The New Sudoku Players' Forum

[laidback_tyke]

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.

I've been following most of what's been posted earlier in this thread, but the above is where I hit the brick wall. As I've been using up to now a version of what I've just discovered is called "carpet-bombing", the reason for "R2C6 must be a 1 or 6" is not at all clear to me. My grid shows that R2C6 could be 1,6 or 7, so what stops it being a 7?
Please could you explain?

I'd begun to suspect that there must be deeper logical techniques needed when I hit a brick wall in the Times Friday The 13th puzzle, but I've only just discovered that there are codified methods (with their own names, like "x-wing" and such). Is this what's implied by the use of the term "obvious" later in the thread?

[Animator]

Look at row 2. What cells can have the number 1 or the number 6? there are only two such cells...

[laidback_tyke]

Look at row 2. What cells can have the number 1 or the number 6? there are only two such cells...

OK, I see that principle (thanks for that - I wasn't familiar with it before).
However, R2C2 can also be 6 (or 2,5,9), which would leave R2C1 & R2C6 to have a 1 in them.
We could have 2 or 5 in R2C1, R2C2 and R2C8 ... and so on.

Is this a case of having to chase it down, repeatedly applying the same principle to eliminate possibilities?

[whohe]

Look at row 2. What cells can have the number 1 or the number 6? there are only two such cells...

OK, I see that principle (thanks for that - I wasn't familiar with it before).
However, R2C2 can also be 6 (or 2,5,9), which would leave R2C1 & R2C6 to have a 1 in them.
We could have 2 or 5 in R2C1, R2C2 and R2C8 ... and so on.

Is this a case of having to chase it down, repeatedly applying the same principle to eliminate possibilities?

Hi there.

R2C2 can't be a 6 because Column 1 must have a 6 in it, and the 6 can only go in R1C1 or R2C1, therefore no other cell in the top left block of 9 could be a 6.

[Guest]

Look at row 2. What cells can have the number 1 or the number 6? there are only two such cells...

OK, I see that principle (thanks for that - I wasn't familiar with it before).
However, R2C2 can also be 6 (or 2,5,9), which would leave R2C1 & R2C6 to have a 1 in them.
We could have 2 or 5 in R2C1, R2C2 and R2C8 ... and so on.

Is this a case of having to chase it down, repeatedly applying the same principle to eliminate possibilities?

Hi there.

R2C2 can't be a 6 because Column 1 must have a 6 in it, and the 6 can only go in R1C1 or R2C1, therefore no other cell in the top left block of 9 could be a 6.

Ahh, yes, now that's coming in from a different direction - thanks for the reminder not to rely on only one principle.
I'd still like to pin down the "1-6 pair" thing. Is this a hidden thing? I'll have to do some more thinking about this before I make any more input

... and I thought I was good at logic! I need to get some sleep (not staying up until 0200 hrs working on "fiendish" sudoku!)

Thanks folks, for the stimulation

[Pappocom]

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.

That's correct. The topic you were thinking of is probably http://forum.enjoysudoku.com/viewtopic.php?t=211

- Wayne

[Guest]

the puzzle that you first posted, was that the original one or was it after you had filled in some boxes. because i make progress and fill up 7 3x3 matrices but there is an altercation between the bottom left and middle matrices. if this was after your input, then there is a possibility that you have made an error. i suspect the error originates in row 7th from the bottom.

[Time^-]

The original would be nice indeed...

[joolslee]

Kwasi, I think the original poster had partially completed the puzzle and was asking for help to move forward. I have finished the puzzle and it checks correctly.

[scrose]

The original would be nice indeed...

The original is in the daily sudoku archive.

[Guest]

Hi Dan.
History: I saw a puzzle in the paper and thought about how it should be solved. Generally a cell can only contain valid values that have not been used acording to the rules. Can I create a spreadsheet to display the valid values for each cell? Well it took a day between work and a few wrong turns to create the tool.
Testing: So I grabbed your hard to solve puzzle, put the initial values in, and just as I expected not one cell showed a single valid value only doubles. That is many cells showed the value could be 2 or 5 for example.
Solved: So I jumped in and selected one of the 2 possible values. Must have picked the right one, as then single valid values for other cells were then presented.
Result: In about 5 minutes I had the answer with very little mind activity, excluding the day it took to design the spreadsheet.

639.254.781
158.796.324
274.831.965

493.182.657
517.649.238
826.375.149

341.567.892
765.928.413
982.413.576

Thanks

[Animator]

You eliminated a cell by trial and error, that is, you guessed a cell and tried it...

That's not how it is usually done... I only fill in a cell if there are enough clues to elimante the other candidates.

[Guest]

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.

That's correct. The topic you were thinking of is probably http://forum.enjoysudoku.com/viewtopic.php?t=211

- Wayne

G'day all

The hint function at the Daily Sudoku should give you the next logical move. In the case of the very hard puzzle from last Sat, it appears to screw up a give you a number which you can't get just yet (leading to a flood of emails from confused people). I've only just worked out why, and will be attempting to fix at the w/e!

Best wishes

Sam

[Guest]

Hi,

Here is the beginning of the final solution to the problem:

All references are 'Row, Column'

Because of the 6 in 4,7 and 5,4 and 7,5; the 6 in the top-right 3x3 square must be in the first column. This means that the only places in the second row that can be either a one or a six are 2,1 and 2,6 (i.e. no longer 2,2 no longer). These are two possible numbers can only go in two cells so no other numbers can go in these two cells. 2,6 was one of only two options for the 7 in the top-middle square. Therefore the 7 goes in 2,4

Afterwards:
7 in 7,6 (only place in column, row and square)
9 in 8,4 (only place in column)
8 in 8,6 (only place in row and square)
8 in 3,4 (only place in row)
5 in 7,4 (only possible number in cell)
2 in 8,5 (only possible number in cell)

And afterwards it is essentially trivial / normal.

BTW, I have written a 'sudoku buddy' programme in Excel that allows you to solve Sudokus on screen. It is complete with notes sections for each cell and there are 7 potential levels of help from none at all to doing all the obvious things for you. You can solve the hardest puzzles on level 6 or 7 by allowing the buddy to do the normal calculations while you focus on the harder strategies (like those needed to solve this puzzle). I usually use level 2 which just points out repeated number errors alone. It is available on http://www.sudokubuddy.co.uk but the website is not properly sorted yet. I'd love some people to test using it - e-mail me on andrew@sudokubuddy.co.uk to get a copy (don't worry about the charge on the website - I want some other people to start using it before I'm realistically expecting any money!).

Cheers,


Andrew

[Guest]

6-3-9 | 2-5-4 | 7-8-1
1-5-8 | 7-9-6 | 3-2-4
2-7-4 | 8-3-1 | 9-6-5
___________________________
4-9-3 | 1-8-2 | 6-5-7
5-1-7 | 6-4-9 | 2-3-8
8-2-6 | 3-7-5 | 1-4-9
___________________________
3-4-1 | 5-6-7 | 8-9-2
7-6-5 | 9-2-8 | 4-1-3
9-8-2 | 4-1-3 | 5-7-6

This is the solution.

Each row and column must have at least one number in them between 1 and 9. The sum of number from 1 to 9 is 45.
(It is safe to say that if you require one number in a row or column, then deduct the sum of numbers in that row or column from 45)
Also use each cluster of 3x3 to add numbers where you have 1 space remaining within it.

I use a spreadsheet for this

Take any row that requires the least numbers to complete it, which in this case row 6 and columns 7 & 8.
This is where I guessed a bit.
In R5C7 I placed a 2. That meant that R7 needed only 1 number to complete it - a 5 at R9C7.
With only 1 block space at R4C8, the only number available was a 5
That gave me one number required to complete Column 8 (R2C8).
The only number available to R3C9 is a 5

You should have Columns 7 and 8 complete and be well on your way to solving the rest of the puzzle.