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Ron Moore

Joined: 13 Aug 2006
Posts: 72
Location: New Mexico

Posted: Thu Dec 14, 2006 7:37 pm    Post subject: 11 Dec 2006 Nightmare: X-Cycle

In the 11 Dec Nightmare, this is the position after eliminations from basic techniques, a naked "279" triple in r123c1, and a naked "178" triple in r789c9:
 Code: .----------------.-------------------.-----------------. | 29   1    8    | 7     *35     6   | 259  #359   4   | | 79   3    4    | 158    58     2   | 1579  5789  6   | | 27   6    5    | 138    4      9   | 127  -378  *23  | :----------------+-------------------+-----------------: | 4    28   123  | 238    9      378 | 6     137   5   | |*135  9    7    | 2-356  2-356 -34  | 8     1-34 #23  | | 6    258  23   | 2358   1      3478| 27    347   9   | :----------------+-------------------+-----------------: | 15   4    1269 | 268    2678   18  | 3     59    78  | |#135  25   12-39| 4      2-378  138 | 59    6     78  | | 8    7   *36   | 9     #36     5   | 4     2     1   | '----------------'-------------------'-----------------'

Here we have an X-cycle for digit "3." An X-cycle is an Alternating Inference Chain (AIC) for a single digit, i.e., a chain with alternating links of strong and weak inference for that digit, which loops back to its starting point. The cells in the cycle are alternately marked with "*" and "#". If we begin to traverse the cycle starting at r1c5 and proceeding to r1c8, the links from "*" to "#" are links of strong inference (meaning at least one of the cells must be "true" for digit "3") and links from "#" to "*" are links of weak inference (meaning at least one of the cells must be "false" for digit "3"). In Eureka notation:

(3): r1c5 = r1c8 - r3c9 = r5c9 - r5c1 = r8c1 - r9c3 = r9c5 - r1c5.

The existence of the cycle shows that the links of weak inference are in fact links of strong inference as well, so that "3" can be eliminated from any cell seeing both end nodes of any side of the cycle. (Note this is the same idea as with an X-wing -- an X-wing being the simplest form of an X-cycle.)

Many of the eliminations for digit "3" which are found one by one by the Sudocue solver are taken care of in one blow by the X-cycle.

Following up these eliminations, and making use of a naked triple of "259" in r8c237, we come to this position:
 Code: .---------------.------------------.--------------. | 29   1    8   | 7     35     6   | 259  359  4  | | 79   3    4   | 158   58     2   | 1579 5789 6  | | 27   6    5   | 138   4      9   | 127  78   23 | :---------------+------------------+--------------: | 4    28   1   | 238   9      378 | 6    37   5  | | 3-5  9    7   |*26+5 *26+5   4   | 8    1    23 | | 6    258  23  | 2358  1      378 | 27   4    9  | :---------------+------------------+--------------: |#15   4    269 |*26+8 *26+78 #18  | 3    59  #78 | | 13   25   29  | 4     78     138 | 59   6    78 | | 8    7    36  | 9     36     5   | 4    2    1  | '---------------'------------------'--------------'

From here, the quickest way to proceed seems to be a uniqueness argument based on the threatening deadly pattern of "26" cells in r57c45 (marked with "*"). All cells have at least one surplus candidate. We can avoid the deadly pattern only by placing a "5" in one of r5c45, or by placing one of the surplus digits "7" or "8" in one of [Edit: r7c45]. In the former case, (5)r5c1 is immediately eliminated; in the latter, the ALS r7c169 (marked with "#") becomes locked with "5" forced into r7c1, which again eliminates (5)r5c1. In AIC form:

(5=26)r5c45 - UR - (26=(7or8))r7c45 - (178=5)r7c169 => r5c1 <> 5.

The rest is easy after this elimination.

Last edited by Ron Moore on Thu Dec 21, 2006 12:27 am; edited 1 time in total
Myth Jellies
Hooked

Joined: 04 Apr 2006
Posts: 42

Posted: Sat Dec 16, 2006 8:33 pm    Post subject:

Hey, Ron.

I look forward to reading your posts. They are the interesting observations and solutions that I appreciate.

 Code: .---------------.------------------.--------------. | 29   1    8   | 7     35     6   | 259  359  4  | | 79   3    4   | 158   58     2   | 1579 5789 6  | | 27   6    5   | 138   4      9   | 127  78   23 | :---------------+------------------+--------------: | 4    28   1   | 238   9      378 | 6    37   5  | | 3-5  9    7   |*26+5 *26+5   4   | 8    1    23 | | 6    258  23  | 2358  1      378 | 27   4    9  | :---------------+------------------+--------------: |#15   4    269 |*26+8 *2678  #18  | 3    59  *78 | | 13   25   29  | 4    *78     138 | 59   6   *78 | | 8    7    36  | 9     36     5   | 4    2    1  | '---------------'------------------'--------------'

There is also that r78c59 UR which can either kill a 78 in your UR (an additional step compared to yours) or if you just want to combine it with your UR to form a MUG you can eliminate the same 5 with a similar slightly shorter chain. The asterisks and plus signs denote the MUG.

Perhaps we are going to have to figure out a way to post simultaneously so that people won't think we are the same person
Ron Moore

Joined: 13 Aug 2006
Posts: 72
Location: New Mexico

Posted: Thu Dec 21, 2006 12:58 am    Post subject:

(Initial post edited for minor correction.)

 Myth Jellies wrote: Perhaps we are going to have to figure out a way to post simultaneously so that people won't think we are the same person

I don't really think there's much danger of that yet. But I will admit to having learned from your responses to my posts, especially your first one (here). It may not seem like much, but the added insight there really helped to firm up some of my thinking which hadn't totally gelled. More recently I have been aware of some of the alternatives you've given, such as the "unique corner" elimination of (78)r7c5 you mention above. I thought this might detract a bit from the primary argument. You're right, though, maybe it's time for me to read up on MUG's.
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