Vanhegan Fiendish October 11

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Vanhegan Fiendish October 11

Postby ArkieTech » Fri Oct 11, 2013 1:27 am

Code: Select all
 *-----------*
 |.5.|1.3|.9.|
 |9..|546|..3|
 |..4|.9.|6..|
 |---+---+---|
 |63.|.5.|.89|
 |...|...|...|
 |89.|.2.|.64|
 |---+---+---|
 |..3|.1.|8..|
 |1..|438|..6|
 |.8.|6.5|.3.|
 *-----------*


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dan
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Re: Vanhegan Fiendish October 11

Postby Leren » Fri Oct 11, 2013 2:40 am

Code: Select all
*-----------------------------------------------------*
| 27   5    6     | 1    8    3     | 4    9    2-7   |
| 9    27   8     | 5    4    6     |d127  127  3     |
| 3    1    4     | 2    9    7     | 6    5    8     |
|-----------------+-----------------+-----------------|
| 6    3    12    | 7    5    4     | 12   8    9     |
| 25   4    1257  | 8    6    9     | 3    127 B1257  |
| 8    9    57    | 3    2    1     |c57   6    4     |
|-----------------+-----------------+-----------------|
| 57   6    3     | 9    1    2     | 8    4   a57    |
| 1    27   59    | 4    3    8     |b59   27   6     |
| 4    8    29    | 6    7    5     | 129  3    12    |
*-----------------------------------------------------*

M Wing Type 7 A or B (57):

(7=5) r7c9 - r8c7 = (5-7) r6c7 = r2c7 => - 7 r1c9; or

(7=5) r7c9 - r5c9 = (5-7) r6c7 = r2c7 => - 7 r1c9; stte

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Re: Vanhegan Fiendish October 11

Postby tlanglet » Fri Oct 11, 2013 1:04 pm

Digits 5 & 7 seem to offer a variety of choices. I initially spotted one of Leren's post but decided to find an alternate. Here is a messy one using the digits 5 & 7.
Code: Select all
 *-----------------------------------------------------------*
 |c27    5     6     | 1     8     3     | 4     9    b27    |
 | 9     27    8     | 5     4     6     | 127   127   3     |
 | 3     1     4     | 2     9     7     | 6     5     8     |
 |-------------------+-------------------+-------------------|
 | 6     3     12    | 7     5     4     | 12    8     9     |
 | 25    4     1257  | 8     6     9     | 3     12-7 *57=12 |
 | 8     9     57    | 3     2     1     |*57    6     4     |
 |-------------------+-------------------+-------------------|
 |d57    6     3     | 9     1     2     | 8     4     57    |
 | 1    e27    59    | 4     3     8     | 59   f27    6     |
 | 4     8     29    | 6     7     5     | 129   3    a12    |
 *-----------------------------------------------------------*

ANS(75=12)r6c7,r5c9
1: (75) r6c7,r5c9 => r5c8<>7
2: (12)r5c9 => LS(12)r59c9-(2=7)r1c9-r1c1=r7c1-r8c2=7r8c8 => r5c8<>7
I did not know how to notate this as a single statement with the "Almost" component forming a LS. Any suggestions?

I then looked for a solution not involving digits 5 & 7.
Code: Select all
 *-----------------------------------------------------------*
 | 27    5     6     | 1     8     3     | 4     9     27    |
 | 9    a27    8     | 5     4     6     |*127   127   3     |
 | 3     1     4     | 2     9     7     | 6     5     8     |
 |-------------------+-------------------+-------------------|
 | 6     3     12    | 7     5     4     |*12    8     9     |
 | 25    4     1257  | 8     6     9     | 3     127   1257  |
 | 8     9     57    | 3     2     1     | 57    6     4     |
 |-------------------+-------------------+-------------------|
 | 57    6     3     | 9     1     2     | 8     4     57    |
 | 1    b27    59    | 4     3     8     | 59    27    6     |
 | 4     8    c29    | 6     7     5     |d9-12  3     12    |
 *-----------------------------------------------------------*

ANS(12=7)r42c7-(7=2)r2c2-r8c2=(2-9)r9c3=9r9c7 => r9c7<>12

Ted
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Re: Vanhegan Fiendish October 11

Postby JC Van Hay » Fri Oct 11, 2013 2:35 pm

tlanglet wrote:ANS(75=12)r6c7,r5c9
1: (75) r6c7,r5c9 => r5c8<>7
2: (12)r5c9 => LS(12)r59c9-(2=7)r1c9-r1c1=r7c1-r8c2=7r8c8 => r5c8<>7
I did not know how to notate this as a single statement with the "Almost" component forming a LS. Any suggestions?
There are different "equivalent" ways to justify -7r5c8 :
1. Forcing Chain :
If r5c8=7->r8c2=7=r1c1,r1c9=2,r9c9=1; r5c9=5; r6c7 is empty :=> -7r5c8
This can be fully justified by writing down the following "Triangular Matrix" :
Code: Select all
7r5c8
7r8c8=7r8c2
      7r7c1=7r1c1
            7r1c9=2r1c9
                  2r9c9=1r9c9
7r5c9=============2r5c9=1r5c9=5r5c9
7r6c7=========================5r6c7
2. The Triangular Matrix may be written without any reference of the target :
Code: Select all
7r8c8=7r8c2
      7r7c1=7r1c1
            7r1c9=2r1c9
                  2r9c9=1r9c9
7r5c9=============2r5c9=1r5c9=5r5c9
7r6c7=========================5r6c7
In this case, theory shows that the first column is a derived strong set : 7r8c8=7r5c9=7r6c7 :=> -7r5c8
3. Finally, the Triangular Matrix may be rewritten as a Transfer Matrix :
Code: Select all
7r8c8=7r8c2
      7r7c1=7r1c1
            7r1c9=2r1c9
                  2r9c9=1r9c9
                  2r5c9=1r5c9=5r5c9=7r5c9
                              5r6c7=7r6c7
that is, in Eureka notation :

7r8c8=7r8c2-7r7c1=7r1c1-(7=2)r1c9-NP(12)r59c9=NP(57)r6c7,r5c9 :=> -7r5c8


Note : This "AIC" is readable from left to right and from right to left !

JC
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Re: Vanhegan Fiendish October 11

Postby Marty R. » Fri Oct 11, 2013 3:59 pm

Code: Select all
+------------+-------+--------------+
| 27 5  6    | 1 8 3 | 4   9   27   |
| 9  27 8    | 5 4 6 | 127 127 3    |
| 3  1  4    | 2 9 7 | 6   5   8    |
+------------+-------+--------------+
| 6  3  12   | 7 5 4 | 12  8   9    |
| 25 4  1257 | 8 6 9 | 3   127 1257 |
| 8  9  57   | 3 2 1 | 57  6   4    |
+------------+-------+--------------+
| 57 6  3    | 9 1 2 | 8   4   57   |
| 1  27 59   | 4 3 8 | 59  27  6    |
| 4  8  29   | 6 7 5 | 129 3   12   |
+------------+-------+--------------+

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M-Wing (7=5)r7c9-r5c9=(5-7)r6c7=r2c7=>r1c9<>7

M Wing Type 7 A or B


Leren, is there someplace that lists and defines these various types of M-Wings?
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Re: Vanhegan Fiendish October 11

Postby daj95376 » Fri Oct 11, 2013 4:51 pm

Marty R. wrote:Leren, is there someplace that lists and defines these various types of M-Wings?

Leren would probably answer ...

ronk deleted his original postings on the M-Wing and M-Ring definitions and exemplars. However, StrmCkr reposted them once discussions surfaced about the various types.

Look here
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Re: Vanhegan Fiendish October 11

Postby tlanglet » Sun Oct 13, 2013 1:42 am

JC Van Hay wrote:
tlanglet wrote:ANS(75=12)r6c7,r5c9
1: (75) r6c7,r5c9 => r5c8<>7
2: (12)r5c9 => LS(12)r59c9-(2=7)r1c9-r1c1=r7c1-r8c2=7r8c8 => r5c8<>7
I did not know how to notate this as a single statement with the "Almost" component forming a LS. Any suggestions?
There are different "equivalent" ways to justify -7r5c8 :
1. Forcing Chain :
If r5c8=7->r8c2=7=r1c1,r1c9=2,r9c9=1; r5c9=5; r6c7 is empty :=> -7r5c8
This can be fully justified by writing down the following "Triangular Matrix" :
Code: Select all
7r5c8
7r8c8=7r8c2
      7r7c1=7r1c1
            7r1c9=2r1c9
                  2r9c9=1r9c9
7r5c9=============2r5c9=1r5c9=5r5c9
7r6c7=========================5r6c7
2. The Triangular Matrix may be written without any reference of the target :
Code: Select all
7r8c8=7r8c2
      7r7c1=7r1c1
            7r1c9=2r1c9
                  2r9c9=1r9c9
7r5c9=============2r5c9=1r5c9=5r5c9
7r6c7=========================5r6c7
In this case, theory shows that the first column is a derived strong set : 7r8c8=7r5c9=7r6c7 :=> -7r5c8
3. Finally, the Triangular Matrix may be rewritten as a Transfer Matrix :
Code: Select all
7r8c8=7r8c2
      7r7c1=7r1c1
            7r1c9=2r1c9
                  2r9c9=1r9c9
                  2r5c9=1r5c9=5r5c9=7r5c9
                              5r6c7=7r6c7
that is, in Eureka notation :

7r8c8=7r8c2-7r7c1=7r1c1-(7=2)r1c9-NP(12)r59c9=NP(57)r6c7,r5c9 :=> -7r5c8


Note : This "AIC" is readable from left to right and from right to left !

JC


JC,
I appreciate your detailed response to my query. I am not familiar with either "Triangular Matrix" or "Transfer Matrix" but I was able to follow the logic involved. I did not try to find an alternate solution to remove 7 from r5c8, but was wondering how to use Eureka notation to my original ANS() premise. Maybe one day I will pursue moving to the next level of Sudoku fundamentals.

Thanks, Ted
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