## Number Groups and "very hard"

Advanced methods and approaches for solving Sudoku puzzles
I see in Sudoku that solving techniques have names.
Is there a name for this combination?

I'm still learning

Posts: 19
Joined: 07 June 2005

Thanks for all the helpful replies. With Scrose'sfirst post concentrating on row 2 i could immediately see 2 groups of 456/1379 which allowed 7 to be removed from r2c5 and allow 7 to be inserted in r8c5.
Arnie

Posts: 49
Joined: 19 May 2005

joolslee wrote: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)

Thanks Jules. I felt a bit like the boys were picking on the old lady lol. I'm sure they wern't really, it was just a bit of a shock when I saw both replys asking me to explain!

Does this help folks?

In box six my candidates were
from top to bottom C8 and C9 respectivly: 458,458,148, and 256,256,16. My thinking was 4 and 8 must be in column 8 therefore I had just to determine the other number. If in column 9 the cells R4 and R5 were numbers 2 and 6 I didn't have enough numbers to go round therefore those cells must contain 2 and 5 which left 1 and 6 remaining. I knew from the candidates number 6 couldn't be in column 8 so it must be number 1. I had however, looked at other boxes and rows to I think qualify it other ways because like I say not having done many at this level I try to be careful.
Roz

Posts: 34
Joined: 25 May 2005

BadCujo wrote:I see in Sudoku that solving techniques have names.
Is there a name for this combination?

Some people call it a disjoint subset. Others simply refer to it as a triplet.
scrose

Posts: 322
Joined: 31 May 2005

Roz wrote:I felt a bit like the boys were picking on the old lady lol. I'm sure they wern't really

Roz, I'm really sorry if I sounded like I was picking on you! That was certainly unintended. My apologies, again.

I'll go through your explanation (thanks very much for providing it!) step-by-step.

Roz wrote:In box six my candidates were from top to bottom C8 and C9 respectivly: 458,458,148, and 256,256,16. My thinking was 4 and 8 must be in column 8 therefore I had just to determine the other number. If in column 9 the cells R4 and R5 were numbers 2 and 6

So our block 6 looks like this.

Code: Select all
`9 .458 .2567 .458 .2563 14.8 1..6`

By placing 2 and 6 in r4c9 and r5c9, block 6 now looks like this.

Code: Select all
`9 .458 .27 .458 .63 14.8 16`

We can remove the candidate 6 from r6c9, leaving the 1 behind. That means we can remove the candidate 1 from r6c8.

Code: Select all
`9 458 27 458 63 4.8 1`

Roz wrote:I didn't have enough numbers to go round

This is where I disagree with you. It is still possible to fill column 8 of block 6 as (top to bottom) 458, 548, 584, or 854.

Roz wrote:therefore those cells must contain 2 and 5 which left 1 and 6 remaining

Ahhh, so that's how you thought you could remove the candidate 5's from column 8 of block 6 and place the 1 in r6c8. Now I understand how you reasoned this.

Do you understand why I think you made a (very small) mistake? By placing the 261 in column 9 of block 6, we are left with the following.

Code: Select all
`9 458 27 458 63 4.8 1`

As I mentioned above, there are still four separate ways to complete column 8 of block 6. Because of that, placing 261 in column 9 of block 6 doesn't actually allow us to conclude that 256 are the correct values for column 9 of block 6.

Does that make sense?

Fixed: mistake that Roz pointed out.
Last edited by scrose on Thu Jun 16, 2005 6:56 pm, edited 1 time in total.
scrose

Posts: 322
Joined: 31 May 2005

Thanks Scrose . To be honest, I'm really tired just now and can't be bothered to read and digest lol. I'll read it again tommorow, when I hopefully feel less brain dead .
Roz

Posts: 34
Joined: 25 May 2005

I will just reply quickly with this. I havn'tt read your reply fully but noticed this discrepancy. It could be tnhat I didn't explain my candidates properly.

Box six looked like this

9 458 256
7 458 256
3 148 16
Roz

Posts: 34
Joined: 25 May 2005

Roz, you explained your candidates properly, but I accidently left the candidates 2 and 5 out of r5c9 when I drew the block. I have corrected my post. Thank you for pointing that out!

As it turns out, it doesn't affect my explanation about where I think your error is. I'll let you digest it tomorrow morning!
scrose

Posts: 322
Joined: 31 May 2005

Roz wrote:Thanks Jules. :D I felt a bit like the boys were picking on the old lady lol. I'm sure they wern't really, it was just a bit of a shock when I saw both replys asking me to explain!

That was, ofcourse, unintended.

I asked for more explenation for two reasons:
a) you saw something we missed/you discovered a new pattern,
b) you made a mistake when solving it, which is a good oppurtunity to give some real help/explanation why it is wrong (and therefor help in everyone's learning process).

Roz wrote:If in column 9 the cells R4 and R5 were numbers 2 and 6 I didn't have enough numbers to go round therefore those cells must contain 2 and 5 which left 1 and 6 remaining. I knew from the candidates number 6 couldn't be in column 8 so it must be number 1.

So what you tried was that 2 and 6 had to go in r4c9 and r5c9. This would leave only the value 1 in r6c9. So you tried it correct?

But by trying it you reached a point where you had a cell without candidates. (somewhere else in the grid I guess in column 3 (that's where I found such a cell).

Therfor you decided that the numer 1 cannot go r6c9, correct?

If that is what you did, then you used Trial and Error...

(A small note though, when you got to that point you can easily fill in the number 5 in box 9, then you don't need the group of 4 numbers of column 8)
Animator

Posts: 469
Joined: 08 April 2005

BadCujo wrote: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) ?

As Scose says this is correct but I think there is a fractionally easier way and that is to look at the two outer cells at each end.

We have (379) (379) (1379) (1379)

Four numbers and four cells, if any of these numbers appears elsewhere in row #2 other than these four cells they can be eliminated.

Fractionally easier but the same result.
MCC

Posts: 1275
Joined: 08 June 2005

will someone explain where the game came from cause i don't really understand.
Guest

Posts: 312
Joined: 25 November 2005

air16 wrote:will someone explain where the game came from cause i don't really understand.

Visit the wikipedia entry for the history of sudoku.

scrose

Posts: 322
Joined: 31 May 2005

air16 wrote:will someone explain where the game came from cause i don't really understand.

Try looking under General Puzzle, there are a couple of threads there.
MCC

Posts: 1275
Joined: 08 June 2005

MCC wrote:We have (379) (379) (1379) (1379) ... Fractionally easier but the same result.

To-may-to, to-mah-to, right? When it comes to pairs, triples, etc., there is usually at least two ways to look at the row/column/block. BadCujo found the hidden triple, and you found the naked quad.
scrose

Posts: 322
Joined: 31 May 2005

Thanks Animator and Scrose. I understand now, I'll stick to the tried and tested ways in future!

Roz
Roz

Posts: 34
Joined: 25 May 2005

PreviousNext