The New Sudoku Players' Forum

[White Knight]

I am tearing my hair out after 90 mins of trying to solve this V Hard from the program. I have been doing these puzzles for the past couple of months on and off (I'm not a total adict yet!), and gradually working my way up to VHards. I think that I have grasped pairs, triples and Xwing, but it still does not seem to get me any further than you see below. Any help would be appreciated.

**8 7** ***
7** *3* *45
3*1 **2 *76

2** 5** *34
**4 *** 7**
*9* **1 **8

*** 9** 6*7
47* *6* **9
*** **7 4**

[Animator]

Row 6 is really intersting... and after that box 6 is intresting.

Take a look at both of them...

[White Knight]

OK, lets look at row 6 first; the candidates are :-
[56]9[367][2346][2479]1[25][256]8

are you saying that [56],[25],[256] are a triple? Or that [56],[256] and/or[25][256] are doubles; in which case my understanding is flawed (not unlikely!!)

Otherwise I'm afraid that I will have to ask to to spell it out in words that a simpleton like me can understand

[Animator]

[56], [256], [25] is indeed a triple...

So take a close look at box 6... and remember this: where can the 2 go in row 6?

[White Knight]

OK, so my understanding of triples was wrong!
I assumed that they were used to get rid of options within the three cells, not in the rest of the row/column/box.
So having eliminated some of the 2s in row 6, it becomes obvious that the only available options for a 2 in row six are the two in box six. Therefore the other 2s in box 6 row 5 can be eliminated.
I'll go walk the dog and ponder this for half an hour, then try again!

Just to clarify triples, can any extras within each of the three cells be eliminated, or doe the fact that there are extras within them rule the cells out from tripledom?

[Animator]

More numbers in the cells makes this triplet invalid... although it depends...

But in this case there are really two triplets.
First note, your list of candidates include 9 in column 5, this ofcourse is incorrect.

Removing that will make it that there are only three places where the numbers 3, 4 and 7 can go... and therefor you can elminante the numbers too...

It all depends on how you look at it :)

And keep on looking on box 6. you're on the right track... You can fill in a number... (after establishing the thing about the 2's, which you just did)

[White Knight]

Thanks for pointing out the "9" error. Senility must be setting in!

Yes, having eliminated the extra 2s from box6, I can put in the 1 and then the 9 in that box, then everything else unravels nicely . Thanks for all your help.

However I am Intrigued by your comments "there are really two triplets" and the "... although it depends... " ! Would you care to elaborate

[White Knight]

Yes, I've just redone the puzzle, and I think that you are talking of the second triple as the 3,4,7 in row 6. However, by the time I get to this, there seem to be no other 3,4,7s in the row to eliminate!

[Animator]

Well, if those are the only cells that can have those three numbers then it doesn't matter if there are more in there... (which is the case with the 3, 4, 7 one)...

If they can occur on other places aswell, such as with the 2, 5, 6, then an extra number in the cell will make it an invalid triple...

[White Knight]

Yes, I think I have it. There seem to be two flavours of valid triplet.
If I can give some examples to sum up :-
(* indicates a cell which does not contain any of the three triplet numbers)

TYPE 1 Triplets in three clear cells, with additional instances in other cells

[123][*][*][12][23][*][245][359][167] is a valid triplet and reduces to

[123][*][*][12][23][*][45][59][67] by deleting triplet numbers from other cells


TYPE 2 Triplets in three mixed cells, with no additional instances in other cells

[1234][*][*][1235][236][*][*][*][*] is a valid triplet and reduces to

[123][*][*][12][23][*][*][*][*] by deleting extraneous numbers from the triplet cells


INVALID TYPE Triples in three mixed cells with additional instances in other cells

[1234][*][*][1235][236][*][245][359][167] Does not reduce.

Is this correct. If so its a great leap forward for me!

[White Knight]

Thinking about it more, I see that it must be a valid technique for any number of candidates in the same number of cells, (1 in 1, 2 in 2, 3 in 3 etc., although 4 and above probably get a bit academic) right down to the single level where we naturally use it all the time without thinking. Foe example, if there is a cell in a set with a single candidate, then all the other identical candidate numbers in the corresponding row, column and box can be deleted! (Type 1) and if the candidate is the only instance of the number in the set of cells, then any other candidates which may also be in that in the cell can be deleted, leaving it as the sole candidate. (Type 2)

Its great to have a good extensible technique like this, it make programming so much easier!

[Animator]

It is correct.

Also note that if you find one such group (wheter it is one number, two, three, ... is irrelevant) you will always find another one.

This means that you should maximum look at 4 numbers.

[White Knight]

Thanks for the tip. I will watch out for that over the next few puzzles