Advanced methods and approaches for solving Sudoku puzzles

I've got this far and and am totally stuck, despite much time looking for a logical next step. Can someone suggest one please?

7*1 239 64*
6*2 145 **7
*49 867 *1*

4** *5* *7*
9** *8* *61
*6* *2* *53

*9* *18 724
*** 47* 5**
*74 59* 1*6

tnx, Mike
MikeF

Posts: 11
Joined: 07 September 2005

[Edit: Argh, I missed the boat with this post!
My hint did not lead anywhere ... and I realize
now that Simple Sudoku is very well known ...

There is a naked pair in the 1st column.

Let me recommend a program to you called Simple Soduku,
which is available free here:

http://www.angusj.com/sudoku/

In just a few seconds, I pasted your puzzle into the
program, clicked on the Hint button, and
the program reported the hint I wrote above.
It's a great program!

Cheers,
Nick
Last edited by Nick67 on Sat Sep 17, 2005 4:48 pm, edited 1 time in total.
Nick67

Posts: 113
Joined: 24 August 2007

From the naked pair I can solve this by testing candidates 3,5 in r3,c1 and from there through boxes 2 and 3 and back to 1.

From this I deduce r3,c1 can't be 5 otherwise r1,c2 would also be 5. Is this what is called forcing chain(or just T&E)?

Is there another way of doing it eg Xwing of 2s in r8 and r9 that doesn't require T&E?
emm

Posts: 987
Joined: 02 July 2005

I'm sorry, I can't quite see how you were able to
deduce that r3,c1 is not 5. Could you please
explain once more?

Thanks,
Nick

P.S.

For the record, I took this route:
I eliminated 3 as a candidate
from r8,c1 and r9,c1, based on the naked pair.
Next, I eliminated 8 as a candidate from r4,c2 and r8,c2.
This is because 8 appears as a candidate in box 1 only in c2.

Next, I eliminated 2 as a candidate from r8,c2, because
2 appears as a candidate in box 2 only in c2.

Similarly, I then eliminated 3 from r2,c8, because
3 appears as a candidate in box 9 only in c8.

But then ... I'm stuck. Actually, I confess that I
used the Hint Button in Simple Sudoku to find all
of the above. And when I reach this point, and
click the button again, the program reports
"No hint available".

So ... a lot of help I am!
Nick67

Posts: 113
Joined: 24 August 2007

Hmmm! Strangely, I can't figure out my own reasoning not even with all the brilliant hieroglyphics in the margin. I expect I made a mistake that resulted in the 3 being placed in r3,c1 which unlocked the puzzle.

Well if SS can't solve it I may as well pick up my pencil and move on.

PS : I can't believe I missed the locked candidates I was so busy looking for something complicated!
emm

Posts: 987
Joined: 02 July 2005

I wonder if there is indeed a good reason for
placing 3 in r3,c1? As you suggested, it
is exactly the right move! But why?
Maybe you were on the right track when
you suggested there might be a forcing
chain somewhere?

Just by the way, there is a nice description of
forcing chains here:

http://www.simes.clara.co.uk/programs/sudokutechnique7.htm

... but I can't figure out how to find the starting point
of a forcing chain in our puzzle.
Nick67

Posts: 113
Joined: 24 August 2007

A 4 at R6C6 and it all falls apart aswell - might be an easier place to start looking for a chain.

stuartn
stuartn

Posts: 211
Joined: 18 June 2005

I also wonder if this puzzle can be
solved using one of the
(I'm guessing that the solution
requires more than simple coloring
techniques, since SS doesn't give any hints.
I also tried simple coloring manually ...)

BTW, apologies to MikeF for not recognizing
earlier that this is a tough puzzle!
Nick67

Posts: 113
Joined: 24 August 2007

stuartn wrote:A 4 at R6C6 and it all falls apart aswell - might be an easier place to start looking for a chain.
stuartn

I only found a rather long chain for that:
r1c2=8->r2c2=3->r8c2=1->r4c2<>1->r6c1=1->r6c6=4
r1c2=5->r1c9=8->r3c9=5->r3c7=2->r5c7=4->r5c6=3->r6c6=4
Wolfgang

Posts: 208
Joined: 22 June 2005

Many thanks for for all the replies. I guess that, apart from forcing chains, there is no other logical approach here that might work. The various naked pairs -- like the 35s in col 1 -- don't seem to help further reduce candidates in a way that solves the puzzle. Personally, I hate forcing chains or any T&E type approach.
MikeF

Posts: 11
Joined: 07 September 2005

Do you consider xy-chain an T&E approach? If yes, don't read further.

Starting r6c3, [78]-[81]-[14]-[43]-[37] => r6c4<>7 => r6c4=9 => r5c4=7 and r6c3=7

Starting r6c1, [18]-[82]-[23]-[34]-[41] non-repetitive => r8c1<>8, r4c6<>3 and r8c6<>3

Starting r8c2, [31]-[12]-[26]-[63] => r7c1<>3 => r7c1=5

There are so many xy-chains, but these 3 are sufficient to solve the puzzle. There is another non-repetitive one after the 9 and 7s are fixed if you would like to find it as an exercise.
Jeff

Posts: 708
Joined: 01 August 2005

That's pretty impressive Jeff. No way could I spot all that with mere pencil and paper in hand. Thanks very much for outpointing.

Mike
MikeF

Posts: 11
Joined: 07 September 2005

MikeF

Actually, xy-wing, xy-chain, xyz-wing and xyz-chains are much easier to spot than x-wing, swordfish, turbot fish and colours (some people may disagree), techniques I am sure you would consider logical. This is because the formers don't require any filtering and are therefore most suitable for direct pencil/rubber on newspaper application. It all comes down to practice as to how many of these patterns you could spot in a puzzle.

As to forcing chains, it can be T&E, but then again can be not T&E at all. It hinges on whether you have a system to identify these chains. Forcing chain itself is not T&E, but usually the process of finding these chains can be T&E. Using a bilocation/bivalue combination plot, coupled with a set of proven rules, I can identify forcing chains without T&E. It is a time consuming process, but in return you would enjoy full satisfaction for being able to solve the puzzle logically.
Last edited by Jeff on Sat Sep 10, 2005 2:08 am, edited 1 time in total.
Jeff

Posts: 708
Joined: 01 August 2005

I found that Nick70's coloring technique is also a
good way to solve this puzzle.
Here is the puzzle after adding candidates and
doing a few standard reductions:

Code: Select all
` 7     58    1    |     2     3     9    |    6     4    586     38    2    |     1     4     5    |    389   89   735    4     9    |     8     6     7    |    23    1    25----------------------------------------------------------4     123  38    |     369   5    136   |    289   7    2899     235  357   |     37    8    34    |    24    6    118    6    78    |     79    2    14    |    489   5    3----------------------------------------------------------35    9    356   |     36    1    8     |    7     2    4128  13    368   |     4     7    236   |    5     389  8928   7     4     |     5     9    23    |    1     38   6 `

Next, let's add coloring, starting with cell (6,1).
(I'll place colors directly below candidates.)
There are many conjugate pairs, so many of
the candidates can be colored.

Code: Select all
`7     58    1    |     2     3     9    |    6     4    58 6     38    2    |     1     4     5    |    389   89   735    4     9    |     8     6     7    |    23    1    25----------------------------------------------------------4     123  38    |     369   5    136   |    289   7    289      BB                          A9     235  357   |     37    8    34    |    24    6    1      A                           AB         BA   18    6    78    |     79    2    14    |    489   5    3AB                                BA         B----------------------------------------------------------35    9    356   |     36    1    8     |    7     2    4128  13    368   |     4     7    236   |    5     389  89B    AB28   7     4     |     5     9    23    |    1     38   6`

There are 2 candidates with the B color in cell (4,2),
so B must be false. Therefore A must be true.

So, every candidate with the A color is the correct
candidate for its respective cell, and every candidate
with the B color can be eliminated, leading to
this:

Code: Select all
`7     58    1    |     2     3     9    |    6     4    58 6     38    2    |     1     4     5    |    389   89   735    4     9    |     8     6     7    |    23    1    25----------------------------------------------------------4     3    38    |     369   5    1     |    289   7    2899     2    357   |     37    8    3     |    4     6    11     6    78    |     79    2    4     |    89   5    3----------------------------------------------------------35    9    356   |     36    1    8     |    7     2    428    1    368   |     4     7    236   |    5     389  8928   7     4     |     5     9    23    |    1     38   6 `

From here the solution is straightforward.
[/quote]

Thanks to Nick70 for this coloring technique,
which is described here:

http://www.setbb.com/phpbb/viewtopic.php?t=77&sid=f1a9110279421a8b358d722f515d3729&mforum=sudoku
Nick67

Posts: 113
Joined: 24 August 2007

post cancelled
Last edited by Jeff on Sat Sep 10, 2005 12:26 pm, edited 1 time in total.
Jeff

Posts: 708
Joined: 01 August 2005

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