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ravel
Joined: 21 Apr 2006
Posts: 536
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 Posted: Mon Dec 03, 2007 4:44 pm Post subject: A wapati |
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Always a good place to find challenging puzzles is wapati's thead, which moved from the players to the programmers forum. To this one (Dec 01) he noted "Sue de coq", i needed a lot of coloring (strong links, ER) plus x,xy-wing.
| Code: | . 7 .|. . .|1 . .
6 . 9|. . 4|5 . .
. 8 .|9 3 .|. . .
-----+-----+-----
. . 1|. 9 .|. . 2
. . 8|3 . 1|4 . .
. 9 .|. 8 .|. . .
-----+-----+-----
4 6 .|. 7 .|. 5 .
. . .|. . .|6 . .
. . .|1 . .|. . 3 |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Mon Dec 03, 2007 6:16 pm Post subject: |
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don't know about sue de coq (didn't get around to really study it, shame).
For me it started out with some coloring eliminations ("2" from r3c6, r9c6, "7" from r4c7)
Then - in this position
| Code: |
+--------------------------+--------------------------+--------------------------+
| 25 7 4 | 2568 256 2568 | 1 3 9 |
| 6 3 9 | 27 1 4 | 5 278 78 |
| 1 8 25 | 9 3 57 | 27 46 46 |
+--------------------------+--------------------------+--------------------------+
| 357 4 1 | 567 9 567 | 38 678 2 |
| 257 25 8 | 3 256 1 | 4 9 567 |
| 2357 9 6 | 4 8 257 | 37 1 57 |
+--------------------------+--------------------------+--------------------------+
| 4 6 23 | 28 7 2389 | 289 5 1 |
| 89 1 2357 | 258 245 23589 | 6 2478 478 |
| 89 25 257 | 1 2456 5689 | 2789 2478 3 |
+--------------------------+--------------------------+--------------------------+
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there is this useless xy-wing pivot r3c6 27-57-52 which means either r3c3 or r2c4 must be 2.
From r3c3=2 we get r3c7=7, r6c7=3, r4c7=8
From r2c4=2 we get r7c4=8
thus r7c7<>8 (the whole thing is an xy chain linking r7c4 to r4c7).
From there an xy-wing 57-27-25 pivot r2c4 removes some "5"s.
Currently I am stuck here, but I am sure there is something that can be done either with "5"s or "2"s - such nice long chains ...
| Code: |
+--------------------------+--------------------------+--------------------------+
| 25 7 4 | 268 256 2568 | 1 3 9 |
| 6 3 9 | 27 1 4 | 5 278 78 |
| 1 8 25 | 9 3 57 | 27 46 46 |
+--------------------------+--------------------------+--------------------------+
| 357 4 1 | 567 9 567 | 38 678 2 |
| 257 25 8 | 3 256 1 | 4 9 567 |
| 2357 9 6 | 4 8 257 | 37 1 57 |
+--------------------------+--------------------------+--------------------------+
| 4 6 23 | 28 7 2389 | 29 5 1 |
| 89 1 2357 | 25 245 239 | 6 2478 478 |
| 89 25 257 | 1 2456 69 | 2789 2478 3 |
+--------------------------+--------------------------+--------------------------+
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Mon Dec 03, 2007 8:57 pm Post subject: |
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nataraj, I took your advice and looked at the 2's
I was thinking finned x-wing all the way, but its not now that I look at it.
being horrible at chains, have no idea what this is. I found it interesting however:
r9c7=2
makes 2 in r7c3, r8c3 a pointing pair, -2 r3c3
that means 2 goes in r3c7, that is a contradiction. 2 can't go in r9c7. |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Mon Dec 03, 2007 9:23 pm Post subject: |
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in the meantime (and while I had Photoshop open) I couldn't resist and did some Medusa coloring (starting with those 2/5 pairs in the upper left)
And lo and behold ...
if red is true, the fives in r1c1 and r3c6 make r4c4=5
but if green is true then r4c4=7,
which means r4c4<>6 and r1c4=6.
That takes us a long way I believe. |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Mon Dec 03, 2007 9:35 pm Post subject: |
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| storm_norm wrote: | nataraj, I took your advice and looked at the 2's
I was thinking finned x-wing all the way, but its not now that I look at it.
being horrible at chains, have no idea what this is. I found it interesting however:
r9c7=2
makes 2 in r7c3, r8c3 a pointing pair, -2 r3c3
that means 2 goes in r3c7, that is a contradiction. 2 can't go in r9c7. |
Norm, I think your reasoning is quite correct.
Nothing horrible about the chains, I will try to write it as a grouped AIC/nice loop
(I've never done this, so please, guys, correct me if I'm wrong)
The discontinuity must be at r9c7 (two weak links) and it "sees" both r9 (box 7) and r3c7
2: -(r9c23)=(r78c3)-r3c3=r3c7- => r9c7<>2 |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Mon Dec 03, 2007 9:40 pm Post subject: |
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| hmm, I guess it would help if I found those relationships that actually move the puzzle ahead, he he. |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Mon Dec 03, 2007 9:42 pm Post subject: |
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| storm_norm wrote: | | hmm, I guess it would help if I found those relationships that actually move the puzzle ahead, he he. |
Every contribution counts, let's eliminate them buggers ... 
No, but seriously. I think spotting a connection (even if it only removes one single candidate) in one of these difficult puzzles is worth more than solving a load of easy ones by rote. |
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ravel
Joined: 21 Apr 2006
Posts: 536
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| Posted: Mon Dec 03, 2007 11:53 pm Post subject: |
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Great, nataraj, i love to watch, how you attack this puzzle.
Since i read about Sue de coq, i tried to spot it (watching pairs), but i missed it. After asking SudoCue (dont have it under linux), i saw, it was not on my solving path. It made the same elimination as you with the Medusa deduction: r4c4<>6.
In your second grid above there is an ER in 2 with strong link in row 3 to eliminate 2 in r9c7. A kite then takes out 2 from r7c4, placing 8 there and in r1c6. Then an ER (strong link in row 1) takes out 5 from r5c5 to get here:
| Code: | +--------------------------+--------------------------+--------------------------+
| 25 7 4 | 268 256 8 | 1 3 9 |
| 6 3 9 | 27 1 4 | 5 278 78 |
| 1 8 25 | 9 3 #57 | 27 46 46 |
+--------------------------+--------------------------+--------------------------+
| 357 4 1 | 5-67 9 #567 | 38 678 2 |
| 257 25 8 | 3 #26 1 | 4 9 567 |
| 2357 9 6 | 4 8 #257 | 37 1 57 |
+--------------------------+--------------------------+--------------------------+
| 4 6 23 | 8 7 239 | 29 5 1 |
| 89 1 2357 | 25 245 239 | 6 2478 478 |
| 89 25 257 | 1 2456 69 | 789 2478 3 |
+--------------------------+--------------------------+--------------------------+
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A 2 or 6 in in box 5 and a 5 or 7 in column 6 make the marked cells invalid.
After all it is harder to find the Sue de coq than to solve the puzzle  |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Tue Dec 04, 2007 3:44 am Post subject: |
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ravel,
Congratulations on finding the Sue de Coq! I couldn't. Who knew it was so path-dependent.
nataraj,
As ravel points out, that R9C7 <2> elimination is a basic ER with the strong link in R3. The Eureka notation, with the R3 strong link notation condensed ("(2=2)"), is:
(2)R9C7-(2)R9C23=(2)R789C3-(2=2)R3C37-(2)R9C7; R9C7<>2
Notice that all three Rows must be included on the right side of the ER strong link ("(2)R9C23=(2)R789C3"). This is so that the AIC is valid in the reverse direction. Since it is a strong inferential link position, it is okay for both sides to have the possibility of being true, which would be the case if the ERI at R9C3 were to be true.
Also, nice work on those Medusa red <5>s!
I solved the puzzle without any Medusa (though I was tempted!). Just lots of ERs and XY Wings and a couple of Finned X-Wings and Kites and an XYZ Wing or two, if I'm not mistaken. It was a rather lengthy slog. |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Tue Dec 04, 2007 6:50 am Post subject: |
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Thank you ravel and Asellus for your help and encouragement!
The ER - yes, I see it now. Must have been a case of not seeing the wood for the trees. I stared at the plot for a long time but there are so many "2"s in the position...
Also thanks for pointing out the reverse direction issue with my chain, but I am not happy with the solution:
| Quote: | | all three Rows must be included on the right side of the ER strong link ("(2)R9C23=(2)R789C3") |
because it includes r9c3 twice. Are you sure this is valid?
Maybe this
(2)R9C7-(2)R9C2=(2)R789C3-(2=2)R3C37-(2)R9C7; R9C7<>2
will work? Actually, the "reverse" direction is much better suited to the ER interpretation where one "comes from" the strong link in r3.
Continuing, I found the same kite ravel did (plus the removal of 2 from r7c6)
| Code: |
+·····+·····+·····+
·* ·o o o· ·
· \ · · ·
· \ ·*-------* ·
· \ · · / ·
· *-------* ·
+·····+·····+|····+
· · ·| ·
· · ·| ·
·o * · * ·| ·
· | · \ ·| ·
·*---------*·| ·
+··|··+·····+|····+
· | o·o o·* ·
· | · · ·
· | o·o o o· o ·
· | · · ·
· * o· o · o ·
+·····+·····+·····+
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, which opened an x-wing r37 and got rid of 2 in r897c3 (we seem to be caught in that same loop forever ...)...
... and xy-wing (27-25-57) pivot r3c3 ... (some basic stuff) ...
... a kite to remove 5 from r4c1 ...
... and we're home free.
Seems that Norm's ER really cracked the puzzle ... "Eureka !"
(and I did not meet the famous Sue so she's probably going to remain a mystery to me for some time...) |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Tue Dec 04, 2007 1:01 pm Post subject: |
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| nataraj wrote: | Are you sure this is valid?
Maybe this
(2)R9C7-(2)R9C2=(2)R789C3-(2=2)R3C37-(2)R9C7; R9C7<>2
will work? |
"Sure" might be too strong a word; but I'm fairly confident. Your alternate notation is still asymmetric, structured differently in the two directions. I like symmetry. So, the choice is to include the ERI cell in both groups or to exclude it from both groups. I vote for inclusion. There's no logical problem that I can see: remember, the link in this case is between the cell groups, not the individual cells. If the first group is false, the second group must be true. The individual cells aren't important.
By the way, you might note that when the ERI cell is occupied (by the ER candidate digit), the ER can only be used at a strong link position. However, an ER without the candidate in the ERI cell can be used at both strong and weak link positions.
| nataraj wrote: | | (and I did not meet the famous Sue so she's probably going to remain a mystery to me for some time...) |
Just to make sure you (or others) realize... the thing ravel marked with # in his grid is the "famous Sue" de Coq technique. It's been described as "disjoint overlapping locked sets" or something like that. (Don't quote me!) Any way that you solve those four cells, <5> and <7> can't be anywhere else in C6 (so, had there been any others, they would have been eliminated) and <2> and <6> can't be anywhere else in Box 5. Alas, the only elimination was that single <6>.
I'm not very good at spotting them, though I try. It doesn't help that they don't seem to come up very often. |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Tue Dec 04, 2007 3:25 pm Post subject: |
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Thanks for the clarification, Asellus, regarding the grouped chain. I'll have to think about that for a while tonight.
As for meeting Sue
| Quote: | | the thing ravel marked with # in his grid is the "famous Sue" de Coq |
I thought so. Only it seems to me that after the kite in "2" ravel wrote about, we took a slightly different path and I don't remember ever coming to the position ravel posted. I'll have to meditate on that diagram as well ... |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Tue Dec 04, 2007 10:05 pm Post subject: |
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asellus wrote:
| Quote: | | As ravel points out, that R9C7 <2> elimination is a basic ER with the strong link in R3. The Eureka notation, with the R3 strong link notation condensed ("(2=2)"), is |
I really need practice on these ER's. everytime I see finns or that pattern with candidates lined up on one side of a box, then the three other sides of the "x-wing" then I can also look for the ER. I just naturally look for x-wings rather than an ER pattern. |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Tue Dec 04, 2007 10:24 pm Post subject: |
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| storm_norm wrote: | | I just naturally look for x-wings rather than an ER pattern. |
X-Wings, Skyscrapers, Kites, Turbot Fish, ERs and more all just exploit strong (conjugate) links. So, with experience ones perspective can (and perhaps should) shift from looking for a specific technique to looking for the strong links and then scanning for how to exploit them. For instance, if two strong links are parallel, do one or both ends line up? Then: Skyscraper/X-Wing. If perpendicular, to two ends share a Box: Kite. Does one end line up with an ER? Can I join them up into a color chain/cluster? See how it works?
nataraj's dot grids on the side are a good way to help one do this. |
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