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boweasel
Joined: 15 May 2006 Posts: 5
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Posted: Mon May 15, 2006 8:34 pm Post subject: Black Belt Sudoku (2nd degree) - Frank Longo |
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Picked up his book the other day, and am having difficulty with this puzzle. Is there a technique beyond guessing I can use to solve it?
Code: | 2 3 6/7 4 8 6/7 9 1/5 1/5
4/9 4/9 5 3 1 2 6/8 7 6/8
6/8 1/8 1/6/7 9 6/7 5 4 3 2
4/5/9 4/9 3/8 1/5 2 4/8 1/3/5 6 7
4/5 6 2 1/5/7 3 4/7 1/5 8 9
1 7 3/8 5/6 9 6/8 2/3/5 2/4/5 4/5
7 2 9 8 5/6 3 5/6 1/4 1/4
3 1/5/8 1/6 2/6 4 9 7 2/5 6/8
6/8 5/8 4 2/6/7 5/6/7 1 2/5/6/7 9 3 | [/ |
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Marty R.
Joined: 12 Feb 2006 Posts: 5770 Location: Rochester, NY, USA
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Posted: Mon May 15, 2006 9:19 pm Post subject: |
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In row 9, the only possibilities for "8" are in box 7, therefore, the "8" cannot be in r8c2. That solves r8c9, since that is the only remaining "8" in that row. That will help you solve more cells, but I don't know that that's enough to complete the puzzle. The "7" can't be in r9c7, as that number is already solved in that column, but that's of little help.
There is a unique rectangle in rows 2 and 4, columns 1 and 2. I don't know you or what techniques you're familiar with, but Keith wrote an excellent primer on rectangles here:
http://www.dailysudoku.co.uk/sudoku/forums/viewtopic.php?t=744
I think using that rectangle to solve a cell might break the puzzle open.
Good luck. |
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Guest
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Posted: Tue May 16, 2006 12:05 am Post subject: |
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Somehow I knew I would screw up this entry...
Row 9 col 7 should read 2/5/6/8. I hope that's the only typo I made.
And why do I get the msg 'sorry but this username has already been taken'?
Code: | 2 3 6/7 4 8 6/7 9 1/5 1/5
4/9 4/9 5 3 1 2 6/8 7 6/8
6/8 1/8 1/6/7 9 6/7 5 4 3 2
4/5/9 4/9 3/8 1/5 2 4/8 1/3/5 6 7
4/5 6 2 1/5/7 3 4/7 1/5 8 9
1 7 3/8 5/6 9 6/8 2/3/5 2/4/5 4/5
7 2 9 8 5/6 3 5/6 1/4 1/4
3 1/5/8 1/6 2/6 4 9 7 2/5 6/8
6/8 5/8 4 2/6/7 5/6/7 1 2/5/6/8 9 3
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David Bryant
Joined: 29 Jul 2005 Posts: 559 Location: Denver, Colorado
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Posted: Tue May 16, 2006 12:15 am Post subject: There are a couple of XY-Wings |
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Code: | 2 3 6/7 4 8 6/7 9 1/5 1/5
4/9 4/9 5 3 1 2 6/8 7 6/8
6/8 1/8 1/6/7 9 6/7 5 4 3 2
4/5/9 4/9 3/8 1/5 2 4/8 1/3/5 6 7
4/5 6 2 1/5/7 3 4/7 1/5 8 9
1 7 3/8 5/6 9 6/8 2/3/5 2/4/5 4/5
7 2 9 8 5/6 3 5/6 1/4 1/4
3 1/5/8 1/6 2/6 4 9 7 2/5 6/8
6/8 5/8 4 2/6/7 5/6/7 1 2/5/6/7 9 3 |
There are a couple of fairly obvious XY-Wings, also, at r8c4.
-- r8c4 = 2 ==> r8c8 = 5
-- r8c4 = 6 ==> r6c4 = 5
Therefore r6c8 <> 5.
-- r8c4 = 2 ==> r8c8 = 5
-- r8c4 = 6 ==> r7c5 = 5
Therefore r7c7 <> 5, and we must have r7c7 = 6.
There's also a coloring argument that allows one to eliminate the "6" at r9c4 -- if r3c1 = 6, then r6c4 = 6; and if r3c1 <> 6, then r9c1 = 6. Either way, there can't be a "6" at r9c4.
Unfortunately, these aren't enough to solve the puzzle. For that you either need to fall back on the "UR" that Marty pointed out, or use some form of a forcing chain. After following up on the XY-Wings there's a cute "double-implication chain" lurking in r6c9 ...
A. r6c9 = 5 ==> r6c4 = 6.
B. r6c9 = 4 ==> r6c8 = 2 ==> r8c4 = 2 ==> r9c4 = 7 ==> {1, 5} pair in column 4 ==> r6c4 = 6.
Setting r6c4 = 6 cracks the puzzle. dcb
Last edited by David Bryant on Tue May 16, 2006 3:01 pm; edited 1 time in total |
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Marty R.
Joined: 12 Feb 2006 Posts: 5770 Location: Rochester, NY, USA
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Posted: Tue May 16, 2006 12:56 am Post subject: |
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Quote: | Therefore r6c7 <> 5, and we must have r6c7 = 6. |
Just a minor typo, David. I think you meant r7c7. |
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David Bryant
Joined: 29 Jul 2005 Posts: 559 Location: Denver, Colorado
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Posted: Tue May 16, 2006 3:00 pm Post subject: Oops! I made a mistake. |
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You're right, Marty -- it should be r7c7. Thanks for pointing that out. I've edited the original post. dcb |
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Guest
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Posted: Wed May 17, 2006 1:49 am Post subject: |
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Wow... unique rectangles! I never would have thought of using a technique based on the principle that the Sudoku puzzle solution must be unique. It just never would've occurred to me.
Using UR enabled me to easily solve the puzzle. Thanks, Marty.
And David, I appreciate your input, although your sentence:
"There's also a coloring argument that allows one to eliminate the "6" at r9c4 -- if r3c1 = 6, then r6c4 = 6; and if r3c1 <> 6, then r9c1 = 6. Either way, there can't be a "6" at r9c4."
is over my head. I mean if I set r3c1 = 6 I have go through a significant number of cells before I finally get to where r6c4 = 6. I'm sure there's simply something I'm not grasping.... |
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David Bryant
Joined: 29 Jul 2005 Posts: 559 Location: Denver, Colorado
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Posted: Wed May 17, 2006 4:00 pm Post subject: Answering Boweasel's questions |
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Boweasel wrote: | ... why do I get the msg 'sorry but this username has already been taken'? |
I think that you've registered. Now you just need to "Log In" to be "Boweasel" on this board. If you don't do that, and you post a message as a Guest, the system won't let you use the registered name.
Oh -- you may find it handy to set the "Log me in automatically" check box on the index page (bottom right -- scroll down to find it). That way you won't have to log in each time you visit the forum.
Boweasel wrote: | ... if I set r3c1 = 6 I have go through a significant number of cells before I finally get to where r6c4 = 6. I'm sure there's simply something I'm not grasping.... |
Yes, you need to learn about "binary chains". You can run a search on the word "coloring" in this forum to find quite a few articles on the subject. For now I'll just illustrate how "coloring" works on this puzzle.
Code: | 2 3 6/7+ 4 8 6/7- 9 1/5 1/5
4/9 4/9 5 3 1 2 6/8 7 6/8
6/8= 1/8 1/6/7 9 6/7 5 4 3 2
4/5/9 4/9 3/8 1/5 2 4/8 1/3/5 6 7
4/5 6 2 1/5/7 3 4/7 1/5 8 9
1 7 3/8 5/6- 9 6/8+ 2/3/5 2/4/5 4/5
7 2 9 8 5/6 3 5/6 1/4 1/4
3 1/5/8 1/6 2/6 4 9 7 2/5 6/8
6/8~ 5/8 4 2/6/7 5/6/7 1 2/5/6/8 9 3 |
There are two binary chains on the "6"s at this stage in the puzzle. Notice that there are only two ways to enter a "6" in column 1 -- at r3c1, or at r9c1. I've marked these two cells "=/~". This is the first binary chain.
There are also only two ways to enter a "6" in row 1, two ways to enter a "6" in column 6, and two ways to enter a "6" in row 6. The four cells r1c3, r1c6, r6c6, and r6c4 constitute the second binary chain -- I've marked these "+/-" in the grid above.
To rule out the "6" at r9c4 we reason as follows.
-- If r9c1 = 6 then r9c4 can't be a 6.
-- If r9c1 is not a "6" then r3c1 must be a "6". But then r1c3 cannot be 6 (same 3x3 box as r3c1), r1c6 must be a 6 (only spot left in row 1), r6c6 cannot be a 6 (because of 6 at r1c6), and r6c4 must be a 6 (only spot left in row 6). But r6c4 = 6 ==> r9c4 is not 6.
I hope that makes sense -- "coloring" is a powerful technique. dcb |
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boweasel
Joined: 15 May 2006 Posts: 5
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Posted: Wed May 17, 2006 8:32 pm Post subject: |
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Yes. thanks David. It does make sense, although there certainly seems to be more of an educated 'trial and error' quality to coloring than to other logic techniques such as XY Wings, and it might be that very quality that makes me unlikely to use it unless all else fails. |
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David Bryant
Joined: 29 Jul 2005 Posts: 559 Location: Denver, Colorado
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Posted: Wed May 17, 2006 10:16 pm Post subject: "X-Wing" and "swordfish" are coloring |
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Boweasel wrote: | ... there certainly seems to be more of an educated 'trial and error' quality to coloring than to other logic techniques such as XY Wings ... |
Maybe so. I don't quite agree, though.
The rather standard techniques known as "X-Wing" and "swordfish" are in fact just special cases of coloring, as the following diagrams demonstrate. Do those techniques strike you as "trial and error"? dcb
Code: | X-Wing Swordfish
. . . . . . . . . . . . . . . . . .
. . . . . . . . . . S+ . . . . S- . .
. . . X+ . . X- . . . . . . . . . . .
. . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . S- . . S+ . .
. . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . .
. . . X- . . X+ . . . S- . S+ . . . . .
. . . . . . . . . . . . . . . . . . |
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ravel
Joined: 21 Apr 2006 Posts: 536
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Posted: Thu May 18, 2006 3:04 pm Post subject: |
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IMO (multiple) coloring is a systematic way to eliminate candidates. Just follow the well defined rules (color all strong linked cells for a candidate and make the eliminations at the end - if there are any)
So there is not more trial&error than for xy-wings: go to all bivalue cells and try to find appropriate wings. |
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Lone Cactus
Joined: 12 Jan 2009 Posts: 3 Location: Ranch en BCS Mexico
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Posted: Mon Jan 12, 2009 4:10 pm Post subject: |
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This is my first time every to post anything. I am working through Longo's book, Black Belt Sudoku. I am currently stumped with this one, Black Belt 2nd Degree Longo #76
If anyone could give me an idea what to try next I would appreciate it.
I have gotten this far:
[code]
+------------+----------------+------------+
| 5 3 9 | 68 4 68 | 7 1 2 |
| 4 12 8 | 12 7 5 | 3 6 9 |
| 26 7 126 | 3 9 12 | 8 5 4 |
+------------+----------------+------------+
| 3 8 12 | 9 26 246 | 14 7 5 |
| 9 12 5 | 247 8 247 | 6 34 13 |
| 7 6 4 | 5 1 3 | 9 2 8 |
+------------+----------------+------------+
| 1 5 37 | 2467 26 9 | 24 8 36 |
| 26 4 26 | 18 3 18 | 5 9 7 |
| 8 9 37 | 247 5 247 | 124 34 136 |
+------------+----------------+------------+
10-Q for any help. |
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wapati
Joined: 10 Jun 2008 Posts: 472 Location: Brampton, Ontario, Canada.
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Posted: Mon Jan 12, 2009 4:23 pm Post subject: |
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Your markup has 2 sixes eliminated that should not have been.
Try it from here.
Code: | .------------------.------------------.------------------.
| 5 3 9 | 68 4 68 | 7 1 2 |
| 4 12 8 | 12 7 5 | 3 6 9 |
| 26 7 126 | 3 9 12 | 8 5 4 |
:------------------+------------------+------------------:
| 3 8 12 | 9 26 246 | 14 7 5 |
| 9 12 5 | 247 8 247 | 6 34 13 |
| 7 6 4 | 5 1 3 | 9 2 8 |
:------------------+------------------+------------------:
| 1 5 37 | 2467 26 9 | 24 8 36 |
| 26 4 26 | 18 3 18 | 5 9 7 |
| 8 9 37 | 2467 5 2467 | 124 34 136 |
'------------------'------------------'------------------' |
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arkietech
Joined: 31 Jul 2008 Posts: 1834 Location: Northwest Arkansas USA
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Posted: Mon Jan 12, 2009 4:27 pm Post subject: |
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Lone Cactus wrote: | This is my first time every to post anything. I am working through Longo's book, Black Belt Sudoku. I am currently stumped with this one, Black Belt 2nd Degree Longo #76
If anyone could give me an idea what to try next I would appreciate it.
I have gotten this far:
[code]
+------------+----------------+------------+
| 5 3 9 | 68 4 68 | 7 1 2 |
| 4 12 8 | 12 7 5 | 3 6 9 |
| 26 7 126 | 3 9 12 | 8 5 4 |
+------------+----------------+------------+
| 3 8 12 | 9 26 246 | 14 7 5 |
| 9 12 5 | 247 8 247 | 6 34 13 |
| 7 6 4 | 5 1 3 | 9 2 8 |
+------------+----------------+------------+
| 1 5 37 | 2467 26 9 | 24 8 36 |
| 26 4 26 | 18 3 18 | 5 9 7 |
| 8 9 37 | 247 5 247 | 124 34 136 |
+------------+----------------+------------+
10-Q for any help. |
There only one place for a 6 in row 9
The puzzle still has problems. |
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Lone Cactus
Joined: 12 Jan 2009 Posts: 3 Location: Ranch en BCS Mexico
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Posted: Mon Jan 12, 2009 7:19 pm Post subject: |
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You are correct I took out the 6's in error,but unfortunately I am still stuck. Any hints or suggestions for my next move...
Code:
.------------------.------------------.------------------.
| 5 3 9 | 68 4 68 | 7 1 2 |
| 4 12 8 | 12 7 5 | 3 6 9 |
| 26 7 126 | 3 9 12 | 8 5 4 |
:------------------+------------------+------------------:
| 3 8 12 | 9 26 246 | 14 7 5 |
| 9 12 5 | 247 8 247 | 6 34 13 |
| 7 6 4 | 5 1 3 | 9 2 8 |
:------------------+------------------+------------------:
| 1 5 37 | 2467 26 9 | 24 8 36 |
| 26 4 26 | 18 3 18 | 5 9 7 |
| 8 9 37 | 2467 5 2467 | 124 34 136 |
'------------------'------------------'------------------'
The only way I can figure out to solve it is my picking an options in one of the cells and seeing if it works, but there must be some method I can apply to eliminate more candidates.
Lone Cactus/Jane |
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wapati
Joined: 10 Jun 2008 Posts: 472 Location: Brampton, Ontario, Canada.
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Posted: Mon Jan 12, 2009 8:16 pm Post subject: |
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There is a short way to finish using a Unique Rectangle.
Code: | .------------------.------------------.------------------.
| 5 3 9 | 68 4 68 | 7 1 2 |
| 4 12 8 | 12 7 5 | 3 6 9 |
|*26 7 1-2-6| 3 9 12 | 8 5 4 |
:------------------+------------------+------------------:
| 3 8 12 | 9 26 246 | 14 7 5 |
| 9 12 5 | 247 8 247 | 6 34 13 |
| 7 6 4 | 5 1 3 | 9 2 8 |
:------------------+------------------+------------------:
| 1 5 37 | 2467 26 9 | 24 8 36 |
|*26 4 *26 | 18 3 18 | 5 9 7 |
| 8 9 37 | 2467 5 2467 | 124 34 136 |
'------------------'------------------'------------------' |
There is a much longer way using several methods.
One available right now is a skyscraper on 4s.
Code: | .------------------.------------------.------------------.
| 5 3 9 | 68 4 68 | 7 1 2 |
| 4 12 8 | 12 7 5 | 3 6 9 |
| 26 7 126 | 3 9 12 | 8 5 4 |
:------------------+------------------+------------------:
| 3 8 12 | 9 26 *246 |*14 7 5 |
| 9 12 5 | 27-4 8 247 | 6 34 13 |
| 7 6 4 | 5 1 3 | 9 2 8 |
:------------------+------------------+------------------:
| 1 5 37 |*2467 26 9 |*24 8 36 |
| 26 4 26 | 18 3 18 | 5 9 7 |
| 8 9 37 | 2467 5 267-4| 124 34 136 |
'------------------'------------------'------------------' |
Also available is this remote pair on 12. You could also use ER for this elimination. Marked about the same with hinge at r2c6. It is also a W-wing.
Code: | .------------------.------------------.------------------.
| 5 3 9 | 68 4 68 | 7 1 2 |
| 4 *12 8 |*12 7 5 | 3 6 9 |
| 26 7 126 | 3 9 *12 | 8 5 4 |
:------------------+------------------+------------------:
| 3 8 12 | 9 26 246 | 14 7 5 |
| 9 *12 5 | 247 8 47-2 | 6 34 13 |
| 7 6 4 | 5 1 3 | 9 2 8 |
:------------------+------------------+------------------:
| 1 5 37 | 2467 26 9 | 24 8 36 |
| 26 4 26 | 18 3 18 | 5 9 7 |
| 8 9 37 | 2467 5 2467 | 124 34 136 |
'------------------'------------------'------------------' |
A little cleanup and now use sue de coq, xy-chain or ALS do do the marked eliminations. They all act the same here.
Code: | .---------------.-----------------.-----------------.
| 5 3 9 | 68 4 68 | 7 1 2 |
| 4 12 8 | 12 7 5 | 3 6 9 |
| 26 7 126 | 3 9 12 | 8 5 4 |
:---------------+-----------------+-----------------:
| 3 8 12 | 9 26 246 | 14 7 5 |
| 9 12 5 | 27 8 47 | 6 34 13 |
| 7 6 4 | 5 1 3 | 9 2 8 |
:---------------+-----------------+-----------------:
| 1 5 37 | 47-6 *26 9 |*24 8 *36 |
| 26 4 26 | 18 3 18 | 5 9 7 |
| 8 9 37 | 467 5 267 | 12-4 *34 136 |
'---------------'-----------------'-----------------' |
Finally we can use either an ALS or a chain, both using the marked candidate squares and having the same result.
Code: | .---------------.---------------.---------------.
| 5 3 9 | 68 4 68 | 7 1 2 |
| 4 12 8 | 12 7 5 | 3 6 9 |
| 26 7 126 | 3 9 12 | 8 5 4 |
:---------------+---------------+---------------:
| 3 8 12 | 9 #26 246 | 14 7 5 |
| 9 12 5 |#27 8 47 | 6 34 13 |
| 7 6 4 | 5 1 3 | 9 2 8 |
:---------------+---------------+---------------:
| 1 5 37 |*47 2-6 9 | 24 8 36 |
| 26 4 26 | 18 3 18 | 5 9 7 |
| 8 9 37 |*467 5 267 | 12 34 136 |
'---------------'---------------'---------------' |
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Lone Cactus
Joined: 12 Jan 2009 Posts: 3 Location: Ranch en BCS Mexico
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Posted: Wed Jan 14, 2009 11:36 pm Post subject: |
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Working on Longo 2nd Degree Back Belt, #80.
Have gotten to this point:
Code: |
+----------------+-----------------+----------------+
| 3679 2 3679 | 137 137 4 | 8 5 679 |
| 34679 49 8 | 237 2357 357 | 1 479 4679 |
| 47 1 5 | 6 9 8 | 2 47 3 |
+----------------+-----------------+----------------+
| 4689 3 469 | 1478 1467 167 | 5 2 4789 |
| 5 48 46 | 9 3467 2 | 37 3478 1 |
| 2 7 1 | 348 345 35 | 39 6 489 |
+----------------+-----------------+----------------+
| 1 45 347 | 2347 8 9 | 6 37 257 |
| 3789 589 379 | 1237 12367 1367 | 4 3789 25789 |
| 34789 6 2 | 5 347 37 | 379 1 789 |
+----------------+-----------------+----------------+ |
Draw/Play says the next logical step is a 5 in R7C2. Maybe I am just missing something obvious, but I can't see how you get the reduction to the lone 5 in R7C2. Can someone tell me what to look for. |
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keith
Joined: 19 Sep 2005 Posts: 3355 Location: near Detroit, Michigan, USA
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Posted: Thu Jan 15, 2009 12:22 am Post subject: |
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Lone Cactus wrote: | Working on Longo 2nd Degree Back Belt, #80.
Have gotten to this point:
Code: |
+----------------+-----------------+----------------+
| 3679 2 3679 | 137 137 4 | 8 5 679 |
| 34679 49 8 | 237 2357 357 | 1 479 4679 |
| 47 1 5 | 6 9 8 | 2 47 3 |
+----------------+-----------------+----------------+
| 4689 3 469 | 1478 1467 167 | 5 2 4789 |
| 5 48 46 | 9 3467 2 | 37 3478 1 |
| 2 7 1 | 348 345 35 | 39 6 489 |
+----------------+-----------------+----------------+
| 1 45 347 | 2347 8 9 | 6 37 257 |
| 3789 589 379 | 1237 12367 1367 | 4 3789 25789 |
| 34789 6 2 | 5 347 37 | 379 1 789 |
+----------------+-----------------+----------------+ |
Draw/Play says the next logical step is a 5 in R7C2. Maybe I am just missing something obvious, but I can't see how you get the reduction to the lone 5 in R7C2. Can someone tell me what to look for. |
I have no idea about R7C2.
There are pairs in C6 and C9. Then, look at <9> in R6.
In the resulting grid, there are two X-wings and an XYZ-wing.
Keith |
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Asellus
Joined: 05 Jun 2007 Posts: 865 Location: Sonoma County, CA, USA
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Posted: Thu Jan 15, 2009 11:54 pm Post subject: |
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I agree with Ravel about coloring: no trial and error is involved. The example above is "multi-coloring" because it involves more than one color "cluster", in this case two clusters: =/~ and +/-. The two clusters "see" each other in Box 1 such that + and = cannot both be true. (Some call this weak link connection the "bridge" between the two clusters.) This, in turn, means that - and ~ <6>s cannot both be false. So, any <6> that can see both a - <6> and a ~ <6> can be eliminated, such as the <6> at r9c4.
There is no need to assume that any particular <6> is true or false. |
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