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sett.sudoku 04082017

 
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Ajò Dimonios



Joined: 01 May 2017
Posts: 339
Location: Sassari Italy

PostPosted: Mon Aug 07, 2017 12:06 pm    Post subject: sett.sudoku 04082017 Reply with quote

Hi everyone

Code:

+-------+-------+-------+
| 7 . 5 | . . 8 | . . . |
| . 9 . | 1 . 3 | 5 . . |
| 4 . . | . 7 . | . 3 . |
+-------+-------+-------+
| . 1 . | . . . | . 5 3 |
| . . 3 | . . . | 7 . . |
| 9 7 . | . . . | . 1 . |
+-------+-------+-------+
| . 6 . | . 3 . | . . 2 |
| . . 9 | 6 . 1 | . 7 . |
| . . . | 8 . . | 4 . 1 |
+-------+-------+-------+

Play this puzzle online at the Daily Sudoku site

Ciao a Tutti
Paolo
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JC Van Hay



Joined: 13 Jun 2010
Posts: 494
Location: Charleroi, Belgium

PostPosted: Mon Aug 07, 2017 4:05 pm    Post subject: Reply with quote

15 singles; LC(6B1, 8B9)
The solutions of the 2 exclude {2r2c3, 2r3c7, 2r5c8, 2r8c2}
Code:
+----------------+-----------------+----------------+
| 7    3    5    | 249   6   8     | 1    249  49   |
| 268  9    68   | 1     24  3     | 5    248  7    |
| 4    28   1    | 259   7   259   | 689  3    689  |
+----------------+-----------------+----------------+
| 26   1    246  | 2479  8   24679 | 269  5    3    |
| 5    248  3    | 249   1   2469  | 7    489  4689 |
| 9    7    2468 | 3     5   246   | 268  1    468  |
+----------------+-----------------+----------------+
| 1    6    47   | 457   3   457   | 89   89   2    |
| 28   48   9    | 6     24  1     | 3    7    5    |
| 3    5    27   | 8     9   27    | 4    6    1    |
+----------------+-----------------+----------------+
2L2C5 -> 1 solution
2L8C5 -> 7 singles; XWings(2C26, 8R26) -> NP(49)r5c48 -> LC(9C9) -> XYWing(249)r1c49.r5c4 -> r5c8=9; 0 solution
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Ajò Dimonios



Joined: 01 May 2017
Posts: 339
Location: Sassari Italy

PostPosted: Mon Aug 07, 2017 5:09 pm    Post subject: Reply with quote

Hi everyone

Code:

+--------------+---------------+----------------+
| 7   3   5    | 249  6  8     | 1    249  49   |
| 268 9   268  | 1    24 3     | 5    248  7    |
| 4   28  1    | 259  7  259   | 2689 3    689  |
+--------------+---------------+----------------+
| 26  1   246  | 2479 8  24679 | 269  5    3    |
| 5   248 3    | 249  1  2469  | 7    2489 4689 |
| 9   7   2468 | 3    5  246   | 268  1    468  |
+--------------+---------------+----------------+
| 1   6   47   | 457  3  457   | 89   89   2    |
| 28  248 9    | 6    24 1     | 3    7    5    |
| 3   5   27   | 8    9  27    | 4    6    1    |
+--------------+---------------+----------------+

Play this puzzle online at the Daily Sudoku site

My solution after basics.

1) AIC (2)R9C3=(2)R9C6-(2)R8C5=(2)R2C5-(2)R2C3=>-(2)R2C3;
2) [(2)R3C7-(2)R3C2=(2)R2C1-(2=4)R2C5 and (2)R3C7-(2)R46C7=(2)R5C8-(2)R5C2=(2)R8C2-(2=4)R8C5]=>contradiction two 4 singles in column 5=>-2R3C7.
3) [(4)R4C4-(4)R456C6=(4)R7C6-(4)R7C3 and (4-7)R4C4=(7)R7C4-(7)R7C3]=>contradiction R7C3=Ø=>-4R4C4.
4) [(7-4)R4C6=(4)R4C3-(4)R7C3 and (7)R4C6-(7)R9C6=(7)R9C3-(7)R7C3] ]=>contradiction R7C3=Ø=>-7R4C6.
5) R4C4=7
6) [(2-8)R8C2=(8)R8C1-(8)R2C1; (2)R8C2-(2=4)R8C5-(4=2)R2C5-(2=6)R2C1 and (2)R8C2-(2)R3C2=(2)R2C1-(2=6)R4C1]=>contradiction two 6 singles in column 1=>-2R8C2.
7) [(2)R6C3-(2=6)R3C1; (2-8)R6C3=(8)R2C3-(8)R2C1 and (2-8)R6C3=(8)R5C2-(8=2)R3C2-(2=6)R2C1]=> contradiction two 6 singles in column 1=>-2R6C3.
8 ) [(8)R6C7-(8)R6C3=(8)R5C2-(8)R3C2 and (8)R6C7-(8)R6C3=(8-6)R2C3=(6-8)R2C1=(8-2)R3C2=>contradiction R3C2=Ø =>-(8)R6C7.
9) [(2)R1C4-(2)R2C5=(2)R8C5-(2)R8C1; (2)R1C4-(2)R3C46=(2-8)R3C2-(2)R5C2; (2)R1C4-(2)R5C4=(2-6)R5C6=(6)R5C9-(6)R46C7=(6-8)R3C7=(8)R3C9-(8)R56C9=(8)R5C8-(8)R5C2=(8)R8C2-(8)R8C1]=>contradiction R8C1=Ø=>-2R1C4
10) R2C8=2;
11) [(4)R2C5-(4=9)R1C4; (4)R2C5-(4)R2C8=(4)R5C8-(4)R5C4 and (4-2)R2C5=(2)R3C46-(2)R3C2=(2)R5C2-(2=9)R5C4]=>contradiction two singles 9 in column 4=>-4R2C5=>solution stte.

Ciao a Tutti
Paolo
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JC Van Hay



Joined: 13 Jun 2010
Posts: 494
Location: Charleroi, Belgium

PostPosted: Mon Aug 07, 2017 5:36 pm    Post subject: Reply with quote

or the following simpler 2-steps solution determined by the previous 1-step solution:
Code:
+--------------------+--------------------+------------------+
| 7      3      5    | (249)  6     8     | 1    9(24)  49   |
| 26(8)  9      6(8) | 1      2(4)  3     | 5    (248)  7    |
| 4      -8(2)  1    | 259    7     259   | 689  3      689  |
+--------------------+--------------------+------------------+
| 26     1      246  | 2479   8     24679 | 269  5      3    |
| 5      48(2)  3    | (249)  1     2469  | 7    89(4)  4689 |
| 9      7      2468 | 3      5     246   | 268  1      468  |
+--------------------+--------------------+------------------+
| 1      6      47   | 457    3     457   | 89   89     2    |
| 28     48     9    | 6      24    1     | 3    7      5    |
| 3      5      27   | 8      9     27    | 4    6      1    |
+--------------------+--------------------+------------------+
8r2c13 8r2c8
       2r2c8 2r1c8
       4r2c8 4r1c8 4r5c8
       4r2c8             4r2c5
             2r1c4       4r1c4 9r1c4
                   4r5c4       9r5c4 2r5c4
                                     2r5c2 2r3c2 -> 8r2c13==2r3c2 => r3c2=2; LC(2R5, 8B1)
+--------------------+-----------------+----------------------+
| 7     3     5      | 249  6   8      | 1      249    49     |
| 68    9     68     | 1    24  3      | 5      24     7      |
| 4     2     1      | 59   7   59     | 689    3      689    |
+--------------------+-----------------+----------------------+
| 26    1     246    | 479  8   4679   | (269)  5      3      |
| 5     4(8)  3      | 249  1   49(26) | 7      49(8)  489(6) |
| 9     7     468(2) | 3    5   46     | 68(2)  1      468    |
+--------------------+-----------------+----------------------+
| 1     6     47     | 457  3   457    | 8(9)   (89)   2      |
| (28)  4(8)  9      | 6    24  1      | 3      7      5      |
| 3     5     7-2    | 8    9   7(2)   | 4      6      1      |
+--------------------+-----------------+----------------------+
2r6c3 2r6c7
2r9c6       2r5c6
            6r5c6 6r5c9
      2r4c7       6r4c7 9r4c7
                        9r7c7 9r7c8
                              8r7c8 8r5c8
                                    8r5c2 8r8c2
                                          8r8c1 2r8c1 -> 2r6c3==2r9c6==2r8c1 => -2r9c3; stte
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JC Van Hay



Joined: 13 Jun 2010
Posts: 494
Location: Charleroi, Belgium

PostPosted: Mon Aug 07, 2017 7:08 pm    Post subject: Reply with quote

Hi Paolo,

your
step 6 is included in step 2;
step 8 is to be corrected, for example : {8C3, 6R2, (62)r4c1, 2C7} -> -8r6c7;
step 10 : r1c8=2.

JC
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Ajò Dimonios



Joined: 01 May 2017
Posts: 339
Location: Sassari Italy

PostPosted: Mon Aug 07, 2017 9:23 pm    Post subject: Reply with quote

Hi JC

Sorry JC but I can not understand why Step 6 is already included in step 2 (Perhaps because , differed from you, I started from after basics). After inserting 7 in R4C4 and eliminating for locked candidates of 2 in R5C8, in cell R8C2 2 is still a possible candidate along with 4 and 8. In my opinion it is necessary to prove that 2 in R8C2 is false. Sorry but I also do not understand why step 8 needs to be corrected. The demonstration that 8 in R6C7 is false seems to me correct.

Ciao a Tutti
Paolo
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JC Van Hay



Joined: 13 Jun 2010
Posts: 494
Location: Charleroi, Belgium

PostPosted: Mon Aug 07, 2017 11:45 pm    Post subject: Reply with quote

Hi Paolo,

your step 2 : the solutions of your 5 constraints {(24)r28c5, 2C2B26} effectively exclude {2r3c7, 2r8c2}. To wit : 2r2c5 -> 2r3c2 , while 4r2c5 -> 2r8c2, EmptyRectangle(2C2, 2B6). Therefore 2r3c7 and 2r8c2 are excluded in both cases.

your step 8 contains in the second part the simultaneous implications +8r2c3 and +8r3c2[instead of +2r3c2 !]

JC
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Ajò Dimonios



Joined: 01 May 2017
Posts: 339
Location: Sassari Italy

PostPosted: Tue Aug 08, 2017 11:01 am    Post subject: Reply with quote

Hi JC

Sorry for the distraction in step 10 (clearly R1C8=2) and error of the second part of step 8 . Replace the whole step with
8 ) [(8)R6C7-(8)R6C3=(8)R5C2-(8)R3C2 and (8-2)R6C7 =(2)R4C7-(2)R4C13=(2)R5C2-(2)R3C2=>contradiction R3C2=Ø =>-(8)R6C7.=>contradiction R3C2=Ø =>-(8)R6C7.
As for Step 6 included in Step 2, I simply work step by step. Deleting 2 in R3C7 does not automatically involve eliminating 2 in R8C2, to eliminate candidate 2 in R8C2 after eliminating 2 in R3C7 you need the fin swordfish of the 2 in line 3,5 and 9 (fin R9C3 = 2). As you have shown with a more global view, R3C7=2 and R8C2=2 are biunivoca, when one is True the other is also true and viceversa ,[(2)R3C7-(2)R3C2; (2)R3C7-(2)R12C8=(2)R5C8-(2)R5C2=(2)R8C2 ; (2)R8C2-(2)R3C2 and (2)R8C2-(2)R8C5=(2)R2C5-(2)R3C46=(2)R3C7]. In practice when R3C7 = 2 => R8C2 = 2 and when R8C2 = 2 => R3C7 = 2. I consider it another step.
Ciao a Tutti
Paolo


Last edited by Ajò Dimonios on Tue Aug 08, 2017 4:06 pm; edited 1 time in total
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Ajò Dimonios



Joined: 01 May 2017
Posts: 339
Location: Sassari Italy

PostPosted: Tue Aug 08, 2017 2:20 pm    Post subject: Reply with quote

Hi JC


JC Van Hay wrote:

Quote:
2r2c5 -> 2r3c2 , while 4r2c5 -> 2r8c2, EmptyRectangle(2C2, 2B6)


Reading your post I would like to understand why you consider this a single step. It seems to me that you can apply the EmpTyRectangle after eliminating the 2 in R8C2, which happens with 2r2c5 -> 2r3c2, while 4r2c5 -> 2r8c2, otherwise there are three 2 in column 2 and the Emptyrectangle is not applicable.

Ciao a Tutti
Paolo
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JC Van Hay



Joined: 13 Jun 2010
Posts: 494
Location: Charleroi, Belgium

PostPosted: Tue Aug 08, 2017 4:25 pm    Post subject: Reply with quote

Hi Paolo,

Using your constraints in step 2, one can write the following mixed block matrix :
Code:
|2r46c7 2r5c8|
|2r3c2  2r5c2| 2r8c2
              &2r8c5 4r8c5      &
              &      4r2c5 2r2c5&
              &2r3c2       2r2c1& -> 2r8c5==2r3c2 => -2r8c2, 2r46c7==2r3c2 => -2r3c7
Note :
1. the ER between the symbols |
2. the AIC between the symbols &
2. 4C5 is optional -> AIC=ER

In Eureka notation : ER(2B6, 2r35*c2)=*2r8c2-[(2=4)r8c5-(4=2)r2c8-2r2c1=2r3c2] -> 2r8c5==2r3c2 => -2r8c2 -> ER true => -2r3c7

Conclusion : step 2 is of course a combined 2 steps move as it is, in fact, made up of 2 "Empty rectangles" : 2B1C5 and 2C2B6. But in any case, it is up to the player to determine or not all the exclusions by the solutions of a set of constraints as in your step 2. In both cases, it is a single step. Otherwise, what would be the case of a continuous chain\network, a double exocet, ... where tenths of exclusions are possible !

JC
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Ajò Dimonios



Joined: 01 May 2017
Posts: 339
Location: Sassari Italy

PostPosted: Wed Aug 09, 2017 11:03 am    Post subject: Reply with quote

Hi JC

Thank you for your excellent and precise explanation.

Anyway, I'm of the opinion that a single step is that of an advanced technique along with one or more basic techniques. If that was not the case, it would be absurd that always with a single step that includes many advanced techniques, one can get the solution.I think that when using more advanced techniques, each application represents a step. I have not found in this example any basics technique that proves that once proved to be false R3C7 = 2 attest to the falseness of R8C2 = 2 or vice versa.

Ciao a Tutti
Paolo
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Ajò Dimonios



Joined: 01 May 2017
Posts: 339
Location: Sassari Italy

PostPosted: Wed Aug 09, 2017 2:23 pm    Post subject: Reply with quote

Hi JC

JC Van Hay Wrote:
Quote:
In Eureka notation : ER(2B6, 2r35*c2)=*2r8c2-[(2=4)r8c5-(4=2)r2c8-2r2c1=2r3c2] -> 2r8c5==2r3c2 => -2r8c2 -> ER true => -2r3c7




Confirmation of what has been said in my previous post also from your writing in Eureka notation easily shows that there are two logical processes of which one is completely independent. The first conclusion (2R8C5 = = 2R3C2 => 2R8C2) depends only on the final AIC [2 = 4) R8C5- (4 = 2) R2C5-2R2C1 = 2R3C2] which is certainly true irrespective of the strong initial binding of the chain. In fact, only to comply with the logical process, first demonstrate that the AIC is true and after, as a consequence of a second logical process, I demonstrate the truth of the ER.

Ciao a Tutti
Paolo
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