The New Sudoku Players' Forum

[yorm]

I'm stuck on another one. If someone could please help me understand the technique used to make progress I'd be grateful.

Here's the original:

Code: Select all

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


Here's what I've been able to accomplish:

Code: Select all

*-----------------+-----------------+-----------------*
|   139  6    8   |   14   5    2   |   49   7    13  |
|   123  124  127 |   6    17   9   |   48   5    138 |
|   159  14   157 |   147  8    3   |   2    149  6   |   
|-----------------+-----------------+-----------------|
|   4    12   6   |   157  127  15  |   3    8    9   |
|   8    5    9   |   3    4    6   |   1    2    7   |
|   7    3    12  |   19   129  8   |   5    6    4   |   
|-----------------+-----------------+-----------------|
|   1256 7    3   |   159  19   145 |   4689 149  258 |
|   156  8    145 |   2    3    7   |   469  149  15  |
|   125  9    1245|   8    6    145 |   7    3    125 |   
*-----------------------------------------------------*


Thanks!

- Oh, by the way, this is from "Su Doku for Dummies" - puzzle 186 (tough)

[TKiel]

Yorm,

There is a naked quad in row 7 (or a hidden triple, whichever you prefer), which leads to an x-wing. Neither directly solve the puzzle, though.

Tracy

[QBasicMac]

Yorm,

There is a naked quad in row 7 (or a hidden triple, whichever you prefer), which leads to an x-wing. Neither directly solve the puzzle, though.

Tracy

Triple 268 in row 7 leads to c1=26, c7=68 c9=28
X-Wing r4c4-r7c6 leads to r9c6=14

That gets here

Code: Select all

+----------------+---------------+---------------+
| 139  6    8    | 14   5    2   | 49   7    13  |
| 123  124  127  | 6    17   9   | 48   5    138 |
| 159  14   157  | 147  8    3   | 2    149  6   |
+----------------+---------------+---------------+
| 4    12   6    | 157  127  15  | 3    8    9   |
| 8    5    9    | 3    4    6   | 1    2    7   |
| 7    3    12   | 19   129  8   | 5    6    4   |
+----------------+---------------+---------------+
| 26   7    3    | 159  19   145 | 68   14   28  |
| 156  8    145  | 2    3    7   | 469  149  15  |
| 125  9    1245 | 8    6    14  | 7    3    125 |
+----------------+---------------+---------------+


Errk! You're right. That didn't help much.

Mac

[re'born]

Here is an ugly path towards a solution:

(9,3)2 > (7,1)6 which creates a deadly pattern ([89],[19])<15>. Therefore, (9,3)!2. From here, looking at 2's, we easily conclude (2,1)!2.

(8,7)4 > (2,7)8 & (1,7)9 which creates a deadly pattern ([12], [19])<13>. Therefore, (8,7)!4. From here, looking at 4's, we easily conclude (3,8)!4.

Now a long series of xy-wings and multicoloring will solve the puzzle.

I sincerely hope to see a much better solution soon, as I am going nuts trying to find one.

[Carcul]

Code: Select all

 *-----------------------------------------------------------*
 | 139   6     8     | 14    5     2     | 49    7     13    |
 | 123   124   127   | 6     17    9     | 48    5     138   |
 | 159   14    157   | 147   8     3     | 2     149   6     |
 |-------------------+-------------------+-------------------|
 | 4     12    6     | 157   127   15    | 3     8     9     |
 | 8     5     9     | 3     4     6     | 1     2     7     |
 | 7     3     12    | 19    129   8     | 5     6     4     |
 |-------------------+-------------------+-------------------|
 | 26    7     3     | 159   19    145   | 68    14    28    |
 | 156   8     145   | 2     3     7     | 469   149   15    |
 | 125   9     1245  | 8     6     14    | 7     3     125   |
 *-----------------------------------------------------------*


[r7c9](-2-[r7c1]-6-[r8c1])-2-[r8c9|r9c9]{=(Almost Unique Rectangle:r8c1/r8c9/r9c1/r9c9)=2=[r9c1]-2-[r9c3]}-1-[r7c8|r8c8]=1=[r3c8]=9=[r3c1]=5=[r3c3](=7=[r3c4]-7-[r4c4])=7=[r2c3]=2=[r6c3](-2-[r4c2]-1-[r4c4])-2-[r6c5]=(Almost Unique Rectangle:r6c4/r6c5/r7c4/r7c5)=2|5=[r7c4]-5-[r4c4]

from which we can conclude that r7c9=2 would make r4c4 an empty cell. So, we must have r7c9<>2 and the puzzle is solved.

Carcul

[Neunmalneun]

Can't see why this is an ugly path. Seems logic and elegant to me. But I am still wondering how you found the 9 and the 3 in R23C6. Could not find a solution for these candidates.

[re'born]

Carcul,

Maybe it is the 1:30 in the morning talking, but that solution is frickin' beautiful! I worked all night on the plethora of AUR's and never thought to employ two of them. You are my hero.

[re'born]

Neunmalneun,

I think there must be a typo in your post. R2C6 is a given and R3C6 comes easily with singles.

[ronk]

(9,3)2 > (7,1)6 which creates a deadly pattern ([89],[19])<15>. Therefore, (9,3)!2. From here, looking at 2's, we easily conclude (2,1)!2.

Starting with QBasicMac's candidates: after the AUR deduction, there remains a disjoint x-wing and swordfish in 2s.

Code: Select all

+----------------+---------------+---------------+
| 139  6    8    | 14   5    2   | 49   7    13  |
| 13   124  127  | 6    17   9   | 48   5    138 |
| 159  14   157  | 147  8    3   | 2    149  6   |
+----------------+---------------+---------------+
| 4    12   6    | 157  127  15  | 3    8    9   |
| 8    5    9    | 3    4    6   | 1    2    7   |
| 7    3    12   | 19   129  8   | 5    6    4   |
+----------------+---------------+---------------+
| 26   7    3    | 159  19   145 | 68   14   28  |
| 156  8    145  | 2    3    7   | 469  149  15  |
| 125  9    145  | 8    6    14  | 7    3    125 |
+----------------+---------------+---------------+

How did you then conclude (2,1)!2? [edit: Sorry, I "read" that as (2,2)!2.]

[ravel]

How did you then conclude (2,1)!2?

Box elimination, i suppose.
This is a very nice puzzle and i like the way, rep'nA solves it, 2 AUR's, 3 xy-wings. I did not finish with coloring, but a 5-cell bivalue chain, which was easy to spot.

[ronk]

How did you then conclude (2,1)!2?

Box elimination, i suppose.

If you mean line-box interaction, I don't see it ... and Angus' Simple Sudoku provides no hint at that stage either.

This is a very nice puzzle and i like the way, rep'nA solves it, 2 AUR's, 3 xy-wings.

I like rep'nA's solution too, and that's why I'm trying to follow each step.

[ravel]

If you mean line-box interaction, I don't see it ...

Hm, after the elimination of the 2 in r9c3 with the AUR there are only two 2's in box 7 (r7c1 and r9c1), which kill the one in r2c1.

[ronk]

If you mean line-box interaction, I don't see it ...

Hm, after the elimination of the 2 in r9c3 with the AUR there are only two 2's in box 7 (r7c1 and r9c1), which kill the one in r2c1.

Duh! Sorry to waste your time, and think I'll get another cup of coffee. I was reading that as r2c2<>2, which is a next step ... a non-trivial step ... I'd worked out.

Code: Select all

*139  6    8    | 14   5    2    |*49   7    13
 123 *124  127  | 6    17   9    |*48   5    138
*159  14  *157  |*147  8    3    | 2    149  6
----------------+----------------+---------------
 4   *12   6    |*157 *127  15   | 3    8    9
 8    5    9    | 3    4    6    | 1    2    7
 7    3    12   | 19   129  8    | 5    6    4
----------------+----------------+---------------
 26   7    3    | 159  19   145  |*68   14   28
 156  8    145  | 2    3    7    |*469  149  15
 125  9    1245 | 8    6    14   | 7    3    125

A pure bilocation chain (not to be confused with a pure strong inference chain):

r2c2-2-r4c2=2=r4c5=7=r4c4-7-r3c4=7=r3c3=5=r3c1=9=r1c1-9-r1c7=9=r8c7=6=r7c7=8=r2c7=4=r2c2 implying r2c2<>2
... and with two subsequent xy-wings, this solves the puzzle. But I like the solution by rep'nA better.

[ravel]

r2c2-2-r4c2=2=r4c5=7=r4c4-7-r3c4=7=r3c3=5=r3c1=9=r1c1-9-r1c7=9=r8c7=6=r7c7=8=r2c7=4=r2c2

Nice, and i think it can be made a bit shorter after r1c1-9-
r1c1-9-r1c7-4-r2c7=4=r2c2
(but maybe it is no pure bilocation chain then ? I dont know that - ah, i guess this is it)

Hi Carcul,
please can you wrap the long chains - the window is getting too wide, so we have to shift around then.

[re'born]

Ravel and Ron,

Thanks for taking the time to go through my deductions in detail. Ravel, I've been trying to find your 5 cell bivalue chain that solves the puzzle. I came up with the following:

Starting from just after the 2 AUR's, an xy-wing shows that (3,4)!1 and (1,9)!1. After cleaning up a little, another xy-wing shows that (2,[23])!1. A third xy-wing shows that (6,5)!1. This leaves us with...

Code: Select all

 

 *--------------------------------------------------*
 | 19   6    8    | 14   5    2    | 49   7    3    |
 | 3    24   27   | 6    17   9    | 48   5    18   |
 | 159  14   157  | 47   8    3    | 2    19   6    |
 |----------------+----------------+----------------|
 | 4    12   6    | 157  127  15   | 3    8    9    |
 | 8    5    9    | 3    4    6    | 1    2    7    |
 | 7    3    12   | 19   29   8    | 5    6    4    |
 |----------------+----------------+----------------|
 | 26   7    3    | 159  19   145  | 68   14   28   |
 | 156  8    145  | 2    3    7    | 69   149  15   |
 | 125  9    145  | 8    6    14   | 7    3    125  |
 *--------------------------------------------------*

 


Now (6,4)1 > (6,5)9 > (7,5)1 > (2,5)!1 > (1,4)1, a contradiction. Therefore (6,4)9 and this solves the puzzle.

Is this what you had in mind? In principle, I like this solution better than the one by coloring. In practice, I am much better at finding coloring deductions than finding forcing chains.