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.