The New Sudoku Players' Forum

[claudiarabia]

Code: Select all

. 5 . . . . . 7 .
9 . . . . . . . 3
. . 8 1 . 2 6 . .
7 . . . . . . . 2
. 1 . 3 . 7 . 4 .
2 . . . 9 . . . 1
. . 2 6 . 8 5 . .
4 . . . . . . . 9
. 8 . . . . . 2 .

SE 6-8 X-Wing, xy-wing, 4 bi-directional Y-cycles, BUG Type 2 in the End.

Happy solving!

Claudia

[claudiarabia]

Code: Select all

6 1 . . 4 . . . .
. 9 8 . 1 7 . . .
. . . 8 . . . . .
8 . . . . 9 . 7 .
. . 7 2 . . . 4 5
. . . . 3 . 2 . .
. . 6 . 9 . . 1 .
1 . . . . . . 6 9
. 7 . . . 3 . . 4


SE 6-6
just want to see if this nasty empty page will disappear now.

Claudia

[re'born]

Claudia, well done turning someone else's mistake into something that kept me entertained for far too long. I love the puzzles where you end up with a nearly cleaned up board and then you can apply tons of different techniques to solve it. ...

The sudoku on the wikipedia-page on the top right is very easy and has sooo many initial clues. After removing some clues I got a fantastic sudoku with 22 clues only. It is to solve with three xy-wings and a bi-directional Y-cycle.

Code: Select all

5 3 . . 7 . . . .
. . . 1 . 5 . . .
. 9 8 . . . . 6 .
8 . . . 6 . . . .
4 . . 8 . 3 . . 1
. . . . 2 . . . .
. 6 . . . . 2 8 .
. . . 4 . 9 . . 5
. . . . . . . 7 .


Happy solving

Claudia

Very cool Claudia! It also solves in one step with a semi-remote naked pair (nee Y-wing style) on <13> that implies r8c5<>3.

[re'born]

Code: Select all

6 1 . . 4 . . . .
. 9 8 . 1 7 . . .
. . . 8 . . . . .
8 . . . . 9 . 7 .
. . 7 2 . . . 4 5
. . . . 3 . 2 . .
. . 6 . 9 . . 1 .
1 . . . . . . 6 9
. 7 . . . 3 . . 4


SE 6-6
just want to see if this nasty empty page will disappear now.

Claudia

Here is a one step "simple" 3D-coloring solution (simple in that it only uses two colors):
[r7c1]=3=[r5c1]=9=[r5c7]-9-[r3c7]-4-[r2c7]=4=[r2c1]-4-[r7c1], => r7c1<>4.

[claudiarabia]

Code: Select all

2 . . . 9 . 5 . .
. 7 . . . 3 . . .
. . . 4 . . . 6 .
. . 4 . . 9 . . 2
9 . . . . . . . 8
5 . . 6 . . 7 . .
. 6 . . . 7 . . .
. . . 5 . . . 4 .
. . 3 . 8 . . . 1


This sudoku has various techniques, among them Unique Rectangle Typ 3, every common kind of wing and Turbot fish.

Happy solving

Claudia

[claudiarabia]

Today I had my sudoku-storming phase. I just typed in some numbers very fast and stopped until the solution was valid. This came out. It is to solve with one bi-directional cycle consisting of two numbers and one sky-craper and one turbot-fish.

Code: Select all

. . . . 7 . . . 2
. 5 . . . . . 9 .
6 . 3 . . 5 . . 7
8 . . . . 9 4 . .
. 1 . . . . . . .
. . . 2 . . . 6 .
. 9 . . . 4 . . 1
7 . . 6 . . 2 . .
. . 5 . . . . . 8


Happy solving

Claudia

[re'born]

Today I had my sudoku-storming phase. I just typed in some numbers very fast and stopped until the solution was valid. This came out. It is to solve with one bi-directional cycle consisting of two numbers and one sky-craper and one turbot-fish.

Code: Select all

. . . . 7 . . . 2
. 5 . . . . . 9 .
6 . 3 . . 5 . . 7
8 . . . . 9 4 . .
. 1 . . . . . . .
. . . 2 . . . 6 .
. 9 . . . 4 . . 1
7 . . 6 . . 2 . .
. . 5 . . . . . 8


Happy solving

Claudia

I found a bicycle (your hint that it contained two numbers was surprisingly helpful), which led me to the following one-step solution (found with 3D-coloring):
[r2c1]-2-[r7c1]=2=[r7c3]=8=[r1c3]=9=[r1c1]=1=[r2c1], =>[r2c1]<>2.

Edit: Or perhaps you might enjoy the following proof of the same deduction that I found using the methods of my recent post. One almost has a simple xy-chain (perhaps it should be called a finned xy-chain ):

Code: Select all

 *--------------------------------------------------------------------*
 | 19     48*    489    | 3      7      6      | 5      148    2      |
 | 12-    5      7      | 148    1248   128    | 3      9      6      |
 | 6      24(8)% 3      | 9      12     5      | 18     148    7      |
 |----------------------+----------------------+----------------------|
 | 8      237    26     | 157    1356   9      | 4      12     35     |
 | 2359   1      2469   | 4578   34568  378    | 78     28     359    |
 | 359    347    49     | 2      13458  1378   | 178    6      359    |
 |----------------------+----------------------+----------------------|
 | 23*    9      28     | 578    358    4      | 6      37     1      |
 | 7      38*    1      | 6      9      38     | 2      5      4      |
 | 4      6      5      | 17     123    1237   | 9      37     8      |
 *--------------------------------------------------------------------*


except for the 8 in r3c2. But as this sees the other two 8's in the chain, all is good and we can still eliminate 2 from r2c1. As an ALS xz-rule application, it would be A={2,3} on r7c1, B={2,3,4,8} on r138c2, x=3, z=2.

[claudiarabia]

Could it be qualified ?

Code: Select all

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



Reborn posted a very good solution. I have only a shaky hypothesis.
Here is the pencilmark-grid before the forcing chain:

Code: Select all

 *-----------------------------------------------------------*
 | 59    56    8     | 13    7     13    | 2     46    49    |
 | 67    1     4     | 89    2     89    | 56    3     57    |
 | 37    29    23    | 4     5     6     | 1     79    8     |
 |-------------------+-------------------+-------------------|
 | 19    24    7     | 1259  6     1249  | 8     25    3     |
 | 24    3     5     | 27    8     247   | 9     1     6     |
 | 189   89    6     | 1259  3     129   | 4    *257   27    |
 |-------------------+-------------------+-------------------|
 | 38    48   *213   | 6     9     5     | 7     4-2   12    |
 | 46    7     9     | 23    1     23    | 56    8     45    |
 | 25    56    12    | 78    4     78    | 3    *269  *219   |
 *-----------------------------------------------------------*


I propose a BUG-lite type 2 solution. Regardless of the wild middle box and r6c1 we can see that in r7c3, r9c98 and r6c8 we have 4 cells with the candidate 2 as the third candidate in the cell, which is showing up at least three times in it's row, column and box. All these 2s are seen by the 2 in r7c8. After eliminating the 2 in r7c8 you can set single numbers and the puzzle is solved. I went not that deep into BUG-Lite-Theorie but for me it seems logical that, when most cells in the puzzle are bivalual and when so many candidate-buddies as third candidate in a cell see the same number in one cell, this seen candidate can be removed. I will observe this phänomenon if it is only a special coincidence in this puzzle or if one can generalize.

With best regards

Claudia

[re'born]

Could it be qualified ?

Code: Select all

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



Reborn posted a very good solution. I have only a shaky hypothesis.
Here is the pencilmark-grid before the forcing chain:

Code: Select all

 *-----------------------------------------------------------*
 | 59    56    8     | 13    7     13    | 2     46    49    |
 | 67    1     4     | 89    2     89    | 56    3     57    |
 | 37    29    23    | 4     5     6     | 1     79    8     |
 |-------------------+-------------------+-------------------|
 | 19    24    7     | 1259  6     1249  | 8     25    3     |
 | 24    3     5     | 27    8     247   | 9     1     6     |
 | 189   89    6     | 1259  3     129   | 4    *257   27    |
 |-------------------+-------------------+-------------------|
 | 38    48   *213   | 6     9     5     | 7     4-2   12    |
 | 46    7     9     | 23    1     23    | 56    8     45    |
 | 25    56    12    | 78    4     78    | 3    *269  *219   |
 *-----------------------------------------------------------*


I propose a BUG-lite type 2 solution. Regardless of the wild middle box and r6c1 we can see that in r7c3, r9c98 and r6c8 we have 4 cells with the candidate 2 as the third candidate in the cell, which is showing up at least three times in it's row, column and box. All these 2s are seen by the 2 in r7c8. After eliminating the 2 in r7c8 you can set single numbers and the puzzle is solved. I went not that deep into BUG-Lite-Theorie but for me it seems logical that, when most cells in the puzzle are bivalual and when so many candidate-buddies as third candidate in a cell see the same number in one cell, this seen candidate can be removed. I will observe this phänomenon if it is only a special coincidence in this puzzle or if one can generalize.

With best regards

Claudia

I don't yet see the logic behind your deduction, but I'll keep thinking about it. In the meantime, from your point there is a uniqueness deduction that, while it doesn't solve the puzzle, does reduce it to down to an xy-wing. Observe the famous UR meets X-wing in r79c39<12> with strong links on the 1's. This implies that r7c3, r9c9<>1.

[ronk]

I have only a shaky hypothesis.
Here is the pencilmark-grid before the forcing chain:

Code: Select all

 *-----------------------------------------------------------*
 | 59    56    8     | 13    7     13    | 2     46    49    |
 | 67    1     4     | 89    2     89    | 56    3     57    |
 | 37    29    23    | 4     5     6     | 1     79    8     |
 |-------------------+-------------------+-------------------|
 | 19    24    7     | 1259  6     1249  | 8     25    3     |
 | 24    3     5     | 27    8     247   | 9     1     6     |
 | 189   89    6     | 1259  3     129   | 4    *257   27    |
 |-------------------+-------------------+-------------------|
 | 38    48   *213   | 6     9     5     | 7     4-2   12    |
 | 46    7     9     | 23    1     23    | 56    8     45    |
 | 25    56    12    | 78    4     78    | 3    *269  *219   |
 *-----------------------------------------------------------*


I propose a BUG-lite type 2 solution. Regardless of the wild middle box and r6c1 we can see that in r7c3, r9c98 and r6c8 we have 4 cells with the candidate 2 as the third candidate in the cell, which is showing up at least three times in it's row, column and box.Claudia

Sorry, but those four cells do not comprise a BUG-Lite. Indeed, your deduction based on those four cells is not even valid.

Why? It fails backtesting! Placing the eliminated candidate instead doesn't cause a contradiction somewhere in the set of cells that allegedly "causes" the elimination.

[Ocean]

Thanks JPF and claudiarabia for the several nice contributions during my vacation!!

Will try to categorize and label the puzzles, when I get some more time (will be away from home pretty much the next couple of weeks also.) I don't have an automatic tool to do this, and while Sudoku Explainer can be of great help, it's still a bit time-consuming, and requires a proper "mood of thinking". Thanks to re'born and ronk for providing solutions and comments to the puzzles!

[claudiarabia]

Code: Select all

. . . . . . . . .
. . 8 3 . . . 5 .
. 7 . . 6 . 2 . 4
9 . . . . 8 . . 3
. 6 . . . . . 1 .
2 . . . . 9 . . 6
. 4 . . 2 . 9 . 7
. . 6 1 . . . 8 .
. . . . . . . . .


SE 7.0
This happy fish has stranded in your bi-cycle-thread. It is to solve by X-wing, xy-wing, UR Type 4 and a bi-directional cycle.

Claudia

[claudiarabia]

wandering through my thread I re-discovered this nice 19-clue sudoku made by JPF. His "planetary system" can be solved with x-wing, 2xy-wings, xyz-wing, 2 turbotfishes, a unique rectangle type 4 and on bi-directional y-cycle.

Code: Select all

. . . . 1 2 3 . .
. . . 4 . . . . .
. . 5 . . . . . 6
. 2 . . . . . . 7
3 . . . 8 . . . 9
7 . . . . . . 6 .
4 . . . . . 2 . .
. . . . . 9 . . .
. . 9 6 7 . . . .


SE 6.8



Claudia

[claudiarabia]

Code: Select all

. . 3 . . . . . .
. 6 . . . . 4 3 1
2 . . . . . . 7 .
. 4 9 1 . 3 . 6 .
7 . . . 5 . . . 2
. 2 . 9 . 7 . 1 .
. . . 2 . . 6 . .
. . . 3 1 . 2 . 8
. . . . . 9 . . .


Very assymmetric but next to a bi-directionyl y-cycle you will find a wxyz-wing here too.

Claudia

[claudiarabia]

Code: Select all

. . . . 6 2 . . .
. . . 4 . . 9 . .
. . 2 . . . . 1 .
1 . . 9 . . . . .
3 . . . 8 6 5 . .
4 . . . 2 . . 7 .
. 8 . 5 . . . . 6
. . 6 . . . . 4 .
. . . . . 1 3 . .
 

SE 7.2

This Sudoku can be solved by some triplets and in the end a bi-directional cycle (with a pair as node when starting in a certain cell). It eliminates 4 candidates leaving a BUG Type 1.

Happy solving

Claudia