The New Sudoku Players' Forum

[yessam]

Can someone solve this and tell me the technique,I found this to be very tough

XXX XXX X9X
X7X 4X5 XXX
X9X X36 2XX

3XX 15X 7XX
X69 XXX 51X
XX1 X4X XX9

XX4 56X X8X
XXX 9X4 X5X
X2X XXX XXX

[MCC]

yessam check that you've copied the numbers out correctly.

The grid is not symmetrical, you appear to have a number missing at:

(r6c6).

Edit: I just checked it out it's rated a valid hard on pappocom.

yessam you can place a 6 in box 5.
Also a 9 in box 2, then a 9 in box 5.

[yessam]

checked again,no mistakes except r4c4 is also blank not 1 (1 is what I had figured out while solving)

what you mean by 'not symmetrical'??

[MCC]

yessam check out my previous reply which I've edited.

yessam how did you come to place the 1 in r4c4 before doing the 9's first?
After placing the 9's the 1 resolves itself in r4c4.

Without the 1 you placed, the grid is symmetrical.


What do I mean by not symmetrical; The grid is non-symmetrical about an axis of rotation, i.e., it does not retain its symmetry after being rotated.

If you rotate the grid around cell r5c5 by 180 degrees ( in the plane of the page) so that box 1 becomes box 9 and box 9 becomes box 1, then all cells that contained a number in the grid before rotation will contain a number after rotation (albeit not the same number).

In your grid you have for box 5:

Code: Select all

15.
...
.4.

If you rotate this by 180 degrees you'll have:

Code: Select all

.4.
...
.51

Cell r4c4 has a 1 before rotation but is empty after rotation so therefore the grid is not symmetrical.

[yessam]

Thanks for clarifying,as for as 1 is concerned I got apart from 1 the foll right

r2c5 9
r4c6 9
r6 c4 6
r4c4 1

thats it only 4 I could get,all the others have multiple options and no apaprant techniques are working!
I hope you can figure it out better and let me know

[simes]

You have two locked candidates in block 2 and row 1, and then another in block 1 and column 1, then another in block 4 and column 1.

Simon

[QBasicMac]

Take your original puzzle (with r4c4=space)
Then r2c5=9 r4c6=9 r4c4=1 r6c4=6

After that, it is just a matter of endless pencilmark elimination by simple techniques. The only problem you could have is making a mistake.

Example pencilmark elimination: r1c56 contain the only 1's in box 2 therefore 1 can be eliminated as a candidate in r1c12 and r1c79.

Also: r23c1 means 1 can be eliminated from r789c1.

r1c456 means 2 can be eliminated from r1c13

It goes on and on with no technique required other than noticing locked candidates and singles like I described above.

The trick is being very careful and accurate.

Mac

[iRob]

symmetrical... If you rotate the grid around cell r5c5 by 180 degrees ( in the plane of the page) so that box 1 becomes box 9 and box 9 becomes box 1, then all cells that contained a number in the grid before rotation will contain a number after rotation (albeit not the same number)

Oh, how fascinating! Let me ask, as I'm a relative newb to the theory of Sudoku, is this sort of symmetry a requirement for a unique solution sudoku? If so, is there a proof for this?

Thanks!

[PaulIQ164]

No, it's by no means necessary. It's just aesthetics, the same way crosswords tend to be symmetrical. There's no reason for it - it just looks nice.

[Bigtone53]

No, it's by no means necessary. It's just aesthetics, the same way crosswords tend to be symmetrical. There's no reason for it - it just looks nice.

As someone said earlier, you are envisioning rotational symmetry which as a crossword solver, I like. However, there are a number of Sudokus published which go for refective symmetry about either an axis or a diagonal. Typically, these have a cluster of numbers on one side/corner, moving towards large blank spaces at the other. Personally, I find these less pleasing.