V2_00 Easter Monster summary

 

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That path, dated march 2010 has been solved using the revised version of the solver in construction.

 

This is as usual a “solver path”, very progressive, so not really something a player would like.

 

I guess the “final path” in that version will be shorter after I have introduced deadly UR pattern.

 

On top of the tagging, the main rules in that version are:

 

· SK loop AAHS/AC2 are kept all along the process. In fact, as the main eliminations in that loop are finished in step 2, all loop effects are there from that point.

· Contrary to the previous version, derivation of weak links is authorized in level one of the tagging (maximum 2 successive derivations) and choices are activated.

· The complementary tagging has been improved and is now very active

· As soon a derived weak link has been printed, the solver is free to reuse it in next steps.

· All nice loops giving no immediate elimination generate equivalences for the next steps (same for complementary tagging.

 

As a matter of facts, that path is based mainly on SK loop effects and accumulation of equivalences generated by nices loops and complementary tagging.

 

Clearing sequences are relatively short, but are requesting a precise registration of equivalences generated

 

Here is a summary of the path followed

 

=============================================

Step 1 identification of the SK loop, clearing of the loop.

r5c6=4

Step 2:

A key one establishing strong links all along the loop after 2 super candidates have been invalidated

 

Followed by nice loops giving equivalences

 

8r2c1==8r1c7    8r2c3==8r3c9  4r9c2==4r1c3 4r7c2==4r3c1

 

Step 3:

 

That step publishes several basic derived weak links reused later

1r2c5=>1r8c3

1r3c6=>1r6c23

1r4c8=>1r8c5 

1r5c2=>1r3c8

1r6c2=>6r2c9   

1r6c3=>1r23c6

1r7c6=>7r4c8   

1r8c4=>1r5c8 

2r4c8=>2r2c5 2r4c8=>2r5c4 2r4c8=>2r7c6 

2r8c4=>2r5c2 2r8c4=>2r23c6 2r8c4=>2r6c7

6r1c6=>6r6c12   

6r2c5=>6r8c1   

6r2c6=>1r7c2   

6r4c8=>6r8c5 6r4c8=>2r5c8 6r4c8=>1r5c2 6r4c8=>1r78c4.n   

6r5c2=>6r1c8

6r6c1=>6r12c6  

6r6c2=>1r2c7

6r8c4=>2r8c5 6r8c4=>6r5c8 6r8c4=>1r2c5 6r8c4=>1r4c78     

 

Makes 2 eliminations<1r2c6>  <1r6c7>

 

And find 2 equivalences 6r4c1==6r1c4   1r3c4==1r4c3

 

Steps 4;5: provide new equivalences

I summarize all equivalences found at the end of these steps

 

1r3c4==1r4c3

1r3c6==1r6c3

1r4c5==1r8c4

1r5c2==1r7c4==6r1c6

1r6c2==1r7c6==6r1c4==6r4c1

2r2c5==6r2c6==6r6c9

4r1c3==4r9c2

4r3c1==4r7c2

6r1c6==6r6c1 

6r1c8==6r8c1  Linking tightly boxes 3 and 7

6r2c5==6r4c9

6r5c2=>6r9c4     

8r2c1==8r1c7

8r2c3==8r3c9 

 

The main tagging within the sk loop looks like that.

In each AAHS/AC2 of the loop red groups for a strong link, blue groups as well.

 

1    47c8  3458 |3567  389 5678 |3489 36a9  2   

2c38 9     37C8 |4     1A26 267c|1a38 5     36A8

3458 2C48  6    |1235  389 1258 |7    1A39  3489

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

2468 5     1478 |9     126 3    |1A28 1267d 678 

389  126   389  |126   7   4    |3589 126   3589

2369 1267C 1379 |8     5   126  |239  4     3679

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

7    1a48  4589 |1235  348 1258 |6    2d39 3459

456a 3     1A45 |1267D 126 9    |2D45 8     457d

4589 46A8  2    |3567  348 5678 |3459 37D9  1   

 

Step 6: eliminates <4r3c1;4r7c2> <2r4c8>

Step 7: eliminates <8r7c3> <6r8c4> <8r1c2>  <2r2c5;6r6c9;6r2c6>

Step 8: eliminates  <6r4c5> <2r6c6>

    And extend an equivalence 1r5c2==1r7c4==6r1c6==6r4c8

Steps 9 10 :create new equivalences

1r4c3== 6r6c2

1r6c3==6r5c2

1r8c5==8r4c7

2r8c4==2r5c2

2r3c6==7r9c4

         eliminate  <7r1c6> <2r4c1> <3r6c3>  <2r7c4> <9r6c3>

 

Steps 11 12 13:

 

Extend equivalence 6r1c6==6r6c1==7r4c9 

Add equivalences

8r4c9==1r7c6 extending an existing equivalence set

1r4c8==7r4c3

1r78c4==8r4c13

7r4c9==8r5c9

8r5c7==2r4c7

3r7c9==5r9c7

 

Eliminates  <4r9c7>   <3r2c7> <8r2c1;8r1c7> <3r3c1> <3r1c7> <9r7c9>

              <3r1c3;8r2c3;3r2c9;8r3c9> all linked to the SK loop

After eliminations, the SK loop looks like this

 

1    47c   458  |3567  389 568  |49   36a9  2   

2c3 9     37C  |4     1A26 27c |1a8  5     6A8

58   2C48  6    |1235  389 1258 |7    1A39  349

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

468  5     1478 |9     12  3    |1A28 167d  678 

389  126   389  |126   7   4    |3589 126   3589 

2369 1267C 17   |8     5   16   |239  4     379

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

7    1a8   459  |135   348 1258 |6    2d39 345

456a 3     1A45 |127D 126 9    |2D45 8     457d

4589 46A8  2    |3567  348 5678 |359  37D9  1   

 

Next step eliminates first the “C” tag of that PM

Which is enough to make the remaining situation “simple”

 

In the same step, the solver eliminates

<7r4c3;1r4c8;7r6c9> 

<9r3c8>

<1r6c2;1r7c6;6r1c4;6r4c1;8r4c9;6r5c8>

<3r7c8>

<8r5c1>

<9r5c7>

<5r9c1>

 

After basic moves, including a UR, the puzzles restarts here

 

1   7   458 |35   389 568 |49  369 2   

2   9   3   |4    16  7   |18  5   68  

58  48  6   |1235 389 258 |7   13  349 

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

48  5   148 |9    2   3   |18  67  67  

39  12  89  |6    7   4   |35  12  3589

369 26  7   |8    5   1   |239 4   39  

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

7   18  459 |135  348 258 |6   29  345 

456 3   145 |27   16  9   |245 8   457 

489 468 2   |357  348 568 |359 379 1   

 

This is a situation any skill player can solve.

 

 

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