The New Sudoku Players' Forum

[PaulIQ164]

I was doing a puzzle (Times Book 4 #83 as it happens), and I reached this position:

Code: Select all

 12¦84 ¦3 7
 93¦  1¦4 
 4 ¦  3¦  1
---+---+---
1 9¦534¦ 7
 5 ¦   ¦134
   ¦17 ¦  9
---+---+---
   ¦617¦925
5 1¦3  ¦746
 76¦4 5¦813

Now, there are three cells unfilled in row 8, and three cells unfilled in box 8. In both cases, those cells must be 2, 9 and 8. Two of the cells are in row 8 and box 8 (r8c5, r8c6), so the remaining cell in row 8 (r8c2) and the remaining cell in box 8 (r9c5) must contain the same number, either 2 or 9 or 8. There's a 9 already in column 2, so r8c2 can't be 9, and there's a 8 already in row 8, so r9c5 can't be 8. But they also both have to be the same thing, so if they can't be 8 and they can't be 9, both those cells must be 2s. So I put them in and the rest of the puzzle fell apart. Hurrah!

But what I can't decide is whether this technique is ever truly useful. You see in this case, I could have placed the 9 in box 7 (only one place it can go), and form that placed the 2 similarly, and it'd have done the same thing for me, with only standard techniques. So can anyone clever work out if this is a legitimate tactic, or just a consequence that will sometimes arise from things you can do anyway?

[bennys]

Eliminating the 8 from R8C2 based on the fact that the 8 in box 8 must be in R8 is standard no need any advance technic.

[Shazbot]

That's an interesting idea. I'll keep it in mind as I play and see if it's helpful when other techniques fail. I think it could possibly be useful, but need to work with it a few times to tell.

[PaulIQ164]

Eliminating the 8 from R8C2 based on the fact that the 8 in box 8 must be in R8 is standard no need any advance technic.

Yes, that's true. I think it's essentially a reworking of that. Probably.

[stuartn]

Isn't R1C1 forced to be a 6? - not sure I see where the new strategem comes in.

stuartn

[PaulIQ164]

Yes, it is. I'm not saying the 'new' technique is the only thing you can do (it couldn't possibly be - it's a Pappocom puzzle, and they only use a certain set of techniques of which this is not one), but it is something you can do.

[r.e.s.]

Bob Harris's laws of leftovers (explained in his tutorials on sudoku with irregular boxes) can also be applied here -- nothing really new, just a different way of looking at the picture: If two units intersect, in each of these units the set of "leftovers" (cells in the unit but not in the intersection) contains the same set of digits.

Here are rows 7, 8, 9, of the candidate grid for PaulIQ164's puzzle ...


[2007/04/05: Updated link address.]

BLUE is the intersection of Row 8 and Box 8. YELLOW and RED are the two sets of "leftovers". The set of digits in YELLOW is the same set of digits in RED, thus eliminating the 8 from '28' in r8c2 and the 9 from '29' in r9c5; consequently, r8c2=r9c5=2.

The method generalises: If a chosen set of cells is extended in two different ways to make a collection of whole units each way, then the set of added cells in each of the two ways must contain the same set of digits; alternatively, ...

If a chosen collection of whole units is modified into a different collection of whole units by adding some cells and removing others, then the set of cells added and the set of cells removed must contain the same set of digits.

[Lardarse]

A picture is worth a thousand words, and that picture elegantly describes both this situation, and also the technique that needs to be used. Looks useful, if you can remember to use it...

[stuartn]

This can also be looked at as ' the only 8's in the middle box (of the partial puzzle) are in R2 - therefore there can be no other possibility for 8 elsewhere in the row.

The same sort of thing applies to the 9's - the only 9's in R2 (of the partial puzzle) are in the middle box - therefore there can be no 9's elsewhere in the middle box.

hope this helps!

stuartn

[Pi]

You don't need that tecnhique for the picture example.

The only nine candidates for the middle row are in the same box and so there is no way that th othe box could be a nine and so must be a 2

[stuartn]

Didn't I just say that? -

[Cec]

Didn't I just say that? -

Yes - and I actually find Stuartn's explanation easier to understand.

This scenario simply shows the technique of applying 'Locked Candidates (1)' where the candidate 8's in r8 are locked (restricted) in box 8 which therefore eliminates any other candidate 8 in this row. This leaves 2 to fill r8c2 and hence candidate 9 (also a 'hidden' single in box 7) as the only possibility to fill cell r9c1. This then leaves candidate 2 as the only possibility for cell r9c5 in box8.

Cec

[Pi]

Sorry i think i may have been typing my message at the time when the other was posted

[cho]

Sorry i think i may have been typing my message at the time when the other was posted

Then you should get a typing tutor or a better ISP. There appears to be four hours between the two posts.

The 'simplest' solving technique is surely the lone nine in the left hand box (forced by the other two nines in the left stack and the nine in R7). I think the discussion was more about techniques than finding a solution.

cho

[NorwegianViking]

... so the remaining cell in row 8 (r8c2) and the remaining cell in box 8 (r9c5) must contain the same number, either 2 or 9 or 8. There's a 9 already in column 2, so r8c2 can't be 9, and there's a 8 already in row 8, so r9c5 can't be 8.

Paul, I think you have a typo: "and there's a 8 already in row 8, so r9c5 can't be 8". I suppose it should read: "and there's a 8 already in row 9, so r9c5 can't be 8".

Anyway, thanks for posting your discovery.