The New Sudoku Players' Forum

[Guest]

Today's Sudoku at the Daily Sudoku is listed as "very hard," but I think it should be called "nigh impossible!" Between my wife and I we've restarted 3 times after realizing we'd gotten to an impossible state. Here's what I've gotten so far but I'm thoroughly stuck:

– 3 – . 2 – – . 7 8 1
– – 8 . – – – . 3 – 4
– 7 4 . – 3 – . 9 6 –

– – – . 1 8 – . 6 – 7
– 1 7 . 6 – 9 . – 3 8
8 – – . 3 7 5 . 1 4 9

– 4 1 . – 6 – . 8 9 –
7 – – . – – – . 4 1 –
9 8 – . 4 1 3 . – 7 –

In desperation I took a hint and the solver said that R1C1 is a 6. A 6?! How did it come up with that?

[Guest]

Today's Sudoku at the Daily Sudoku is listed as "very hard," but I think it should be called "nigh impossible!" Between my wife and I we've restarted 3 times after realizing we'd gotten to an impossible state. Here's what I've gotten so far but I'm thoroughly stuck:

– 3 – . 2 – – . 7 8 1
– – 8 . – – – . 3 – 4
– 7 4 . – 3 – . 9 6 –

– – – . 1 8 – . 6 – 7
– 1 7 . 6 – 9 . – 3 8
8 – – . 3 7 5 . 1 4 9

– 4 1 . – 6 – . 8 9 –
7 – – . – – – . 4 1 –
9 8 – . 4 1 3 . – 7 –

In desperation I took a hint and the solver said that R1C1 is a 6. A 6?! How did it come up with that?

I guess it depends what you mean by "logic".

Clearly R1C1 can only be a 5 or a 6, and a 6 must be in either R1C1 or R2C1. If you try R1C1 with a 5, then you get stuck pretty quickly.

[Guest]

If you try R1C1 with a 5, then you get stuck pretty quickly.

Can you explain what you mean by "stuck pretty quickly"? Do you mean stuck as in it doesn't help you get any farther, or stuck as in you get into an impossible situation?

A 5 in R1C1 means that there's only two 5's on the board, and as far as I can see (which isn't very far with these things), it doesn't lend itself to finding many more squares.

[whohe]

If you try R1C1 with a 5, then you get stuck pretty quickly.

Can you explain what you mean by "stuck pretty quickly"? Do you mean stuck as in it doesn't help you get any farther, or stuck as in you get into an impossible situation?

A 5 in R1C1 means that there's only two 5's on the board, and as far as I can see (which isn't very far with these things), it doesn't lend itself to finding many more squares.

Sorry - "stuck" as in you get into an impossible situation quite quickly.

As presented, there are 37 squares left to find, of these -

19 have only 2 possible correct entries, 15 have 3, and the remaing 3 have 4 (yes - I spent a bit of time last night checking this manually!).

I'm not sure of the correct terminology, but as far as I can see, there are no instances of {(x/y);(x/y);(x/y/z)} triplets forcing z, and neither are there any instances of {(w/x/y);(w/x/y);(w/x/y);(w/x/y/z)} quads forcing z. Therefore, I think I'm right in saying that the only way to solve it is to pick one of the {x/y}'s and follow the conclusions around the board.

[joolslee]

What candidates are you showing for Row 2? It is pretty straightforward filling in R2C4

[whohe]

What candidates are you showing for Row 2? It is pretty straightforward filling in R2C4

Is it? I've got 5,7,9. What am I missing?

[Guest]

I had same candidates for that cell but it's more about the candidates for R2C1 and R2C6 - these are the only possible cells for numbers 1 & 6, by deleting other candidates for those cells you are left with only one option for R2C4

[whohe]

I had same candidates for that cell but it's more about the candidates for R2C1 and R2C6 - these are the only possible cells for numbers 1 & 6, by deleting other candidates for those cells you are left with only one option for R2C4

I don't quite follow you.

I see what you're saying about the 1&6 in row 2, but would you mind explaining a little further how that directly leaves just a single option for R2C4?

/edit

Sorry - got you. Regardless of whether or not R2C8 is a 2 or a 5, R2C4 is forced to a 7.

What I don't understand is why the "hint" that was given was a 6 in R1C1, when surely the first logical possibility for the puzzle is a 7 in R4C2?

[joolslee]

The only options for 7 were in R2C4 and R2C6, but R2C6 must be a 1 or 6 so you can put 7 in R2C4.

[joolslee]

Sorry I'm not sure about the hint - I don't use them and don't know whether they are based on the next logical move or just an extra random clue.

[whohe]

The only options for 7 were in R2C4 and R2C6, but R2C6 must be a 1 or 6 so you can put 7 in R2C4.

D'oh - I missed the obvious as I was so caught up in the detail!

Still doesn't explain why the "hint" given was R1C1 (unless whatever program is doing the hinting doesn't provide them in a logical order - which would seem pretty stupid).

[joolslee]

Lol ... I often can't see for looking myself. I agree the hint should be based on a logical move but can't figure that out with the 6 you were given.

[whohe]

Lol ... I often can't see for looking myself. I agree the hint should be based on a logical move but can't figure that out with the 6 you were given.

Nice to know I'm not the only one!

Fancy taking a quick peak at Sydney Guest's problem? I initially thought it was similar (but that was under the assumption that R1C1 in this one was the next "logical" step), but I'd be interested to see if you can spot anything I've missed.

[joolslee]

Would love to help but absolutely must drag myself away from SuDoku'ing for a few hours. Starting to wonder if I need therapy

[Guest]

Still doesn't explain why the "hint" given was R1C1 (unless whatever program is doing the hinting doesn't provide them in a logical order - which would seem pretty stupid).

In one of the forums, not sure which one, pappocom has replied to a similar message saying that a hint is random, no logic involved.
It's up to the hintee if they want to accept the hint.