The New Sudoku Players' Forum

[Animator]

Hmm, strange... in that grid I don't see an immediate possibility to fill in r6c7 and/or r9c5... (but note that the X-wing and the other moves are possible at this point too)

[Guest]

MCC, can you post your current grid? it will be easier for us to tell you what move you are missing... yes I could start from scratch too but then there is the risc that I make the same move as Duncan without thinking about it... (and thus without posting the reason why)

Here is where I'm at the moment.

*27|8**|*5*
*83|*5*|791
5**|7**|*2*
**8|***|5*6
*1*|*4*|*8*
7*5|*8*|4**
*5*|**9|***
869|*1*|24*
*7*|**8|*6*

I've managed to solve the placement of 4 in (6,7) by using a x-wing of 7's in cells (5,6)(5,9)(8,6)(8,9) this eliminates the 7 in (7,9) leaving the possibles in column 9 of: 3,4 in (1,9), 3,4,8 in (3,9) and 3,8 in (7,9) you can now eliminate the 4 in (6,9) leaving (6,7) the only position for a 4.

[Animator]

Yes, but that requires the X-wing and column 9... And the question is did or did he not see that one?

If both the X-Wing and Column 9 were seen by Duncan, then his only problem was Column 6... (as in, only one place for the numbers 5 and 7 after the X-wing) and this is enough to solve the entire puzzle...

Duncan, do you still have the pencilmarks you used when you were stuck (if you used pencilmarks that is)?

[su_doku]

I've managed to solve the placement of 4 in (6,7) by using a x-wing of 7's in cells (5,6)(5,9)(8,6)(8,9) this eliminates the 7 in (7,9) leaving the possibles in column 9 of: 3,4 in (1,9), 3,4,8 in (3,9) and 3,8 in (7,9) you can now eliminate the 4 in (6,9) leaving (6,7) the only position for a 4.[/quote]


I dont see how you eliminate with certainty the 4 in (6,9) just because you had a possible 4 in (1,9) and (3,9) since you could have a 4 in (1,7) and (3,7)

[Guest]

I've managed to solve the placement of 4 in (6,7) by using a x-wing of 7's in cells (5,6)(5,9)(8,6)(8,9) this eliminates the 7 in (7,9) leaving the possibles in column 9 of: 3,4 in (1,9), 3,4,8 in (3,9) and 3,8 in (7,9) you can now eliminate the 4 in (6,9) leaving (6,7) the only position for a 4.

I dont see how you eliminate with certainty the 4 in (6,9) just because you had a possible 4 in (1,9) and (3,9) since you could have a 4 in (1,7) and (3,7)[/quote]


Su-doku there are three cells (1,9)(3,9)(7,9) and three numbers 3 and 4 and 8, these three numbers must fit these three cells, if you place a 4 in (6,9) then that leaves the numbers 3 and 8 to fit three cells.
Hope this is clear, if not just ask.

[su_doku]

Yes MCC I see that and a further vision:

since the 3 of column 9 can therefore not be in (9,8) and (9,9) it follows that the 3 of row 8 is in (8,4) or (8,6) and the 3 of row 9 must be in (9,1). Hence the 2 in (9,5). Should be straightforward from now

[Duncan]

Yes, but that requires the X-wing and column 9... And the question is did or did he not see that one?

If both the X-Wing and Column 9 were seen by Duncan, then his only problem was Column 6... (as in, only one place for the numbers 5 and 7 after the X-wing) and this is enough to solve the entire puzzle...

Duncan, do you still have the pencilmarks you used when you were stuck (if you used pencilmarks that is)?

No I don't still have the PMs. I remember the 4 in r6c7 was the last number I got if my memory serves me correctly it was because the 3into3 with the 348 ment that the 4 in col 9 had to be in block 3 so it left nowhere else for the 4 to go in block6 but in r6c7.

I rubbed out my PMs because I went back to scratch on the empty cells to see if I'd missed something silly.
I'm still a bit miffed that I missed the 5 and 7 2into2 in col6

I only use PMs for the Vhards if I can manage it.

Duncan

[Duncan]

I'll like to ask Duncan as to his thinking behind the placement of a 4 in (6,7) and a 2 in (9,5)?

I've been on this, starting from scratch, for about an hour and I cannot see how he reached his conclusions.

I see from your latter posts you got the 4 worked out.

I'd have to start the thing again to tell you how I got the 2 in 9,5 but then I run the risk of having to kill myself. Did you get it yet?

Duncan

[Guest]

I echo MCC having started from scratch - here's the current grid


*27|8**|*5*
*83|*5*|791
5**|7**|*2*
**8|***|5*6
*1*|*4*|*8*
7*5|*8*|***
*5*|**9|***
869|*1*|24*
*7*|**8|*6*

The set (3,4,8) is found in (1,9),(3,9), and (7,9). So they can be removed from the other cells in column 9.

This results in (5,7) being found in (8,6) and (8,9). So 5 is removed from (8,4) leaving only 3 in (8,4). This was pointed out by Duncan, but not incorporated in the latest grid.

Code: Select all

*27|8**|*5*
*83|*5*|791
5**|7**|*2*
**8|***|5*6
*1*|*4*|*8*
7*5|*8*|***
*5*|**9|***
869|31*|24*
*7*|**8|*6*


[Guest]

The 3 in (8,4) also gave me 2 in (9,5), also seen in earlier posts. I finally figured out the 4 in (6,7) a bit behind here...

Code: Select all

*27|8**|*5*
*83|*5*|791
5**|7**|*2*
**8|***|5*6
*1*|*4*|*8*
7*5|*8*|4**
*5*|**9|***
869|31*|24*
*7*|*28|*6*


[Guest]

(Another coffe break, lots of them today)

I only see two possible solutions for the remaining 7s, so I tried 7 in (5,6). This gave me a contradition very soon, so I concluded that 7 in (5,6) was not an option. Leaving only 5 in that position, I got further.

Code: Select all

*27|8**|*5*
*83|*5*|791
5**|7**|*2*
**8|*7*|5*6
*1*|*45|*87
7*5|*8*|4*2
*5*|469|*7*
869|317|245
*7*|528|*69


Now I'm stuck again!

[Animator]

Take a close look at the number 3...

[su_doku]

(Another coffe break, lots of them today)

I only see two possible solutions for the remaining 7s, so I tried 7 in (5,6). This gave me a contradition very soon, so I concluded that 7 in (5,6) was not an option. Leaving only 5 in that position, I got further.

Code: Select all

*27|8**|*5*
*83|*5*|791
5**|7**|*2*
**8|*7*|5*6
*1*|*45|*87
7*5|*8*|4*2
*5*|469|*7*
869|317|245
*7*|528|*69


Now I'm stuck again!

Jonas I respectfully disagree - I dont see the contradiction you refer to: inserting (and I am not happy doing that as it isn't a logical leap) the 7 in (5,6) in fact solves it beautifully. Unless of course inserting the 5 in (5,6) also leads to a solution in which case the uniqueness theory.....

[su_doku]

Inserting 5 into (5,6) eventually leads to a contradiction when you reach (5,3)

However I have seen the logical leap without inserting either 5 or 7 into (5,6). It involves inserting the 6 into (6,6) using the knowledge of (1,6), (2,6), (3,6) and (4,6)

[Guest]

Jonas I respectfully disagree - I dont see the contradiction you refer to: inserting (and I am not happy doing that as it isn't a logical leap) the 7 in (5,6) in fact solves it beautifully. Unless of course inserting the 5 in (5,6) also leads to a solution in which case the uniqueness theory.....

su_doku, trying the other 7 solution also lead me into a dead end...so obviously I made an error somewhere, which I blame on my pen - doing this on paper takes a while. And of course, I cannot tell on which branch I went wrong. My paper with 3x3x3x3x3x3 little squares is now useless.

I took that step because I was out of ideas of new unambiguous paths.