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

Advanced methods and approaches for solving Sudoku puzzles

### Re: bi-directional cycles

claudiarabia wrote:Here the simple example of a bi-directional cycle with the minimum number of 4 cells in which 2 numbers form the cycle in their special constellation. Our cycle numbers, the 1 and the 3 must be without other numbers in the r3c7 and r7c3 so that the cycle can work.

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`.  .  .  .  .  .  .  .  . .  .  .  .  .  .  .  .  ..  1x . -1  .  . 13  .  ..  .  .  .  .  .  .  .  ..  .  .  .  .  . -3  .  ..  .  .  .  .  .  .  .  ..  13 .  .  .  . 3x  .  ..  .  .  .  .  .  .  .  ..  .  .  .  .  .  .  .  .`

You move at first from r3c2 in clockwise direction and secondly from the same position counterclockwise. The numbers 1 and 3 are occuring only as I wrote in the respective columns and rows. In every direction you have only one defined way to go. The two ways in each direction are slung together and they form the cycle with the same cells everytime. It is not important which numbers are in addition in the cells r3c2 or r7c7. These additional numbers are the x. There are only eliminations outside of the cycle-cells, never inside. If you consider the X-Wing a simple bi-directional x-cycle, you know what I mean. There are also bi-directional cycles which involve boxes but here we have only 2 rows and 2 columns where there are strong links of 1 and 3 combined together in this cycle. Our aim is to eliminate the 1 in r3c4 and the 6 in r5c7.
1. If you start in r3c2 placing the Nr 1 inside. If you go clockwise, the 1 in r3c2 deletes the 1 in r3c7 and the 1 in r3c4 too. Because there is left a 3 in r3c7 only, we place the 3 here and this 3 eliminates the 3 in r5c7 and r7c7. Because there are only two 3s in r7, the eliminated 3 in r7c7 confirmes the 3 in r7c2. The 3 here eliminates the 1 and the 1 occures again in our starting cell r3c2. End first round.
2. If you start again in r3c2 assuming there is no 1 there going counterclockwise you see that the 1 has to be placed now in r7c2. The 1 there eliminates the 3 there and by this elimination the 3 in r7c7 is confirmed by this confirmation the 3 in r5c7 is eliminated again. By placement of the 3 in r7c7 the 3 in r3c7 is also eliminated and the 1 appears to be left alone and this 1 in r3c7 confirms our starting position of no 1 in r3c2 and eliminates in the second turn the 1 in r3c4. The cycle is closed in both directions and the eleminations of 1 in r3c4 and 6 in r5c7 are valid and final.

For the manual solvers, looking for the above pattern should be fairly straightforward and can be coupled with looking for the pattern in Steke K's recent thread.
re'born

Posts: 551
Joined: 31 May 2007

### second results

rep'nA wrote:
Just to check whether I've figured this bi-directional cycle business out, is this what you had in mind:
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` *--------------------------------------------------------------------* | 78     4      9      | 1678-  1678-  2      | 5      18     3      | | 3      57     1      | 57(8)  4      9      | 6      2      (78)   | | 57-8   6      2      | 3    157-8-   (78)   | 4      189    (7)89  | |----------------------+----------------------+----------------------| | 24     9      8      | 267    267    467    | 13     35     15     | | 45     15     3      | 189    189    48     | 2      7      6      | | 6      12     7      | 12     3      5      | 89     89     4      | |----------------------+----------------------+----------------------| | 279    27     6      | 4      25789  1      | 389    35     2589   | | 129    3      5      | 2689   2689   68     | 7      4      1289   | | 1279   8      4      | 2579   2579   3      | 19     6      1259   | *--------------------------------------------------------------------*`

Eliminations : r1c45<>8, r3c1<>7, r3c5<>7,8

Yes I had these eliminations in mind. You found the explainer-cycle. I oversaw that the placement of the only two digits 78 in r3c6 make it together with r3c9 a strong link of 7 and 8.
I found the following longer cycle:

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` *--------------------------------------------------------------------* | (78)    4     9      | 1678-  1678-  2      | 5      (18)   3      | | 3      (57)   1      | (5)78  4      9      | 6      2      78     | | 57-8    6     2      | 3    (15)7-8-   78   | 4      (1)89  789    | |----------------------+----------------------+----------------------| | 24     9      8      | 267    267    467    | 13     35     15     | | 45     15     3      | 189    189    48     | 2      7      6      | | 6      12     7      | 12     3      5      | 89     89     4      | |----------------------+----------------------+----------------------| | 279    27     6      | 4      25789  1      | 389    35     2589   | | 129    3      5      | 2689   2689   68     | 7      4      1289   | | 1279   8      4      | 2579   2579   3      | 19     6      1259   | *--------------------------------------------------------------------*`
claudiarabia

Posts: 288
Joined: 14 May 2006

### swinging naked pairs as nodes in the bi-directional cycle

You remember perhaps this puzzle from the thread: Sudokus with a rare shape
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`. 3 . . . 5 . 7 . . . 9 . . . . . 8 1 . . 6 . . 2 . . . . 4 . 8 . . . 5 . . . . . 7 . . . 6 . . . . . 3 . . . 7 . . . 2 . 9 . . . 5 . . . . . 1 8 . . 1 . . 4 . . `

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` *-----------* |.3.|..5|.7.| |..9|...|5.8| |15.|6..|23.| |---+---+---| |..4|.8.|..5| |5..|..7|...| |6..|...|3..| |---+---+---| |.71|..2|89.| |..5|...|.21| |8..|1..|45.| *-----------*`

While solving it I came to the following situation:

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` *---------------------------------------------------------------------* | #24     3      68-2   | 2489    1249    5     | 16      7       49  | | #247   #246     9     | 2347    1234    134   | 5      #16      8   | | 1       5       78    | 6       479     489   | 2       3       49  | |-----------------------+-----------------------+---------------------| | #2379   129     4     | 239     8       1369  | 179-6  #16      5   | | 5       1289    23    | 2349    123469  7     | 19-6   1468    #26  | | 6       189    #27    | 459     1459    149   | 3       148    #27  | |-----------------------+-----------------------+---------------------| | 34      7       1     | 345     3456    2     | 8       9       36  | | 349     469     5     | 34789   3469    34689 | 67      2       1   | | 8       269     236   | 1       3679    369   | 4       5       37  | *---------------------------------------------------------------------*`

The new element is a changing pair as a full node of the cycle. In both directions there is a change in the three cells r1c1, r2c12. In clockwise direction, placing the 1 in r2c8 you have a naked pair of 24 in r12c1, making 6 in r2c2. Anticlockwise meanwhile, placing the 6 in r2c8, you will have a naked pair emerging in r1c1 and r2c2 of 24, following the 7 in r2c2 and so on.

Here the cycle in abstraction:

The whole cycle:
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` 24 .-2 . . . . . .247 246 . . . . 16.  . . . . . . . . .  7 . . . . .-6 16.  . . . . . .-6 . 26  . . 27. . . . . 27  . . . . . . . . .  . . . . . . . . .  . . . . . . . . .`

clockwise
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`24 .-2 . . . . . .24 6 . . . . . 1 . . . . . . . . . . 7 . . . . .-6 6 . . . . . . .-6 . 2 . . 2 . . . . . 7 . . . . . . . . . . . . . . . . . . . . . . . . . . .`

anti-clockwise:
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`24 .-2 . . . . . .7 24 . . . . . 6 . . . . . . . . . .-7 . . . . .-6 1 . . . . . . .-6 . 6 . . 7 . . . . . 2 . . . . . . . . . . . . . . . . . . . . . . . . . . .`

It occured several times that I found such cycles and it always worked.
Claudia
Last edited by claudiarabia on Tue Aug 14, 2007 5:02 pm, edited 4 times in total.
claudiarabia

Posts: 288
Joined: 14 May 2006

### Re: swinging naked pairs as nodes in the bi-directional cycl

claudiarabia wrote:
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` *--------------------------------------------------------------*  | #24    3     68-2 | 2489   1249    5     | 16      7     49  |  | #247  #246    9   | 2347   1234    134   | 5      #16    8   |  | 1      5      78  | 6      479     489   | 2       3     49  |  |-------------------+----------------------+-------------------|  | #2379  129    4   | 239    8       1369  | 179-6  #16    5   |  | 5      1289   23  | 2349   123469  7     | 19-6   1468  #26  |  | 6      189   #27  | 459    1459    149   | 3       148  #27  |  |-------------------+----------------------+-------------------|  | 34     7      1   | 345    3456    2     | 8       9     36  |  | 349    469    5   | 34789  3469    34689 | 67      2     1   |  | 8      269    236 | 1      3679    369   | 4       5     37  |  *--------------------------------------------------------------*`

The new element is a changing pair as a full node of the cycle. In both directions there is a change in the three cells r1c1, r2c12. In clockwise direction, placing the 1 in r2c8 you have a naked pair of 24 in r12c1, making 6 in r2c2. Anticlockwise meanwhile, placing the 6 in r2c8, you will have a naked pair emerging in r1c1 and r2c2 of 24, following the 7 in r2c2 and so on.

That's a nice illustration of the power of the ALS (almost locked set) in a continuous loop (cycle). In nice loop (NL) notation ...

r2c8 -1- (ALS B:r4c8=1|26|7=r5c9,r6c9) -7- r6c3 =7= r4c1 -7- (ALS A:r12c2=7|24|6=r2c12) -6- Loop

Due to the continuous loop (cycle), ALS set A contains either 724 or 246. (This is indicated by the order of the digits in the NL expression above.) IOW digits 2 and 4 are locked into set A because of the continuous loop. Any 2s candidate that sees all the 2s in set A may be eliminated.

Ditto for the 4 of set A. Ditto for digits 2 and 6 of ALS set B.
ronk
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