## Continuous (coloring) chains i.e. bi-directional(x)cycles

Advanced methods and approaches for solving Sudoku puzzles

### Continuous (coloring) chains i.e. bi-directional(x)cycles

When solving sudokus, do you regularily use Continuous (coloring) chains i.e. bi-directional(x)cycles?
I think it is more easy to spot a simple forcing chain then these cycles. But nevertheless it is very useful to spot them. In the Explainer I observed that he uses forcing chains to produce such cycles while eliminating clues who disturb the continuity of the cycle. But even if I use sometimes the solving strategies of the SE as a rough guide, I'm never sure if a certain chain I found will create or destroy a cycle. That's why in every forcing chain-sudoku I would never search for cycles.
How do you handle that?
claudiarabia

Posts: 288
Joined: 14 May 2006

Before i look for more complicated chains, i try to combine strong links and should spot the eliminations from bidirectional x-cycles, which are built of strong links only.
I dont look for bidirectional y-cycles explicitely, but sometimes stumble on one, when i look for (extended) xy-chains and come back to the starting number.
ravel

Posts: 998
Joined: 21 February 2006

### bi-directional cycles

ravel wrote:Before i look for more complicated chains, i try to combine strong links and should spot the eliminations from bidirectional x-cycles, which are built of strong links only.
I dont look for bidirectional y-cycles explicitely, but sometimes stumble on one, when i look for (extended) xy-chains and come back to the starting number.

Hi, ravel,

At least one person, looking for cycles like me. I made even a puzzle-collection with sudokus who requires cycles only in addition to pairs, UR's, Swordfisches, X-Wings, XY-Wings etc. avoiding any forcing chains, because they can be laid so arbitrarily throughout the whole sudoku, while cycles are well defined like Swordfishes.

Well, a forcing chain everybody may spot somewhere but to find such a bi-directional cycle is a high and beautiful art in sudoku-solving which doesn't unveil itself to everyone.

You wrote you look for bidirectional X-cycles, which are built of strong links only. Unfortunately until now I produced only few sudokus who expose such cycles and even less examples with such bi-directional x-cycles excluding additional forcing chains.
Most cycles are the simple bi-directional, which for me are the most difficult to spot, especially when they include more then 4 cells. Y-Cycles on the other hand I can detect more easily, but I still miss some eliminations when they produce many. But until now the eliminations I found were enough to solve the puzzle.
Code: Select all
`. . 4 . . . 5 . 2. 7 . 2 . 1 . . .. . . . 9 . . . .7 . 6 . . . . . .. 3 . . . . 1 7 .2 . . 1 . . . . 94 9 . . . 2 . 6 .. . . 9 4 . . . 3. . . . . 7 8 . .`

This one is a sudoku which requires no forcing chains but includes a bi-directional y-cycle as well as a bi-directional x-cycle in addition to other fancy techniques.

Code: Select all
` . . 4 . . . 6 9 . . 8 . . . 6 . . 1 . 1 . . 7 . . . . . 3 . . 8 9 . . . . . 1 . 4 . 2 . . . . . 3 6 . . 4 . . . . . 2 . . 8 . 9 . . 5 . . . 7 . . 7 6 . . . . . . `
I borrowed it from my thread. It requires a Turbot fish and a bi-directional x-cycle only.

regards

Claudia
claudiarabia

Posts: 288
Joined: 14 May 2006

ravel wrote:Before i look for more complicated chains, i try to combine strong links and should spot the eliminations from bidirectional x-cycles, which are built of strong links only.
I dont look for bidirectional y-cycles explicitely, but sometimes stumble on one, when i look for (extended) xy-chains and come back to the starting number.

you can combine strong and weak edges and still be bi-directional
see the writeup of up the math behind my X&Y cycle algorithm
gsf
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### Re: bi-directional cycles

claudiarabia wrote:I made even a puzzle-collection with sudokus who requires cycles only in addition to pairs, UR's, Swordfisches, X-Wings, XY-Wings etc. avoiding any forcing chains, because they can be laid so arbitrarily throughout the whole sudoku, while cycles are well defined like Swordfishes.

Would you be willing to post this collection?

rep'nA
re'born

Posts: 551
Joined: 31 May 2007

### Re: bi-directional cycles

rep'nA"][quote="claudiarabia wrote:I
Would you be willing to post this collection?

rep'nA

Yes, here is a starter:

Code: Select all
`5 . . 4 . 1 . . .. 8 . . . . . 2 .. . 3 . . 6 . . 17 . . . 5 . 9 . .. . 4 . . . . . 2. 3 . 6 . . . 8 .. . 8 . . 7 5 . 3. 7 . . 1 . . . .6 . . . 9 . . . 8`

This very assymmetrical puzzle is to be solved by some pairs and a bi-directional cycle with many cells which eliminates a lot of numbers in the surrounding. I found it stunning how a mere bi-directional cycle can push up the ER Rating to 7,3.

And a 2nd one:
Code: Select all
`. . 8 . . . 4 . .. . . 2 . 8 . . 36 . . . 9 . . . 1. 3 . . . . . 4 .. . 9 . . . 6 . .. . . 7 . 3 . . .. . . . 8 . . . .1 . . . . . . . 77 4 . . . . . 6 5`

This one can be solved with some pairs and 3 bi-directional Y-Cycles. It ends up in a BUG1-Situation. For this interesting constellation I even accepted the asymetrical 3 in r2c9. With sudoku it is like with human beings. The most interesting ones are too often not the perfect symmetric ones. ;-)

Happy solving

Claudia
claudiarabia

Posts: 288
Joined: 14 May 2006

### Re: bi-directional cycles

claudiarabia wrote:
Code: Select all
`5 . . 4 . 1 . . .. 8 . . . . . 2 .. . 3 . . 6 . . 17 . . . 5 . 9 . .. . 4 . . . . . 2. 3 . 6 . . . 8 .. . 8 . . 7 5 . 3. 7 . . 1 . . . .6 . . . 9 . . . 8`

This very assymmetrical puzzle is to be solved by some pairs and a bi-directional cycle with many cells which eliminates a lot of numbers in the surrounding. I found it stunning how a mere bi-directional cycle can push up the ER Rating to 7,3.

can you list the bidirectional cycle
thanks
gsf
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### Re: bi-directional cycles

gsf wrote:can you list the bidirectional cycle

Although I've seen the term a few times, I hadn't realized that SE's bidirectional cycle is the same as continuous loop on this forum.

[edit: I see claudiarabia has the same pencilmarks, so I'll just leave the continuous nice loop expression.]

r1c5=8=r1c7=3=r5c7-3-r5c6-9-r6c6-2-r6c1=2=r3c1-2-r3c5=2=r1c5

Due to the continuous loop, all the weak links become conjugate links.
Last edited by ronk on Sat Apr 21, 2007 2:14 pm, edited 1 time in total.
ronk
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### Re: bi-directional cycles

gsf"][quote="claudiarabia wrote:
Code: Select all
`5 . . 4 . 1 . . .. 8 . . . . . 2 .. . 3 . . 6 . . 17 . . . 5 . 9 . .. . 4 . . . . . 2. 3 . 6 . . . 8 .. . 8 . . 7 5 . 3. 7 . . 1 . . . .6 . . . 9 . . . 8`

can you list the bidirectional cycle
thanks

Of course here it is:
Code: Select all
`   *-------------------------------------------------------------------* | 5      269    2679    | 4      #28     1    |#3678  3679   679    | | 149    8      1679    | 3579   37     *35-9 | 467    2      45679 | | #249   *49-2    3     | 579     #28     6   | 478    4579   1     | |-----------------------+---------------------+---------------------| | 7      126    126     | 138    5      238   | 9      1346   46    | | 8      1569   4       |*17-39  *7-3   #39   |#136    *156-3  2    | | #129    3     *159-2  | 6      4      #29   | 17     8      57    | |-----------------------+---------------------+---------------------| | 149    149    8       | 2      6      7     | 5      149    3     | | 3      7      259     | 58     1      458   | 246    469    469   | | 6      1245   125     | 35     9      345   | 1247   147    8     | *-------------------------------------------------------------------*`

The cycle cells have the route-sign. The cells in which an elimination occurs, have a star. The eliminated numbers come after the numbers to stay and a minus.

The cycle consists of the numbers:

Code: Select all
`.  .  .  .  28 .  38 .  . .  .  .  .  .  .  .  .  . 2  .  .  .  2  .  .  .  ..  .  .  .  .  .  .  .  ..  .  .  .  .  39 3  .  .2  .  .  .  .  29 .  .  . .  .  .  .  .  .  .  .  ..  .  .  .  .  .  .  .  ..  .  .  .  .  .  .  .  .`
Last edited by claudiarabia on Sun Apr 22, 2007 10:58 am, edited 3 times in total.
claudiarabia

Posts: 288
Joined: 14 May 2006

### Re: bi-directional cycles

claudiarabia wrote:The cycle cells have the route-sign. The cells in which an elimination occurs, have a star. The eliminated numbers come after the numbers to stay and a minus.

The cycle consists of the numbers:

Code: Select all
`.  .  .  .  2  .  38 .  . .  .  .  .  .  .  .  .  . 2  .  .  .  2  .  .  .  ..  .  .  .  .  39 3  .  .2  .  .  .  .  29 .  .  . .  .  .  .  .  .  .  .  ..  .  .  .  .  .  .  .  ..  .  .  .  .  .  .  .  ..  .  .  .  .  .  .  .  .`

I'm familiar with cycles based on one value (X cycles) and cycles based on bivalue cells (Y cycles)
but those don't advance the puzzle here
can you bear with me and explain exactly how this cycle is formed, cell by cell
thanks

edit somehow I missed ronk's post and went right to claudiarabia's
I'd still like to be fed this one cycle from first principles
thanks again
edit**2 my questions come from the perspective of a not using nice loop notation
ronk's post cleared up what I'm missing from that perspective
this also gives some insight into what I'm missing in pure X/Y cycles
thanks again**2
gsf
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### Re: bi-directional cycles

claudiarabia wrote:
And a 2nd one:
Code: Select all
`. . 8 . . . 4 . .. . . 2 . 8 . . 36 . . . 9 . . . 1. 3 . . . . . 4 .. . 9 . . . 6 . .. . . 7 . 3 . . .. . . . 8 . . . .1 . . . . . . . 77 4 . . . . . 6 5`

This one can be solved with some pairs and 3 bi-directional Y-Cycles. It ends up in a BUG1-Situation. For this interesting constellation I even accepted the asymetrical 3 in r2c9.

I'm trying hard to spot these bi-directional Y-cycles, but I don't quite have the knack for it yet. My first solution to this puzzle was to note the Sue de Coq in r7c1234,r9c3 that leads to r8c3<>2,3. We then are left with a BUG+4 grid (is it +3 or +4 when two of the candidates are in the same cell?) and it is easy to check that this implies r8c5<>5, solving the puzzle.

I'm not sure I understand the definition of bi-directional Y-cycle. Would the the following be an example?

[r8c6]-2-[r8c8]-9-[r1c8]-2-[r1c2]-9-[r2c1]-4-[r5c1]=4=[r5c5]=1=[r9c5]-1-[r9c4]-9-[r9c6]-2-[r8c6]

implying r8c6<>2, solving the puzzle.

claudiarabia wrote:With sudoku it is like with human beings. The most interesting ones are too often not the perfect symmetric ones. ;-)
Claudia

Funny. That's the pickup line I used when I met my wife.
re'born

Posts: 551
Joined: 31 May 2007

### Re: bi-directional cycles

rep'nA wrote:I'm not sure I understand the definition of bi-directional Y-cycle. Would the the following be an example?

[r8c6]-2-[r8c8]-9-[r1c8]-2-[r1c2]-9-[r2c1]-4-[r5c1]=4=[r5c5]=1=[r9c5]-1-[r9c4]-9-[r9c6]-2-[r8c6]

No, for two reasons. Firstly, it is not "bi-directional" -- according to b]Nicolas Juillerat[/b]'s apparent usage -- since the loop is discontinuous instead of continuous. Personally, I think "bidirectional" is a poorly chosen term because the inferences of a discontinuous loop are also bi-directional.

Secondly, a y-cycle deduction has strong inferences based on bivalue cells only. The elimination cell need not be bivalued, however. Your loop has two strong inferences of an x-cycle (highlighted above). (see last paragraph)

We then are left with a BUG+4 grid (is it +3 or +4 when two of the candidates are in the same cell?) and it is easy to check that this implies r8c5<>5, solving the puzzle.

That confused me for some time too. In the BUG+n notation, the number of candidates in a poly-valued cell is irrelevant. As I was unable to spot that deduction, would you please post the pencilmarks and the BUG+n splitting of candidates TIA

Would someone please explain the original meaning of the term "cycle?" Does it refer to the closed property of loop? Or that inferences must necessarily alternate between strong inference and weak inference? Something else? And where did the term orginate?
ronk
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A Y-cycle is equivalent to an X-cycle, but in the former case the cells are all bivalues and the later they are strong links. Jeff defined X-cycles as:
Jeff wrote:An x-cycle is a cycle in which all cells are linked by a single digit 'x'. It can be of any length. It can be continuous or discontinuous.

Gaby's definition which points to Bob Hanson's original post also shows Y-cycles to be continuous or discontinuous bivalues. I would think that bi-directionality or reversability is independent of the loop being continuous or discontinuous.

Heres a discontinuous Y-cycle which solves claudiarabia's puzzle:
Code: Select all
`6-node XY-chain (r5c6-5-r5c1-4-r6c3-2-r9c3-3-r9c5-1-) => r5c5<>1+---------------+---------------+-------------+|  3  29     8  | 15     7  15  |   4  29  6  || 49   1    45  |  2     6   8  |  59   7  3  ||  6  25     7  |  3     9   4  |  25   8  1  |+---------------+---------------+-------------+| 25   3     1  | 69    25  69  |   7   4  8  || 45*  7     9  |  8  45-1  15* |   6   3  2  ||  8   6    24* |  7    24   3  |   1   5  9  |+---------------+---------------+-------------+| 29  59  2356  | 56     8   7  |  23   1  4  ||  1   8  2356  |  4    35  26  | 239  29  7  ||  7   4    23* | 19    13* 29  |   8   6  5  |+---------------+---------------+-------------+`
Mike Barker

Posts: 458
Joined: 22 January 2006

### Re: bi-directional cycles

ronk wrote:Would someone please explain the original meaning of the term "cycle?" Does it refer to the closed property of loop? Or that inferences must necessarily alternate between strong inference and weak inference? Something else? And where did the term orginate?

with cells (vertices) and links (edges) I would like to see some confluence with graph theory
which provides definitions for path, walk, cycle
along those lines discontinuous cycle is an oxymoron
gsf
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Mike Barker wrote:A Y-cycle is equivalent to an X-cycle, but in the former case the cells are all bivalues and the later they are strong links.

That's a very broad definition of equivalency.

Jeff defined X-cycles as:
Jeff wrote:An x-cycle is a cycle in which all cells are linked by a single digit 'x'. It can be of any length. It can be continuous or discontinuous.

Gaby's definition which points to Bob Hanson's original post ...

Based on that link, it appears Bob Hanson picked up the x-cycle and y-cycle terminology from Glenn S. Fowler -- our very own gsf.
ronk
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