## Number Groups and "very hard"

Advanced methods and approaches for solving Sudoku puzzles

### Number Groups and "very hard"

I read with interest this useful thread on number groups because this is the area I havemost difficulty with.

http://forum.enjoysudoku.com/viewtopic.php?t=410&postdays=0&postorder=asc&start=30

I cannot see the groups necessary to solve this computer generated VH puzzle. I see a pair in box 1 and a triplet in column 7 but nil else...

4** *91 **8
*** 2*8 ***
*1* *53 *6*

**3 1*7 9**
**1 **9 7**
**7 5*2 3**

*64 *15 82*
1** 9*4 6**
5** *26 1*4

I have tried to identify the 2 logical approaches Andrew mentions in his very helpful response in the thread above but cannot take this puzzle furhter.. any suggestion much appreciated
Arnie

Posts: 49
Joined: 19 May 2005

I'm fairly new to very hard myself Arnie, so what I am seeing might not be the best approach, but I think the answer lies in column 8. I say this after studying box 6, and related upper and lower boxes (1,9).
Roz

Posts: 34
Joined: 25 May 2005

The options for column 8 are:

37
1379
6
458
458
148
2
753
739

I cannot see now to take this on... simply "eyeballing" this column suggests there might be 2 groups of458/1379 but I think this is not right because of the 1 and 5. this scenario is where I repeatedly get stuck with very hards . What else do you see with column 8?
Arnie

Posts: 49
Joined: 19 May 2005

Look at column 3, then at row 2, then fill a number in column 5.
scrose

Posts: 322
Joined: 31 May 2005

I think with me it's just a case of looking at things how it seems to work for me. So basically I saw what I think is a group of four (candidates 3,5,7,9) in column 8 rows 1,2,8 and 9, but I like you Arnie, am not too confident with groups, but I'd already determined the central 3 cells must contain a 4 and a 8, so with that added information it worked, enabling me to fill number 1 in C8,R6.
No doubt Scrose approach is much better, but thought I'd attempt to tell you my thought process.
Roz

Posts: 34
Joined: 25 May 2005

Although you managed to find the correct cell for the 1 in column 8, I think your logic is faulty. You cannot eliminate the candidate 1 from r2c8. Consider the following. We have the following candidates for column 8.
{37} {1379} [6] {458} {458} {148} [2] {357} {379}

Even though it is incorrect, suppose we place a five at r4c8. Now we have the following candidates.
{37} {1379} [6] [5] {48} {148} [2] {37} {379}

With the {48} pair, we can eliminate the candidate 1 in r6c8 place a 1 in r2c8. We are left with the following candidates.
{37} [1] [6] [5] {48} {48} [2] {37} {379}

Our list of candidates is still valid (even though the grid will eventually become invalid). This is why you cannot remove the candidate 1s from r2c8.

I might not have explained this as clearly as others can. I'll think about this a little longer and try to explain it a second time.

Updated: fixed an error
Last edited by scrose on Thu Jun 16, 2005 12:23 pm, edited 1 time in total.
scrose

Posts: 322
Joined: 31 May 2005

Okay, here is my second attempt at explaining why you cannot eliminate the candidate 1 from r2c8.

You were looking at the candidates 3,5,7,9 in column 8. According to simes' explanation, some combination of these candidates must only occur is 4 cells in column 8. However, these candidates actually occur in 6 cells because of the candidate 5's in r4c8 and r5c8. That is why you cannot make any eliminations based on the candidates 3,5,7,9.
scrose

Posts: 322
Joined: 31 May 2005

scrose wrote:Okay, here is my second attempt at explaining why you cannot eliminate the candidate 1 from r2c8.

You were looking at the candidates 3,5,7,9 in column 8. According to simes' explanation, some combination of these candidates must only occur is 4 cells in column 8. However, these candidates actually occur in 6 cells because of the candidate 5's in r4c8 and r5c8. That is why you cannot make any eliminations based on the candidates 3,5,7,9.

I agree it isn't the way Simes explains, nor was it mean't to be. It was a mixture of the group technique plus looking at adjacent cells, boxes and rows. I still have however more candidates in the remaining cells than your example. For that row I have (3,5,7,9)(3,5,7,9)(6)(4,8)(4,8)(1)(2)(3,5,7)(3,7,9). I do not beleive my logic is faulty, just maybe a different, and not so refined approach.

I haven't actually looked at the puzzle more than to explain, since last night (haven't had time) but will do later.
Roz

Posts: 34
Joined: 25 May 2005

Can you post all your candidates? (preferably before filling in the number 1 in column 8).

I can see no reason why you removed the number 1 of r2c8...

(You can remove candidates from r8c1 and r8c2 based on the information of column 7)
Animator

Posts: 469
Joined: 08 April 2005

Animator, you beat me to it. I'll post my reply anyway as it explains the eliminations made due to column 7 in a little more detail.

Roz wrote:For that row I have (3,5,7,9)(3,5,7,9)(6)(4,8)(4,8)(1)(2)(3,5,7)(3,7,9).

I think you meant column, right? From my perspective the candidates for column 8 are as follows.
{357} {134579} [6] {458} {458} {148} [2] {357} {379}

Notice that the candidate 4's in column 7 are only in block 3. Thus you can eliminate the candidate 4 from r2c8.
{357} {13579} [6] {458} {458} {148} [2] {357} {379}

Similarily, notice that the candidate 5's in column 7 are only in block 3. Thus you can eliminate the candidate 5's from r1c8 and r2c8.
{37} {1379} [6] {458} {458} {148} [2] {357} {379}

That explains how you can remove the candidate 5's from r1c8 and r2c8 in your list of candidates.

You can also remove the candidate 9 from r1c8 in your list of candidates because there is already a 9 at r1c5.

Your list of candidates is missing the candidate 5's in r4c8 and r5c9. How did you remove those?
scrose

Posts: 322
Joined: 31 May 2005

Eck, I feel like I'm at school lol. ;-)

If I have time I will try to explain later. I know you are both very much better at this and I had already qualified my lack of experience by explaining to Arnie that I haven't done many very hard yet. I just thought in posting last night it might help him a little. I'm begining to wish I hadn't bothered!
Roz

Posts: 34
Joined: 25 May 2005

I'm very sorry if it seems we're being intimidating or like we're ganging up on you!

I'm just trying to understand your reasoning and the process you went through to make the eliminations you did. I'm looking forward to your reply when you have more time! Whatever you do, don't give up on sudoku!
scrose

Posts: 322
Joined: 31 May 2005

Hi Roz, I just wanted to say that your posting IS helpful. It opens up the discussion to show why a particular approach or a misunderstood technique doesn't work ... and sometimes I find that really useful in helping me to gain a better understanding. I see where you're coming from with your strategy and I'm looking forward to see why that can't work to help me understand these triple and quadruple sets more clearly. I'm sure that is the intention of Animator and Scrose, not just to shoot you down :o)
joolslee

Posts: 29
Joined: 05 May 2005

I'm showing the following in row #2

(379) (379) (56) (2) (467) (8) (45) (1379) (1379)

can the combination (56) (467) (45) allow me to eliminate the 7 in the (467) ?