## Got quite a long way then...

Advanced methods and approaches for solving Sudoku puzzles

### Got quite a long way then...

As a result of my being inspired to attempt a V. Hard without any pencil marks, I sat here this morning with a monumental hangover and had a go. I was genuinely surprised at how easy the first twenty-one moves were, but we all know that there are grades within grades. Then... *Bang*. Sudden stop. So I sat for twenty minutes, then gave up and inserted all the pencil marks. *Louder Bang*. No help there. So I really gave up and dubbed the puzzle into Sadman's program and asked it for the next move, and now we get the the crux of the matter...

Can anyone tell me why the next step is to place a 7 in r2c1?

* * * 7 3 9 * * *
* 1 * 5 * 2 * 9 *
* * 9 1 * 8 7 * *
3 * * 8 1 6 9 4 2
9 4 1 2 7 3 * 6 *
8 2 6 9 5 4 * * 7
* * 2 4 9 * 6 * *
6 9 * 3 2 * * 7 *
* * * 6 8 7 * * 9

(I'm expecting I've missed something obvious like a hidden triple, if that's not an oxymoron, but the Stella is still talking to me.)
Karyobin

Posts: 396
Joined: 18 June 2005

Look for an x-wing and make some eliminations.
Then look for a triple in box 1 and make some eliminations.
Then you can place the 7 at r2c1.

scrose

Posts: 322
Joined: 31 May 2005

Nope. I'm assuming the x-wing is in the 8's, but I've checked rows and columns and I can only think I've got a few dodgy pencil marks.
Karyobin

Posts: 396
Joined: 18 June 2005

Karyobin wrote:Nope. I'm assuming the x-wing is in the 8's

Code: Select all
`.8.|...|.8....|...|......|...|...-----------...|...|......|...|......|...|...-----------.8.|...|.8....|...|......|...|...`
angusj

Posts: 306
Joined: 12 June 2005

Oh, for God's sake - nothing wrong with my pencil marks, just my eyes. Still, kind of proves my original thesis - would've been hard to complete that without a few pencil marks.

Cheers peeps.
Karyobin

Posts: 396
Joined: 18 June 2005

Karyobin,
X-Wing found for rows 1 and 7 in columns 2 and 8 for 8 - updating cell(s) (1,3)(1,7)(1,9)(7,9)
block 1: disjoint subset 245 in cells (1,1)(3,1)(1,3) - updating candidates for cell(s) (2,1)(1,2)(3,2)(2,3)
(2,1) = 7 : only possible value for this cell
etc

Or did it give you something different?
Last edited by simes on Sun Dec 11, 2011 10:13 am, edited 1 time in total.
simes

Posts: 324
Joined: 11 March 2005
Location: UK

You mean it explains itself to you? Always knew I should explore these things a little more.

How do you make it do that then?

{Five minutes later}

Of course, I could always just click on 'View solution log' couldn't I?
Karyobin

Posts: 396
Joined: 18 June 2005

### 245 triple?

How can this be Simes? - isn't there at 2 at R7C3?

Stuart
www.brightonandhove.org/sudoku/solver5c.xls
stuartn

Posts: 211
Joined: 18 June 2005

### triples or not....

With the 2 in place at R7C3, the logic goes like this....

The x-box reduces R1C3 to 45
If R1C2 was a 4, R1C3 would be a 5
Which would make R4C3 a 7
Which would mean R2C3 couldn't be a 7

Which would mean that R1C2 would be the only candidate for 7

Which is invalid - so it has to be 7 at R1C2.

Without triples.....;0)

Stuart
stuartn

Posts: 211
Joined: 18 June 2005

### Re: 245 triple?

stuartn wrote:How can this be Simes? - isn't there at 2 at R7C3?

Now you've confused me - who mentioned R7C3? (I didn't even mention R3C7, so it's not a row/column confusion.)
Last edited by simes on Sun Dec 11, 2011 10:13 am, edited 1 time in total.
simes

Posts: 324
Joined: 11 March 2005
Location: UK

### 2 atR7C3

In karyobins post there is a 2 at R7C3 which eliminates a triple including a 2 at R1C3. I'm sure you'll agree.

Stuart
www.brightonandhove.org/sudoku/solver5c.xls
stuartn

Posts: 211
Joined: 18 June 2005

### Re: 2 atR7C3

stuartn wrote:In karyobins post there is a 2 at R7C3 which eliminates a triple including a 2 at R1C3. I'm sure you'll agree.

Well, I agree there's a two at R7C3. Yep, I've just checked again, it's definitely there. To say this eliminates a triple including a 2 at R1C3 though? A 2 in column 3 eliminates another 2 in column 3? hmmm. Isn't that a teeny bit obvious? I must be misunderstanding, so I'm still confused. The only triple I can see is the one my solver found in a previous post.

<shrug>So there are at least two ways to proceed after the XWing. You've found a conclusion from a look-ahead approach, while my solver found the subset 245 in cells (1,1)(3,1)(1,3), and so eliminated candidates from cells (2,1)(1,2)(3,2)(2,3).

But in either case, I'm still missing the significance of:
stuartn wrote:How can this be Simes? - isn't there at 2 at R7C3?
simes

Posts: 324
Joined: 11 March 2005
Location: UK

X-Wing found for rows 1 and 7 in columns 2 and 8 for 8 - updating cell(s) (1,3)(1,7)(1,9)(7,9)
block 1: disjoint subset 245 in cells (1,1)(3,1)(1,3) - updating candidates for cell(s) (2,1)(1,2)(3,2)(2,3)
(2,1) = 7 : only possible value for this cell
etc

You sent this. I've underlined the relevant entry. You also say this...

Well, I agree there's a two at R7C3

How can your ' disjoint subset 245' be possible in R1C3 when as you've agreed, there's a found 2 C3 already? Please enlighten?
stuartn

Posts: 211
Joined: 18 June 2005

### triple

245 568 458
47 1 3478
245 356 9
3 57 57
9 4 1
8 2 6
157 3578 2
6 9 458
145 35 345

This may help make my point - (without resort to sarcasm). Cols 1 - 3.

It's not possible that you missed out the 2 at R7C3 when entering into your solver is it? - heaven forbid. ;0)
stuartn

Posts: 211
Joined: 18 June 2005

### Re: triple

Ahhh... I see.

OK, "disjoint subset" doesn't require that all the members of the subset appear as candidates for all the cells, just that the total number of cells with only subset members as candidates is the same as the number of subset members.

The subset consists of 2, 4 and 5, but the 2 doesn't appear as a candidate for (1,3)

Before applying the XWing, the candidates are
Code: Select all
`{2,4,5} {5,6,8} {4,5,8}{4,7}   {1}     {3,4,7,8}{2,4,5} {3,5,6} {9}`

After applying the XWing, the candidates are
Code: Select all
`{2,4,5} {5,6,8} {4,5}{4,7}   {1}     {3,4,7,8}{2,4,5} {3,5,6} {9}`

After applying the disjoint subset, they become
Code: Select all
`{2,4,5} {6,8} {4,5}{7}     {1}   {3,7,8}{2,4,5} {3,6} {9}`

stuartn wrote:It's not possible that you missed out the 2 at R7C3 when entering into your solver is it? - heaven forbid. ;0)

No, of course not, I used copy'n'paste.
simes

Posts: 324
Joined: 11 March 2005
Location: UK

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