The New Sudoku Players' Forum

[QBasicMac]

Well, T&E works fine. Is there a pattern I missed?

Mac

Solution So Far

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532 481 -9-
768 392 541
914 756 238
197 863 452
845 --7 3--
623 --- 8--
47- -38 925
289 --- 1-3
35- --9 -84


Pencilmarks So Far

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-   -   -   -   -   -   67  -   67
-   -   -   -   -   -   -   -   - 
-   -   -   -   -   -   -   -   - 
-   -   -   -   -   -   -   -   - 
-   -   -   129 12  -   -   16  69
-   -   -   159 14  45  -   17  79
-   -   16  16  -   -   -   -   - 
-   -   -   56  47  45  -   67  - 
-   -   16  126 127 -   67  -   - 


[masb]

According to my solver program (http://www.axcis.com.au/bb/viewtopic.php?t=25) you have to resort to trial and error to find R1C7 is 6 and then it is easy to solve.

[r.e.s.]

According to my solver program (http://www.axcis.com.au/bb/viewtopic.php?t=25) you have to resort to trial and error to find R1C7 is 6 and then it is easy to solve.

That's incorrect. There's a "uniqueness rectangle" using r7c3, r7c4, r9c3 and r9c4, then an xy-wing.
(Alternatively, there's a forcing chain r6c5-4-r8c5-7-r8c8-6-r5c8-1-r6c8 giving r6c5<>1.)
Either way, the solution then follows with only simple moves.

[QBasicMac]

There's a "uniqueness rectangle" using r7c3, r7c4, r9c3 and r9c4, then an xy-wing.

Wow! I finally see a "uniqueness rectangle" that makes sense!

THANKS!

So therefore r8c4=5

But I don't know why you said "then there is an XY-Wing". After the above placement, the puzzle solves immediately.

Mac

[r.e.s.]

Wow! I finally see a "uniqueness rectangle" that makes sense! [...] So therefore r8c4=5

But I don't know why you said "then there is an XY-Wing". After the above placement, the puzzle solves immediately.

Actually, the uniqueness rectangle (which gives r9c4=2) is not really needed, and the xy-wing (on 7's) is enough to get r8c8=6 -- but in either case, it's after the xy-wing that r8c4=5 can be deduced.

PS: Careful with the quotes ... It would be nice if you were to edit your previous post so it doesn't attribute to "masb" something I wrote, as it seems to do.

[QBasicMac]

Actually, the uniqueness rectangle (which gives r9c4=2) is not really needed

Oops, now I am confused again. Just when I "Understood" uniqueness rectangles at long last.

On my workpad, I see

Code: Select all

16   16
 9   56
16  126


And I conclude that therefore the possible patterns are

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1  6                6  1
9  5      or        9  5   
6  1                1  6


And therefore, r8c4=5 which immediately and nicely solves the puzzle.

Now you are saying that is not the uniqueness rectangle after all, as the only value solved by recongizing that pattern is r9c4=2.

*Headache* - Back to the drawing board.

Mac

P.S. Edited post

[Max Beran]

You can lose the 6 in r5c8 by following an xy-chain from r5c9 to r8c8. It then completes eezy-peezy. That's not the only one either. You can eliminate the 7 as per masb with an xy chain - no need for "cheats".

[r.e.s.]

Actually, the uniqueness rectangle (which gives r9c4=2) is not really needed

Oops, now I am confused again. Just when I "Understood" uniqueness rectangles at long last.

The uniqueness rectangle is this part ...

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16 | 16
 . | .
16 | 126


which (given that the puzzle is valid) implies that the '126' reduces to '2'.

[QBasicMac]

The uniqueness rectangle ... implies that the '126' reduces to '2'.

So evidently, my conclusion about the "5" was just a lucky guess, not logically drawn from the facts.

Interesting

Mac

[Max Beran]

There isn't an xy-chain that directly eliminates the 6 from r8c4 but there is the next best thing, an xy-chain that eliminates the 5 from r8c6 (which of course immediately forces the 5 under discussion): 56(r8c4)-67-71-14-45(r6c6).

So, no need to invoke uniqueness here.

Max Beran

[QBasicMac]

So, no need to invoke uniqueness here.

Good, because I REALLY don't understand.

Given
16 16
16 126

Why would we conclude this:
16 16
16 2

Why not also this:
16 16
16 1

which yeilds

1 6
6 1

Seems like for some puzzle somewhere, that would be the correct elimination.

So it is really T&E between 1,2,and 6 as I see it. Meaning the whole concept is bogus.

Mac

[PaulIQ164]

Notice that if your solution were the case, you could swap round those 1s and 6s, and get another valid solution to the puzzle. So while yes, there could be a puzzle somewhere where that would be the elimination, it would have to be a puzzle with multiple solutions. So if you know that the puzzle does have a unique solution, then you can set that cell to be a 2.

[Shazbot]

ah - pipped at the post - onya Paul! But I'm leaving my explanation anyway - says the same thing as Paul's but gives the examples too.

-------------

I'm not really up on the uniqueness rectangles, but you're working on the assumption that the puzzle itself is unique.

Given
16 16
16 126

If you place the 2 somewhere else, and there's nothing to filter down the remaining candidates, you're left with
16 16
16 16

which could by guessing either be narrowed down to
1 6
6 1

or
6 1
1 6

Thus the puzzle has 2 solutions.

In using the uniqueness rectangle you're saying "I believe this puzzle has only ONE solution, so if I get to a point where the elimination of one candidate is going to leave me with TWO solutions, then that candidate must not be eliminated."

Thus if the elimination of the 2 leaves you with a two-solution puzzle, the 2 cannot be eliminated, so must be the correct value for that cell. You're therefore left with
16 16
16 2

This should allow you to eliminate candidate 2s in other parts of the grid, and should eventually lead to being able to finish off those remaining 3 cells giving you only 1 solution - either
1 6
6 2

or
6 1
1 2


Of course if you check the answer they COULD have 1661 or 6116 in those 4 cells, in which case the puzzle was poorly written in the first place and the author didn't ensure a single solution.

So it's probably not T&E in the truest sense - by logic you reach the decision that the only possible candidate to use in that cell to ensure a unique solution, is 2. (Don't we see ALL newer methods and advanced techniques as T&E until we actually understand them and can apply them ourselves?)

[r.e.s.]

Given
16 16
16 126

Why would we conclude this:
16 16
16 2

Why not also this:
16 16
16 1

[...]So it is really T&E between 1,2,and 6 as I see it. Meaning the whole concept is bogus.

No, it's neither T&E nor bogus, given that the puzzle is valid (i.e. that it has a unique solution). ...

It's important to show the box-boundaries -- what we have is this ...

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16 | 16
 . | .
16 | 126

So, if the '126' reduced to '16', then the possibilities would be

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 1 | 6        6 | 1
 . | .        . | .
 6 | 1        1 | 6

But if the puzzle were to solve to either one of those two, then switching the '1' and '6' on the corners of the rectangle would produce another solution, which would contradict the given validity of the puzzle.

For valid puzzles, it's a "theorem" that the occurrence of such a "uniqueness rectangle" pattern allows the reduction to the appropriate candidate (here reducing the '126' to '2') -- and no T&E is involved because we're applying that theorem, not going through the logic to prove it every time.

[QBasicMac]

So, if the '126' reduced to '16', then the possibilities would be

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 1 | 6        6 | 1
 . | .        . | .
 6 | 1        1 | 6


Yes "if the '126' reduced to '16' then.... is bad

But what if the '126' reduced to '12'? That is my mental block.

As I see it, the following possibilities exist in the reduction of pencilmarks in '126': The box reduces to 1, 2, 6, 12, 26. But not 16.

From that list, you select '2'. I guess I imagine that someday we will find a valid puzzle where '1' would be the correct value and '2' is incorrect.

I very clearly understand we cannot have
16 16
16 16
in a one-solution valid puzzle.

But we could certainly have
16 16
16 1

which is the same as
1 6
6 1

Anyway, go to bed and forget me. I doubt anyone can explain what's going on other than to simply repeat the same old mantra.

Thanks for trying, anyway. At least I now know that I don't understand the uniqueness rectangle and that my placement of the 5 was just a lucky guess.

Mac