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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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 Posted: Wed May 10, 2006 5:14 pm Post subject: "Nightmare", April 28 |
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This is my fifth try. Every time I find myself backed into the same corner and it's maddening; I can't learn from my mistakes when I don't know what mistakes I'm making. Maybe they're mechanical, maybe logical. I've used a few techniques that are fairly new to me and if you'd let me know if my reasoning is correct, then I can try to discover mechanical errors.
After basic elimination techniques, this is what I have:
| Code: | ----------------------------------------------------------
|9 358 6 |7 135 13 |2 4 58 |
|1245 257 1247 |29 259 8 |5679 3 569 |
|235 23578 278 |6 2359 4 |579 578 1 |
----------------------------------------------------------
|56 1 47 |3 47 9 |8 56 2 |
|2456 25679 2479 |124 8 17 |1567 1567 3 |
|8 27 3 |5 127 6 |17 9 4 |
----------------------------------------------------------
|7 2369 129 |8 1369 5 |4 126 69 |
|1236 4 1289 |19 13679 137 |1569 12568 5689 |
|16 689 5 |149 1469 2 |3 168 7 |
---------------------------------------------------------- |
I would like to eliminate the "3" in r1c2 based on strong links in cols 1 and 6. Would also like to eliminate the "8" in r1c2 based on a finned X-wing in r3&9, thus solving r1c2 for "5" and r1c9 for "8." Then I would like to eliminate the "3" from r1c5 to avoid the "deadly pattern" which starts with 8-3-8 in r1c9, r5c9 and r5c5, respectively. After the eliminations
therefrom, I arrive at the following:
| Code: | -------------------------------------------------
|9 5 6 |7 1 3 |2 4 8 |
|124 27 1247 |29 259 8 |5679 3 569 |
|23 38 278 |6 259 4 |579 57 1 |
-------------------------------------------------
|56 1 47 |3 47 9 |8 56 2 |
|2456 69 2479 |124 8 17 |567 567 3 |
|8 27 3 |5 27 6 |1 9 4 |
-------------------------------------------------
|7 369 129 |8 369 5 |4 12 69 |
|1236 4 1289 |19 3679 17 |569 128 569 |
|16 689 5 |149 469 2 |3 18 7 |
------------------------------------------------- |
And this brings me back to the same situation as the previous four times. With the three "27" pairs in c2 and r6, I eliminate the "2" from r2c5 to kill the "deadly pattern." Then I do a chain on "2", starting at r2c2, going down, across, and then back up. I end up with the type of chain I inquired about in my thread on simple coloring, i.e., a chain with two "+" in one group.
There are two "+" in r2, at c2 and c4. So the "+" must be changed to "-" and r2c2 and r2c4 must be "7" and "9", respectively. Then a few eliminations in boxes 1 and 2 will yield two "7" in box 1 or two "9" in box 2, depending on the path.
Can anyone spot what I might have done wrong?
Last edited by Marty R. on Wed May 10, 2006 8:08 pm; edited 1 time in total |
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Steve R
Joined: 24 Oct 2005
Posts: 289
Location: Birmingham, England
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| Posted: Wed May 10, 2006 7:17 pm Post subject: "Nightmare", April 28 |
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Marty
The last step is wrong. As I have learned from bitter experience, the deadly pattern is deadly to those who use it directly. There is absolutely nothing wrong with
7…2
2…7
What is wrong (when assuming uniqueness) is the pattern
2 or 7…2 or 7
2 or 7…2 or 7
The difference is that, in the second case, you can juggle the 2s and 7s around to give two different solutions. If 7 had been a candidate for r2c5 you might well have been able to make certain eliminations to avoid the unique rectangle, as the second pattern is known. As it is, I cannot see any progress involving the four cells concerned. There are one or two pairs about, however.
In addition r35c78 looks temptingly rectangular.
Steve |
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David Bryant
Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado
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| Posted: Wed May 10, 2006 8:04 pm Post subject: "Deadly Pattern" must lie in just two houses |
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| Marty R wrote: | | Then I would like to eliminate the "3" from r1c5 to avoid the "deadly pattern" which starts with 8-3-8 in r1c9, r5c9 and r5c5, respectively. |
This step is invalid, Marty.
The "deadly pattern" is only deadly if it leads to multiple solutions. That can only happen if the four corners of the rectangle lie in just two different 3x3 boxes. In this case the corners lie in four different 3x3 boxes.
You can't interchange the "3"s with the "8"s in this situation -- if you did you'd have two "3"s in box 3 and box 5, and two "8"s in box 2 and box 6. That's not a valid solution, so "uniqueness" is no help. dcb
PS There's a "hidden pair" in r2c1 & r2c3. And there's also a "fork" on the digit "2" in columns 2 and 4, which allows you to set r6c2 = 2 ... |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Wed May 10, 2006 8:23 pm Post subject: |
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Steve and David,
Thank you, thank you, thank you! This puzzle was threatening my sanity, if indeed, there is any sanity to begin with.
The thing is, I knew the "deadly pattern" principle applied only when it was in two boxes, yet that never occurred to me while doing the puzzle, and probably wouldn't have if you hadn't pointed it out.
I assume what I did with the strong links and Finned X-Wing was OK, since you didn't comment. Now I will have my sixth go at it and maybe reach an impasse instead of duplicates. 
| Quote: | | In addition r35c78 looks temptingly rectangular. |
It certainly does, but it I don't think there's enough there for me to do anything with it. |
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Steve R
Joined: 24 Oct 2005
Posts: 289
Location: Birmingham, England
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| Posted: Wed May 10, 2006 9:05 pm Post subject: "Nightmare", April 28 |
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Marty
Yes, you were absolutely right until the fraudulent rectangle came up.
I pointed out the other (genuine) rectangle since it is an example of Keith’s recent discovery. R5c78 are conjugates with respect to 7 in the fifth row so one of these two cells must contain 7.
We cannot have a rectangle of (57)s in two boxes, two rows and two columns. Thus r3c7 contains 9 or r5c8 contains 6. In the former case r3c7 does not contain 7. In the latter, r5c7 contains 7 and also eliminates 7 from r3c7.
Steve |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Wed May 10, 2006 11:54 pm Post subject: |
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| Quote: | | We cannot have a rectangle of (57)s in two boxes, two rows and two columns. Thus r3c7 contains 9 or r5c8 contains 6. In the former case r3c7 does not contain 7. In the latter, r5c7 contains 7 and also eliminates 7 from r3c7. |
While I understand the general principle, Steve, I don't understand the start of the reasoning. At my skill level, I would have said that r3c7 must be "9" or r5c7 or r5c8 must be"6." I don't know how the "6" is not considered for r5c7.
By the way, all's well that ends well. I finished the puzzle now that I knew that the "deadly pattern" was only deadly on me. The chain on "2" opened things up considerably, and basic techniques plus two rectangles of the ab-ab-ab-abc type finished things up. |
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Steve R
Joined: 24 Oct 2005
Posts: 289
Location: Birmingham, England
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| Posted: Thu May 11, 2006 2:35 am Post subject: |
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Mea culpa. I failed to spell it out.
In the second position you posted, 6 is excluded from the eighth column in boxes 3 and 9. Accordingly it must occupy that column in box 6.
Steve |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Thu May 11, 2006 4:57 pm Post subject: |
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| Steve R wrote: | Mea culpa. I failed to spell it out.
In the second position you posted, 6 is excluded from the eighth column in boxes 3 and 9. Accordingly it must occupy that column in box 6.
Steve |
I'll take a mea culpa as well. I always look for locked candidates and just happened to miss that one, so that second position I posted shouldn't have shown "6" as a candidate in r5c7. Thanks. |
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