The New Sudoku Players' Forum

[hana somekh]

Hi !

I would really appreciate some help in finishing off the following sudoku logically :

Here is where I got so far :

2 4 * 9 * 1 7 * 5

5 * 9 4 * 7 * * *

* 7 * 8 2 5 9 4 *


4 * * 5 1 3 * 9 7

7 5 * * 4 9 3 1 8

9 * * 7 8 * 5 * 4


8 * 4 * 5 * * 7 9

* * 7 * 9 8 4 5 *

6 9 5 1 7 4 8 3 2

The solution to the puzzle is :

243 961 785
589 437 621
176 825 943

468 513 297
752 649 318
931 782 564

824 356 179
317 298 456
695 174 832

Thanks,
Hana.

[PaulIQ164]

According to the solver I use (http://sudoku.sourceforge.net), the puzzle can't be solved any further without using trial and error (or at least, without using the 'Nishio' technique, which is essentially trial and error. Where did this puzzle come from?

[simes]

It can be solved with either colouring or forcing chains.

Forcing chains:
Either candidate of r2c2 forces r1c5 to be 6
r2c2=6 => r2c5=3 => r1c5=6
r2c2=8 => r1c3=3 => r1c5=6

Colouring 6:
r7c6=6, so r6c6<>6, so r6c8=6, so r4c7<>6
this gives a double exclusion on r7c7, so 6 can be eliminated.

Simes
(BTW, colouring will be in V2.2 of my solver)

[PaulIQ164]

Ah yes, the solver I use doesn't do those methods. Anyway, they're still a bit trial-and-errory, but that's a whole nother thread.

[hana somekh]

Simes,

Thank you very much for the colouring technique solution !

I did try the colouring technique for the digit 6, however, somehow missed that ! Thank you for pointing it out to me !

Hana.

[tso]

According to the solver I use (http://sudoku.sourceforge.net), the puzzle can't be solved any further without using trial and error (or at least, without using the 'Nishio' technique, which is essentially trial and error. Where did this puzzle come from?

Actually, that solver doesn't make the claim the puzzle *can't* be solved without trial and error, only that IT cannot do so. It all depends on what tactics are implemented. Pappocom considers puzzles that require swordfish in the T&E category.

Setting asided that you'll never be convinced that forcing chains aren't a little bit T&E, coloring isn't in the slightest. It may be misleading that Simes gave the *results* of the coloring tactic as a chain. The *results* of x-wing can also be written as a chain.

He could have described it this way:

Code: Select all

  .  .  . | .  6  . | .  6  .
  . -6  . | .  6  . |+6  6  .
  .  . +6 | .  .  . | .  . -6
  --------+---------+--------
  . +6  . | .  .  . |-6  .  .
  .  . -6 |+6  .  . | .  .  .
  .  .  . | .  . -6 | . +6  .
  --------+---------+--------
  .  .  . | .  . +6 |-6  .  .
  .  .  . |-6  .  . | .  . +6
  .  .  . | .  .  . | .  .  .

Pairs of cells are labled + or - if they are the only two in a group that can contain a 6. There is no particular chain required. No matter where you start, an exclusion will be made. In this case, there are two 6's with the same sign in column 7 [correction, thanks Pauliq164] -- they both can't be 6, so they both must NOT be 6. You can fill in ALL the "+" as sixes in one fell swoop.

Forcing chains are a proof, but coloring is a tactic to FIND a proof.

[PaulIQ164]

But you're still making a hypothesis, and rejecting it if it leads to an impossible situation, aren't you? Sorry if I'm not understanding it, I'm afraid I'm not fully read up on colouring.

[simes]

It may be misleading that Simes gave the *results* of the coloring tactic as a chain.

<shrug> I couldn't think how to represent it.

S

[Jeff]

Colouring 6 can be viewed as a turbot fish shown in amber with solid arrow => strong link and dotted arrow => weak link.

There is also a swordfish of 6s shown in pink.

These options can be considered before using double forcing chains.

[Jeff]

There is also an xy-chain, which can be seen without filtering. Same as the turbot fish or colouring 6, the 6 in r7c7 is eliminated.

[hana somekh]

Jeff,

I had found the swordfish, however, thank you very much for pointing out crystal clear that there is indeed a turbot fish in the puzzle !! Extremely helpful


Hana

[hana somekh]

Yet another difficult sudoku which I am stuck with !

Apart from the sudoku puzzle I posted originally which with your help I have found the solution rather easily, there is another one I am trying to solve and would really appreciate some help with :

I have got this far :

* * * 2 * 7 6 * 3
* * 3 6 1 * 4 * 7
* 6 * 4 3 8 5 * 1

3 1 6 * 8 2 * 4 *
2 * * * 6 1 3 * 8
* 8 * 3 4 * 2 1 6

6 7 9 8 * 4 1 3 *
4 * * * 7 3 8 6 *
* 3 * * * 6 * * 4

The solution to the puzzle is :

541 297 683
823 615 497
967 438 521

316 782 945
294 561 378
785 349 216

679 854 132
452 173 869
138 926 754


Hana.

[Nick70]

But you're still making a hypothesis, and rejecting it if it leads to an impossible situation, aren't you?

Not in the slightest.

Sorry if I'm not understanding it, I'm afraid I'm not fully read up on colouring.

If you don't understand it, why do you dismiss it as T&E?

[Karyobin]

Ooohh, a bit of needle creeping in...

You're not Australian are you by any chance Nick?

Were you the 70th Nick on the prison ship?

[PaulIQ164]

Well, I've studied it a bit, and I remain unconvinced. Seems to me you're still trying two possible sets of places for the sixes, and rejecting the one that leads to an error.

And by the way:

there are two 6's with the same sign in box 6 -- they both can't be 6, so they both must NOT be 6.

The two 6s in box 6 have the different signs. The only unit that has two sixes of the same sign that I can see is column 7. Is this what was meant.