The New Sudoku Players' Forum

[vz]

Do I have to resort to trial and error on this puzzle? I hope not.
I also tried naked triples.

000000012
072135086
010620073
700200300
020400800
100000200
000012730
287000140
351000620

Thanks in advance.

[QBasicMac]

That "puzzle" has 952 solutions.

The interesting thing about multiple-solution puzzles is that Trial and Error works so well - No matter what you guess, you win! (Almost).

But conversely, despite there being 952 solutions, there is no non-T&E solution for any of them.

The non-T&E techniques work only on single-solution puzzles.

My advise, forget about getting puzzles from whatever source you have. There are plenty of reliable sources, this site for example.

Mac

P.S. Regarding your question in another thread about how the solver got the two solutions. The answer is that the solver used T&E. No other answer is possible IMHO.

[tso]

Actally, trial and error is not required AND not helpful.

If a puzzle has several hundred solutions, NO method will help you arrive at whatever solution the creator of the puzzle had in mind. There is simply NOTHING you can do to come up with the solution "in the back of the book" short of looking in the back of the book.

I'm thinking of a digit that is greater than 1 and less than 4. What is it? Trial and error will NOT help you get the answer because there is no way for you to know if you have the right answer! You can guess -- but that won't help -- you still don't know if you have the answer without asking me. The puzzle as posed is NOT SOLVABLE. The only real solution is: "The number you are thinking of is either 2 or 3 but I do not have enough information to determine which", which does not require trial and error nor guessing.

The same situation exists with Sudokus that have multiple solutions -- whether it has 2 or 200. A certain number of cells will be solvable, and solving them could just as well be very easy as very difficult. At that point, you could either:

1) Say this puzzle is invalid as it has 2 (or 200 or simply mulitple) solutions.

or

2) Find both (or all 200) solutions.

Again, this could be just as well be easy or difficult. Guessing and trial and error will have no more part to play than in a standard puzzle.


On the other hand, if your goal is to merely fill in the muliple solution Sudoku so that there are no duplications in any row, column or box, you can CHOSE a solution -- you still are not guessing. The puzzle changes to "find a number greater than 1 and less than 4". No guessing is required. Take your pick.

[vz]

Thank you so much for your prompt and very useful reply. I spent some time trying to find the source of the puzzle but could not. It was one of the first I recorded. I now include the source of any puzzle I record.

I am really trying to develop my logical solving skills (X-Chains?, etc). I am afraid some of the vocabulary is a bit beyond me. I have another puzzle that I am also stuck on. I could not find any chain, etc. I hope it is not another flawed puzzle: Would you give it a shot?

497152863
518030042
623804501
031500004
000400130
000301250
370285019
000013087
180040325

[QBasicMac]

If a puzzle has [multiple] solutions, NO method will help you arrive at whatever solution the creator of the puzzle had in mind.

Agreed!

You can get "a" solution via T&E, but not "the" solution.

The only correct T&E answer is a list of ALL solutions. And that is only correct in some sense. As you indicated, the correct solution is really "This puzzle is bogus". They use that in logic a lot.

I have a sheet of paper. On one side it says "The statement on the other side is false". The other side says "The statement on the other side is true".

Question: Is the first statement true?

Answer: The question is bogus. Logic does not address such questions.

Mac

[QBasicMac]

I hope it is not another flawed puzzle: Would you give it a shot?

It is not flawed. Try harder and start a new thread if you are having problems. Simple techniques will suffice. No X-Wings, etc.

Mac

[tso]

Code: Select all

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

a) In box 6, r56c9 must be 6 and 8, so neither r4c7 or r4c8 can be 6 (or 8).

b) Because of (a) there are only two cells in row 4 that can hold a 6. Both are in box 5. Therefore, the other three cells in box 5 are not 6.

c) Since r3c5<>6 (r3c1=6) and r56c5<>6 (from (b)), r4c5 *must* be 6, as it's the last cell in column 5 that can still hold one.

Only singles are required after this. No pencil marks are needed.

[vz]

Thank you both, QBasicMac and tso. I actually saw Q's post first and I went back and solved it. I am a bit ashamed that I missed the 6's restricted to 2 cells in 1 row. Thanks again.