The New Sudoku Players' Forum

[Go Mifune]

I have managed to figure this one out through trial and error but I dont like the feeling of guessing at it can someone please tell me how to solve this without guessing I am losing my mind on this. Thanks

129 538 674
573 ### 189
486 #1# 523
95# 2## 34#
34# #95 #62
268 374 951
835 ### #16
694 ### 23#
712 ##3 #9#

[Go Mifune]

this is the puzzle as it is published\

*2* |5** |*74
*73 |*** |***
4** |*** |5*3
9** |2** |3**
*4* |*9* |*6*
**8 |**4 |**1
8*5 |*** |**6
*** |*** |23*
71* |**3 |*9*

[MCC]

Look at box 7, bottom left.

There is only one for a 2.
Once you placed the 2 then there is only one place for a 4.

Go Mifune, it would help us if you also posted a candidate grid.


MCC

[re'born]

Go Mifune,

I assume that you have reached this point:

Code: Select all

  *-----------------------------------------------------------*
 | 1     2     9     | 5     3     8     | 6     7     4     |
 | 5     7     3     | 46    246   26    | 1     8     9     |
 | 4     8     6     |*79    1    *79    | 5     2     3     |
 |-------------------+-------------------+-------------------|
 | 9     5     17    | 2     68    16    | 3     4     78    |
 | 3     4     17    | 18    9     5     | 78    6     2     |
 | 2     6     8     | 3     7     4     | 9     5     1     |
 |-------------------+-------------------+-------------------|
 | 8     3     5     |*479   24   *279   |-47    1     6     |
 | 6     9     4     | 178   58    17    | 2     3     578   |
 | 7     1     2     | 468   4568  3     | 48    9     58    |
 *-----------------------------------------------------------*


Consider the cells in r3c4, r3c6, r7c4, r7c6. If r7c4 does not equal 4 and r7c6 does not equal 2, then we would get a deadly pattern on <79>. Therefore either r7c4 = 4 or r7c6 = 2. However, by inspection, we also have r7c4=9 or r7c6 = 9. Hence, it is not possible that r7c4 = 7 or that r7c6 = 7. Consequently, r7c7 = 7 and from here the puzzle is solved.

An alternative way to think about this deduction is: If r7c7 = 4, then r7c5 = 2 and so r7c4 does not equal 4 and r7c6 does not equal 2, leaving a deadly pattern.

[re'born]

Go Mifune,

Another path if you don't like unique rectangles is the following chain:

[r8c9]-7-[r8c6]=1=[r4c6]=6=[r4c5]=8=[r4c9]=7=[r8c9] => r8c9 <> 7.

I've never understood the above notation very well (and I have probably stated my chain incorrectly), so let me state it differently using notation that makes sense to me.

(8,9)7 > (8,6)1 > (4,6)6 > (4,5)8 > (4,9)7, a contradiction.

After this step, it should be all singles.

[re'born]

Sorry, but one last solution.

Code: Select all

  *-----------------------------------------------------------*
 | 1     2     9     | 5     3     8     | 6     7     4     |
 | 5     7     3     | 46    246   26    | 1     8     9     |
 | 4     8     6     | 79    1     79    | 5     2     3     |
 |-------------------+-------------------+-------------------|
 | 9     5     17    | 2     68    16    | 3     4     78    |
 | 3     4     17    | 18    9     5     | 78    6     2     |
 | 2     6     8     | 3     7     4     | 9     5     1     |
 |-------------------+-------------------+-------------------|
 | 8     3     5     | 479   24    279   | 47    1     6     |
 | 6     9     4     | 178  *58    17    | 2     3    *578   |
 | 7     1     2     | 468  *4568  3     | 48    9    *58    |
 *-----------------------------------------------------------*


The cells in r8c5, r8c9, r9c5, r9c9 form a unique rectangle coinciding with an x-wing on the 5's (see this thread for more details) and so we can eliminate 5 from r8c9 and r9c5.

The downside of this solution is that it still requires another advanced step (assuming that you continue to avoid the previous two solving methods I described) before it yields to singles. The upside is that when you applying coloring at this point, it is far more effective than before (coloring on 8 will eliminate candidates from 5 cells, 3 of which are now naked singles).

[Carcul]

I've never understood the above notation very well

The correct notation is the following:

[r8c9]-7-[r8c6]-1-[r4c6]-6-[r4c5]-8-[r4c9]-7-[r8c9], => r8c9<>7.

Regards, Carcul

[re'born]

I've never understood the above notation very well

The correct notation is the following:

[r8c9]-7-[r8c6]-1-[r4c6]-6-[r4c5]-8-[r4c9]-7-[r8c9], => r8c9<>7.

Regards, Carcul

Thanks Carcul. I'll have to go back and read the notation post again. I always thought that you could use =x= when the two cells form a conjugate pair with respect to x.

[Carcul]

Code: Select all

 *-----------------------------------------------------------*
 | 1     2     9     | 5     3     8     | 6     7     4     |
 | 5     7     3     | 46    246   26    | 1     8     9     |
 | 4     8     6     | 79    1     79    | 5     2     3     |
 |-------------------+-------------------+-------------------|
 | 9     5     17    | 2     68    16    | 3     4     78    |
 | 3     4     17    | 18    9     5     | 78    6     2     |
 | 2     6     8     | 3     7     4     | 9     5     1     |
 |-------------------+-------------------+-------------------|
 | 8     3     5     | 479   24    279   | 47    1     6     |
 | 6     9     4     | 178   58    17    | 2     3     578   |
 | 7     1     2     | 468   4568  3     | 48    9     58    |
 *-----------------------------------------------------------*


[r8c6]=1=[r4c6]=6=[r4c5]-6-[r279c5]=(AUR: r2789c5/r89c9)=2,4,6|7=[r8c9]-7-[r8c6], => r8c6<>7 which solves the puzzle.

Carcul

[Sped]

Code: Select all

 *-----------------------------------------------------------*
 | 1     2     9     | 5     3     8     | 6     7     4     |
 | 5     7     3     | 46    246   26    | 1     8     9     |
 | 4     8     6     | 79    1     79    | 5     2     3     |
 |-------------------+-------------------+-------------------|
 | 9     5     17    | 2     68    16    | 3     4     78    |
 | 3     4     17    | 18    9     5     | 78    6     2     |
 | 2     6     8     | 3     7     4     | 9     5     1     |
 |-------------------+-------------------+-------------------|
 | 8     3     5     | 479   24    279   | 47    1     6     |
 | 6     9     4     | 178   58    17    | 2     3     578   |
 | 7     1     2     | 468   4568  3     | 48    9     58    |
 *-----------------------------------------------------------*


The XY chain

5-(r8c5)-8-(r4c5)-6-(r4c6)-1-(r4c3)-7-(r4c9)-8-(r9c9)-5

eliminates the 5s in r8c9 and r9c5 and allows the puzzle to be solved.

Would the correct notation be:

[r8c9]-5-[r8c5]-8-[r4c5]-6-[r4c6]-1-[r4c3]-7-[r4c9]-8-[r9c9]-5-[r8c9], => r8c9<>5
[r9c5]-5-[r8c5]-8-[r4c5]-6-[r4c6]-1-[r4c3]-7-[r4c9]-8-[r9c9]-5-[r9c5], => r9c5<>5

??

[lb2064]

..........Another path if you don't like unique rectangles is the following chain:

[r8c9]-7-[r8c6]=1=[r4c6]=6=[r4c5]=8=[r4c9]=7=[r8c9] => r8c9 <> 7.

I've never understood the above notation very well (and I have probably stated my chain incorrectly), so let me state it differently using notation that makes sense to me......

Actually, I beleive this chain does work but leads to r8c6 <> 7.

Using the same notation and cells that u have but rewritting it, we have:

[r8c6]=1=[r4c6]=6=[r4c5]=8=[r4c9]=7=[r8c9]-7-[r8c6] => r8c6 <> 7

[Sped]

Actually, I beleive this chain does work but leads to r8c6 <> 7.

Using the same notation and cells that u have but rewritting it, we have:

[r8c6]=1=[r4c6]=6=[r4c5]=8=[r4c9]=7=[r8c9]-7-[r8c6] => r8c6 <> 7

Code: Select all

 *-----------------------------------------------------------*
 | 1     2     9     | 5     3     8     | 6     7     4     |
 | 5     7     3     | 46    246   26    | 1     8     9     |
 | 4     8     6     | 79    1     79    | 5     2     3     |
 |-------------------+-------------------+-------------------|
 | 9     5     17    | 2     68    16    | 3     4     78    |
 | 3     4     17    | 18    9     5     | 78    6     2     |
 | 2     6     8     | 3     7     4     | 9     5     1     |
 |-------------------+-------------------+-------------------|
 | 8     3     5     | 479   24    279   | 47    1     6     |
 | 6     9     4     | 178   58    17    | 2     3     578   |
 | 7     1     2     | 468   4568  3     | 48    9     58    |
 *-----------------------------------------------------------*


The chain, in my preferred notation, is this:

7-(r8c6)-1-(r4c6)-6-(r4c5)-8-(r4c9)-7

It starts on r8c6 and meanders to r4c9. It allows the exclusion of 7s in cells that see both the start of the chain at r8c6 and the end of the chain at r4c9. r8c9 sees both r8c6 and r4c9 so it can have its 7 excluded.

The more traditional way of expresing the chain is this:

[r8c9]-7-[r8c6]-1-[r4c6]-6-[r4c5]-8-[r4c9]-7-[r8c9], => r8c9<>7

r8c6 is one of the ends of the chain, not a cell to have its 7 excluded.

[lb2064]

.....r8c6 is one of the ends of the chain, not a cell to have its 7 excluded.

Not necessarily true Sped. For a discontinuous nice loop chain that meets the Theorem 5 requirements (which is what I have written) you do not need both cells at the end to have the -7 link between the cells at each end.

At the beginning of the chain I wrote the [r8c6]=1=[r4c6] => [r8c6] <> 1 => [r8c6] is 7 thus forming the contradiction with the end of the chain.

I know it's not a common (traditional) type of chain to write but it does meet the Nice Loop notation as specified by Jeff in his reference to Theorem 5 here

[Sped]

Not necessarily true Sped. For a discontinuous nice loop chain that meets the Theorem 5 requirements (which is what I have written) you do not need both cells at the end to have the -7 link between the cells at each end.

At the beginning of the chain I wrote the [r8c6]=1=[r4c6] => [r8c6] <> 1 => [r8c6] is 7 thus forming the contradiction with the end of the chain.

I know it's not a common (traditional) type of chain to write but it does meet the Nice Loop notation as specified by Jeff in his reference to Theorem 5 here

Heh. I obviously have some work to do before I get a handle on these loops.

I'll be studying the reference you gave. When I looked at it before my eyes sort of glazed over. It seems to make more sense on second look though.

[Go Mifune]

thanks for all of your help guys.
MCC I am new to this but as soon as I figure out how to post the puzzles that way I will, Cheers!