The New Sudoku Players' Forum

[wcedge]

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

Thanks,

Charlie

[moggymidge]

Hi Charlie,

Hope the following gets you started:-


The 6 in column 7 has to be either of the ones in box 6, otherwise there wouldn't be a 6 in the column, therefore exclude the 6's in column 7 rows 2 and 3.

In box 3 two cells contain a 2 and a 4 exclusively, exclude the 2's from other cells in box 3.

In box 2 the 2's have to be in row 1 otherwise that row would be without 2's, so exclude all other 2's in the box.


good luck - Moggy

[ravel]

But (according to susser) i fear its getting worse then, you will need more advanced techniques (also an xy-wing does not solve it).

[QBasicMac]

What do you mean "solve by hand"?

I hope you mean you have prepared a sheet like this:

Code: Select all

+------------------+-----------------+---------------+
| 3469  8     3469 | 26   246   5    | 1    7   269  |
| 1469  1249  7    | 8    1246  3    | 246  5   269  |
| 146   124   5    | 7    1246  9    | 246  3   8    |
+------------------+-----------------+---------------+
| 1579  159   19   | 369  8     16   | 367  2   4    |
| 147   6     148  | 23   5     124  | 378  9   137  |
| 2     3     1489 | 69   7     146  | 68   18  5    |
+------------------+-----------------+---------------+
| 8     7     369  | 1    2369  26   | 5    4   23   |
| 1356  15    136  | 4    236   2678 | 9    18  1237 |
| 1349  149   2    | 5    39    78   | 378  6   137  |
+------------------+-----------------+---------------+


That way you can start reducing candidates.

On your worksheet, you can follow moggymidges advice to erase the 6's in column 7, rows 2 and 3. Also the 2's in r12c9. Right? Finally, the 2's in box 2, giving

Code: Select all

+------------------+-----------------+---------------+
| 3469  8     3469 | 26   246   5    | 1    7   69   |
| 1469  1249  7    | 8    146   3    | 24   5   69   |
| 146   124   5    | 7    146   9    | 24   3   8    |
+------------------+-----------------+---------------+
| 1579  159   19   | 369  8     16   | 367  2   4    |
| 147   6     148  | 23   5     124  | 378  9   137  |
| 2     3     1489 | 69   7     146  | 68   18  5    |
+------------------+-----------------+---------------+
| 8     7     369  | 1    2369  26   | 5    4   23   |
| 1356  15    136  | 4    236   2678 | 9    18  1237 |
| 1349  149   2    | 5    39    78   | 378  6   137  |
+------------------+-----------------+---------------+


Is this the kind of work you want to do? Personally, I dislike pencilmarks and get very easy puzzles to solve so I can just stare at the puzzle and fill in squares. But your puzzle is tougher.

The only thing I can add is to point out that r89c6 have the only 78's in box 8, so the 26 can be removed from r8c6.

There may be other stuff, but I can't see it.

Mac

[Cec]

Hi Charlie,
To find out more about posting to the forum including how to post a grid, how cells are defined and other suggested terminology then click on --> Here

".. I dislike pencilmarks .."

I know what you mean Mac. However I notice in the "Sticky" thread that posting "pencilmarks" is mentioned which could be confusing because this is what you have posted though I would prefer it were called a "candidate" grid.
Cec

[moggymidge]

QBasicMac wrote

What do you mean "solve by hand"?

I presume that Charlie means he's one of us who prefer the book/pencil/paper approach to solving rather than using a Sudoku software package of some description.

Each to his own - if you want to dig deep and analyse each puzzle to within an inch of it's life then fair enough - and believe me some of the advanced solving methods that can be applied easily to pencil and paper have been a god send and I applaud those contributors on this forum who have the brain power work out and come up with such things

But some of us are 'old school' and there can't be anything more satisfying (sudoku wise) than completing a 'hard' or a 'fiendish' puzzle in a book/newspaper using only a pencil and eraser with pencil marks (candidates) scribbled all over the place.

I really do believe that the best puzzles are 'solved by hand' even if it does soemtimes take days. Anyway, I know this thread 'man v machine etc. etc.' has already been discussed elsewhere so I'll stop now.

Moggy

[Carcul]

Code: Select all

 *-----------------------------------------------------------*
 | 3469  8     3469  | 26    246   5     | 1     7     69    |
 | 1469  1249  7     | 8     146   3     | 24    5     69    |
 | 146   124   5     | 7     146   9     | 24    3     8     |
 |-------------------+-------------------+-------------------|
 | 1579  159   19    | 369   8     16    | 367   2     4     |
 | 147   6     148   | 23    5     124   | 378   9     137   |
 | 2     3     1489  | 69    7     146   | 68    18    5     |
 |-------------------+-------------------+-------------------|
 | 8     7     369   | 1     2369  26    | 5     4     23    |
 | 1356  15    136   | 4     236   78    | 9     18    1237  |
 | 1349  149   2     | 5     39    78    | 378   6     137   |
 *-----------------------------------------------------------*


[r1c4]{(-6-[r2c5|r3c5])(-6-[r1c1|r1c3])-6-[r1c9]-9-[r1c1|r1c3]-4-[r2c2|r3c2]=(Almost Unique Pattern: r2c257/r3c257)=4|9=[r2c2]=2=[r3c2]=1=[r2c1|r3c1]-1-[r5c1]}=2=[r5c4](=3=[r4c4]-3-[r4c7])-2-[r5c6]=2=[r7c6]-2-[r7c9](-3-[r5c9|r9c9])-3-[r9c6|r9c7]-7-[r9c9]-1-[r5c9](-7-[r5c1])-7-[r4c7]-6-[r4c6]-1-[r5c6]-4-[r5c1]

which means that, if r1c4=6 then r5c1 would be an empty cell - a contradiction. So, r1c4 cannot be "6" and that solve the puzzle.

Carcul

[ronk]

[r1c4]{(-6-[r2c5|r3c5])(-6-[r1c1|r1c3])-6-[r1c9]-9-[r1c1|r1c3]-4-[r2c2|r3c2]=(Almost Unique Pattern: r2c257/r3c257)=4|9=[r2c2]=2=[r3c2]=1=[r2c1|r3c1]-1-[r5c1]}=2=[r5c4](=3=[r4c4]-3-[r4c7])-2-[r5c6]=2=[r7c6]-2-[r7c9](-3-[r5c9|r9c9])-3-[r9c6|r9c7]-7-[r9c9]-1-[r5c9](-7-[r5c1])-7-[r4c7]-6-[r4c6]-1-[r5c6]-4-[r5c1]

which means that, if r1c4=6 then r5c1 would be an empty cell - a contradiction. So, r1c4 cannot be "6" and that solve the puzzle.

Yikes! I don't mean to burst your balloon, but how is that an appropriate response to a first time poster?

[Neunmalneun]

Next step could be this one: There is an AUR (24) in R23C27. R23C2 must contain a "2", so either of the cells can't be "4". Eliniaiton of 4 in R23C2 leaving a single 4 in R9C2.

[Carcul]

Hi Ronk.

Thanks for sharing your opinion. Here is a simpler solution:

1. [r6c4]=9=[r4c4]=3=[r4c7]=6=[r6c7]-6-[r6c4], => r6c4<>6.

2. [r6c6]=4=[r5c6]=2=[r7c6]-2-[r7c9](-3-[r9c9])-3-[r9c6|r9c7]-7-[r9c9]-1-[r8c8]=1=[r6c8]-1-[r6c6], => r6c6<>1.

3. [r5c9]-1-[r5c6]=1=[r4c6](-1-[r4c2])-1-[r4c3]-9-[r4c2]-5-[r8c2]-1-[r8c8]=1=[r6c8]-1-[r5c9], => r5c9<>1.

4. [r5c9]=7=[r9c9]-7-[r9c7]-3-[r7c9]-2-[r7c6]=2=[r5c6]-2-[r5c4]-3-[r5c9], r5c9<>3 and that solve the puzzle.

Regards, Carcul

[ronk]

Here is a simpler solution:

Thanks, that's much easier to understand IMO. Personally, I think expressing the multiple inferences as almost-locked-sets might help a little too. For example ...

[r6c6]=4=[r5c6]=2=[r7c6]-2-[r7c9](-3-[r9c9])-3-[r9c6|r9c7]-7-[r9c9]-1-[r8c8]=1=[r6c8]-1-[r6c6], => r6c6<>1

... could be written ...

r6c6=4=r5c6=2=r7c6-2-r7c9-3-(ALS:r9c679=3|1=r9c9)-1-r8c8=1=r6c8-1-r6c6, => r6c6<>1.

P.S. I stripped the square brackets, as I've never seen them as useful.

[ravel]

After Neunmalneun's UR (type 4) you have:

Code: Select all

 *-----------------------------------------------------------*
 | 3469  8     3469  | 26    246   5     | 1     7     69    |
 | 1469  129   7     | 8     146   3     | 24    5     69    |
 | 146   12    5     | 7     146   9     | 24    3     8     |
 |-------------------+-------------------+-------------------|
 | 1579  159   19    | 369   8     16    | 367   2     4     |
 | 147   6     148   | 23    5     124   | 378   9     137   |
 | 2     3     1489  | 69    7     146   | 68    18    5     |
 |-------------------+-------------------+-------------------|
 | 8     7     369   | 1     2369  26    | 5     4     23    |
 | 1356  15    136   | 4     236   78    | 9     18    1237  |
 | 139   4     2     | 5     39    78    | 378   6     137   |
 *-----------------------------------------------------------*


The next easy step is an xy-wing (from r4c6=16 either r4c3 or r6c4 is 9)
r6c3=9 => r4c3=1 => r4c6=6 => r6c4=9, so r6c4=9

Then the following chain solves the puzzle:
r9c9=1 (=> r9c1<>3) => r8c3=9 (=> r8c5<>9 => r9c5=9 => r9c5<>3)
=> r4c3=1 => r4c6=6 => r7c6=2 => r7c9=3 => r9c79<>3
Because there is no 3 left for row 9 then, r9c9=1.

A nice loop notation:
r9c9-1-r8c3(-9-r8c5=9=r9c5)-9-r4c3-1-r4c6-6-r7c6-2-r7c9-3-r9c79