Puz 160 page 1

 

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98.7.....76....5....5.......9..4..3...85..6.......2..1..98..7......3..2......1..4

 

In that first PM, we see the exocet location and the SK loop location

In red are the direct eliminations

 

9      8     1234 |7      1256  3456  |1234   146    236   

7      6     1234 |12349  1289  3489  |5      1489   2389  

1234   1234  5    |123469 12689 34689 |123489 146789 236789

-----------------------------------------------------------

1256   9     1267 |16     4     678   |28     3      2578  

1234   12347 8    |5      179   379   |6      479    279   

3456   3457  3467 |369    6789  2     |489    45789  1     

-----------------------------------------------------------

123456 12345 9    |8      256   456   |7      156    356   

14568  1457  1467 |469    3     45679 |189    2      5689  

23568  2357  2367 |269    25679 1     |389    5689   4     

 

Now the same PM after eliminations and a small change in colours in the sk loop to highlight the 2 belts

 

69r46c4 =>24r89c4 =>56r7c56 =>13r7c89 =>89r89c7 =>24r46c7 =>79r5c89 =>13r5c56 loop

13r46c4 =>79r5c56 =>24r5c89 =>89r46c7 =>13r89c7 =>56r7c89 =>24r7c56 =>69r89c4 loop

Also are underlined the cells r5c1 and r7c1 that could generate UR’s with the base of our exocet r3c12

 

9     8    1234 |7    1256  3456  |1234 146    236   

7     6    1234 |1234 1289  3489  |5    1489   2389  

1234  1234 5    |1234 689   689   |1234 6789   6789

-----------------------------------------------------

1256  9    1267 |16   4     678   |28   3      578   

1234  1234 8    |5    179   379   |6    479    279   

3456  3457 3467 |369  6789  2     |489  5789   1     

-----------------------------------------------------

1234  1234 9    |8    256   456   |7    156    356   

14568 1457 1467 |469  3     5679  |189  2      5689  

23568 2357 2367 |269  5679  1     |389  5689   4   

 

Just consider the digits 1234 of the exocet

if we have 2 common digits for the red and blue cells in columns  4 et 7 the exocet rule gives no digit possible in r3c4 r3c7

 

The main effect for the next step is that the solution can not be all red and can’t be all blue.

In each minirow or mini column, we have one blue and one red

 

BTW, this is, whatever is the way to prove it (but it must be proven) something I always verified in a SK loop

 

Also, it’s not possible to have the same digit 1234 red or blue (sk loop) in columns 4 and 7, the exocet would have no solution

 

 

Here, we continue combining both patterns

 

Cells r3c12 r5c12 r7c12 must not show an UR pattern

This give restrictions for any start in r3c12 r1c7 r4c2 (exocet). 

What we show with values in r1c7 r4c2

This uses the results in bold underlined:

 

r1c7 r4c2

 1     2   UR in r3c12 r7c12

 2     1   UR in r3c12 r5c12 12r3c12 killed

 3     4   UR in r3c12 r7c12

 4     3   UR in r3c12 r5c12 34 r3c12 killed

 1     4   UR in r3c12 r7c12

 4     1   UR in r3c12 r5c12 14 r3c12 killed

 2     3   UR in r3c12 r5c12

 3     2   UR in r3c12 r7c12 23 r3c12 killed

 

Are remaining as possible 13 r3c12 and 24 r3c12

But here, we don’t have the UR invalid pattern.

 

 

We continue keeping in mind the three constraints in red and using it as aften as possible.

 

I follow during some steps eliminations proposed by the solver

The process is more or less expansion in dynamic mode (in Sudoku Explainer terminology)

with the above constraints

 

<9>r8c6

  9r8c6 =>9r6c4;3r5c6; 7r5c5 => 7r8c6 - 9r8c6

 

Then

<9>r9c8

 

      9r9c8 => 9r6c7;2r4c7

            => 9r8c4;2r9c4 not possible for the exocet

Then

<8>r4c9

 

 8r4c9 => 9r6c7;4r6c8;7r6c9=>7r3c8 – 89r3c8 forces 9r2c8

                           => 8r89c7 forces 8r2c8

 

9     8    1234 |7    1256  3456  |1234 146    236   

7     6    1234 |1234 1289  3489  |5    1489   2389  

1234  1234 5    |1234 689   689   |1234 6789   6789

-----------------------------------------------------

1256  9    1267 |16   4     678   |28   3      57    

1234  1234 8    |5    179   379   |6    479    279   

3456  3457 3467 |369  6789  2     |489  5789   1     

-----------------------------------------------------

1234  1234 9    |8    256   456   |7    156    356   

14568 1457 1467 |469  3     567   |189  2      5689  

23568 2357 2367 |269  5679  1     |389  568    4   

 

Next step proposed by my solver at that point is to show <8>r9c7

 

The path is not as short as the others and likely many possibilities exist.

Here the  path proposed by  the solver.

 

<8>r9c7

 

8r9c7 => 12 column 7 forces 34 column 4 (from our constraints)

         =>3r6c4 ;  <3>r6c2 ; <3>r6c3

      => 8r6c8;8r4c6,5r4c9=>7r5c9;8r4c6;=>7r6c5;7r8c6; <5>r8c6

      => 8r8c1;                                         <5>r8c1

      => 9r8c9;                                        <5> r8c9

          => (set) 5r8c2; <5>r6c2  

          => 7r6c5;      <7>r6c2

          =>       (set)4r6c2               <4>r6c3

      => 2r7c5;3r7c9 => 14r7c12=>67r8c3

      (8r4c6;5r4c9) => 7r4c3    => 7r8c3 =>  4r6c3

   

                                                                                                                           

 

Then becoming easier

 

<7>r5c6

7r5c6 =>1r5c5;6r4c4;8r4c6;2r4c7 => 3r4c4;9r9c7 => 49 r8c4 not valid

 

<2>r9c1 ( may be not necessary for key steps)

 

2r9c1 => 4r8c4;2r4c7=>9r8c79;3r9c7;1r4c4 =><12>r4c3

      => 8r9c8; 8r8c1=>6r46c1 <6>r4c3

                  =>5r6c8;7r4c9 <7>r4c3    not valid r4c3 empty   

 

                

9     8    1234 |7    1256  3456  |1234 146    236   

7     6    1234 |1234 1289  3489  |5    1489   2389  

1234  1234 5    |1234 689   689   |1234 6789   6789

-----------------------------------------------------

1256  9    1267 |16   4     678   |28   3      57    

1234  1234 8    |5    179   39    |6    479    279   

3456  3457 3467 |369  6789  2     |489  5789   1     

-----------------------------------------------------

1234  1234 9    |8    256   456   |7    156    356   

14568 1457 1467 |469  3     567   |189  2      5689  

3568  2357 2367 |269  5679  1     |39   568    4   

 

<4>r6c2  ( may be not necessary for key steps)

 

4r6c2 =><7>r89c3 ; 2r4c7 ;8r4c6 ;7r8c6 <7>r8c2 =>7r9c2

    =><4>r6c4 ;4r8c4 => (4r13c3 exocet 13r3c12)

        =><2>r9c4 ;<2>r9c3                     =>2r9c2 not valid

<5>r9c1  ( may be not necessary for key steps)

 

5r9c1=>8r9c8;9r79c7;5r4c9=> 6r8c9              <6>r1c9

     =>5r7c8 1r9c7 9r9c7 9r8c4 4r6c7

     Now pair 23 in r13c7                      <23>r1c9 not valid                 

 

then

 

<3>r9c1  ( may be not necessary for key steps)

 

3r9c1 => 9r9c7 1r8c7 9r8c4 2r8c9 4r6c7 3r6c4      <3>r6c3 <4>r6c7

      => <3>r6c2  8r8c1 5r8c2 <5>r6c2 => 7r6c2    <7>r6c3

      => 6r46c1                                   <6>r6c3   not valid 

 

  

Then an interesting step clearing one sub scenario for the exocet    

 

We clear XX (<1>r3c4 <3>r3c7) (same sub scenario for the pair 24 in r3c12)  

 

XX=> 24r3c12 13r13c3 2r1c79                   <12>r1c5                

  =>3r6c4  9r9c7 9r8c4 2r9c4 2r13c7 <23>r1c9  <6>r1c5

  => 4r7c6 <5>r7c6

  => 13r13c3(exocet) <3>r9c3 <3>r9c7 => 3r9c2

      =>5r8c12 <5>r8c6 => 5r79c5              <5>r1c5 not valid

    

 

From that point, I leave the solver path and go directly to action on the exocet or on the SK loop

 

 

9     8    1234 |7    1256  3456  |1234 146    236   

7     6    1234 |1234 1289  3489  |5    1489   2389  

1234  1234 5    |234  689   689   |124  6789   6789

-----------------------------------------------------

1256  9    1267 |16   4     678   |28   3      57    

1234  1234 8    |5    179   39    |6    479    279   

3456  357  3467 |369  6789  2     |489  5789   1     

-----------------------------------------------------

1234  1234 9    |8    256   456   |7    156    356   

14568 1457 1467 |469  3     567   |189  2      5689  

68    2357 2367 |269  5679  1     |39   568    4   

 

at that point we try to establish

 

XX  13r3c12   1r4c2 3r1c7 not valid

 

To make it easier, we do first evident steps

Just applying the rules for exocet and sk loop

The PM specific to that scenario comes to

 

9     8    24   |7    256   3456  |3   146    26   

7     6    24   |1   289   3489  |5    1489   289  

13    13  5    |24   689   689   |24   6789   6789

-----------------------------------------------------

1256  9    1267 |6    4     78    |8    3      57    

1234  1234 8    |5    1     9     |6    79     2     

3456  357  3467 |3    78    2     |4    579    1     

-----------------------------------------------------

1234  1234 9    |8    56    4     |7    56     3    

14568 1457 1467 |9    3     567   |1    2      568  

68    2357 2367 |2    567   1     |9    568    4 

 

XX=>7r4c6;5r4c9;79r56c8 ;8r6c5

  => 68 r3c68 =>9r3c5

  =>1r1c8 4r2c8 2r2c3

    No possibility in r2c5 so XX is not valid

 

We still have valid 

13r3c12 3r4c2 1r1c7

24r3c12  (2 sub scenarios)

 

 

 

9     8    1234 |7    1256  456   |124  46     236   

7     6    1234 |234  1289  489   |5    489    2389  

1234  1234 5    |234  689   689   |124  6789   6789

-----------------------------------------------------

256   9    267  |1    4     678   |28   3      57    

124   124  8    |5    79    3     |6    479    279   

3456  357  3467 |69   6789  2     |489  5789   1     

-----------------------------------------------------

234   234  9    |8    256   456   |7    1      56   

14568 1457 1467 |469  3     567   |89   2      5689  

68    257  267  |269  5679  1     |3    568    4   

 

We do the same with XX 24r3c12 2r1c7 4r3c4

 

9     8    3    |7    1     56    |  4      6   

7     6    1    |4    2     89    |5    89     3   

24    24  5    |3    689   689   |1    6789   6789

-----------------------------------------------------

256   9    267  |1    4     678   |8    3      57    

124   124  8    |5    79    3     |6    79     2      

3456  357  3467 |69   6789  2     |4    5789   1     

-----------------------------------------------------

234   234  9    |8    56    4     |7    1      56   

14568 1457 1467 |69   3     567   |89   2      5689  

68    257  267  |2    5679  1     |3    568    4   

 

XX  => 9r9c5 5r1c6 5r7c5 6r7c9 conflict with r1c9

 

And with XX 24r3c13  4r1c7 2r3c4

 

Again the PM reduced after evidence

 

9     8    3    |7    1     5     |4    6      2     

7     6    1    |  89    4     |5    89     3  

24    24   5    |3    689   689   |1    789    789

-----------------------------------------------------

256   9    267  |1    4     678   |2    3      57    

124   124  8    |5    79    3     |6    4      79   

3456  357  3467 |69   6789  2     |89   5789   1     

-----------------------------------------------------

234   234  9    |8    2     56    |7    1      56   

14568 1457 1467 |4    3     567   |89   2      5689  

68    257  267  |69   5679  1     |3    568    4   

 

   After evidence      <14>r8c3

XX => 6r7c6;8r7c6      <7>r8c3

   => 6r8c9            <6>r8c3  scenario not valid

 

The last valid scenario for the exocet is

 

13r3c12 1r1c7 3r2c4

 

9     8    24  |7    256   456   |  46     3   

7     6    24  |3    1     489   |5    489    289  

13    13  5   |24   689   689   |24   6789   6789

-----------------------------------------------------

256   9    67  |1    4     678   |28   3      57    

124   124  8   |5    79    3     |6    479    279   

456   57   3   |69   6789  2     |489  5789   1     

-----------------------------------------------------

234   234  9   |8    256   456   |7    1      56   

4568  457  1   |469  3     567   |89   2      5689  

68    257  67  |269  5679  1     |3    568    4   

 

 

To finish the work in the sk loop, we have to show

 

XX <4>r8c4 and associated candidates

 

9     8    24  |7    256   456   |  46     3   

7     6    24  |3    1     489   |5    489    289  

13    13  5   |24   689   689   |24   6789   6789

---------------------------------------------------

256   9    67  |1    4     678   |28   3      57    

124   124  8   |5    79    3     |6    479    279   

456   57   3   |69   6789  2     |489  5789   1     

-----------------------------------------------------

234   234  9   |8    256   456   |7    1      56   

4568  457  1   |469  3     567   |89   2      5689  

68 257  67  |269  5679  1     |3    568    4   

 

Xx => 4r3c7 6r1c8               <6>r1c56

   => 2r4c7 8r4c6 8r3c5     <6>r3c5  6r3c6 <6>r78c6

   =>    2r7c5                              <6>r7c5

   =>    2r2c9 8r8c9 <5>r89

   => 6r7c9 5r4c9 7r4c3 6r9c3               <6>r9c56

   With <6>r8c4 direct, all ‘6’ in box 8 have disappeared

 

 

And we go to the end with singles.