Code: Select all
 *-----------------------------------------------------------*
 | 28-   5     9     | 123   128   7     | 24%   6     34*   |
 | 267   4     1     | 5     26    369   | 27    8     39*   |
 | 2678  27%   3     | 249   2468  469   | 1     79*   5     |
 |-------------------+-------------------+-------------------|
 | 1     9     27    | 27    3     8     | 5     4     6     |
 | 237   237   4     | 6     127   5     | 8     179   79    |
 | 5     8     6     | 147   9     14    | 3     17    2     |
 |-------------------+-------------------+-------------------|
 | 4     137   5     | 1379  17    139   | 6     2     8     |
 | 9     6     8     | 147   5     2     | 47    3     147   |
 | 237   1237  27    | 8     1467  1346  | 9     5     147   |
 *-----------------------------------------------------------*


[r3c2]-7-[r3c8]-9-[r2c9]-3-[r1c9]-4-[r1c7], => r1c1<>2, which (along with a naked pair down the line) solves the puzzle. Incindentally, one could also view this xy-chain as an application of the ALS xz-rule, with A={2,7,9} on r3c28, B = {2,3,4,9} on r1c79, r2c9, x = 9, z = 2. Cool puzzle, Claudia.
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Posts: 551
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Postby JPF » Mon Jun 04, 2007 3:48 pm

An attempt to be qualified for this thread...

Code: Select all
 . . . | . . . | . . .
 . . 1 | 2 . 3 | 4 . .
 . 5 . | 4 . 6 | . 7 .
-------+-------+-------
 . 6 8 | . . . | 2 5 .
 . . . | . . . | . . .
 . 3 5 | . . . | 1 9 .
-------+-------+-------
 . 4 . | 7 . 1 | . 6 .
 . . 2 | 8 . 4 | 9 . .
 . . . | . . . | . . .


JPF
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