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Earl
Joined: 30 May 2007
Posts: 677
Location: Victoria, KS
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 Posted: Sat Mar 13, 2010 3:43 pm Post subject: March 13 DB |
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The March 13 DB is tempting.
A one-step solution: an xy-wing (267) is tempting, but does not solve. An xy-chain, eliminating the (1) in R4C1 solves.
Earl
of the Chain Gang
| Code: |
+-------+-------+-------+
| . . . | . . 9 | 7 3 . |
| . 2 1 | . 6 . | 5 4 . |
| . 3 . | . . . | . . . |
+-------+-------+-------+
| . . 3 | 8 . . | . . . |
| 5 . . | 6 2 7 | . . 3 |
| . . . | . . 3 | 4 . . |
+-------+-------+-------+
| . . . | . . . | . 6 . |
| . 4 2 | . 3 . | 1 5 . |
| . 8 5 | 1 . . | . . . |
+-------+-------+-------+
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Play this puzzle online at the Daily Sudoku site
Last edited by Earl on Sat Mar 13, 2010 7:09 pm; edited 1 time in total |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sat Mar 13, 2010 6:00 pm Post subject: |
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After basics:
| Code: | +----------------+----------------+----------------+
| 8 5 6 | 24 14 9 | 7 3 12 |
| 7 2 1 | 3 6 8 | 5 4 9 |
| 4 3 9 | 25 7 125 | 26 8 126 |
+----------------+----------------+----------------+
| 12A 67 3 | 8 145 14C | 26 9 2567 |
| 5 9 4 | 6 2 7 | 8 1 3 |
| 12 67 8 | 9 15 3 | 4 27 2567 |
+----------------+----------------+----------------+
| 3 1 7 | 245 8 245 | 9 6 24 |
| 9 4 2 | 7 3 6 | 1 5 8 |
| 6 8 5 | 1 9 24B | 3 27 247 |
+----------------+----------------+----------------+ |
ABC is a flightless XY-wing. Any cell that sees A and B is not 2. Coloring on 2 starting in B:
| Code: | +----------------+----------------+----------------+
| 8 5 6 | 24 14 9 | 7 3 12b |
| 7 2 1 | 3 6 8 | 5 4 9 |
| 4 3 9 | 25 7 125 | 26B 8 126b |
+----------------+----------------+----------------+
| 12A 67 3 | 8 145 14C | 26b 9 -2567b|
| 5 9 4 | 6 2 7 | 8 1 3 |
|1-2 67 8 | 9 15 3 | 4 27B 2567b|
+----------------+----------------+----------------+
| 3 1 7 | 245 8 245 | 9 6 24B |
| 9 4 2 | 7 3 6 | 1 5 8 |
| 6 8 5 | 1 9 24B | 3 27b 247b |
+----------------+----------------+----------------+ |
Solving R6C1 solves the puzzle.
Keith |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sat Mar 13, 2010 6:16 pm Post subject: |
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| Code: | ABC is a flightless XY-wing. Any cell that sees A and B is not 2. Coloring on 2 starting in B:
Code:
+----------------+----------------+----------------+
| 8 5 6 | 24 14 9 | 7 3 12b |
| 7 2 1 | 3 6 8 | 5 4 9 |
| 4 3 9 | 25 7 125 | 26B 8 126b |
+----------------+----------------+----------------+
| 12A 67 3 | 8 145 14C | 26b 9 -2567b|
| 5 9 4 | 6 2 7 | 8 1 3 |
|1-2 67 8 | 9 15 3 | 4 27B 2567b|
+----------------+----------------+----------------+
| 3 1 7 | 245 8 245 | 9 6 24B |
| 9 4 2 | 7 3 6 | 1 5 8 |
| 6 8 5 | 1 9 24B | 3 27b 247b |
+----------------+----------------+----------------+ |
Not surprisingly, I don't follow the coloring. However, r4c1 can be transported to r3c7, taking out the 2 from r3c6. After a few eliminations therefrom, r9c6 can be transported to r7c9, taking out the 2 from r1c9. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sat Mar 13, 2010 7:09 pm Post subject: |
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| Quote: | | Not surprisingly, I don't follow the coloring. |
Code: If B in R9C6 is true (=2), b is not true.
Step 1: If R9C6 is B, R8C9 is B, forming pincers on R4C9.
Step 2: Then R3C7 is B.
Step 3: Then R6C8 is B, forming pincers on R6C1.
Keith |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sat Mar 13, 2010 8:29 pm Post subject: |
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| Thanks, that's easy enough, I just got distracted by the lower case b's. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sat Mar 13, 2010 9:28 pm Post subject: |
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| Quote: | | r4c1 can be transported to r3c7 |
So can R9C6!
Last edited by keith on Sat Mar 13, 2010 10:41 pm; edited 1 time in total |
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arkietech
Joined: 31 Jul 2008
Posts: 1834
Location: Northwest Arkansas USA
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| Posted: Sat Mar 13, 2010 9:50 pm Post subject: |
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This AIC led to a one step solution: | Code: | *-----------------------------------------------------------*
| 8 5 6 | 24 14 9 | 7 3 12 |
| 7 2 1 | 3 6 8 | 5 4 9 |
| 4 3 9 | 25 7 125 | 26 8 126 |
|-------------------+-------------------+-------------------|
| 12 67 3 | 8 145 14 | 26 9 2567 |
| 5 9 4 | 6 2 7 | 8 1 3 |
| 12 67 8 | 9 15 3 | 4 27 2567 |
|-------------------+-------------------+-------------------|
| 3 1 7 | 245 8 245 | 9 6 24 |
| 9 4 2 | 7 3 6 | 1 5 8 |
| 6 8 5 | 1 9 24 | 3 27 247 |
*-----------------------------------------------------------*
(2=4)r7c9-r7c4=(4-2)r1c4=r1c9-r3c7=r4c7-(2=1)r4c1-(1=4)r4c6-(4=2)r9c6;
r7c46,r9c89<>2
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sun Mar 14, 2010 7:51 am Post subject: |
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| keith wrote: | | Quote: | | r4c1 can be transported to r3c7 |
So can R9C6! |
I can augment the coloring:
| Code: | +----------------+----------------+----------------+
| 8 5 6 | 24B 14 9 | 7 3 12b |
| 7 2 1 | 3 6 8 | 5 4 9 |
| 4 3 9 | 25b 7 125b | 26B 8 126b |
+----------------+----------------+----------------+
| 12AB 67 3 | 8 145 14C | 26b 9 -2567b|
| 5 9 4 | 6 2 7 | 8 1 3 |
|1-2b 67 8 | 9 15 3 | 4 27B 2567b|
+----------------+----------------+----------------+
| 3 1 7 | 245b 8 245b | 9 6 24B |
| 9 4 2 | 7 3 6 | 1 5 8 |
| 6 8 5 | 1 9 24B* | 3 27b 247b |
+----------------+----------------+----------------+ |
Code: If B in R9C6* is true (=2), cells marked b are not true, cells marked B are true.
If you can extend (color) one pincer (here, B*), such that if it is true the other pincer (here, A) is also true, then both are true.
This grouped coloring should be applied with care. Note that in this case, for single digit coloring, B* false does not imply true nor false for A, and neither true nor false for A allows any conclusion on B*.
Keith
(And, yes, you are up late at night just waiting to read this factoid.)  |
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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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| Posted: Sun Mar 14, 2010 11:24 pm Post subject: |
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What to me is interesting in Dan's AIC is that the pincers for 2 are at
r7c9 and r9c6 and the proposition is:
"If r7c9 is set to 4 (or NOT 2) then r9c6 is 2." So 2 will be in at least one of those cells and so both 2s in r9c89 are toast. End of.
The "sudophilosophic" moment comes when you do a quick inspection and say:
If r7c9 is 4 then r7c45 can not be 4 So r9c6 is 4 and no 2's are eliminated.
A conundrum to the casual observer. |
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arkietech
Joined: 31 Jul 2008
Posts: 1834
Location: Northwest Arkansas USA
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| Posted: Mon Mar 15, 2010 2:22 am Post subject: |
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| Mogulmeister wrote: | What to me is interesting in Dan's AIC is that the pincers for 2 are at
r7c9 and r9c6 and the proposition is:
"If r7c9 is set to 4 (or NOT 2) then r9c6 is 2." So 2 will be in at least one of those cells and so both 2s in r9c89 are toast. End of.
The "sudophilosophic" moment comes when you do a quick inspection and say:
If r7c9 is 4 then r7c45 can not be 4 So r9c6 is 4 and no 2's are eliminated.
A conundrum to the casual observer. |
(2=4)r7c9-r7c4=(4-2)r1c4=r1c9-r3c7=r4c7-(2=1)r4c1-(1=4)r4c6-(4=2)r9c6;
r7c46,r9c89<>2
The logic above says that one or both of the cells r7c9 or r9c6 is a 2.
if r7c9 is a 2 all other row 7 2's are toast. All other box 9 2's are also history.
if r9c6 is a 2 other row 9 2's are gone and other box 8 2's are outta here.
As your logic indicates a casual observer would have to say neither can be 4. Isn't this fun? 
it would be better to say:
(2=4)r7c9-r7c4=(4-2)r1c4=r1c9-r3c7=r4c7-(2=1)r4c1-(1=4)r4c6-(4=2)r9c6;
r7c9,r9c6=2 or <>4 |
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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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| Posted: Mon Mar 15, 2010 6:29 am Post subject: |
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Indeed Dan r7c9 and r9c6 can never be 4, but to the casual observer who only operates in boxes 8 and 9, starting with the "not 2" argument is the interesting conundrum:
Setting r7c9 to 4 (or NOT 2) puts a 4 in r9c6
Setting r9c6 to 4 (or NOT 2) puts a 4 in r7c9
Neither case appears to eliminate the 2s in r9c89. |
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arkietech
Joined: 31 Jul 2008
Posts: 1834
Location: Northwest Arkansas USA
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| Posted: Mon Mar 15, 2010 7:20 am Post subject: |
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| Mogulmeister wrote: | Indeed Dan r7c9 and r9c6 can never be 4, but to the casual observer who only operates in boxes 8 and 9, starting with the "not 2" argument is the interesting conundrum:
Setting r7c9 to 4 (or NOT 2) puts a 4 in r9c6
Setting r9c6 to 4 (or NOT 2) puts a 4 in r7c9
Neither case appears to eliminate the 2s in r9c89. |
Shows you what "plugging" in numbers will do.  |
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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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| Posted: Mon Mar 15, 2010 10:39 am Post subject: |
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Indeed.  |
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Steve R
Joined: 24 Oct 2005
Posts: 289
Location: Birmingham, England
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| Posted: Mon Mar 15, 2010 7:13 pm Post subject: |
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Keith’s grid may be written as:
| Code: | +----------------+------------------+----------------+
| 8 5 6 | 24 14 9 | 7 3 12 |
| 7 2 1 | 3 6 8 | 5 4 9 |
| 4 3 9 | 25+ 7 25+1 | 26 8 126 |
+----------------+------------------+----------------+
| 12 67 3 | 8 145 14 | 26 9 2567 |
| 5 9 4 | 6 2 7 | 8 1 3 |
| 12 67 8 | 9 15 3 | 4 27 2567 |
+----------------+------------------+----------------+
| 3 1 7 | 25+4 8 25+4 | 9 6 24 |
| 9 4 2 | 7 3 6 | 1 5 8 |
| 6 8 5 | 1 9 24 | 3 27 247 |
+----------------+------------------+----------------+ |
This version brings out the UR in r37c46. If r3c6 contains 1, r4c6 contains 4. If r3c6 does not contain 1, r7c46 contains 4. So 4 may be eliminated from r9c6 and the puzzle is solved.
Steve |
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