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Victor
Joined: 29 Sep 2005
Posts: 207
Location: NI
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 Posted: Mon Mar 17, 2008 5:51 pm Post subject: Bouncing ER |
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M3980837 (160)
| Code: |
+---------------+-------------+-----------------+
| 236 37 67 | 26 5 48 | 14 148 9 |
| 269 4 1 | 3 267 289 | 57 28 258 |
| 29 5 8 | 79 1 249 | 347 234 6 |
+---------------+-------------+-----------------+
| 18 18 3 | 79 27 29 | 6 5 4 |
| 4 6 9 | 1 8 5 | 2 7 3 |
| 7 2 5 | 4 3 6 | 8 9 1 |
+---------------+-------------+-----------------+
| 156 179 467 | 8 246 3 | 1459 1246 25 |
| 13568 138 46 | 256 9 7 | 1345 123468 258 |
| 3568 389 2 | 56 46 1 | 3459 3468 7 |
+---------------+-------------+-----------------+
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Play this puzzle online
You'll see that the 2 in r3c6 goes - W-wing or short XY-chain. After that? ...
(a) Can you spot a bouncing ER? - only one extra line, but still a first for me.
(b) However, that's not necessary for the solution. This is a 1-move sudoku really, which I'm embarrassed to say took me ages to spot. Hopefully you'll be quicker. |
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Steve R
Joined: 24 Oct 2005
Posts: 289
Location: Birmingham, England
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| Posted: Sun Apr 06, 2008 12:52 pm Post subject: |
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OK, Victor, I give up.
I did manage to solve the puzzle using the fact that r2c1 has two conjugates in the second row. One, with respect to 6, is r2c5; the other, with respect to 9, is r2c6. So r2c5 contains 6 or r2c6 contains 9. Either places 7 in r3c4 and reduces the puzzle to singles.
I don’t know if this is the trick you had in mind. However I’m intrigued by your “bouncing ER,” useless though it may be. Would you care to elaborate?
Steve
Last edited by Steve R on Sun Apr 06, 2008 5:27 pm; edited 1 time in total |
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ravel
Joined: 21 Apr 2006
Posts: 536
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| Posted: Sun Apr 06, 2008 3:02 pm Post subject: |
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Suppose, you mean 7 in r3c4.
I had a longer way for the same: From the w-wing one of r3c1 and r4c6 must be 2.
r3c1=2 => r1c4=2 => r7c5=2
==> r4c5<>2 |
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Victor
Joined: 29 Sep 2005
Posts: 207
Location: NI
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| Posted: Sun Apr 06, 2008 4:26 pm Post subject: |
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I'd forgotten about this! I do love your smart solution, Steve.
| Code: |
+---------------+-------------+-----------------+
| 236A 37 67 | 26B 5 48 | 14 148 9 |
| 269 4 1 | 3 267 289 | 57 28 258 |
| 239C 5 8 | 79 1 49 | 347 234 6 |
+---------------+-------------+-----------------+
| 18 18 3 | 79 27 29D | 6 5 4 |
| 4 6 9 | 1 8 5 | 2 7 3 |
| 7 2 5 | 4 3 6 | 8 9 1 |
+---------------+-------------+-----------------+
| 156 179 467 | 8 2#46 3 | 1459 12*46 25 |
| 13568 138 46 | 2#56 9 7 | 1345 123468 258 |
| 3568 389 2 | 56 46 1 | 3459 3468 7 |
+---------------+-------------+-----------------+
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(a) ER. The #d 2s both see the target elimination *, the upper one directly, the lower via boxes 2 & 3. (In my small experience, it's unusual for an ER of any kind not to be solved easily in some other way, which seems to be the case here.)
(b) I'll expand a little for the benefit of anyone reading this who is unfamiliar with skyscrapers, etc. Here's a skyscraper in 2s.
| Code: |
+-------+-------+
|A2x. . | . .B2y|
| . . . | . . . |
|2z . . | . . . |
+-------+-------+
| . . . | . . . |
|C2w. . | .D2v. |
| . . . | . . . |
+-------+-------+
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x,y... are any old numbers. The 2s in A-B & in C-D are conjugate (no others in those rows).
One way to explain it is to start with either B or D not being 2. If B is not 2, then A is. And therefore C is not, and therefore D is. And it's bi-directional: D <> 2 => B =2.
So (at least) one of B & D is 2, and thus we can remove any 2s that they can both 'see'.
Of course once one understands this, one just looks for two strong links connected at one end by a weak link. (I suspect that it's so automatic for some people that they just think of it as colouring, even though strictly speaking that uses all-conjugate links - in this case there would be no other 2s in column 1.)
Note that this depends on the fact that the 2s in the two rows are strongly linked: if one is false the other MUST be true. We do NOT need them to be precisely conjugate, which has the additional implication that if one is true then the other is false.
Now think of the ends of an XY-chain / W-wing / skyscraper / etc. These ends are strongly linked: if one is false, the other is true. They're not conjugate (one being true does NOT imply that the other is false - might both be true), but we don't need that.
Here's the punchline: we could replace one/both of AB/CD by strong link(s) such as between the ends of an XY-chain to get a sort of pseudo-skyscraper.
In the puzzle, the 2s in C & D are strongly linked: ends of a W-wing. And the 2s in A & B are strongly linked: they're conjugate. And so we can eliminate any 2 seen by both B & D, in this case r2c6.
(Purists might complain that this is an unnecessary idea, and that the whole construct can be seen as an AIC; or that you can find a smart alternative as Steve & Ravel have done. That's true - but some of us like our patterns to be bite-sized.) |
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Steve R
Joined: 24 Oct 2005
Posts: 289
Location: Birmingham, England
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| Posted: Sun Apr 06, 2008 5:31 pm Post subject: |
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Victor
Wow! No wonder I didn’t twig what you had in mind. Thanks for the explanation and the kind remark.
How many hinges? Should it be three?
Box 2: row 2 and column 4
Box 3: row 2 (shared with box 2) and column 8
Box 8: row 7 and column 4 (shared with box 2)
Conclusion: eliminate from the row 7 column 8 intersection.
Actually box 8 is a something of a shady character because its column and row are not well defined. Perhaps it’s twins plus a conjugate. Good find, anyway.
ravel
You’re right, of course. Thanks for the correction, now edited into the original post.
Steve |
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Victor
Joined: 29 Sep 2005
Posts: 207
Location: NI
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| Posted: Sun Apr 06, 2008 7:05 pm Post subject: |
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You are working from a grid with the 2 gone from r3c6? Two hinges then, in b2 & b3. When I first read up about ERs I found an explanation in another forum which made a big deal out of it, but actually it's a pretty simple idea - no technical knowledge required, no need to even think about empty rectangles per se. Here, you can just say that r8c4 = 2 => b2r2 = 2 => r3c8 = 2 => * <>2.
PS I'm about to post a puzzle with an ER that uses 3 hinges. |
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ravel
Joined: 21 Apr 2006
Posts: 536
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| Posted: Sun Apr 06, 2008 7:26 pm Post subject: |
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| ravel wrote: | From the w-wing one of r3c1 and r4c6 must be 2.
r3c1=2 => r1c4=2 => r7c5=2
==> r4c5 |
How could i miss that ? Already r3c1=2 => r1c4=2 eliminates 2 in r23c6. Thanks Victor. |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Sun Apr 06, 2008 9:56 pm Post subject: |
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| Victor wrote: | | In my small experience, it's unusual for an ER of any kind not to be solved easily in some other way, which seems to be the case here. |
It's not the case here, either. That ER elimination is also a Finned Swordfish in r138.
By the way... you need to lose the <3> in r3c1 from the grid in your second post in order for your Skyscraper/W-Wing thing to make sense. |
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Victor
Joined: 29 Sep 2005
Posts: 207
Location: NI
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| Posted: Sun Apr 06, 2008 11:06 pm Post subject: |
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Asellus says: | Quote: | | That ER elimination is also a Finned Swordfish in r138. |
Indeed! (However, I did say 'solved easily'!)
| Quote: | .. you need to lose the <3> in r3c1 from the grid in your second post in order for your Skyscraper/W-Wing thing to make sense.
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Indeed again. Actually it isn't really there (pointing pr in b3r3), & wasn't in my original grid - just one of my silly mistakes. |
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