## Solving without pencilmarks

Advanced methods and approaches for solving Sudoku puzzles
7b53 wrote:

With/without pencil marks

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 . | . . . `

Pat

Posts: 4056
Joined: 18 July 2005

i distinctly remember that 7b53 solved the above puzzle ( joel64 # 191 # 2 )
and posted a response — which has now vanished

Pat

Posts: 4056
Joined: 18 July 2005

### Re: Solving without pencilmarks

The logs show that 7b53 deleted their own post back at the end of November.

JasonLion
2017 Supporter

Posts: 642
Joined: 25 October 2007
Location: Silver Spring, MD, USA

### Re: Solving without pencilmarks

yes, I deleted the post myself. hope didn't spoiled anything.
7b53
2012 Supporter

Posts: 156
Joined: 01 January 2012
Location: New York

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 `

Pat

Posts: 4056
Joined: 18 July 2005

### Re: Solving without pencilmarks

Pat ; two pairs will do for this puzzle.
not sure your tricky part is...
7b53
2012 Supporter

Posts: 156
Joined: 01 January 2012
Location: New York

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

Pat

Posts: 4056
Joined: 18 July 2005

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

Pat

Posts: 4056
Joined: 18 July 2005

### Re: Solving without pencilmarks

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.
7b53
2012 Supporter

Posts: 156
Joined: 01 January 2012
Location: New York

### Re: Solving without pencilmarks

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.
Noumenon

Posts: 6
Joined: 25 August 2010

### Re: Solving without pencilmarks

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?

Posts: 4
Joined: 09 December 2013

### Re: Solving without pencilmarks

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.

Posts: 4
Joined: 09 December 2013

### Re: Solving without pencilmarks

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.

Posts: 4
Joined: 09 December 2013

### Re: Solving without pencilmarks

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.
RW
2010 Supporter

Posts: 1010
Joined: 16 March 2006

### Re: Solving without pencilmarks

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.|
*-----------*