The New Sudoku Players' Forum

[Led_Zeppelin]

Hey all. First time post here:
OK, maybe I've been looking at this puzzle for too long... I can't seem to figure it out. I've gotten pretty close, but can't find the next step... anyone care to help? Thanks!

Code: Select all

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


[emm]

To move ahead with the puzzle you need to understand the principle of locked candidates.

Start by looking at which rows the 7s can go in across boxes 4, 5 & 6.

You can learn from an expert about this technique - and many more - at Simple Sudoku.

Click on the blue words for a fountain of knowledge.

[ronk]

... the next step. .. anyone care to help?

If you post a candidate list similar to the following, someone can give you specific help.

Code: Select all

 *--------------------------------------------------------------------*
 | 7      5      4      | 3      8      1      | 9      6      2      |
 | 9      2      1      | 6      47     47     | 3      8      5      |
 | 3      6      8      | 25     9      25     | 1      7      4      |
 |----------------------+----------------------+----------------------|
 | 45     3      6      | 1      2      8      | 457    459    79     |
 | 2      1      9      | 457    6      457    | 457    3      8      |
 | 8      47     57     | 9      457    3      | 2      1      6      |
 |----------------------+----------------------+----------------------|
 | 45     8      57     | 2457   1      24579  | 6      459    3      |
 | 1      479    3      | 8      457    6      | 457    2      79     |
 | 6      479    2      | 457    3      4579   | 8      459    1      |
 *--------------------------------------------------------------------*

[Lardarse]

Some locked candidates let you put a 7 in box 2, and then some colouring and an x-wing will help you finish the puzzle.

[QBasicMac]

Some locked candidates let you put a 7 in box 2, and then some colouring and an x-wing will help you finish the puzzle.

Locked candidate 7 in box 4: remove 7 from r6c5
Locked candidate 7 in box 5: remove 7 from r5c7
Locked candidate 5 in box 7: remove 5 from r7c468
Locked candidate 7 in box 9: remove 5 from r8c25
Hidden single 7 at r2c5
Naked single 4 at r2c6
Leads to the marks below, by my calculation.

I can't see the X-Wing you mention.

Looks like time for T&E!
r9c8=4 leads to an invalid solution
r9c8=9 leads to an invalid solution
Therefore, r9c8=5
(Easy after that)

Mac

Code: Select all

754  381  962
921  674  385
368  -9-  174

-36  128  ---
219  -6-  -38
8--  9-3  216

-8-  -1-  6-3
1-3  8-6  -2-
6-2  -3-  8-1


Code: Select all

+-------------+--------------+--------------+
| 7   5    4  | 3    8   1   | 9    6    2  |
| 9   2    1  | 6    7   4   | 3    8    5  |
| 3   6    8  | 25   9   25  | 1    7    4  |
+-------------+--------------+--------------+
| 45  3    6  | 1    2   8   | 457  459  79 |
| 2   1    9  | 457  6   57  | 45   3    8  |
| 8   47   57 | 9    45  3   | 2    1    6  |
+-------------+--------------+--------------+
| 45  8    57 | 247  1   279 | 6    49   3  |
| 1   49   3  | 8    45  6   | 457  2    79 |
| 6   479  2  | 457  3   579 | 8    459  1  |
+-------------+--------------+--------------+


[ronk]

Looks like time for T&E!

Simple coloring is required to proceed beyond that point. If you look at candidate 4, you should find a chain of conjugate pairs, meaning rows, cols, or boxes with only two instances of candidate 4. If one of these candidates is FALSE, the other must be TRUE.

The chain is r8c5(T) to r6c5(F) to r5c4(T) to r5c7(F). If r8c5 is FALSE, then r5c7 is TRUE. [edit: Conversely, if r5c7 is FALSE, then r8c5 is TRUE.] Candidate 4 at intersection of r8 and c7 (r8c7) may be eliminated.

[Lardarse]

Locked candidate 7 in box 4: remove 7 from r6c5
Locked candidate 7 in box 5: remove 7 from r5c7
Locked candidate 5 in box 7: remove 5 from r7c468
Locked candidate 7 in box 9: remove 5 from r8c25
Hidden single 7 at r2c5
Naked single 4 at r2c6
Leads to the marks below, by my calculation.

I can't see the X-Wing you mention.

Code: Select all

+-------------+--------------+--------------+
| 7   5    4  | 3    8   1   | 9    6    2  |
| 9   2    1  | 6    7   4   | 3    8    5  |
| 3   6    8  | 25   9   25  | 1    7    4  |
+-------------+--------------+--------------+
| 45  3    6  | 1    2   8   | 457  459  79 |
| 2   1    9  | 457  6   57  | 45   3    8  |
| 8   47   57 | 9    45  3   | 2    1    6  |
+-------------+--------------+--------------+
| 45  8    57 | 247  1   279 | 6    49   3  |
| 1   49   3  | 8    45  6   | 457  2    79 |
| 6   479  2  | 457  3   579 | 8    459  1  |
+-------------+--------------+--------------+


The X-Wing comes after the colouring. First you have our old friend the Turbot Fish:

The chain is r8c5(T) to r6c5(F) to r5c4(T) to r5c7(F). If r8c5 is FALSE, then r5c7 is TRUE. Conversely, if r8c5 is FALSE, then r5c7 is TRUE. Candidate 4 at intersection of r8 and c7 (r8c7) may be eliminated.

Locked candidates remove the 4 at r4c8.
The X-Wing finally appears, and removes the 4 in r9c2.
Then this happens:


All of those blue squares have their 5 removed. The green squares then become 5 and the puzzle finally gives up its secrets.

[QBasicMac]

If r8c5 is FALSE, then r5c7 is TRUE.
Conversely, if r8c5 is FALSE, then r5c7 is TRUE.

Conversely? LOL! But I get your point. You meant to say this:

If r8c5=4 then r8c7<>4
If r8c5<>4 then r6c5=4, r4c1=4, r5c7=4, r8c7<>4
Thus r8c7<>4

Thanks. I see that that eliminates a candidate, exposing the X-Wing at r68c25, which leads to elimination of 4 in r9c2. In addition it exposes the locked candidate 4 in r79c8 which means r4c8<>4

So using similar logic next, I guess I could say
r7c1=5 >> r6c3=5 >> r8c5=5 >> c9c8=5 >> r4c8<>5
r8c1<>5 >> r8c3=5 >> r4c1=5 >> r4c8<>5
therefore r4c8<>5

Which leads directly to the solution.

I presume you guys all think all this is easier or somehow better than T*E.

Fine.

Mac

[ronk]

If r8c5 is FALSE, then r5c7 is TRUE.
Conversely, if r8c5 is FALSE, then r5c7 is TRUE.

Conversely? LOL! But I get your point. You meant to say this:

If r8c5=4 then r8c7<>4
If r8c5<>4 then r6c5=4, r4c1=4, r5c7=4, r8c7<>4
Thus r8c7<>4

No, I meant to say what I said, but didn't make a very good explanation of one of the coloring rules.

Picture the chain as alternately colored Blue and Green. We don't know which color will end up being true, only that one of them will be. When one of them isn't true, the other one must be true. If Blue is false, Green is true. Conversely, if Green is false, Blue is true.

So if any candidate can "see" ... by looking around in its box, and in its row, and in its column ... both a Blue somewhere ... and a Green somewhere else ... that candidate can be eliminated ... because one of those colors is representing true.

Once the grid is colored, you don't have to follow the chain again.

[ronk]

So using similar logic next, I guess I could say
r7c1=5 >> r6c3=5 >> r8c5=5 >> c9c8=5 >> r4c8<>5
r8c1<>5 >> r8c3=5 >> r4c1=5 >> r4c8<>5
therefore r4c8<>5

Correct! Or you can color the chain (as Lardarse showed), and end up with ...
r4c1(B)-r7c1(G)-r7c3(B)-r6c3(G)-r6c5(B)-r8c5(B)-r8c7(G)-r9c8(B)-r4c8(B)

Now either Blue(B) or Green(G) represents the eventual TRUE. Noting that r4c1 and r4c8 are in the same row, and knowing that there can't be two TRUEs in the row, we conclude that Blue must represent FALSE. So we can eliminate candidate 5 from *all* cells colored Blue ... not just r4c8.

[QBasicMac]

Hi, ronk,

Well, here it is on separate lines for easy comparison:
If
1) =========== r8c5 is FALSE, then r5c7 is TRUE.
Conversely, if
2) =========== r8c5 is FALSE, then r5c7 is TRUE.


I failed to see that 1 and 2 can be converse.

In fact, I still don't see it.

Mac

[ronk]

I failed to see that 1 and 2 can be converse.

In fact, I still don't see it.

OK, OK, I made a typo ... and then missed by own typo. But, frankly, I wasn't looking for a typo ... because you were missing my point about coloring. But then, I guess, my typo didn't help your understanding ... or lack thereof.