The New Sudoku Players' Forum

[ludadon]

Hi,
I need help with my Sudoku. Its to hard for me.

Thats it. The diagonals are also includet, maybe that could be a help.

[QBasicMac]

If I copied it correctly, you are out of luck. Puzzle has multiple solutions, i.e. logic will not lead to any of them. T&E will find plenty.

Mac

Code: Select all

53-  ---  -97
---  -7-  ---
---  -3-  ---

--1  ---  8--
4--  -2-  --6
--3  ---  7--

---  -1-  ---
---  -9-  ---
81-  -6-  -54


[ludadon]

No, there has to be just one solution. My teacher knows the answer, but he said i should try to find it on my own.
I think you could find a way if you include the diagonals. Otherwise you get over 1000 solutions, I tried it out.

[ronk]

Puzzle has multiple solutions, i.e. logic will not lead to any of them.

So you're saying, even with the extra requirements -- that each diagonal contains each digit 1 thru 9 exactly once -- there are still multiple solutions. Can you show us just two?

I haven't found even one yet.

[Steve R]

The puzzle has a unique solution but I could only find only it by guessing that r1c5 contains 4. Could this have been omitted from the starting grid in error? If so, think X-Wing to solve the amended version.

Steve

[ravel]

The diagonals are also includet, maybe that could be a help.

Since it seems to be a Sudoku-X, you should try to post it in the sudoku variants forum (or try to find a sudoku-x solver) - for me as a sudoku-X newbie it is too hard.

[Sped]

Here's a solution. I don't know if it's the only one.

I got it by guessing a few cells and concentrating on solving the diagonals.

Code: Select all

 *-----------*
 |53.|...|.97|
 |...|.7.|...|
 |...|.3.|...|
 |---+---+---|
 |..1|...|8..|
 |4..|.2.|..6|
 |..3|...|7..|
 |---+---+---|
 |...|.1.|...|
 |...|.9.|...|
 |81.|.6.|.54|
 *-----------*


 *-----------*
 |532|148|697|
 |186|972|435|
 |749|536|182|
 |---+---+---|
 |271|659|843|
 |498|327|516|
 |653|481|729|
 |---+---+---|
 |925|714|368|
 |364|895|271|
 |817|263|954|
 *-----------*


[ludadon]

Oh very good, you got it. I dont know if there are more solutions but i'll tell you when I know more if you want. If you'll find out how to get the solution without guessing digits, please tell me.Thx a lot!

[Crazy Girl]

Assuming there is a 4 in R1C5, to be symmetrical (from Sped's solution), managed to place most of the 4's by the fact that 4 in B6C8 and B5forward diagonal, as indicated by - sign, and get this situation:

Code: Select all

-----------------+------------------+-------------------|
5     3    268   | 1268   4  1268   | 126  9      7     |
1269  689  2689  | 125689 7  125689 | 4    136    12356 |
12679 4    6789  | 125689 3  125689 | 156  1268   1258  |
-----------------+------------------+-------------------|
2679  2679  1    | 3679   5  3469-  | 8    234-   239   |
4     5789  5789 | 1379   2  1379   | 1359 13     6     |
269   2569  3    | 1469-  8  169    | 7    124-   1259  |
-----------------+------------------+-------------------|
23679 25679 569  | 23478  1  234578 | 369  23678  2389  |
2347  56    4    | 23578  9  23578  | 1236 13678  1238  |
8     1     279  | 237    6  237    | 239  5      4     |
-----------------+------------------+-------------------|

Then noticed that there is colouring that can be done on the Forward Diagonal on 3 as indicated below, since 3 only occurs twice on the diagonal and R4C8 see's both of these cells so can't be a 3.

Code: Select all

-----------------+------------------+--------------------|
5     3    268   | 1268   4  1268   | 126  9      7      |
1269  689  2689  | 125689 7  125689 | 4    **136  *12356 |
12679 4    6789  | 125689 3  125689 | 156  1268   1258   |
-----------------+------------------+--------------------|
2679  2679  1    | 3679   5  *3469  | 8    -234   239    |
4     5789  5789 | 1379   2  1379   | 1359 13     6      |
269   2569  3    | 1469   8  169    | 7    124    1259   |
-----------------+------------------+--------------------|
23679 25679 569  | 23478  1  234578 | 369  23678  2389   |
2347  56    4    | 23578  9  23578  | 1236 13678  1238   |
8     1     279  | 237    6  237    | 239  5      4      |
-----------------+------------------+--------------------|

This technique can also be applied to the 1's on the backward diagonal to eliminate 1 from R6C8 to leave us with a naked {24} pair in B6C8, and the following situation:

Code: Select all

-----------------+------------------+-------------------|
5     3    268   | 1268   4  1268   | 126  9      7     |
1269  689  2689  | 125689 7  125689 | 4    136    12356 |
12679 4    6789  | 125689 3  125689 | 156  168    1258  |
-----------------+------------------+-------------------|
2679  2679  1    | 3679   5  3469   | 8    24     39    |
4     5789  5789 | 1379   2  1379   | 1359 13     6     |
269   2569  3    | 1469   8  169    | 7    24     159   |
-----------------+------------------+-------------------|
23679 25679 569  | 23478  1  234578 | 369  3678   2389  |
2347  56    4    | 23578  9  23578  | 1236 13678  1238  |
8     1     279  | 237    6  237    | 239  5      4     |
-----------------+------------------+-------------------|

Sped,
does this help at all with what you've had to guess in solving the puzzle.

Ludadon,
Is there meant to be a number in R1C5 for the puzzle to be symmetrical, and if you find out anything more please let us know.

[Sped]

Assuming there is a 4 in R1C5, to be symmetrical (from Sped's solution), managed to place most of the 4's by the fact that 4 in B6C8 and B5forward diagonal, as indicated by - sign, and get this situation:

Code: Select all

-----------------+------------------+-------------------|
5     3    268   | 1268   4  1268   | 126  9      7     |
1269  689  2689  | 125689 7  125689 | 4    136    12356 |
12679 4    6789  | 125689 3  125689 | 156  1268   1258  |
-----------------+------------------+-------------------|
2679  2679  1    | 3679   5  3469-  | 8    234-   239   |
4     5789  5789 | 1379   2  1379   | 1359 13     6     |
269   2569  3    | 1469-  8  169    | 7    124-   1259  |
-----------------+------------------+-------------------|
23679 25679 569  | 23478  1  234578 | 369  23678  2389  |
2347  56    4    | 23578  9  23578  | 1236 13678  1238  |
8     1     279  | 237    6  237    | 239  5      4     |
-----------------+------------------+-------------------|

Then noticed that there is colouring that can be done on the Forward Diagonal on 3 as indicated below, since 3 only occurs twice on the diagonal and R4C8 see's both of these cells so can't be a 3.

Code: Select all

-----------------+------------------+--------------------|
5     3    268   | 1268   4  1268   | 126  9      7      |
1269  689  2689  | 125689 7  125689 | 4    **136  *12356 |
12679 4    6789  | 125689 3  125689 | 156  1268   1258   |
-----------------+------------------+--------------------|
2679  2679  1    | 3679   5  *3469  | 8    -234   239    |
4     5789  5789 | 1379   2  1379   | 1359 13     6      |
269   2569  3    | 1469   8  169    | 7    124    1259   |
-----------------+------------------+--------------------|
23679 25679 569  | 23478  1  234578 | 369  23678  2389   |
2347  56    4    | 23578  9  23578  | 1236 13678  1238   |
8     1     279  | 237    6  237    | 239  5      4      |
-----------------+------------------+--------------------|

This technique can also be applied to the 1's on the backward diagonal to eliminate 1 from R6C8 to leave us with a naked {24} pair in B6C8, and the following situation:

Code: Select all

-----------------+------------------+-------------------|
5     3    268   | 1268   4  1268   | 126  9      7     |
1269  689  2689  | 125689 7  125689 | 4    136    12356 |
12679 4    6789  | 125689 3  125689 | 156  168    1258  |
-----------------+------------------+-------------------|
2679  2679  1    | 3679   5  *3469  | 8    24     39    |
4     5789  5789 | 1379   2  1379   | 1359 13     6     |
269   2569  3    | 1469   8  169    | 7    24     159   |
-----------------+------------------+-------------------|
23679 25679 569  | 23478  1  234578 | 369  3678   2389  |
2347  56    4    | 23578  9  23578  | 1236 13678  1238  |
8     1     279  | 237    6  237    | 239  5      4     |
-----------------+------------------+-------------------|

Sped,
does this help at all with what you've had to guess in solving the puzzle.

Ludadon,
Is there meant to be a number in R1C5 for the puzzle to be symmetrical, and if you find out anything more please let us know.

Crazy Girl..

It's funny. I did the exact same steps. The colors on the diagonals and the naked pair. But I 'm stuck.

Actually, since I'm totally new to this variant, solving by bifurcation was pretty satisfying. I forget exactly what path I took, but I looked at diagonals with candidates in just two cells and selected one at random, continued solving and backtracked if there was a contradiction. I think I had to backtrack twice.

Anyway, at least I showed there is a solution. And the fact is, to all but suduko snobs like us, bifurcation is considered a perfectly legitimate solving method.

Casual sudoku players probably don't call it "bifurcation", but they use it on hard puzzles all the time and are justifiably satisfied when they've filled in all the cells correctly.

I would like to see a real solution, though.

[Steve R]

I used a rather different trick at the position posted by Crazy Girl.

There are only two cells which can admit 5 in row 6, r6c2 and r6c9. The same applies to column 7, r3c7 and r5c7. As box 6 can hold only one 5, one of r6c2 and r3c7 contains 5. Thus 6 can be entered in r8c2.

Only a pair is then needed to resolve the grid.

Steve

[ravel]

Is there meant to be a number in R1C5 for the puzzle to be symmetrical, and if you find out anything more please let us know.

Since i did not manage to get even one number in 2 hours, i wondered, if it is unique and added the diagonals to my solver.
It is unique without the 4, but obviously very tough at the beginning.
2 guesses solve it with singles:
r1c5=4 and then r5c8=1

[Crazy Girl]

Since the puzzle was set as homework, and there is a unique solution according to the teacher, I decided to add the R1C5 clue from Sped's answers, to make it look symmetrical and that maybe the teacher had forgotten to add this clue to the homework.

Would like to see the teacher's solution to see how he solved it with LOGIC only.

[ludadon]

Today I told him this solution and wanted to know the way he solved it, but hes a strict teacher and he said my homework was to find a way to solve the Sudoku and not to show him the Solution by guessing digits.
Hmm....I really want to know who the hell made this Sudoku and how to solve it. I hope I will soon get the answer .

[ravel]

As i said, i am a sudoku X newbie with no practice in special techniques using the diagonal. But i am very sure that the first steps to a solution are pretty hard.

My program with only singles implemented for the diagonals (no special sudoku X techniques) and using no unique rectangles (which are not valid, when diagonal cells are involved) needs at least 12 brute force eliminations for it. A brute force step means that you can eliminate the number, because you get a contradiction using normal techniques (box/line interaction, x-wing, tuples) without guessing, but it can be a very long way to this contradiction. Compare: for the hardest "normal" sudoku i know, it needs 11 steps.

Here are two 12 step eliminations, that lead to a solution:
r4c8<>3, r8c8<>1, r1c7<>1, r8c7<>1, r2c7<>1, r3c8<>1, r3c3<>6, r8c8<>3, r3c3<>7, r4c4<>3, r3c3<>8, r2c2<>6
r7c3<>6, r8c8<>1, r1c7<>1, r8c7<>1, r2c2<>6, r2c7<>1, r3c8<>1, r3c8<>2, r7c8<>2, r3c3<>6, r4c4<>3, r3c8<>6

So i also would be very interested in your teachers solution. If he does not show it, i assume he does not have a solution, but a sadistic vein

PS: i tested 3 other X-sudokus, one rated hard somewhere was solveable with singles only, one discussed in the variants thread (1 step) and the 13-clue sudoku X of gfroyle (4 steps) - would recommend it as homework for your teacher .
So this one has good chances for the top 10.