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March 13 DB

 
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Earl



Joined: 30 May 2007
Posts: 625
Location: Victoria, KS

PostPosted: Sat Mar 13, 2010 3:43 pm    Post subject: March 13 DB Reply with quote

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 . . | . . . |
+-------+-------+-------+

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: 3344
Location: near Detroit, Michigan, USA

PostPosted: Sat Mar 13, 2010 6:00 pm    Post subject: Reply with quote

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

PostPosted: Sat Mar 13, 2010 6:16 pm    Post subject: Reply with quote

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: 3344
Location: near Detroit, Michigan, USA

PostPosted: Sat Mar 13, 2010 7:09 pm    Post subject: Reply with quote

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

PostPosted: Sat Mar 13, 2010 8:29 pm    Post subject: Reply with quote

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: 3344
Location: near Detroit, Michigan, USA

PostPosted: Sat Mar 13, 2010 9:28 pm    Post subject: Reply with quote

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

PostPosted: Sat Mar 13, 2010 9:50 pm    Post subject: Reply with quote

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: 3344
Location: near Detroit, Michigan, USA

PostPosted: Sun Mar 14, 2010 7:51 am    Post subject: Reply with quote

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.) Wink
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Mogulmeister



Joined: 03 May 2007
Posts: 703

PostPosted: Sun Mar 14, 2010 11:24 pm    Post subject: Reply with quote

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

PostPosted: Mon Mar 15, 2010 2:22 am    Post subject: Reply with quote

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? Very Happy

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: 703

PostPosted: Mon Mar 15, 2010 6:29 am    Post subject: Reply with quote

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

PostPosted: Mon Mar 15, 2010 7:20 am    Post subject: Reply with quote

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. Sad
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Mogulmeister



Joined: 03 May 2007
Posts: 703

PostPosted: Mon Mar 15, 2010 10:39 am    Post subject: Reply with quote

Indeed. Smile
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Steve R



Joined: 24 Oct 2005
Posts: 289
Location: Birmingham, England

PostPosted: Mon Mar 15, 2010 7:13 pm    Post subject: Reply with quote

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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