## Mauricio's Perla Grandisima

Advanced methods and approaches for solving Sudoku puzzles

### Mauricio's Perla Grandisima

Mauricio's Perla Grandisima (very very very large pearl)

Code: Select all
` *-----------* |1..|9..|7..| |.2.|.6.|...| |..3|..5|..8| |---+---+---| |4..|..3|.8.| |...|6..|..2| |..7|.5.|9..| |---+---+---| |3..|...|..1| |.9.|5..|...| |..8|.2.|.4.| *-----------*    SE 10.0`

In this SOLUTION I am assuming some familiarity with my recent solutions in this forum. If you have none, then I suggest you do not start your exploration of these realms with this puzzle.

We are confronted immediately by an extremely difficult first step according to SE, as stated by Mauricio and confirmed by my step (1) with its 5 sub-sub-nets. This I conjecture is because:

a) The weakness or necessarily most easy access lies in the 2s, AND
b) The 2s have only one true member of influence: 2f4, AND
c) 2f4 is not conjugate to anything, and so can not easily be forced. (There are no Candies as yet).

But step (1) will create a Candy, a conjugate cell to f4, in d4. This gives us Candies 1d4 and 8f1. So no more sub-sub-nets will be needed. And we'll get more candies when we eliminate the unwanted 2s at h3, h7 and g8.

(1) ?2c4... (?9e1... ((?6k1...??))-6k1... ((?7k1...??))-7k1... ((?1k4...??))-1k4... ((?7k2...??))-7k2... ((?3h9...??))-3h9...??)-9e1... (?3h9...??)-3h9... (?4h5...??)-4h5...?? -2c4, 2a6.

(2) ?1d4... (?9e1...quickly...??)-9e1...?? -1d4. The first bivalue cell appears on the scene.

(3) ?8f1...?? -8f1...

(4) ?2h7... (?6a9...)-6a9, (?6a8...??)-6a8...?? -2h7.

(5) ?2g8... (?9e1...??)-9e1... (?5k2...??)-5k2, (?5g2...??)-5g2, (?5e1...??)-5e1...?? -2g8.

(6) ?2h3...?? -2h3.

The 2s chain is now complete. Candies still to be eliminated are: 14c7, 56g3, 37h8.

(7) ?3h8...?? -3h8.

(8) ?4c7...?? -4c7.

(9) ?7h8...?? -7h8.

(10) ?1c7...?? -1c7.

(11) ?5g3...?? -5g3.

(12) ?6g3...?? -6g3.

All candies gone. Time now to force either 6d7, 7d4, 6f1, 4g3 or 6h8. 7d4 appears to be the easiest, needing us to eliminate the other two 7s in row d. In accordance with previous experience with this strategy, we can then expect the immediate and total collapse of the puzzle.

(13) ?7d5... (?8e6...??)-8e6... (?9b1...??)-9b1... (?7b1...??)-7b1... (?8e2...??)-8e2...?? -7d5...

(14) ?7d9... (?1d2...??)-1d2, (?6d2...??)-6d2...?? -7d9 and all singles to End.
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Last edited by gurth on Tue Dec 19, 2006 8:13 am, edited 1 time in total.
gurth

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