Ocean's Christmas present for [b]gsf[/b]

Advanced methods and approaches for solving Sudoku puzzles

Ocean's Christmas present for [b]gsf[/b]

Postby gurth » Sat Dec 23, 2006 9:55 am

Ocean's Christmas present for gsf

Code: Select all
# gsfr: "no-solution" (or more correct: no rating)
# ER: 10.7
 *-----------*
 |...|..1|.2.|
 |3..|.4.|5..|
 |...|6..|..7|
 |---+---+---|
 |..2|...|..6|
 |.5.|.3.|.8.|
 |4..|...|9..|
 |---+---+---|
 |9..|..2|...|
 |.8.|.5.|4..|
 |..1|7..|...|
 *-----------*

A glittering acievement, Ocean. And a wonderful Xmas present for me too, the chance to practise and develop my strategies on it.


Note: This solution is designed to be followed using the Simple Sudoku program.
"..." means proceed as far as SS allows, following the hints given by SS in the order given.
This will facilitate rapid checking of this solution.

"?" introduces a move that will be disproved by contradiction. (These moves are only introduced when SS grinds to a halt, saying "no hint available".) In order to "play" this move, you will have to turn off the "block invalid moves" feature of the program. Then insert the move and follow all hints given until you see a contradiction. Once you find the contradiction, you must retrace (withdraw in reverse order) all moves made since the disproved move, by clicking the "undo" arrow repeatedly. Then "correct" the disproved move. EG if this move was "?3f6", then REMOVE the 3 at f6, because you have proved the 3 at f6 false.

SOLUTION:

(1) ...

(2) ?4d6... (?5f4...??)-5f4... (?5f8...((?2f4...??))-2f4, ((?2k1...??))-2k1...??)-5f8... (?5k9...??)-5k9... (?5k1...??)-5k1... (?1d5...??)-1d5... (?1g5...??)-1g5...?? -4d6.

(3) ?7d8... (?2f4...??)-2f4, (?2k9...??)-2k9, (?2k1...??)-2k1... (?2k2...??)-2k2...?? -7d8.

(4) ?5d8... (?2f4...??)-2f4, (?2k9... ((?3a4...??))-3a4...??)-2k9, (?2k1... ((?3a4...??))-3a4...??)-2k1... (?2k2... ((?3a4...??))-3a4...??)-2k2... (?3a4...??)-3a4...?? -5d8...

(5) ?1d8... (?2f4...??)-2f4, (?2k9...??)-2k9, (?2k1...??)-2k1... (?2k2...??)-2k2...?? -1d8.

(6) ?3a9...?? -3a9.

(7) ?8a9...?? -8a9.

(8) ?1g4... (?4a2...??)-4a2... (?4c8...??)-4c8...?? -1g4.

(9) ?8g4... (?4a2...??)-4a2, (?4c8... ((?8b6...??))-8b6...??)-4c8... (?9h6...??)-9h6, (?6g5...??)-6g5, (?9h9...??)-9h9... (?7f3...??)-7f3, (?6f3...??)-6f3, (?3f3...??)-3f3...?? -8g4.

(10) ?2k9... (?4e4...??)-4e4... (?4g2...??)-4g2... (?4k2...??)-4k2...?? -2k9.

(11) ?2k1... (?4e4...??)-4e4... (?4k2...??)-4k2...?? -2k1.

(12) ?2c2... (?4e4...??)-4e4... (?7g2...??)-7g2...?? -2c2.

(13) ?2f4... (?4e4...??)-4e4... (?5g9...??)-5g9... (?5g8...??)-5g8...?? -2f4...

(14) ?2c1...?? -2c1...

(15) ?4g2...?? -4g2...

(16) ?4a9...?? -4a9 and singles to End.
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hey gurth check this out.

Postby StrmCkr » Wed Dec 27, 2006 8:46 am

removed
Last edited by StrmCkr on Sat Dec 13, 2014 6:29 am, edited 3 times in total.
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Re: hey gurth check this out.

Postby ronk » Wed Dec 27, 2006 12:46 pm

StrmCkr wrote:when looking at the chain of 2's where 2 yes's appear the x is overlapped and cannot be 2 by two different directions.

Code: Select all
.n.|...|...|
.2(yes).| not a 2(no)..|...|
nx.|.2(yes).|...|


this tells me the x is not a 2.

And why is it not possible for the "yes" and "no" positions to be reversed?
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Re: hey gurth check this out.

Postby gsf » Wed Dec 27, 2006 4:59 pm

StrmCkr wrote:R7c2 cannot = 4 as
r7c4 <4> - r8c9 <not 4> - r9c2 <4> (overlapping yes's @ r7c2)

I'm having trouble following the logic of your first assertion (and didn't look at the rest)
how does r8c9 w.r.t. 4 come into play when r8c7=4 is a clue?
what about r7c4!=4 which by strong links => r9c2!=4 which leaves r7c23 inconclusive w.r.t. 4?
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Postby StrmCkr » Wed Dec 27, 2006 9:46 pm

removed
Last edited by StrmCkr on Sat Dec 13, 2014 6:30 am, edited 4 times in total.
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Postby ravel » Sun Dec 31, 2006 7:34 pm

Another solution with contradiction chains and subchains using the techniques: singles, locked candidates, locked sets (and maybe x-wing).
So r2c2<>1(r3c8<>1,r2c6<>7,r7c2<>7) means: try r2c2=1, it leads to a grid, where r3c8=1 is tried. This leads to a contradiction and 1 is eliminated from r3c8. Later r2c6=7 is tried and eliminated, later r7c2=7. Then a contradiction arises and 1 can be eliminated from r2c2.

1. r2c2<>1(r3c8<>1,r2c6<>7,r7c2<>7)
2. r2c3<>9
3. r4c2<>9(r6c2<>3,r1c7<>6),(gives the first number r5c3=9)
4. r5c9<>1
5. r8c9<>2 (=> r8c1=2)
6. r3c2<>1 (=> r3c1=1)
7. r1c1<>7
8. r5c7<>2 (=> r9c7=2)
9. r5c4<>1 (=> r5c7=1)
10.r1c1<>8 (=> r4c1=8, r5c1=7, r5c6=6)
11.r7c5<>6 (singles from here)

Happy new year.
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Postby gurth » Thu Jan 04, 2007 8:35 am

ravel,
Many thanks for adding your solution to my thread!

I am finding your work very inspiring. To see what you can do with your computer reveals unsuspected possibilities. It is for ME to study these miraculously revealed possibilities, to analyse them, and try to fathom exactly WHAT makes them possible. Then I hope to get closer to such concise results by my own power, as of course I don't use a computer at all in my solutions, except to deal with the elementary stuff. Just a slight extension of automatised fingermarks, I would say, in the context of this 10-11.2 level.

leon,
I miss your customary contributions, as they are just as instructive.
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Postby wapati » Thu Jan 04, 2007 9:36 am

gurth wrote:ravel,
Many thanks for adding your solution to my thread!


Wow, my sister needs to be in a home.

This really reminds of that fact.
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