Down Under Upside Down - a Sudoku puzzle

Post puzzles for others to solve here.

Postby tarek » Tue Sep 23, 2008 4:40 am

Following recommendations in the HOTLINKING Thread, I started using tinypic.com .... I haven't encountered any problems with that service YET:) .

Why not try that one udosuk ?

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Postby RW » Tue Sep 23, 2008 5:43 am

My first solution was essentially the same as elevens second solution. I now found the solution path that needs XY-wing. Next, let's see if we can get rid of this XY-wing...:)

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Postby udosuk » Tue Sep 23, 2008 6:46 am

I've replaced the image by a tinypic one.

Pat, the symbol mapping between the pic & the text is straight forward, as [6,8,9] stays unchanged, 99 becomes 1, 66 becomes 3, 86 because 7, 98 becomes 4, 68 becomes 2, 89 becomes 5.

tarek, I still prefer imageshack more, for the reason I stated in that thread and also that tinypic requires javascript while imageshack doesn't.

eleven & susume, hopefully you can persist in your quests to find the xy-wing (or better) solving path. I guarantee it's satisfying!:)

RW, congrats and I hope you can find a better solution. I'm onto other projects now, let's see if there will be any new development by the weekend...
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Postby susume » Wed Sep 24, 2008 8:12 pm

In the corners I have not exactly an XY wing, but two overlapping wings of
XY - XYZ - X'Z , where X' is the mirror value of X and the ends of the wing are in mirror cells. I got a couple of eliminations from this but not enough to crack the puzzle.

I'm noticing some properties of emeralds --probably discovered by someone else in 2006.:) Each cell can, in effect, "see" its mirror cell (or its 3 mirrors if you have 4-way symmetry). If a' and a" are symmetrical values, a' in cell e and any value other than a" in e's mirror can't both be true:
Code: Select all
(a')e' - (~a")e"


Cells in r5, c5, or box 5 can see their mirrors both in the usual way and by symmetry, which is sort of like a box-line interaction. Example: a' in r5c1 NAND (~a") in r5c9 (same symmetry inference given above), and a' in r5c1 NAND a" in all the rest of r5 (the symmetry-line interaction).
Code: Select all
  (a')r5c1 - (~a")r5c9
  (a')r5c1 - (a")r5c2345678
  (a")r5c9 - (~a')r5c1
  (a")r5c9 - (a')r5c2345678


I've got symmetric values 6 and 9 both bi-located in row 5, so they each see their own mirror cell and their conjugate's mirror cell; it seems I ought to be able to squeeze an extra inference out of that but haven't nailed it down yet.
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Postby RW » Wed Sep 24, 2008 10:45 pm

susume, have you found the eel yet:?:

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Postby susume » Thu Sep 25, 2008 10:39 pm

RW, thanks for the hint - hadn't heard of eels before but have now read about your emerald seafood (didn't I guess 2006?). Now I've found:
Code: Select all
Emerald eel 89 in r8c46 => r2c46<>89, HS r2c3=89
xy-wing (8|98)r2c1, (8|99)r3c3, (98|99)r9c3 => r9c1<>98, NS r9c1=68
Singles to end


But in order to get the strong links for the eel I first have to use the forcing thing in r19c19 (I can't really reduce it to a wing - just "whichever candidate is in r1c1, 89 is forced in r1|9c9, so r8c9<>89"). I don't think this is simpler than the nice loop I used earlier, so I haven't yet found a truly simple path.
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Postby eleven » Fri Sep 26, 2008 3:25 am

I'd like to know either, if you possibly look at this first step as a xy-wing (r1c9=5 -> r9c1=2, r1c9=7 -> r19c1=24).
I now stared at this grid for an hour again, but all i found out was that 4 can be eliminated from r5c6 with coloring.
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Postby udosuk » Fri Sep 26, 2008 8:04 am

Thank you very much for all the participation guys, all have done well, and many have come close to working out the shortest solution!:)

However, it's time to reveal the answer. I think it's best to explain in the 4-pic series as below:

1. RW calls it EEL, I call it EUR...

2. Another EEL/EUR but different working mechanism, which leads to an immediate hidden single...

3. A very simple xy-wing...

4. After a hidden single, a box-line intersection, 2 naked singles, 3 hidden singles and a cascade of 49 naked singles...

BTW you guys can still beat me by finding a solving path that doesn't need that xy-wing (with simpler moves).:idea:
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Postby RW » Fri Sep 26, 2008 8:29 am


...:?: ...

No I don't call it an Eel, neither can I see how that relates to your EUR (which was later redefined as NEP). Perhaps you could explain?

Your second picture was an Eel, but once again I don't see any relation to your EUR...

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Postby udosuk » Fri Sep 26, 2008 9:31 am

RW wrote:No I don't call it an Eel, neither can I see how that relates to your EUR (which was later redefined as NEP). Perhaps you could explain?

Your second picture was an Eel, but once again I don't see any relation to your EUR...

My bad, the 1st move isn't an EEL. I should have mentioned EUR only.

My EUR works a bit like UR, in the sense that you foresee the prospect of some rectangles turning into illegal patterns and what ensuring eliminations you can make.

In my first pic, you must foresee easily that r19c19 can't form the EUR {86,98}, thus 68 of c1 is locked @ r19c1 and 89 of c9 is locked @ r19c9.

In my 2nd pic, you must foresee easily that r2c46 can't form half of the EUR {68,89}. But since 68 of r2 is locked @ r2c46, r2c46 can't contain 89. This is equivalent to one of the UR types (no time to research).

Both moves are direct consequence of my EUR, but only the 2nd one can be interpreted as your EEL.:idea:
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Postby RW » Fri Sep 26, 2008 3:56 pm

udosuk wrote:In my first pic, you must foresee easily...

Any particular reason why this must be done "easily":?::D

Your logic is of course true, but I don't like the term "EUR" for two reasons:

1. It has nothing to do with uniqueness.
2. You do not need to find a rectangle, a pair is enough. A simple definition: Two cells A and B, where B can see both A and the symmetrical opposite of A, can not contain two symmetrically opposite digits.

In your first step of your puzzle, all we need to notice is that r19c1 only has two symmetrically opposite candidatess plus one extra candidate. This extra candidate may be eliminated from all cells that can see both cells r19c1. Then we may immediately do the symmetrically opposite elimination in c9.

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Postby udosuk » Fri Sep 26, 2008 11:06 pm

RW wrote:Any particular reason why this must be done "easily":?::D

Because I specifically claim my solving path is nothing harder than xy-wing. I have to make sure the readers agree my first 2 steps are easier than an xy-wing.:idea::D

RW wrote:Your logic is of course true, but I don't like the term "EUR" for two reasons:

1. It has nothing to do with uniqueness.

You are so wrong here. It has everything to do with uniqueness. In fact, the whole emerald concept wouldn't work if uniqueness is not guaranteed (no time to draw up a counter-example but hopefully you can see this from elsewhere).

That said, I definitely remember I've changed the full name to Emerald Unlocking Rectangle in the thread you cited. (It is of course an intended pun to the currency unit.:) )

RW wrote:2. You do not need to find a rectangle, a pair is enough. A simple definition: Two cells A and B, where B can see both A and the symmetrical opposite of A, can not contain two symmetrically opposite digits.

In your first step of your puzzle, all we need to notice is that r19c1 only has two symmetrically opposite candidatess plus one extra candidate. This extra candidate may be eliminated from all cells that can see both cells r19c1. Then we may immediately do the symmetrically opposite elimination in c9.

Yep, all this is true. But still, the whole premise of the logic involves 4 cells, which form a rectangle. In my 1st step, I made use of all 4 cells because I needed both sides of the elimination to expose the hidden single in the 2nd step. This can of course be expressed as 2 separate moves (one for r19c1, one of r19c9). I think it would look clumsy to explain pictorially if I just perform the move on the left and then repeat the elimination on the right using symmetry.

In my 2nd step, I only needed the eliminations at the top half so I totally skipped the operations at the bottom half. Perhaps I should call this HEUR (Half Emerald Unlocking Rectangle). It's your choice (and a perfectly valid one) if you prefer to view this as a "pair" instead of a "half rectangle".:)

As a matter of fact, I can think of some other names for this technique, e.g. ERO (Emerald Rectangle Operation) and HERO (Half Emerald Rectangle Operation). All this naming business, of course, is just for leisure purposes.:D
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Postby RW » Sat Sep 27, 2008 12:29 am

udosuk wrote:
RW wrote:1. It has nothing to do with uniqueness.

You are so wrong here. It has everything to do with uniqueness. In fact, the whole emerald concept wouldn't work if uniqueness is not guaranteed

You are right that in a symmetrical sudoku puzzle like this one it is an uniqueness technique. But in a true Emerald puzzle (like those posted by Gurth, where symmetry is an extra given constraint) this is not an uniqueness technique.

udosuk wrote:This can of course be expressed as 2 separate moves (one for r19c1, one of r19c9). I think it would look clumsy to explain pictorially if I just perform the move on the left and then repeat the elimination on the right using symmetry.

It doesn't need to be expressed as two separate moves, because when you make the eliminations in c1, the symmetrical eliminations in c9 are a direct result of this. This is the beauty of symmetry techniques. You never need to think twice about the symmetrical cells. If you find a complex forcing chain, it always eliminates both the target candidate and the symmetrically opposite candidate. If you look for a rexctangle, you only need to look for the pair then you know the rest of the rectangle will also be there.

udosuk wrote:It's your choice (and a perfectly valid one) if you prefer to view this as a "pair" instead of a "half rectangle".:)

I do prefer the pair, because I think or goal when defining techniques should be to find the minimum amount of information required to make each elimination.

udosuk wrote:As a matter of fact, I can think of some other names for this technique, e.g. ERO (Emerald Rectangle Operation) and HERO (Half Emerald Rectangle Operation). All this naming business, of course, is just for leisure purposes.:D

You're right! Naming is not the most important issue here, so get back to work and post some more puzzles!!:D

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Postby udosuk » Sat Sep 27, 2008 1:55 am

RW wrote:You are right that in a symmetrical sudoku puzzle like this one it is an uniqueness technique. But in a true Emerald puzzle (like those posted by Gurth, where symmetry is an extra given constraint) this is not an uniqueness technique.

I see your point. The technique may or may not necessarily be associated with uniqueness, depending on how the puzzle is set out. So my usage of "unlocking" should settle the issue. But I think someone like ronk will probably associate this technique with uniqueness no matter what, because of the similarity to UR.:)

RW wrote:
udosuk wrote:This can of course be expressed as 2 separate moves (one for r19c1, one of r19c9). I think it would look clumsy to explain pictorially if I just perform the move on the left and then repeat the elimination on the right using symmetry.

It doesn't need to be expressed as two separate moves, because when you make the eliminations in c1, the symmetrical eliminations in c9 are a direct result of this. This is the beauty of symmetry techniques. You never need to think twice about the symmetrical cells. If you find a complex forcing chain, it always eliminates both the target candidate and the symmetrically opposite candidate. If you look for a rexctangle, you only need to look for the pair then you know the rest of the rectangle will also be there.

In an intuitive manner every move you perform you can immediately follow with the symmetrical move (unless it's on r5c5). But for my solving path above I'm trying to do the least amount of work before the final "cascade of naked singles". And I also don't want to use any unnecessary technique (e.g. you can notice I didn't bother with the "Gurth's symmetrical placement" at r5c5 which is one of the easiest move to do at the start).

A challenge for you - can you find another solving path, making use of nothing harder than xy-wing, that ends up with a cascade of naked singles of 50 or more?:?:

RW wrote:
udosuk wrote:It's your choice (and a perfectly valid one) if you prefer to view this as a "pair" instead of a "half rectangle".:)

I do prefer the pair, because I think or goal when defining techniques should be to find the minimum amount of information required to make each elimination.

"Pair" has less information than "half rectangle" only in number of letters I think. A "half rectangle" means a "pair of corners of a rectangle". Note the word "rectangle" here also has an implicit meaning that the rectangle must be centred at r5c5, and this implicit meaning can't be easily interpreted with the word "pair".:idea:

RW wrote:Naming is not the most important issue here, so get back to work and post some more puzzles!!:D

I did post quite a few puzzles lately, but it's a matter of whether you guys like to tackle the challenges! For example, I'm still waiting for someone to reply on these puzzles.:idea:
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Postby eleven » Sat Sep 27, 2008 5:49 am

It is really annoying, that both of you did not mention, that you did need my first step in your solution paths.
So we looked for something, which was not there - how could we know, that you have a strange feeling, what is as easy as an xy-wing ? Otherwise it would have been very easy to find the second step.
(btw not to place a single at the beginning is just stupid for me.)

As a revenge here is an unfair challenge for you:
Code: Select all
 *-----------------------*
 | . 9 6 | . . 5 | . 1 . |
 | . . . | 3 . . | . . . |
 | 2 . . | . . 6 | 5 . 9 |
 |-------+-------+-------|
 | . . 5 | . . . | 6 . . |
 | 9 4 2 | . . . | 3 . . |
 | . . 3 | . . . | 1 9 4 |
 |-------+-------+-------|
 | . 2 . | . . 4 | . . 1 |
 | 5 . . | 8 . 2 | 4 . 7 |
 | . 8 4 | 7 . 1 | . 6 . |
 *-----------------------*

Nothing harder than xy-wing needed (no xy-chain).
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