The New Sudoku Players' Forum

[richardm]

As a matter of maximizing my pleasure from sudoku I have not read any account of how to solve this puzzle. To date I have managed to solve all Superior and Super Fendish puzzles by use of direct logic - exept one.

I have heard that guessing is not required, but I wonder what is meant by guessing? Does this mean the one never has to resort to reductio ad absurdum methods? I.e. assume a vale and show that it leads to a contradiction thus eliminating the assumed value as a possibility.

As I mentioned above I have always found a way to avoid this type of logic except in one case.

I have also found that one or two puzzles have paris linked in such a way that toggling one of the pairs between the two possible positions shows that some other potential value can never be selected. There is no guessing involved in this method but it's very close to redicutio ad absurdum that I wonder whether this method is also not necessary in a well formed puzzle.

Finally I sometimes spot certain arrangement of a given digit where there's usually a diagnoal pair in a 3x3 square such that selecting one of the two in the diagonal then the other will also who that some possibilites can never be selected. Hence these are eliminated. This is similat to the case above but involves only on digit. Again, is this logic permitted or does it come under the category of guessing?

I am thinking of puzzles published in the Times and Sunday Times.

Richard

[udosuk]

Richard, perhaps you would consider to post the actual puzzle you can't solve by "logic"... Otherwise people have no idea what you're talking about...

[richardm]

OK, here is an example taken from Sunday Times' Superior 53 6th August 2006

Startting position:

3.. ... .1.
..1 3.. .9.
... ..7 3..

67..3..5.
8.. .9. ..1
.4. .2. .86

..6 4.. ...
.2. ..8 5..
.1. ... ..9

This readily reduces to:

3.. ... .1.
..1 3.. .9.
... ..7 3..

679 831 .5.
852 694 731
143 725 986

..6 4.. 1..
.2. ..8 5.3
.18 ..3 ..9

At this point I can readily sole the puzzle if I can prove that cell d1=2 (I'm using chess board notation: a1 is bottom left and i9 is top right).

By examining the two bottom left 3x3 squares one can see fairly easily that 9, 2 and 5 are inter-related. Furthermore cell d1 can only take the values 2 or 5.

At this point I can employ a reductio ad absudum argument (i.e. trial and error) by first assuming d1=5, showing that this leads to a1=5 which is a contradiction. Hence d1=2.

Once d1=2 is established, the puzzle soves very quickly.

I am disappointed in having to resort to this type of argument. So I am hoping that there is a direct method that doesn't require my testing the supposition that d1=5.

If what I have done is indeed the intended crucial step then I am very disappointed with this puzzle. Since, the type of argument employed can be used as a lazy man's way to get out of any tight spot.

Hope that explains what I am getting at?[/code]

[daj95376]

Your puzzle solves with Naked/Hidden Singles and two Naked Triples. You need to start fresh and maybe it'll work for you this time. Good Luck.

[udosuk]

From your position:

Code: Select all

 *-----------*
 |3..|...|.1.|
 |..1|3..|.9.|
 |...|..7|3..|
 |---+---+---|
 |679|831|.5.|
 |852|694|731|
 |143|725|986|
 |---+---+---|
 |..6|4..|1..|
 |.2.|..8|5.3|
 |.18|..3|..9|
 *-----------*

 *--------------------------------------------------------------------*
 | 3      689    457    | 259    4568   269    | 2468   1      24578  |
 | 2457   68     1      | 3      4568   26     | 2468   9      24578  |
 | 2459   689    45     | 1259   14568  7      | 3      246    2458   |
 |----------------------+----------------------+----------------------|
 | 6      7      9      | 8      3      1      | 24     5      24     |
 | 8      5      2      | 6      9      4      | 7      3      1      |
 | 1      4      3      | 7      2      5      | 9      8      6      |
 |----------------------+----------------------+----------------------|
 | 579   *39     6      | 4      57     29     | 1      27    *278    |
 | 479    2      47     | 19     167    8      | 5      467    3      |
 | 457    1      8      | 25     567    3      | 246    2467   9      |
 *--------------------------------------------------------------------*

There are hidden singles in r7c2 and r7c9 ("b3" & "i3" according to your chess-like system)...

Then a few box-line interactions (locked candidates) and a naked triple solves the puzzle easily...

No need for any guess work...

[richardm]

Thankis. You've restored my faith in these puzzles. Let me take another look at this one.

[Pat]

and a naked triple

hey udosuk -- as usual, you like the "naked" trio -- i prefer the "hidden" trio in the same row --

[udosuk]

hey udosuk -- as usual, you like the "naked" trio -- i prefer the "hidden" trio in the same row --

Hey, I'm an Aussie bloke... We all prefer "naked" stuffs rather than "hidden" stuffs...

As for trios or duos... The more the merrier!

(Think about the Shane Warne "naked trio" photos in the tabloids... )

Happy New Year!

[daj95376]

There are hidden singles in r7c2 and r7c9 ("b3" & "i3" according to your chess-like system)...

Then a few box-line interactions (locked candidates) and a naked triple solves the puzzle easily...

No need for any guess work...

If you check, the Box-Line interactions can be replaced by one Naked Triple. I've run into this often enough, that my solver's hierarchy now has Naked Triples before Locked Candidates.

[richardm]

Code: Select all

b3 and i3 I forgot to copy into my posting. But yes, the triple in row g makes the solution fall out quite quickly. Don know why I didn't see it.

Back to my original questoin about what constitutes a guess.  The following puzzle came from American Way the in-flight mag. Don't know whether this puzzle originates from pappocom, appologes if it doesn't. The starting position is:

+---+---+---+

|641|382|579| 9

|2..|174|6..| 8

|8..|.9.|412| 7

+---+---+---+

|51.|.3.|92.| 6

|.26|.1.|35.| 5

|..4|.2.|1.7| 4

+---+---+---+

|462|853|791| 3

|1..|267|84.| 2

|...|941|2..| 1

+---+---+---+

 abc def ghi


Which quickly reduces to:

+---+---+---+

|641|382|579| 9

|2..|174|6..| 8

|8..|.9.|412| 7

+---+---+---+

|51.|.3.|92.| 6

|.26|.1.|35.| 5

|..4|.2.|1.7| 4

+---+---+---+

|462|853|791| 3

|1..|267|84.| 2

|...|941|2..| 1

+---+---+---+

 abc def ghi

So how to proceed from here?

The key seems to be to eliminate 8 from h4, which you do by looking at the permutations in the two 3x3 squares to the left. Permuting the possible unknowns shows that 8 can only be in row 4 in b4 or f4. But isn't this type of trial-and-error logic tantermount to guessing,


Similarly one often encounters opposite row/column pairs and deduces that these are also column/row pairs. E.g


8.. a   b   | c  8... d    | e  f   g   |
... ... ... | ... ... ... | ... ... ... |
... ... ... | ... ... ... | ... ... ... |

... ... ... | ... ... ... | ... ... ... |
... ... ... | ... ... ... | ... ... ... |
... ... ... | ... ... ... | ... ... ... |

8.. h   i   | j   8... k  | m   n   p   |
... ... ... | ... ... ... | ... ... ... |
... ... ... | ... ... ... | ... ... ... |


a - p are values that don not contain 8. The two rows with 8 have them in the columns

Once can immediately deduce that 8s only appear in the two columns. How do you prove this? The logic needed is to assume 8 appears elsewhere in, without loss of generality, the first column then that forces the second possibility to be chosen for the two rows containing 8, but these happen to be in the same column which can't be allowed.  The method of proof known as reductio ad absurdum (i.e trial and error, choose an alternative hypothesis and show it implies a contradiction).

We employ this theorem implicitly in many of the superior puzzles but the underlying logic employs guesswork. What makes this more legitimate than using the first method I used to solve superior 53 or the AA puzzle?
 


[re'born]

richardm,

Here is what I assume your pencil mark grid would look like at the place you got stuck.

Code: Select all

.---------------.---------------.---------------.
| 6    4    1   | 3    8    2   | 5    7    9   |
| 2    59*  59* | 1    7    4   | 6    38   38  |
| 8    37   37  | 56   9    56  | 4    1    2   |
:---------------+---------------+---------------:
| 5    1    78  | 47   3    68  | 9    2    468 |
| 79   2    6   | 47   1    89  | 3    5    48  |
| 39   38   4   | 56   2    5689| 1    68   7   |
:---------------+---------------+---------------:
| 4    6    2   | 8    5    3   | 7    9    1   |
| 1    359* 359*| 2    6    7   | 8    4    35- |
| 37   3578 3578| 9    4    1   | 2    36   356 |
'---------------'---------------'---------------'


An easy solution is obtained using Unique Rectangles. Either r8c2 or r8c3 is a 3, else one gets a deadly pattern in r28c23. Therefore, r8c9 = 5 and after this a naked pair and singles will solve the puzzle.

[richardm]

Very interesting. The possibility of considering combinations that lead to non-unique solutions had not occurred to me. I agree, it's a more direct method than the one I used to solve this puzzle.

Thanks for sharing this.

Richard

[udosuk]

Richard, the Uniqueness assumption techniques are not accepted by everyone, and one of the main condition you need is some way to verify the puzzle has a unique solution in the first place, such as using a brute-force solver... But it's indeed a brilliant shortcut for advanced level puzzles...

The American Way puzzle from you is surely quite challenging... You have mistakenly copy the same position twice... I think you forgot to post the original starting position...

From this position:

Code: Select all

 *-----------------------------------------------------------*
 | 6     4     1     | 3     8     2     | 5     7     9     |
 | 2     59    59    | 1     7     4     | 6     38    38    |
 | 8     37    37    | 56    9     56    | 4     1     2     |
 |-------------------+-------------------+-------------------|
 | 5     1    *78    | 47    3    -68    | 9     2     468   |
 |*79    2     6     | 47    1    *89    | 3     5     48    |
 | 39    38    4     | 56    2     5689  | 1     68    7     |
 |-------------------+-------------------+-------------------|
 | 4     6     2     | 8     5     3     | 7     9     1     |
 | 1     359   359   | 2     6     7     | 8     4     35    |
 | 37    3578  3578  | 9     4     1     | 2     36    356   |
 *-----------------------------------------------------------*

You can solve it with an advanced move called "xy-wing":
r4c6 cannot be 8, otherwise r4c3=7 and r5c6=9, and we cannot fill in r5c1
Therefore r4c6=6 and the rest are all singles

[daj95376]

richardm,

Here is what I assume your pencil mark grid would look like at the place you got stuck.

Code: Select all

.---------------.---------------.---------------.
| 6    4    1   | 3    8    2   | 5    7    9   |
| 2    59*  59* | 1    7    4   | 6    38   38  |
| 8    37   37  | 56   9    56  | 4    1    2   |
:---------------+---------------+---------------:
| 5    1    78  | 47   3    68  | 9    2    468 |
| 79   2    6   | 47   1    89  | 3    5    48  |
| 39   38   4   | 56   2    5689| 1    68   7   |
:---------------+---------------+---------------:
| 4    6    2   | 8    5    3   | 7    9    1   |
| 1    359* 359*| 2    6    7   | 8    4    35- |
| 37   3578 3578| 9    4    1   | 2    36   356 |
'---------------'---------------'---------------'


An easy solution is obtained using Unique Rectangles. Either r8c2 or r8c3 is a 3, else one gets a deadly pattern in r28c23. Therefore, r8c9 = 5 and after this a naked pair and singles will solve the puzzle.

rep'nA,

If I'm not mistaken, your UR will also eliminate the 3s in [r9c123]. This leads to a solution in singles.

[re'born]

rep'nA,

If I'm not mistaken, your UR will also eliminate the 3s in [r9c123]. This leads to a solution in singles.

daj,

You are absolutely correct. Thanks for pointing that out.

rep'nA