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Captain Pete
Joined: 09 Jun 2007
Posts: 55
Location: Oley, PA
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 Posted: Thu Nov 15, 2007 3:57 pm Post subject: Menneske Super Hard - Help please |
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I have never solved one of these super-hard puzzles. Can someone please help me with the next number on this one?
| Code: |
+----------+--------+------------+
| 4 56 3 | 8 15 2 | 69 79 167 |
| 29 59 7 | 6 15 4 | 3 8 12 |
| 28 1 68 | 9 3 7 | 56 4 256 |
+----------+--------+------------+
| 5 3 2 | 7 4 9 | 1 6 8 |
| 7 8 9 | 3 6 1 | 2 5 4 |
| 6 4 1 | 5 2 8 | 7 3 9 |
+----------+--------+------------+
| 89 2 68 | 4 7 3 | 569 1 56 |
| 3 67 4 | 1 9 5 | 8 2 67 |
| 1 79 5 | 2 8 6 | 4 79 3 |
+----------+--------+------------+
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Play this puzzle online at the Daily Sudoku site |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Thu Nov 15, 2007 4:57 pm Post subject: |
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| Code: | +-------------+-------------+-------------+
| 4 56 3 | 8 15 2 | 69 79 167 |
| 29 59 7 | 6 15 4 | 3 8 12 |
| 28 1 68# | 9 3 7 | 56@ 4 256%|
+-------------+-------------+-------------+
| 5 3 2 | 7 4 9 | 1 6 8 |
| 7 8 9 | 3 6 1 | 2 5 4 |
| 6 4 1 | 5 2 8 | 7 3 9 |
+-------------+-------------+-------------+
| 89 2 68# | 4 7 3 | 569%1 56@ |
| 3 67 4 | 1 9 5 | 8 2 67 |
| 1 79 5 | 2 8 6 | 4 79 3 |
+-------------+-------------+-------------+ |
The W-wing <56> @ has the link on <6> #, so you can eliminate <5> in %.
Then, you have a Bug+1: R1C9 must be <6>.
Keith |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Thu Nov 15, 2007 4:59 pm Post subject: Re: Menneske Super Hard - Help please |
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| Captain Pete wrote: | I have never solved one of these super-hard puzzles. Can someone please help me with the next number on this one?
| Code: |
+----------+--------+------------+
| 4 56 3 | 8 15 2 | 69 79 167 |
| 29 59 7 | 6 15 4 | 3 8 12 |
| 28 1 68 | 9 3 7 | 56 4 256 |
+----------+--------+------------+
| 5 3 2 | 7 4 9 | 1 6 8 |
| 7 8 9 | 3 6 1 | 2 5 4 |
| 6 4 1 | 5 2 8 | 7 3 9 |
+----------+--------+------------+
| 89 2 68 | 4 7 3 | 569 1 56 |
| 3 67 4 | 1 9 5 | 8 2 67 |
| 1 79 5 | 2 8 6 | 4 79 3 |
+----------+--------+------------+
|
Play this puzzle online at the Daily Sudoku site |
There's a neat little XY-Wing with coloring. The Wing is pivoted in r1c2, 56-69-59. It doesn't remove anything. But start a chain on 9 from r2c2>>r2c9>>r9c8 and the new pincers now take out the 9 from r1c8 and r7c7. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Thu Nov 15, 2007 5:07 pm Post subject: |
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OR:
This is a BUG+3. The Deadly Pattern is:
| Code: | +-------------+-------------+-------------+
| 4 56 3 | 8 15 2 | 69 79 17 |
| 29 59 7 | 6 15 4 | 3 8 12 |
| 28 1 68 | 9 3 7 | 56 4 25 |
+-------------+-------------+-------------+
| 5 3 2 | 7 4 9 | 1 6 8 |
| 7 8 9 | 3 6 1 | 2 5 4 |
| 6 4 1 | 5 2 8 | 7 3 9 |
+-------------+-------------+-------------+
| 89 2 68 | 4 7 3 | 59 1 56 |
| 3 67 4 | 1 9 5 | 8 2 67 |
| 1 79 5 | 2 8 6 | 4 79 3 |
+-------------+-------------+-------------+ |
To avoid the DP, one of R1C9, R3C9, and R7C7 must be <6>. Which solves R3C7 and R7C9 as <5>, resulting in the previous BUG+1.
OR:
R37C79 are a Type-6 UR on <56>. R3C9 and R7C7 cannot be <5>, which leads once again the the BUG+1.
Keith |
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Johan
Joined: 25 Jun 2007
Posts: 206
Location: Bornem Belgium
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| Posted: Thu Nov 15, 2007 5:13 pm Post subject: |
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Pete,
To avoid the <56> DP in R37C79, there are 3 possible solutions :
1. R3C9=2
2. R7C7=9
3. Or both R3C9=2 AND R7C7=9
Suppose R7C7=9 => R1C7=6 => R3C7=5 => R3C9=2, which means
<2>can only reside in R3C9, solving the puzzle. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Thu Nov 15, 2007 5:14 pm Post subject: |
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AND:
If you don't like the BUG+1, coloring on <6>, or the finned X-wings on <6> will take out <6> in R1C7 and R3C9, solving it.
Keith |
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Captain Pete
Joined: 09 Jun 2007
Posts: 55
Location: Oley, PA
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| Posted: Thu Nov 15, 2007 6:46 pm Post subject: |
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Thanks to all who replied. Some of these procedures are beyond my experience level, so bear with me.
Marty R: I see the pincers in your XY-chain, but I can't follow the chain. If it begins on 9 in R2C2, you indicate that the next set in your chain is at R2C9. To continue the chain from R2C2, I think I should be looking for a 5, and R2C9 doesn't contain a 5. Am I missing something?
Keith: I like your discovery of the UR. I need to study the DP a bit more because I don't follow your logic there.
Johan: Brilliant!
Coloring: Is there a method to narrowing the coloring candidates? Nine candidates in nine blocks seems to someone new to coloring to provide too many options. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Thu Nov 15, 2007 7:06 pm Post subject: |
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| Quote: | | Marty R: I see the pincers in your XY-chain, but I can't follow the chain. If it begins on 9 in R2C2, you indicate that the next set in your chain is at R2C9. To continue the chain from R2C2, I think I should be looking for a 5, and R2C9 doesn't contain a 5. Am I missing something? |
Pete,
You can see that the pincers (which don't eliminate anything) in the original XY-Wing are on 9, i.e., one or the other must be =9. My little chain from r2c2-->r9c2-->r9c9 is based on strong links on 9. So if r2c2 =9, then r9c8 must be =9. Thus, we can now say that either r9c8 or r1c7 must be =9 and that new pincer combination allows the eliminations.
Hope I was able to make that clear enough. |
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Captain Pete
Joined: 09 Jun 2007
Posts: 55
Location: Oley, PA
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| Posted: Thu Nov 15, 2007 8:55 pm Post subject: |
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Marty,
Thanks for the help. I think I have it.
Pete |
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re'born
Joined: 28 Oct 2007
Posts: 80
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| Posted: Fri Nov 16, 2007 12:33 am Post subject: |
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| keith wrote: | OR:
This is a BUG+3.
To avoid the DP, one of R1C9, R3C9, and R7C7 must be <6>. Which solves R3C7 and R7C9 as <5>, resulting in the previous BUG+1.
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For the exact same reason that you can eliminate 6 from r3c7 and r7c9, you can also eliminate 6 from r1c7, and then the puzzle solves with singles.
| keith wrote: |
OR:
R37C79 are a Type-6 UR on <56>. R3C9 and R7C7 cannot be <5>, which leads once again the the BUG+1.
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Or you can go the long way and avoid the BUG+1 at the end:
(2)r3c9 = (9)r7c7 - (9=6)r1c7 - (6=5)r1c2 - (5=9)r2c2 - (9=2)r2c1, => r2c9<>2, solving the puzzle. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Fri Nov 16, 2007 2:43 am Post subject: Original puzzle? |
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Pete,
This turns out to be a very good example! Can you please post the original puzzle? The starting point?
Thank you,
Keith |
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Earl
Joined: 30 May 2007
Posts: 677
Location: Victoria, KS
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| Posted: Fri Nov 16, 2007 3:29 pm Post subject: Difficult |
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Keith,
I see how coloring or a finned X-wing on <6> eliminates the 6 from R3C9, but I dont grasp how it also clears R1C7. I had to use an x-y chain to clear the 8 from R3C3.
Earl |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Fri Nov 16, 2007 6:10 pm Post subject: |
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Earl,
You have to see both eliminations first:
If R1C7 is the "fin", it eliminates <6> in R3C9. If R3C9 is the fin, it eliminates <6> in R1C7. So, you can eliminate <6> in both of them.
If you eliminate just one of them, you then have a regular X-wing that eliminates the other.
Keith |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Fri Nov 16, 2007 7:12 pm Post subject: |
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Like Earl, I fail to see the second x-wing elimination.
With 6 present in r7c9 ("56"), there are three sixes in r7.
Where is the x-wing used to take out the 6 in r1c7?
After the r3c9 elimination I went for the BUG+2 pattern which solved the puzzle, but now I was trying to do it the other way.
Finned other-something maybe? |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Fri Nov 16, 2007 8:19 pm Post subject: |
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| re'born wrote: | Or you can go the long way and avoid the BUG+1 at the end:
(2)r3c9 = (9)r7c7 - (9=6)r1c7 - (6=5)r1c2 - (5=9)r2c2 - (9=2)r2c1, => r2c9<>2, solving the puzzle. |
It seems to me there is a short-cut:
(2)R3C9=(9)R7C7, but
(2=6)R3C9-(6=7)R8C9-(7=9)R8C8
Thus, R3C9=2 (since R7C7 and R8C8 can't both be <9>).
For those who may be puzzled by (2)R3C9=(9)R7C7, it is the strong inferential link caused by the {56} Deadly Pattern: R3C9 must be <2> and/or R7C7 must be <9>. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Fri Nov 16, 2007 9:31 pm Post subject: |
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Here is the BUG+1 situation:
| Code: | +-------------+-------------+-------------+
| 4 56@ 3 | 8 15 2 | 69# 79 167@|
| 29 59 7 | 6 15 4 | 3 8 12 |
| 28 1 68 | 9 3 7 | 5 4 26% |
+-------------+-------------+-------------+
| 5 3 2 | 7 4 9 | 1 6 8 |
| 7 8 9 | 3 6 1 | 2 5 4 |
| 6 4 1 | 5 2 8 | 7 3 9 |
+-------------+-------------+-------------+
| 89 2 68 | 4 7 3 | 69 1 5 |
| 3 67@ 4 | 1 9 5 | 8 2 67@ |
| 1 79 5 | 2 8 6 | 4 79 3 |
+-------------+-------------+-------------+ |
Take a look at the rectangle R18C29, @. It is almost an X-wing on <6> in the rows, except for R1C7 #, and in the columns, except for R3C9 %.
The finned X-wing: ("true" means has a candidate <6>.)
In the rows: Either the X-wing @ is true, or the fin # is true. Either way, % is not true.
In the columns: Either the X-wing @ is true, or the fin % is true. Either way, # is not true.
Each fin eliminates the other. This kind of situation is quite common, I believe. Check your finned fish carefully before making an elimination!*
In this case, if you make either one of the finned eliminations, the result is a true X-wing which makes the second elimination.
Keith
* The same rule applies to Unique Rectangles. Often, a UR is of more than one type. But, the eliminations for one type may destroy the other. Compare my Type-6 UR (above in this thread) with Johan's much more careful (and effective) reasoning. |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Fri Nov 16, 2007 9:58 pm Post subject: |
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Thx, keith for the diagram.
I was so fixed on the r37c37 x-wing that I didn't see the r18c29 one. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Fri Nov 16, 2007 10:15 pm Post subject: |
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(I apologize if I was confusing people by talking about the BUG+1 situation, when only the BUG+3 was posted.)
| Quote: | | Coloring: Is there a method to narrowing the coloring candidates? Nine candidates in nine blocks seems to someone new to coloring to provide too many options. |
Pete,
| Code: | +-------------+-------------+-------------+
| 4 56a 3 | 8 15 2 | 69a 79 167 |
| 29 59 7 | 6 15 4 | 3 8 12 |
| 28 1 68A | 9 3 7 | 5 4 26a |
+-------------+-------------+-------------+
| 5 3 2 | 7 4 9 | 1 6 8 |
| 7 8 9 | 3 6 1 | 2 5 4 |
| 6 4 1 | 5 2 8 | 7 3 9 |
+-------------+-------------+-------------+
| 89 2 68a | 4 7 3 | 69A 1 5 |
| 3 67A 4 | 1 9 5 | 8 2 67a |
| 1 79 5 | 2 8 6 | 4 79 3 |
+-------------+-------------+-------------+
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In this case it is kind of obvious that <6> is the candidate for coloring. Take a look at the number of unsolved cells:
<1> 4
<2> 4
<3> 0
<4> 0
<5> 4
<6> 9
<7> 6
<8> 4
<9> 8
<7> is a swordfish, <9> is a jellyfish, each using all the unsolved cells, so there are no eliminations.
0 and 4 unsolved cells are similarly uninteresting. Leaving <6>.
Label all the cells that are connected by strong links alternately as "A' or "a". Either "A" or "a" is true. Clearly, "a" is false, look at R1 and C9.
Keith |
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