The New Sudoku Players' Forum

[carol]

Here is a puzzle.
Please help me to know what I should do next.
Thanks very much!

4xx xx6 5x1
3xx xxx 9xx
x5x 8xx x2x

xxx 134 xxx
xx4 682 3xx
xxx 759 xxx

x4x x78 69x
xx6 xxx xx7
897 46x 1x2

Carol

[bennys]

Code: Select all

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

+-------------------+-------------------+-------------------+
| 4     278   289   | 239   29    6     | 5     378   1     |
| 3     12678 128   | 25    124   157   | 9     4678  468   |
| 1679  5     19    | 8     149   137   | 47    2     346   |
+-------------------+-------------------+-------------------+
| 25679 2678  2589  | 1     3     4     | 278   5678  5689  |
| 1579  17    4     | 6     8     2     | 3     157   59    |
| 126   12368 1238  | 7     5     9     | 248   1468  468   |
+-------------------+-------------------+-------------------+
| 125   4     1235  | 235   7     8     | 6     9     35    |
| 25    23    6     | 2359  129   135   | 48    48    7     |
| 8     9     7     | 4     6     35    | 1     35    2     |
+-------------------+-------------------+-------------------+

we cant have R1C8=3 and R9C6=3 because then we will not have 3 in C4.
But  R1C8=3 => R9C8=5 => R9C6=3.
So we get R1C8<>3 and then its easy.

[carol]

That's interesting. Thank you.
What is the name of the technique you used to solve the puzzle?
I used the Susser program, and it needed to use Nishio cycles to solve this one.

[emm]

Simple Sudoku calls it colouring.

Since two of the same colour (-) are in the same group then they are the incorrect ones => eliminate all 3-

Code: Select all

  .   .   .  |  3+ .  .  |  .  3-   . 
  .   .   .  |  .  .  .  |  .   .   . 
  .   .   .  |  .  .  3- |  .   .  3+
-------------+-----------+------------
  .   .    . |  .  .  .  |  .   .   . 
  .   .    . |  .  .  .  |  .   .   . 
  .   3    3 |  .  .  .  |  .   .   . 
-------------+-----------+------------
  .   .   3  |  3  .  .  |  .   .   3- 
  .   3   .  |  3  .  3  |  .   .   . 
  .   .   .  |  .  .  3- |  .   3+  .   

[Cec]

"....we cant have R1C8=3 and R9C6=3 because then we will not have 3 in C4.
But R1C8=3 => R9C8=5 => R9C6=3.
So we get R1C8<>3 and then its easy.

Hi Bennys,
It's probably my diminishing brain cells but I do not understand your above explanations but would like to. For example, I read "we cant have R1C8=3 etc. but then on the next line I read "But R1C8=3 etc. Also what does <> mean?

Any help would be appreciated.

Cec

[Ruud]

Any help would be appreciated

Bennys logic is both simple and elegant:

Premise: When both R1C8=3 and R9C6=3, no candidates for digit 3 are left in column 4. So These cannot both be true.
Assertion: When 3 is placed in R1C8, this forces R9C8 to digit 5, which in turn forces R9C6 to digit 3.

The assertion violates the premise, so the assertion cannot be true. This makes the opposite of the assertion true: R1C8 <> 3 (in English: R1C8 does not have 3 as a candidate)

Ruud.

[QBasicMac]

After processing the locked candidate 1 in box 8 and the naked pair 35 in box 9, you use T&E. Simple!

Try pencilmark 3 in cell r9c8. You easily find this solution
472396581
381527964
659841723
965134278
714682359
238759416
143278695
526913847
897465132

Now, of course, try pencilmark 5 in cell r9c8

That's tougher. I wound up going another level
::Try pencilmark 4 in cell r8c8
Errk! stuck again.
:::Try pencilmark 1 in cell r8c6: That lead to invalid
:::Try pencilmark 5 in cell r8c6
Errk! stuck again.
::::Try pencilmark 1 in cell r7c3: Invalid
::::Try pencilmark 5 in cell r7c3: Invalid Puzzle
::Try pencilmark 8 in cell r8c8: Invalid Puzzle

OK, so the choice of 5 in cell r9c8 was wrong.

Result: The puzzle is valid and has the above solution.

Many easy puzzles to solve (or not solve - same thing)

Mac

[rubylips]

I do not understand your above explanations but would like to.

The following argument is essentially that used by bennys and em - however, you might find it easier to understand as it is presented in chain notation.

Code: Select all

Consider the chain r3c6-3-r1c4-3-r1c8-3-r9c8-3-r9c6.
When the cell r3c6 contains the value 3, so does the cell r9c6 - a contradiction.
Therefore, the cell r3c6 cannot contain the value 3.
When the cell r9c6 contains the value 3, so does the cell r3c6 - a contradiction.
Therefore, the cell r9c6 cannot contain the value 3.
- The moves r3c6:=3 and r9c6:=3 have been eliminated.
The value 5 is the only candidate for the cell r9c6.


Also what does <> mean?

'<>' means 'not equal to'. The symbol is taken from the FORTRAN and BASIC programming languages. Alternatively, you might see '!=', which comes from the C family of programming languages, used to mean the same thing.

[rubylips]

You easily find this solution
472396581
381527964
659841723
965134278
714682359
238759416
143278695
526913847
897465132

So you're guest!

[Cec]

Also what does <> mean?

'<>' means 'not equal to'.. .."

Thanks rubylips - with your above explanations including what this '<>' means and the prior additional help from Ruud I am now able to follow Bennys' chains.

My apologies for a slow reply but have been wading through all your above posts which has been unfamiliar territory for me to date but now trying to learn - up till 1am this morning still trying to absorb it.

With this technique appearing as a case of trying a candidate in a particular cell to see if it works or get a contradiction which then leads to another choice I can see some merit in Mac's alternative of starting from the naked pair 35 in box 9. Cell r9c8 can only be 3 or 5 - if you're lucky you pick 3 and puzzle solved - simple but not sure if it's elegant

Again, I appreciate all of your above help.

Cec

[QBasicMac]

if you're lucky you pick 3 and puzzle solved - simple

No - No matter which one you pick, you must also pick the other to verify there are not multiple solutions. There is no luck - just skill and patience and love of solving puzzles (or proving they can't be solved).

Mac

[Cec]

if you're lucky you pick 3 and puzzle solved - simple

No - No matter which one you pick, you must also pick the other to verify there are not multiple solutions....."

I take your point 'Mac' but personally, once I've solved a puzzle I'd rather not know there were alternative solutions - it kinda' dims the satisfaction of achieving a solution to later have revealed as an invalid puzzle.

Cec

[QBasicMac]

So you're guest!

I thought I had a hide of leather, but that stung!!





















LOL - Just kidding.

Mac