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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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 Posted: Thu Nov 29, 2007 9:55 pm Post subject: Stumped! |
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This one has me stumped.
| Code: | Puzzle: M5006725sh(8)
+-------+-------+-------+
| . 5 2 | . . . | 9 7 . |
| 8 . . | . . . | . . 4 |
| 7 . . | 5 6 2 | . . 8 |
+-------+-------+-------+
| . . 7 | . 2 9 | 1 . . |
| . . 4 | . 5 . | 7 . . |
| . . 1 | 6 7 . | 8 . . |
+-------+-------+-------+
| 2 . . | 9 8 6 | . . 1 |
| 3 . . | . . . | . . 7 |
| . 1 8 | . . . | 6 9 . |
+-------+-------+-------+ |
All I see is a pattern <38> in B245.
Keith |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Thu Nov 29, 2007 10:17 pm Post subject: |
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OK lets see ...
This is where basics get us:
| Code: |
+--------------------------+--------------------------+--------------------------+
| 1 5 2 | 38 4 38 | 9 7 6 |
| 8 6 3 | 17 9 17 | 25 25 4 |
| 7 4 9 | 5 6 2 | 3 1 8 |
+--------------------------+--------------------------+--------------------------+
| 56 38 7 | 348 2 9 | 1 456 35 |
| 69 238 4 | 138 5 138 | 7 26 239 |
| 59 23 1 | 6 7 34 | 8 245 2359 |
+--------------------------+--------------------------+--------------------------+
| 2 7 5 | 9 8 6 | 4 3 1 |
| 3 9 6 | 24 1 45 | 25 8 7 |
| 4 1 8 | 27 3 57 | 6 9 25 |
+--------------------------+--------------------------+--------------------------+ |
There is an xyz-wing (3,4,8) in c4/box5
At this point:
| Code: |
+--------------------------+--------------------------+--------------------------+
| 1 5 2 | 38 4 38 | 9 7 6 |
| 8 6 3 | 17 9 17 | 25 25 4 |
| 7 4 9 | 5 6 2 | 3 1 8 |
+--------------------------+--------------------------+--------------------------+
| 56 38 7 | 348 2 9 | 1 456 35 |
| 69 238 4 | 18 5 138 | 7 26 239 |
| 59 23 1 | 6 7 34 | 8 245 2359 |
+--------------------------+--------------------------+--------------------------+
| 2 7 5 | 9 8 6 | 4 3 1 |
| 3 9 6 | 24 1 45 | 25 8 7 |
| 4 1 8 | 27 3 57 | 6 9 25 |
+--------------------------+--------------------------+--------------------------+ |
it is xy-chains for me ... |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Thu Nov 29, 2007 10:52 pm Post subject: |
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Which is exactly the point I got to.
Keith |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Thu Nov 29, 2007 11:50 pm Post subject: |
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Hi,
I am also a new person to Sudoku and this site, and am definitely still learning.
I looked at the xyz-wing mentioned in box five resulting in the elimination of the <3> in r5c4 and believe I understood how it worked. If I do understand correctly, then it seems to me that the <3> in r6c2 can also be eliminate using the pincer <38> in r4c2.
Comments please.
Ted |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Fri Nov 30, 2007 12:58 am Post subject: |
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Ted,
Hi, and welcome!
The "victim" cells of an XYZ-Wing must be able to "see" all three cells of the wing. In the case here with Pivot R4C4 and Pincers R1C4 and R6C6, the only potential victims are R56C4 (hence the elimination of <3> in R5C4).
I believe you are referring to Pivot R4C4 and Pincers R4C2 and R6C6. The potential victims are R4C56 (where there are no <3>s to eliminate). The <3> at R6C2 cannot be eliminated using this XYZ Wing because it's possible that only R4C4 is <3> and R6C2 cannot "see" R4C4. |
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Johan
Joined: 25 Jun 2007
Posts: 206
Location: Bornem Belgium
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| Posted: Fri Nov 30, 2007 1:10 am Post subject: |
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| Code: |
At this point:
Code:
+--------------------------+--------------------------+--------------------------+
| 1 5 2 | 38 4 38 | 9 7 6 |
| 8 6 3 | 17 9 17 | 25 25 4 |
| 7 4 9 | 5 6 2 | 3 1 8 |
+--------------------------+--------------------------+--------------------------+
| 56 38 7 | 348 2 9 | 1 456 35 |
| 69 238 4 | 18 5 138 | 7 26 239 |
| 59 23 1 | 6 7 34 | 8 245 2359 |
+--------------------------+--------------------------+--------------------------+
| 2 7 5 | 9 8 6 | 4 3 1 |
| 3 9 6 | 24 1 45 | 25 8 7 |
| 4 1 8 | 27 3 57 | 6 9 25 |
+--------------------------+--------------------------+--------------------------+
it is xy-chains for me ... |
There is a 6-cell xy-chain with pincer cells in R6C2 and R2C8, eliminating
<2> in R6C8, but the puzzle isn't solved yet. |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Fri Nov 30, 2007 2:00 am Post subject: |
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Ignoring that XY Chain, I used Medusa and exploited the potential 6-cell{38} Deadly Pattern in Boxes 2, 5 and 4. First, some initial Medusa coloring:
| Code: | +------------------+----------------------+--------------------------+
| 1 5 2 |@38 4 @38 | 9 7 6 |
| 8 6 3 | 1g7r 9 1r7g | 2g5r 2r5g 4 |
| 7 4 9 | 5 6 2 | 3 1 8 |
+------------------+----------------------+--------------------------+
| 56 @38 7 |@34g8 2 9 | 1 4r56 35 |
| 69 @238 4 | 1r8g 5 @1g38 | 7 26 239 |
| 59 23 1 | 6 7 3g4r | 8 24g5 2359 |
+------------------+----------------------+--------------------------+
| 2 7 5 | 9 8 6 | 4 3 1 |
| 3 9 6 | 2g4r 1 4g5r | 2r5g 8 7 |
| 4 1 8 | 2r7g 3 5g7r | 6 9 2g5r |
+------------------+----------------------+--------------------------+ |
If green is false, the only thing to prevent the DP marked @ is the <2> at R5C2. So, this <2> can be colored red and we continue:
| Code: | +----------------+----------------+----------------------+
| 1 5 2 | 38 4 38 | 9 7 6 |
| 8 6 3 | 1g7r 9 1r7g | 2g5r 2r5g 4 |
| 7 4 9 | 5 6 2 | 3 1 8 |
+----------------+----------------+----------------------+
| 5g6r 38 7 | 34g8 2 9 | 1 4r-56g 35 |
| 69 2r38 4 | 1r8g 5 1g38 | 7 26 239 |
| 5r9 2g3r 1 | 6 7 3g4r | 8 -24g5r 2-359 |
+----------------+----------------+----------------------+
| 2 7 5 | 9 8 6 | 4 3 1 |
| 3 9 6 | 2g4r 1 4g5r | 2r5g 8 7 |
| 4 1 8 | 2r7g 3 5g7r | 6 9 2g5r |
+----------------+----------------+----------------------+ |
Notice that we get two red <5>s in R6. This is a Medusa Wrap: all the reds are false and all the greens are true. The puzzle is solved. |
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ravel
Joined: 21 Apr 2006
Posts: 536
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| Posted: Fri Nov 30, 2007 11:57 am Post subject: |
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| Code: | *---------------------------------------------*
| 1 5 2 | 38 4 38 | 9 7 6 |
| 8 6 3 | 17 9 17 | 25 25 4 |
| 7 4 9 | 5 6 2 | 3 1 8 |
|-------------+--------------+----------------|
| 56 38 7 |-348 2 9 | 1 456 #35 |
| 69 238 4 | 18 5 138 | 7 26 239 |
| 59 23 1 | 6 7 #34 | 8 245 2-359 |
|-------------+--------------+----------------|
| 2 7 5 | 9 8 6 | 4 3 1 |
| 3 9 6 | 24 1 #45 | 25 8 7 |
| 4 1 8 | 27 3 @57 | 6 9 @25 |
*---------------------------------------------*
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This looks like a variant of w-wing. Instead of two pairs 35 you only have 35 and pairs 34,45 with the same effect: if the cells are connected with a strong link in 3 or 5, you get an either/or for the other number.
In this grid you can take also 25 in r2c8, 24/45 in row 8 and the strong links 25 in c7, but these do not lead to an elimination. |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Fri Nov 30, 2007 4:47 pm Post subject: |
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| Asellus wrote: | | Ignoring that XY Chain, I used Medusa ... |
Starting from Asellus' first grid and coloring (and not using UR logic this time),
notice that r4c8 cannot be5 (red 4 in cell, green 5 in column)
which makes it possible to color some more candidates: 6 in r4c8 (g) , r5c8 (r), r5c1 (g) and r4c1 (r).
This is the position and it leads directly to a Medusa wrap:
| Code: |
+------------------+----------------------+--------------------------+
| 1 5 2 | 38 4 38 | 9 7 6 |
| 8 6 3 | 1g7r 9 1r7g | 2g5r 2r5g 4 |
| 7 4 9 | 5 6 2 | 3 1 8 |
+------------------+----------------------+--------------------------+
| 56r 38 7 | 34g8 2 9 | 1 4r-56g 35 |
| 6g9 238 4 | 1r8g 5 1g38 | 7 26r 239 |
| 59 23 1 | 6 7 3g4r | 8 24g5 2359 |
+------------------+----------------------+--------------------------+
| 2 7 5 | 9 8 6 | 4 3 1 |
| 3 9 6 | 2g4r 1 4g5r | 2r5g 8 7 |
| 4 1 8 | 2r7g 3 5g7r | 6 9 2g5r |
+------------------+----------------------+--------------------------+
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r9c9 r5 means r4c9 <> 5 => r4c1=5, BUT
r4c1 is r6 thus a contradiction (Medusa wrap), all reds are false, puzzle solved. |
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Earl
Joined: 30 May 2007
Posts: 677
Location: Victoria, KS
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| Posted: Fri Nov 30, 2007 5:03 pm Post subject: Stumped |
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I found a three step xy-chain with pincers (2) at R6C2-R8C7, which pinces nothing, but R8C7 can be extended by coloring to R2C8 which eliminates the 2 in R6C8. Now there is an xy-wing pivoted at R6C8, then another xy-wing pivoted at R4C4, and rest is singles.
Earl |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Fri Nov 30, 2007 7:52 pm Post subject: |
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Hi Asellus,
Thanks for your timely response and the clarification!
In hindsight, I understood that any victim cells must see all three cells in the xyx-wing, but I seem to forget all those details when I am checking the puzzle for a multitude of patterns. Hopefully I will get better at eliminating those "stupid" mistakes with more experience.
Thanks again for your considerations.
Ted |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Fri Nov 30, 2007 8:48 pm Post subject: |
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| tlanglet wrote: | | ... I seem to forget all those details ... |
as we all do, I might add, when it comes to new techniques. I had a collegue once, actually I was in sales and he was the head of quality management, and he said to me:
"One does not assume. One KNOWS."
I've used this advice quite often and it helped a lot. When it comes to sudoku solving, there are many nice methods that help to FIND candidates for exclusion. In and of themselves, those techniques don't prove anything. So, especially for the more complex methods, it helps to go through the proof once more just to make SURE.
In the case of the xyz-wing: WHY does it eliminate the "3" and HOW do we know for sure?
First step: recognize the pattern. Three cells, three common numbers. But not in the same house (otherwise - naked triple). Might be a xy-wing might be a xyz wing might be a "useless" wing might be nothing. How does the xyz wing in boxes 2,4 work?
It works by "case differentiation" - when there are several possibilities and all of them lead to the same result, then the result is true (regardless).
The case differentiation here starts in cell r1c4 with candidates {3,8}.
case 1: r1c4=3. If that's the case, 3 disappears from all cells in col 4 (and row 1), thus r5c4 cannot be 3.
case 2: r1c4=8. In that case, r4c4 is reduced to {3,4} and forms a naked pair (together with r6c6). The naked pair means that 3 (and 4) disappear from all other cells in box 5, notably r5c4.
so, reagrdless of which of the two cases is true, r5c4 cannot be 3.
This whole sequence looks rather long, written out in long sentences. With a little practice it takes just a few seconds. Without practice, maybe 10 secs, half a minute? But if you cannot complete this sequence, you'll NEVER be sure.
So - when you look at r4c2 and you recognize a pattern ("pincer", something with "3", maybe eliminate ...) you've done the first part, the hard part - finding something. The few seconds spent to make sure your wing or whatever really does what you think it does is time well spent.
Same goes for w-wings, xy-wings, xy-chains, empty rectangles, ... (and especially for ANY new technique).
After a few dozens of xyz-wings it'll become routine, I guess ... |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Sat Dec 01, 2007 12:39 pm Post subject: |
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Hi Nataraj,
Your response was very helpful and very encouraging.
At this stage of my "Sudoku development" I find myself trying too many techniques to fast; mostly I need to slow down, check my actions (as you suggested), and learn.
Thanks for your input ......
Ted |
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