The New Sudoku Players' Forum

[Madolite]

take this current hardest puzzle - try solving it with out using a guess and test method ps if you do succumb to trial and error this is a back door size 3 puzzle which requires 3 different clues added to the puzzle at the same time to reach a single solution.

Thanks for underlining my point that using pencil marks important, if not outright necessary.

[eleven]

Can you solve this one with pencilmarks ? It is easy without:

Code: Select all

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

[SpAce]

Can you solve this one with pencilmarks ? It is easy without

I have no idea how to solve it without pm, but with them it wasn't very hard. Such a solution is obviously off-topic here, but mine is below anyway. Wouldn't mind if you enlightened us with the no-pm way!

Hidden Text: Show

1. Grouped X-Loop: 5r4c(4=1) - 5r(5=2)c3 - 5b2p(56=1) - loop => -5 r5c2,r2c12,r57c4
2. AIC: (6=285)r647c7 - (5=8)r7c1 - 8r4c(1=78) => -8 r7c7
3. Kraken Cell (678)r1c7 + Kraken Column 2c1: => -5 r1c4 (*)
4. Loop: (6=7)r1c4 - r1c78 = (7-6)r2c8 = 6r1c78 - loop => -67 r1c12, -289 r2c8; lclste

Code: Select all

(*)

6r1c7 - (6=2)r6c7 - (2)r6c1
||                  ||
||                  (2-1)r2c1 = (1-5)r2c3 = r5c3 - 5r4c(1=4)
||                  ||
||                  (2-5)r4c1 = 5r4c4
||
7r1c7 - r1c1 = (17-2)r26c1 = (2-5)r1c4 = 5r4c4
||
8r1c7 - (8=25)r4c74


=> -5 r1c4

[eleven]

Code: Select all

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

The puzzle is by ssxsssxs, the technique by qiuyanzhe. It is a very nice extension (one of three) of RW's reverse BUG lite.
In columns 5 and 6 (same stack) you can see a triple 124, with only one number in different rows. So it has distinction 1.
Rows 2 and 3 (same band) have the same numbers 3 an 4. If 1 or 5 would be in r2c1 (can't be in r2c8), you would have a triple too, with one extra digit in r3, it would have distinction 1 too.
But this is not possible, if the puzzle is unique. Therefore 1 must be in r2c3, and the puzzle is solved very easy.

And yes, it is very rare, that you can use that technique, but so cool.

[SpAce]

Thanks, eleven! Can't say that I fully understand the technique, but cool it is! I just checked that Hodoku's default solution for this puzzle takes a whopping 23 steps, so it's a pretty effective step indeed (and quite a bit more elegant than Hodoku's one-step net option). I doubt I'll be adding it to my solving arsenal any time soon, though.

[qiuyanzhe]

I have noticed the existence of this topic for long but have never read through the text. Today is my first time browsing them, and as a competitive sudoku player, I have my things to share. There is so much thing so I would just tell a little bit now.

In competition any marking(or try&error, even try-without-error) is allowed, and the only criterium is the solving time. Personally my act is similar to RW's, but I would make some Pencilmarks, but limited to Double candidates, or cells with "unordinary" pencilmark deletions. I learned from grandmasters to mark intersections/pairs between cells, and summarized what deleting a candidate can make. Marking bivalue cells could help discovering (Naked Pairs), (Naked Singles after Intersection/..), (XY-Wing-like chains), or (ab-bc-ca triplets)..

Also I would keep the pencilmarks in bivalue cells ordered. If we have
12 23 13
in a minirow, I would mark
12 23 31,
such that two possibility of all these cells have two possibilities-- all left or all right. If any of their columns has the same property, then ordered bivalue can spread and eliminate some candidates.
This is like branching, and I think it suits competing.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
And I wonder what this topic should be about.
When I first see this topic, instead of "How to solve without marking on the grid", I thought this topic is about "How to discover(and use) informations other than pencilmarks" as pencilmark is not the only form of information. I think this is more open than what we discussed in the beginning.

In 999_Spring's post, the 67-UR becomes the bottleneck of some of the puzzles, while I immediately see that UR is useless. There is a Sue De Coq, which is also what I saw.
SE:UR(with hidden triplet, 4.7, eliminate 3@r5c5)
SE sees the UR as a non-candidate information"r56c6 cannot both be 67", while r3c6's 67 directly shows the same. But it can only utilize such kind of information directly in a UR. And if we change the 67 pair to multivalue "half-ur"(like a 67-75-56) SE uses forcing chains. This confuses me as two problems, one is "why only two-value URs count", and the other is "Why only UR-form non-candidate information can be utilized".
Below is the puzzle (Simplified the UR part).

Code: Select all

+----------------+----------------+----------------+
| 6    2    9    | 3    1    5    | 7    4    8    |
| 5    7    1    | 9    4    8    | 6    3    2    |
| 3    4    8    | 27   26   #67  | 1    5    9    |
+----------------+----------------+----------------+
| 8    5    4    | 12   23   139  | 39   6    7    |
| 2    9    *67  | 4    356  367  | 35   8    1    |
| 1    3    *67  | 57   8    679  | 59   2    4    |
+----------------+----------------+----------------+
| 9    8    2    | 15   35   13   | 4    7    6    |
| 4    1    5    | 6    7    2    | 8    9    3    |
| 7    6    3    | 8    9    4    | 2    1    5    |
+----------------+----------------+----------------+

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
During years of my sudoku career I have used non-candidate methods uncountable number of times, while the first time I discover it was when drawing 6*6 jigsaw sudoku empty grids. I found that some shapes can directly determine equalities.

Code: Select all

o           o     
o o q...--> o a . .   o o       -->   a b
p o o o     a o o o     o o o o     a b o o o o

These equalities can be used to construct jigsaw sudokus that JSudoku even fail to solve logically.

Code: Select all

a a a a b b
c a a b b d
c c b b e d
c c e e e d
c e e f f d
f f f f d d with column 2=1~6

And the first time I used equation in an actual puzzle is like this(when I was 10 and haven't heard of this forum):
Original Puzzle:...........3.....1.8..9..7......6....7....8.....1.3.9.........679..5........4...3 (edit:I cannot find the source anyway.. Bought a book and didn't find the puzzle)

Code: Select all

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

Above is the bottleneck of a 17-cell puzzle, by a UR we can get an 8 in r7c6 but it doesn't break the puzzle. Its bottleneck is a chain on 4(SE7.7).
And here's what I solved it:
According to r5 and c5, (suppose)a=45=r4c4=r5c9, then r3c3=a(hidden single in row), and r9c2=a(hidden single in column), a≠4, a=5.

Such ways can be used in multiple ways, it is hard to tell with a fixed pattern, as another example in a PK is:

Original puzzle: 689.2..3.2..7.9.......3...8.61..4...............8..39.8...4.......5.6..3.7..1.542 (source:oubk.com)
After singles

Code: Select all

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

r9c1=39, hidden single in b4 we have r5c2=r9c1=39, then we get an intersection on 2, so r6c6=7.

It also works on some 999999111/ 999991110-like puzzles. Partial grids solvable by this have various SE ratings up to 8.9 ...
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Still a bit too long.. For things not about specific grids I would post to Coffee Bar: What I See Here before the new year.

[eleven]

A puzzle from 999_spring's list

Code: Select all

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

a in r1c6 must be in r2c8 too, 4 in r1c79
almost x-wing 4r46c67 with fin in r1c7

Code: Select all

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

or 4r1c7, set r6c8=b

Code: Select all

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

then r7c7=b, r4c6=b and r6c6=4
=> r6c1<>4

[eleven]

Another one from there:

Code: Select all

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

a in r1c6 and r2c8 is 2 or 3,

Code: Select all

 +-------+-------+-------+
 | 6 7 5 | 9 8 a | 1 . 4 |
 | 8 - - | 7 5 - | 9 a 6 |
 | x x 9 | - - 6 | 8 5 7 |
 +-------+-------+-------+
 | 7 8 . | 5 . 9 | . 6 2 |
 | 9 5 . | . 6 7 | . 1 8 |
 | x x 6 | 8 * - | 5 7 9 |
 +-------+-------+-------+
 | 5 # 7 | . 9 - | 6 8 1 |
 | - 6 8 | - 7 5 | 2 9 3 |
 | z 9 z | 6 y 8 | 7 4 5 |
 +-------+-------+-------+

almost x-wing ar36c12,
fin r6c5 - r9c5 = r9c13
=> ar7c4

where can b be then ?
not in r2c6 (UR), not in r3c1 (locked)

Code: Select all

 +-------+-------+-------+
 | 6 7 5 | 9 8 a | 1 b 4 |
 | 8 . . | 7 5 - | 9 a 6 |
 | - - 9 | . . 6 | 8 5 7 |
 +-------+-------+-------+
 | 7 8 . | 5 . 9 | . 6 2 |
 | 9 5 . | . 6 7 | . 1 8 |
 | # . 6 | 8 . # | 5 7 9 |
 +-------+-------+-------+
 | 5 . 7 | a 9 # | 6 8 1 |
 | - 6 8 | - 7 5 | 2 9 3 |
 | # 9 z | 6 . 8 | 7 4 5 |
 +-------+-------+-------+

skyscraper br69c1,r67c6 => br7c6
=> r7c46=23

[numpl_npm]

Can you solve this one with pencilmarks ? It is easy without:

Code: Select all

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


This puzzle is not so hard.

With basic techniques only, to next figure.

Code: Select all

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


Either +5r4c1 or +5r4c4, in both case +2r5c8.

If +5r4c1, then ...
+5r1c4 (a), +67r56c4 (b), +2r4c4 (c).
+5r2c3 (d), +1r2c1 (e), +2r6c1 (f),
+7r1c1 (g), +7r2c8 (h), +2r5c8 (i).

Code: Select all

 +-------+-------+-------+
 | g . 4 | a 1 2 | . . . |
 | e . d | 3 . . | 4 h . |
 | 3 . . | 4 . . | 1 5 . |
 +-------+-------+-------+
 | 5 3 6 | c 4 1 | . . 7 |
 | 4 . . | b . . | 3 i . |
 | f . . | b . . | . 4 5 |
 +-------+-------+-------+
 | . 1 7 | . . 4 | . . 6 |
 | . . 2 | 1 . . | . . . |
 | . . 3 | 8 2 . | 9 . . |
 +-------+-------+-------+


If +5r4c4, then ...
+9r7c4 (a), +[67]r1c4 (b), +[67]r2c8 (c), +[67]r3c2 (d), +2r3c9 (e).
6 or 7 in r1c78 (f), so -67r1c12. Then
+267r2c12.r3c2 (g h d), +1r2c3 (i), +1r6c1 (j), +7r2c1 (g),
+6r2c8 (c), +6r6c7 (k), +2r4c1 (l), +2r5c8 (m).

Code: Select all

 +-------+-------+-------+
 | . . 4 | b 1 2 | f f . |
 | g h i | 3 . . | 4 c . |
 | 3 d . | 4 . . | 1 5 e |
 +-------+-------+-------+
 | l 3 6 | 5 4 1 | . . 7 |
 | 4 . . | . . . | 3 m . |
 | j . . | . . . | k 4 5 |
 +-------+-------+-------+
 | . 1 7 | a . 4 | . . 6 |
 | . . 2 | 1 . . | . . . |
 | . . 3 | 8 2 . | 9 . . |
 +-------+-------+-------+


+2r5c8, then
with basic techniques only to the solution.

[eleven]

If +5r4c4, then ...
+9r7c4 (a), +[67]r1c4 (b), +[67]r2c8 (c), +[67]r3c2 (d) ...

Code: Select all

 +-------+-------+-------+
 | . . 4 | b 1 2 | f f . |
 | g h i | 3 . . | 4 c . |
 | 3 d . | 4 . . | 1 5 e |
 +-------+-------+-------+
 | l 3 6 | 5 4 1 | . . 7 |
 | 4 . . | . . . | 3 m . |
 | j . . | . . . | k 4 5 |
 +-------+-------+-------+
 | . 1 7 | a . 4 | . . 6 |
 | . . 2 | 1 . . | . . . |
 | . . 3 | 8 2 . | 9 . . |
 +-------+-------+-------+

why ?

And how many nets did you try until you found this useful one ? What is your way to determine, which starting points are promising ?

[numpl_npm]

If +5r4c4, then ...
+9r7c4 (a), +[67]r1c4 (b), +[67]r2c8 (c), +[67]r3c2 (d) ...

why ?

Code: Select all

 +-------+-------+-------+
 | y y 4 | b 1 2 | f f . |
 | . . . | 3 . . | 4 c . |
 | 3 d . | 4 . . | 1 5 . |
 +-------+-------+-------+
 | . 3 6 | 5 4 1 | . . 7 |
 | 4 . . | x . . | 3 . . |
 | . . . | x . . | . 4 5 |
 +-------+-------+-------+
 | . 1 7 | a . 4 | . . 6 |
 | . . 2 | 1 . . | . . . |
 | . . 3 | 8 2 . | 9 . . |
 +-------+-------+-------+


a not in 267, so bxx=267, and b=6 or b=7 ( b=[67] ).

If b=6 => c=6. If b=7 => c=7. So c=[67].
And 6 or 7 in ff, so 67 both in bff and 67 both not in yy.

If c=6 => d=6. If c=7 => d=7. So d=[67].

how many nets did you try

It is the first trial. In many cases, only one attempt is made.

What is your way to determine, which starting points are promising ?

In this way, continuous guess is necessary.
But there are few starting points where continuous guess can be made.

In this puzzle, 2 seems important.
There are many places where 2 can be placed, but there are many restrictions.

6 and 7 constrain 2. But these are not sufficient constraints.
Looking for other digits, 5 seems effective.

[eleven]

Ah thanks, should have seen the 67's.