ronk wrote: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 Nicolas Juillerat'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.
ronk wrote: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)
ronk wrote: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
.---------------.---------------.---------------.
| 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 14+5 15 | 6 3 2 |
| 8 6 24 | 7 24 3 | 1 5 9 |
:---------------+---------------+---------------:
| 29 59 36+25| 56 8 7 | 23 1 4 |
| 1 8 56 | 4 35- 26 | 39+2 29 7 |
| 7 4 23 | 19 13 29 | 8 6 5 |
'---------------'---------------'---------------'
claudiarabia"][
Let me at first give you a very simple example of the bi-directional cycle as Nicolas Juillerat define it. In the puzzle above the cycle involves not only 8 cells but 4 different numbers too. That justifys the rating of 7,3 in ER.
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.
quote="gsf"][quote="claudiarabia wrote:Of course here it is:
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*-------------------------------------------------------------------*
| 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:
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. . . . 28 . 38 . .
. . . . . . . . .
2 . . . 2 . . . .
. . . . . . . . .
. . . . . 39 3 . .
2 . . . . 29 . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
Mike Barker wrote:Heres a discontinuous Y-cycle which solves claudiarabia's puzzle:
6-node XY-chain (r5c6-5-r5c1-4-r6c3-2-r9c3-3-r9c5-1-) => r5c5<>1
ronk wrote: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.
True...ronk wrote:tarek, you're on the right trail ... but with bivalue cells, the answer can't possibly be limited to one digit.
a c
. / . | . / . | . . .
/ ab* / | / bc* / | / / / b
. / . | . / . | . . .
-----------+-----------+----------
. / . | . / . | . . .
/ ad* / | / cd* / | / / / d
. / . | . / . | . . .
-----------+-----------+----------
. / . | . / . | . . .
. / . | . / . | . . .
. / . | . / . | . . .
r2c2=b=r2c5=c=r5c5=d=r5c2=a=r2c2,
implies r2c2=ab, r2c5=bc, r5c5=cd and r5c2=ad
a c
. * . | . * . | . . .
* ab * | * bc * | * * * b
. * . | . * . | . . .
-----------+-----------+----------
. * . | . * . | . . .
* ad * | * cd * | * * * d
. * . | . * . | . . .
-----------+-----------+----------
. * . | . * . | . . .
. * . | . * . | . . .
. * . | . * . | . . .
r2c2-b-r2c5-c-r5c5-d-r5c2-a-r2c2,
implies r2c1346789<>b, r1346789c5<>c, r5c1346789<>d and r1346789c2<>a
. . . |. . . |. . .
. # x |. . . |. . .
. x . |x . . |. . .
--------------------
. . . |. . . |. . .
. . . |. . . |. . .
. . x |# . . |x . .
--------------------
. . . |x . . |. x .
. . . |. . . |x # .
. . . |. . . |. . .
. . . |. . . |. . .
. x . |. . . |. x .
. . . |. . . |. . .
--------------------
. x . |x . . |. . .
. # . |# . . |. . .
. . . |. . x |. x .
--------------------
. . . |x . x |. # .
. . . |. . . |. . .
. . . |. . . |. . .
ronk wrote:tarek, the question was to everyone, but I was only expecting a response from you, Mike or gsf.
tarek, you're on the right trail ... but with bivalue cells, the answer can't possibly be limited to one digit.
. . . | -3 . . |. . .
. . . | . . . |. . .
. 12 . | 13 . -1 |. . .
----------------------------
. . . | . . . |. . .
. . . | . . . |. . .
. . . | 34 . . |. . .
----------------------------
. 28 . | . -8 . |. 83 .
. . . | 14 . . |13 -3 .
. . . | . . . |. . .
*-------------------------------------------------------------------*
| 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 |
*-------------------------------------------------------------------*
. . . . 28 . 38 . .
. . . . . . . . .
2 . . . 2 . . . .
. . . . . . . . .
. . . . . 39 3 . .
2 . . . . 29 . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .