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keith
Joined: 19 Sep 2005 Posts: 3355 Location: near Detroit, Michigan, USA
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Posted: Fri Nov 23, 2007 2:28 am Post subject: Far too much fun! |
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Another Menneske of the same ilk.
Code: | Puzzle: M5837481sh(14)
+-------+-------+-------+
| . . 4 | . . . | 1 . . |
| 6 . 9 | . 3 . | . . 4 |
| . 1 . | 2 . . | 9 . . |
+-------+-------+-------+
| 7 . . | . . 3 | . . 2 |
| . . . | 8 7 2 | . 4 . |
| . 4 8 | . 5 . | . 9 3 |
+-------+-------+-------+
| . . . | . . 8 | . . 1 |
| . 6 . | 4 . . | . . . |
| . . 1 | . . 7 | . 2 . |
+-------+-------+-------+ |
Happy Thanksgiving!
In Detroit, an inch of wet snow. And watching the Lions lose the football game is such a tradition! At least I have my Beaujolais.
Keith |
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Marty R.
Joined: 12 Feb 2006 Posts: 5770 Location: Rochester, NY, USA
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Posted: Fri Nov 23, 2007 6:27 am Post subject: |
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This puzzle kept me up past my already late bedtime. I must've missed some solution path because it took:
Type 1 UR
Two XY-Wings
Two XY-Wings w' coloring
Three W-Wings w' coloring
Simple coloring
And another XY-Wing w' coloring to finish it off.
Quote: | And watching the Lions lose the football game is such a tradition! |
Hey, they can still go 4-1 to achieve the 10-win promise (boast). |
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keith
Joined: 19 Sep 2005 Posts: 3355 Location: near Detroit, Michigan, USA
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Posted: Fri Nov 23, 2007 3:26 pm Post subject: |
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Marty,
The basics get you to here:
Code: | +-------------------+-------------------+-------------------+
| 358 2 4 | 567 689 569 | 1 35678 5678 |
| 6 78 9 | 157 3 15 | 2 578 4 |
| 358 1 57 | 2 68 4 | 9 35678 5678 |
+-------------------+-------------------+-------------------+
| 7 5 6 | 9 4 3 | 8 1 2 |
| 1 9 3 | 8 7 2 | 56 4 56 |
| 2 4 8 | 16 5 16 | 7 9 3 |
+-------------------+-------------------+-------------------+
| 9 37 57 | 356 2 8 | 4 56 1 |
| 58 6 2 | 4 1 59 | 3 78 789 |
| 4 38 1 | 356 69 7 | 56 2 89 |
+-------------------+-------------------+-------------------+
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There are two XY-wings, taking out <5> in R9C4 and <8> from R3C9.
There is also coloring on <8> (look at R29) that takes out <8> in R8C8 and R13C9.
That reveals another XY-wing which takes out <5> in R3C8. With a bit of cleaning up, you are here:
Code: | +--------------+---------------+--------------+
| 38$ 2 4 | 567 689# 569 | 1 358# 67 |
| 6 78 9 | 157& 3 15 | 2 58@ 4 |
| 358# 1 57 | 2 68# 4 | 9 38$ 67 |
+--------------+---------------+--------------+
| 7 5 6 | 9 4 3 | 8 1 2 |
| 1 9 3 | 8 7 2 | 6 4 5 |
| 2 4 8 | 16 5 16 | 7 9 3 |
+--------------+---------------+--------------+
| 9 37 57% | 35% 2 8 | 4 6 1 |
| 58@ 6 2 | 4 1 59 | 3 7 89 |
| 4 38 1 | 36 69 7 | 5 2 89 |
+--------------+---------------+--------------+
|
The two cells $ must be <3>, because of the Type 6 UR to which they belong.
The two cells @ are a W-wing connected by the chain of <8>'s marked #. The <5> in R8C1 can be extended by coloring (%) to R7C4, so the W-wing takes out <5> in R2C4 &. You are here:
Code: | +-------------+-------------+-------------+
| 3 2 4 | 567 689 569 | 1 58 67 |
| 6 78 9 | 17 3 15 | 2 58 4 |
| 58 1 57 | 2 68 4 | 9 3 67 |
+-------------+-------------+-------------+
| 7 5 6 | 9 4 3 | 8 1 2 |
| 1 9 3 | 8 7 2 | 6 4 5 |
| 2 4 8 | 16 5 16 | 7 9 3 |
+-------------+-------------+-------------+
| 9 37 57 | 35 2 8 | 4 6 1 |
| 58 6 2 | 4 1 59 | 3 7 89 |
| 4 38 1 | 36 69 7 | 5 2 89 |
+-------------+-------------+-------------+ |
This is a BUG+3. If you examine C456, to avoid the deadly pattern you must have R1C4 is <6>, R1C5 is <6>, or R1C6 is <5>. In other words, R1C6 is not <6>, and the puzzle is solved!
Keith |
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Marty R.
Joined: 12 Feb 2006 Posts: 5770 Location: Rochester, NY, USA
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Posted: Fri Nov 23, 2007 5:46 pm Post subject: |
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A couple of comments, Keith.
The BUG+3 reasoning is still something that is beyond me.
I arrived at a different ending position since I don't see the XY-Wing that, with coloring, finished it off for me.
I used the 58 W-Wing. According to my notes, I colored it for an elimination, then further colored it for another elimination.
And that 58 W-Wing--I assume you noticed--is more simply connected by the 8s in column 2. Regardless, I can't follow the chain of 8s, because there are three 8s in row 1, so how are r1c58 connected? |
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keith
Joined: 19 Sep 2005 Posts: 3355 Location: near Detroit, Michigan, USA
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Posted: Fri Nov 23, 2007 7:14 pm Post subject: |
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Marty R. wrote: | And that 58 W-Wing--I assume you noticed--is more simply connected by the 8s in column 2. |
No, I didn't see that!
Marty R. wrote: | Regardless, I can't follow the chain of 8s, because there are three 8s in row 1, so how are r1c58 connected? |
Marty, the UR places a <3> in R1C1 and R3C8. Then, the coloring chain on <8> is R3C1 - R3C5 - R1C5 - R1C8.
Marty R. wrote: | The BUG+3 reasoning is still something that is beyond me. |
Here's a longer explanation:
In the BUG situation, the Deadly Pattern (DP) is such that each candidate appears exacly twice in each row, column and box. Look at the candidates in C4:
5(6)7, 17, 16, 35, 36.
meaning that if R1C4 is <57> you have the DP. If R1C4 is <6>, that breaks up the DP.
Similarly for C5:
(6)89, 68, 69.
and for C6:
(5)69, 15, 16, 59.
Then, check the DP across R1:
5(6)7, (6)89, (5)69, 58, 67.
Each candidate in the DP appears twice in R1.
So, to avoid the DP, R1C4 is <6>, and or R1C5 is <6>, and or R1C6 is <5>. Either of these means R1C6 is not <6>.
Keith |
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Marty R.
Joined: 12 Feb 2006 Posts: 5770 Location: Rochester, NY, USA
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Posted: Fri Nov 23, 2007 9:04 pm Post subject: |
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Thanks Keith, I've printed out the grid and your explanations for further digestion. |
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keith
Joined: 19 Sep 2005 Posts: 3355 Location: near Detroit, Michigan, USA
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Posted: Fri Nov 23, 2007 10:08 pm Post subject: |
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Marty,
When you have mostly two-value cells, and two or three three-value cells, the idea is this:
1. Check some element (row, column, or block) that has only two-value cells. Does each candidate occur exactly twice? If so, good. (If not, there is no BUG.)
2. Find an element that has only one three-value cell. Identify the two candidates that would make the DP, and the one candidate which would kill the DP. If you can do this, good. (If not, there is no BUG.)
3. Do step 2 for the remaining three-value cells, using the DP candidates found in step 2. If you can do this, you have arrived at the DP. All two-value cells, each candidate occurs exactly twice in each element.
4. You now have a list of cells and candidates, any one of which will kill the DP. Only ONE of them is necessarily true. It's up to you to figure out if this is useful information. Each case is different.
Some examples, for two three-value cells, a BUG+2:
a. Cell 1 is X OR Cell 2 is X. Like the pincers of a chain, any cell that sees both cannot be X.
b. Cell 1 is X OR Cell 2 is Y. Cell 1 = X forces Cell 2 = Y. Cell 2 must be Y.
By the way, I don't think this is very obvious. I stared at this thread's puzzle for a long time before I figured out that the BUG killers were saying R1C6 is not <6>.
Best wishes,
Keith |
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Asellus
Joined: 05 Jun 2007 Posts: 865 Location: Sonoma County, CA, USA
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Posted: Wed Nov 28, 2007 10:55 pm Post subject: |
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I'm catching up after the holiday...
Completed this in 5 fairly simple steps without DPs:
1. Skyscraper R29 removes <8> from R13C9 & R8C8.
2. XY Wing, pivot R2C2, removes <5> from R3C89.
3. XY Wing, pivot R9C5, removes <5> from R9C4.
4. W Wing on {38} in B13 removes <3> from R1C8 & R3C1.
5. XY Wing, pivot R9C5, with Pincer R3C5 transported to R8C1 removes <8> from R9C9. |
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re'born
Joined: 28 Oct 2007 Posts: 80
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Posted: Thu Nov 29, 2007 10:37 am Post subject: |
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Here are three not so simple steps that lead to a solution:
1) xy-wing: pivot at r9c5 eliminates 5 from r9c4.
2) finned swordfish: c158/r138 fin r2c8 => r13c9 <> 8
After cleaning up, we get to
Code: | .---------------.---------------.---------------.
| 358 2 4 | 567 689 569 | 1 358 67 |
| 6 78 9 | 157 3 15 | 2 58 4 |
| 358 1 57 | 2 68 4 | 9 358 67 |
:---------------+---------------+---------------:
| 7 5 6 | 9 4 3 | 8 1 2 |
| 1 9 3 | 8 7 2 | 6 4 5 |
| 2 4 8 | 16 5 16 | 7 9 3 |
:---------------+---------------+---------------:
| 9 37 57 | 35 2 8 | 4 6 1 |
| 58 6 2 | 4 1 59 | 3 7 89 |
| 4 38 1 | 36 69 7 | 5 2 89 |
'---------------'---------------'---------------' |
where we do a little 3D-multicoloring (but only using 2 colors) (I started from r9c4) which reveals the following AIC:
3) (6=3)r9c4 - (3=5)r7c4 - (5)r7c3 = (5-7)r3c3 = (7-6)r3c9 = (6)r3c5. This implies r9c5 <> 6, solving the puzzle. |
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