Solving without pencilmarks

Advanced methods and approaches for solving Sudoku puzzles

Postby Pat » Thu Nov 08, 2012 1:24 pm

7b53 wrote:

    With/without pencil marks
    Advantage/disadvantage between them...?

this example should be easy without listing the possibilities in each cell,
tougher for those who do list the possibilities

........1..2..345..6..7.83....3...1...9..1..5.4..58....24....9..586..7..9...4.... # joel64 # 191 # 2

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

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Postby Pat » Tue Jan 08, 2013 11:09 am


    i distinctly remember that 7b53 solved the above puzzle ( joel64 # 191 # 2 )
    and posted a response — which has now vanished
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Re: Solving without pencilmarks

Postby JasonLion » Tue Jan 08, 2013 12:48 pm

The logs show that 7b53 deleted their own post back at the end of November.
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Re: Solving without pencilmarks

Postby 7b53 » Thu Jan 10, 2013 4:16 am

yes, I deleted the post myself. hope didn't spoiled anything.
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Postby Pat » Thu Feb 14, 2013 4:57 pm

the following example is certainly not tough
but is just a bit tricky——

    you may have seen this puzzle,
    it was recently posted at t=30970
2..1...3..35..4.1...6...9.7....1....7..2.5..4....9....5.2...6...1.8..74..7...1..3 # websudoku.com # Evil # 26 cells given

Code: Select all
 2 . . | 1 . . | . 3 .
 . 3 5 | . . 4 | . 1 .
 . . 6 | . . . | 9 . 7
-------+-------+------
 . . . | . 1 . | . . .
 7 . . | 2 . 5 | . . 4
 . . . | . 9 . | . . .
-------+-------+------
 5 . 2 | . . . | 6 . .
 . 1 . | 8 . . | 7 4 .
 . 7 . | . . 1 | . . 3
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Re: Solving without pencilmarks

Postby 7b53 » Sun Feb 17, 2013 5:00 am

Pat ; two pairs will do for this puzzle.
not sure your tricky part is...
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Postby Pat » Sun Feb 17, 2013 1:17 pm

yes, 2 pairs

but i also needed 2 lighter moves
(box\line or line\box)

and the tricky thing (for me)
is where the next cell (c8 2) is solved only by combining 2 branches of the solution-path

    A. c2 duo solves just 1 cell
    but i'm happy as it creates the next move = c8 duo;
    i'm even willing to push on to a 3rd move = (8) r3\b2
    except here it seems to peter out
    and i start thinking perhaps all this was just a red herring

    B. (4) r7\b8 solves just 2 cells
    but i'm happy as it creates the next move = (2) r2\b3
    except here it seems to peter out
    and i must now recall the c8 duo,
    combine moves to solve c8 2
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Postby Pat » Tue Feb 26, 2013 11:19 am

Daily SuDoku (2012.Jun.5) wrote:

    4......581...3..9..7...51.....27.98...7...5...46.91.....16...7..2..8...576......3

    28 cells given [ Play ]
Code: Select all
 4 . . | . . . | . 5 8
 1 . . | . 3 . | . 9 .
 . 7 . | . . 5 | 1 . .
-------+-------+------
 . . . | 2 7 . | 9 8 .
 . . 7 | . . . | 5 . .
 . 4 6 | . 9 1 | . . .
-------+-------+------
 . . 1 | 6 . . | . 7 .
 . 2 . | . 8 . | . . 5
 7 6 . | . . . | . .
too easy perhaps,
but i present it as an example of
solving the next cell by combining branches of the solution-path
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Re: Solving without pencilmarks

Postby 7b53 » Wed Feb 27, 2013 3:57 am

Pat wrote:combining branches of the solution-path

=combining branches of eliminations.

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

   

    r3 "hidden" duo {8,9}
    r5 "hidden" duo {3,8}
    b9 "hidden" duo {2,8} --> (4) r7\b8

    combined to solve c4 4


I like the last one...hidden pair {2,8} at box 9=>r7c7<>4 =>
digit 4 (claiming row 7 at box 8)=>r89c4<>4
resulting r2c4=4

not sure is the same path for pencil-marked grid.
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Re: Solving without pencilmarks

Postby Noumenon » Fri Oct 18, 2013 5:20 pm

I see RW has stuck around; I'm glad to see.
One of my latest tricks for solving tough puzzles without pencil marks is strange, so I thought I'd share it: I am simply not filling in the most obvious cells. Without a lot of heavy analysis on why some puzzles become more difficult to solve after some numerals are filled in correctly, I just went with the idea and it's paying off. I will write in correct numerals only very sparingly, usually only the hardest to get right. (where my logic was so complex, I might forget!) The rest seem to often go faster.
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Re: Solving without pencilmarks

Postby gearss » Sat Dec 14, 2013 2:15 am

Puzzle 1
CODE: SELECT ALL
*-----------*
|.2.|1.8|..5|
|1..|.76|.3.|
|7..|4..|...|
|---+---+---|
|..7|...|2..|
|48.|...|.59|
|..2|...|6..|
|---+---+---|
|...|..9|..1|
|.5.|74.|..6|
|6..|8.1|.7.|
*-----------*


First the obvious ones:
r7c2=7 (B)
r1c7=7 (B)
r6c9=7 (B)
r5c6=7 (B)
r8c3=1 (B)

I can’t see anymore obvious cells by drawing lines, but I can see two pairs: r78c1={2,8} (R) and r23c3={5,8} (B). Next I start checking rows, starting with row 1. Two pairs there also: c1&5={3,9} and c3&8={4,6}. This is enough to solve the next number:

[edit: typo there, the first pair should read r78c1={2,8} (Co) = the two numbers cannot go anywhere else in the column.]

r9c3=9 (Co)
------------------------
How about the skill mentioned by RW? I cannot understand why r9c3=9 (Co)? can someone explain it in detail?
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Re: Solving without pencilmarks

Postby gearss » Sat Dec 14, 2013 2:31 am

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



r3 "hidden" duo {8,9}
r5 "hidden" duo {3,8}
b9 "hidden" duo {2,8} --> (4) r7\b8

combined to solve c4 4


I like the last one...hidden pair {2,8} at box 9=>r7c7<>4 =>
digit 4 (claiming row 7 at box 8)=>r89c4<>4
resulting r2c4=4

not sure is the same path for pencil-marked grid.
---------------
why r89c4<>4? please explain it in detail.
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Re: Solving without pencilmarks

Postby gearss » Sat Dec 14, 2013 2:56 am

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


First I check for unique rectangles (these are often not noticed by puzzle makers and provide easy backdoors), seeing potential deadly patterns 89 in r45c34 and 69 in r56c79, which tells me r5c34 and r56c4<>9. A third one in r12c25 tells me that r2c5<>79.
-------------
why "A third one in r12c25 tells me that r2c5<>79"? please explain it in detail.
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Re: Solving without pencilmarks

Postby RW » Wed Dec 18, 2013 5:01 pm

gearss wrote:why "A third one in r12c25 tells me that r2c5<>79"? please explain it in detail.

Code: Select all
 2 . 5 | 4 . . | 3 6 . 
 . . 6 | . . 3 | . 4 2 
 8 4 3 | 2 6 . | . 9 . 
-------+-------+------
 5 6 . | . 4 2 | 7 1 3 
 . 2 . | . 3a1 | 4 5 . 
a1 3 4 | . 5 . | 2 8 . 
-------+-------+------
 3 . . | . 2 6 | . 7 4 
 4 . 2 |a1 . . | 6 3 . 
 6 5 . | 3 . 4 | 9 2 .

You can see that '1' must go in r12c25 in those two boxes. This means that the other two cells in r12c25 must hold two different digits to avoid the deadly pattern. A 7 or 9 in r2c5 would force the same digit into r1c2, which is impossible due to the uniqueness rule.
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Re: Solving without pencilmarks

Postby gearss » Fri Dec 20, 2013 5:26 am

Can Mr. RW solve the following puzzle step by step in detail? Please donot use the word like "easy to know".
If you can solve it without pencilmark, I beleive millions of player in the world must study your article as textbook.
Best Regards,

*-----------*
|.59|3.6|427|
|...|..2|..3|
|2..|.9.|51.|
|---+---+---|
|7..|1..|.3.|
|3..|.6.|...|
|.6.|923|7.4|
|---+---+---|
|.87|63.|..2|
|9..|2..|...|
|612|4.9|37.|
*-----------*
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