Gurth's Puzzles

Everything about Sudoku that doesn't fit in one of the other sections

Postby RW » Wed Oct 18, 2006 2:02 pm

udosuk wrote:"r9c1<>" what? And sorry I don't see a strong link [r6c9]=8=[r1c9]... When did you eliminate the 8 from r2c9?

Oops, forgot to mention what I eliminated... Should of coure be r9c1<>1. The 8 from r2c9 gets eliminated by locked candidates after the earlier elimination r1c1<>28.
udosuk wrote:Same here! Why is there a strong link [r78c8]=1=[r3c8]? What about the 1 in r2c8?

Ah, mistake when writing the chain. Should be the other way around:
[r2c1]-1-[r3c3]=1=[r3c8]-1-[r78c8]=1=[r8c9]

Silly that I didn't happen to spot the shorter version [r2c1]-1-[r2c9]=1=[r8c9], same as the previous eliminations of candidate 1...

Then to define some emerald seafood:

Note: Digit 5 is very special in Emerald bay and is not affected by these rules.

Emerald Eel
Code: Select all
...|...|...
-..|...|..a
...|...|...
---+---+---
...|...|...
...|...|...
...|...|...
---+---+---
...|...|...
-..|...|..a
...|...|...

If there is a strong link on candidate A between to symmetrically placed cells in a row or column r(x)c(y), r(10-x)c(y), then we may eliminate candidate A from r(x)c(10-y) and r(10-x)(10-y) and it's symmetrical opposite 10-A from r(x)c(y) and r(10-x)c(y).

Emerald Turbot fish
Code: Select all
...|...|..a
...|...|...
...|...|...
---+---+---
...|...|...
...|...|...
..-|...|..a
---+---+---
...|...|...
...|...|...
a.a|...|...

If there is two strong links on digit A that have one end in symmetrically opposite cells r(x)c(y), r(10-x)c(10-y) Then you may eliminate A from all cells that can see both the other cells of the strong links.

I apparently didn't use the full power of the Eel in my solution (second step, r8c1<>1). Same pattern of course also eliminates 1 from r2c1 and makes my last move unnecessary. After both eliminations by the Eel, the puzzle advances to a state where the puzzle solves with an Emerald Turbot fish on digits 2 and 8.

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Postby udosuk » Wed Oct 18, 2006 4:08 pm

Okay, here are the formal definitions of my EURs/EUPs... (On 2nd thought they shouldn't be tagged with "uniqueness" because the logic has nothing to do with whether the solution is unique or not... The eliminations are made because otherwise we cannot achieve a valid "Emerald/symmetrical" state... But since there seems to be another technique called the "ER" I'll keep the "U" but change the word to "Unlocking" since that's pretty much what it does on the puzzles...)

EURs: Emerald Unlocking Rectangles
Code: Select all
.*.|...|.*.
...|...|...
...|...|...
---+---+---
...|...|...
...|...|...
...|...|...
---+---+---
...|...|...
...|...|...
.*.|...|.*.

For all rectangles with corners lying on symmetrical pair of cells (i.e. centred at r5c5):

We must not allow all 4 of them to have the 2 candidates (A,10-A) only, otherwise we'll be forced to have 2 of the same values within a row/column.

Note: unlike URs, these rectangles don't need to involve exactly 2 boxes... The 4 corners can be in 4 different boxes (b1379)...

EUPs: Emerald Unlocking Parallelograms
Code: Select all
...|...|...
...|...|...
...|...|...
---+---+---
.*.|...|*..
...|...|...
..*|...|.*.
---+---+---
...|...|...
...|...|...
...|...|...

For all parallelograms with a pair of horizontal/vertical opposite lines, the other pair of opposite lines lying entirely within 2 corresponding boxes, and opposite vertices located in symmetrical pair of cells:

We must not allow all 4 of them to have the 2 candidates (A,10-A) only, otherwise we'll be forced to have 2 of the same values within a row/column/box.

Perhaps not as common as RW's eels etc, but these don't rely on strong links... Remains to see which will appear more in Gurth's other EC (Emerald Challenge) puzzles...
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Postby RW » Wed Oct 18, 2006 5:03 pm

udosuk wrote:Okay, here are the formal definitions on my EURs/EUPs...

Hmm... IMHO the eliminations made in the puzzle are actually NEPs, Naked Emerald Pairs. One cell has to have one value, the symmetrically opposite has the other value, both values may be eliminated from all cells that can see both cells in the NEP. You could also give a more general description:
Code: Select all
If a bivalue cell C has two symmetrically opposite values A,10-A you may eliminate candidates A and 10-A from all cells that can see both C and it's symmetrical opposite.


I also think the rectangles could be easier defined as
Code: Select all
No digit (except digit 5) may appear in an x-wing formation of 4 symmetrically opposite cells


example of elimination:
Code: Select all
...|.a.|...
...|aaa|...
...|.a.|...
---+---+---
...|.-.|...
...|.-.|...
...|.-.|...
---+---+---
...|.a.|...
...|aaa|...
...|.a.|...

Eliminate a from r456c5


Btw, just realized that my uniqueness argument was wrong, there wouldn't have been 2 symmetrical solutions, but no symmetrical solutions... Actually a good example of the pattern above.

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Postby udosuk » Wed Oct 18, 2006 5:56 pm

RW wrote:Hmm... IMHO the eliminations made in the puzzle are actually NEPs, Naked Emerald Pairs. One cell has to have one value, the symmetrically opposite has the other value, both values may be eliminated from all cells that can see both cells in the NEP...

Geez what a great observation RW!:) OK, so RIP to EURs and EUPs...:( Up come the NEPs!:D

Wait a minute... Perhaps there is still some value to the EUR/Ps...

For example, suppose this is r4:

. . 149 | . 46 . | . 149 .

r4c5 cannot be 4... (can you see why?)

I think this EUP is equivalent to a type 2|3|4 (whatever) UR...
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Emerald Challenge

Postby gurth » Fri Oct 20, 2006 8:19 am

re: Emerald Challenge

GA20Emerald

Udosuk and RW, you've proved me wrong in spades! I eat my words, and my hat.

I'm dazzled by the EUR, EUP, NEP, EE (Eel) and ETF (Turbot Fish). I have to award Acorns to you both for these brilliant inventions.

Obviously, things need to be made tougher already.

The formula n/(10-n), which fixed the central digit at 5 and arranged the other 8 digits in the pairs 19, 28, 37 and 46, is now thrown out of the window. In future, in any puzzle, any digit can be the central digit, and you are not told what the 4 pairings are: you have to discover that for yourself.
The 180-degree symmetry remains unchanged, except that the digit pairings can now change from puzzle to puzzle.

You can try these new rules on GA20 below. There may be some redundant clues - I don't want to make things too tough already. But, of course, there is only one emerald solution.

Code: Select all
 *-----------*
 |6..|752|81.|
 |.7.|.4.|..9|
 |2..|1..|...|
 |---+---+---|
 |1..|4.5|.2.|
 |.8.|...|.9.|
 |...|...|...|
 |---+---+---|
 |...|...|...|
 |...|...|...|
 |...|...|...|
 *-----------*


Udosuk, re your question "r4c5 cannot be 4... (can you see why?) ":
- Because it would create a naked pair [19] in row 5, placing 1 and 9 in spots asymmetrical to each other, whereas the 1 and the 9 must be symmetrical...
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Re: Emerald Challenge

Postby udosuk » Fri Oct 20, 2006 8:29 am

gurth wrote:Udosuk, re your question "r4c5 cannot be 4... (can you see why?) ":
- Because it would create a naked pair [19] in row 5, placing 1 and 9 in spots asymmetrical to each other, whereas the 1 and the 9 must be symmetrical...

What you said would be true if the line was r5. But I said it was r4... Nevermind, these are equally intriguing new eliminations you can make in this exciting new variant!

BTW good on you for introducing the new rules which increase the difficulty considerably... Will try that puzzle later...:)
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Postby tarek » Fri Oct 20, 2006 9:03 am

2 Suggestions......

This (Sum of 10 puzzle) which I've seen in the Superior thread before by Red Ed, should have some other line of symmetry, so that the given cles are symmetrical.......

I'm hoping that with what udosuk & RW presented that we can remove more clues that are redundant........

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Postby RW » Fri Oct 20, 2006 11:23 am

gurth wrote:I don't want to make things too tough already. But, of course, there is only one emerald solution.

Well, they could be much tougher than this, solves with basic sudoku technique and one ESL (Emerald Symmetrical Lock). The starting grid immediately gives away the symmetrical pairing (<89>, r5c2 and r5c8), which advances the puzzle here:
Code: Select all
 *-----------*
 |69.|752|81.|
 |.71|.48|2.9|
 |2.8|1.9|...|
 |---+---+---|
 |1..|495|.28|
 |.8.|...|.9.|
 |9..|.8.|...|
 |---+---+---|
 |...|8..|9..|
 |8..|9..|...|
 |..9|5..|.8.|
 *-----------*

From this we got a new symmetrical pairing (<25>, r1c6 and r9c4) and after basic steps and an empty rectangle we have:
Code: Select all
 *-----------*
 |69.|752|81.|
 |571|.48|2.9|
 |2.8|1.9|.5.|
 |---+---+---|
 |1..|495|.28|
 |482|...|59.|
 |95.|28.|...|
 |---+---+---|
 |.2.|8..|9.5|
 |8.5|9..|..2|
 |..9|52.|.8.|
 *-----------*

Now to define the new technique:

Emerald Symmetrical Lock

If we have strong interference between two symmetrically opposed cells (A,B) in such a way that whatever value is taken by cell A, the same value goes in cell B, then we may eliminate all candidates not present in cell A from r5c5.

This situation will probably mostly occur when box 4 or 6 has row 5 filled and only one empty cell in either of the other rows, similarily for box 2 and 8 with columns. The present stage of the puzzle:
Code: Select all
 *--------------------------------------------------------------------*
 | 6      9      34     | 7      5      2      | 8      1      34     |
 | 5      7      1      | 36     4      8      | 2      36     9      |
 | 2      34     8      | 1      36     9      | 3467   5      3467   |
 |----------------------+----------------------+----------------------|
 | 1      36     367    | 4      9      5      |A367    2      8      |
 | 4      8      2      | 36    -1367   1367   | 5      9      1367   |
 | 9      5     B367    | 2      8      1367   | 13467  3467   13467  |
 |----------------------+----------------------+----------------------|
 | 37     2      46     | 8      1367   13467  | 9      3467   5      |
 | 8      146    5      | 9      367    3467   | 13467  3467   2      |
 | 37     146    9      | 5      2      3467   | 13467  8      13467  |
 *--------------------------------------------------------------------*

Here r4c7 is cell A and whatever value we assign to that cell will also have to appear in it's symmetrical opposite, r6c3. So we know that the only candidates that possibly can be their own opposites, and thereby occupy r5c5, are <367>. We may eliminate 1 from r5c5.

This reveals a hidden single, r7c5=1. The symmetrical opposite to r7c5 (r3c5) may only have two values (36), thus no other digit may appear opposite to a digit 1 and digit one may not appear opposite to any other digits than 3 and 6. Make these eliminations and the puzzle is solved.

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re: GA16 and GA20

Postby gurth » Sun Oct 22, 2006 9:35 am

...sorry, udosuk, I misread your row 4 for 5, as you guessed, later I realised my mistake and saw what you were getting at: two 1s would have been forced into box 4 or 6, and two 9s into the other box.

tarek, thanks for suggestions. RW and udosuk seem to be enjoying these puzzles and are already investing in new techniques, I don't want to make unnecessary changes to invalidate their discoveries already... and I will remove redundant clues as I see they are not necessary.

RW, another very nice technique! Seems this variant is rich in possibilities of deductions and invention. That might make them a bit easier than I was hoping, but worth it because more interesting. I have quite a beauty waiting, but still needs one or two more finishing touches.

If this interest continues or increases I will have to take the trouble to write a better program to produce better emeralds. My present one is too slow at checking for redundancy.
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GA22Emerald

Postby gurth » Mon Oct 23, 2006 8:18 am

GA22Emerald

Code: Select all
 *-----------*
 |..9|71.|.8.|
 |.8.|..3|..2|
 |7..|...|...|
 |---+---+---|
 |8..|...|6..|
 |6..|...|1..|
 |.9.|...|...|
 |---+---+---|
 |...|3..|...|
 |...|...|...|
 |...|5..|...|
 *-----------*


Same rules as for GA20. No redundant clues this time...
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Postby RW » Mon Oct 23, 2006 10:39 pm

gurth wrote:GA22Emerald

Yup, definitely harder than the last one. My solution is so long and ugly that I won't bother to post it now, maybe I can find a shorter later. Found some new interesting techniques though. More on those later.

Thank you for the puzzle Gurth!

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GA25Emerald7

Postby gurth » Thu Oct 26, 2006 9:31 am

GA25Emerald

Code: Select all
    . . 4 9 . . . 2 .
    . 5 . . 8 . 9 . .
    6 . . . . 2 . . .
    4 . . . . . . . .
    . 6 . . . 9 4 . .
    . . 2 . 4 . . 6 .
    . 4 . . 6 . . 9 .
    7 . . . . 4 6 . .
    . . . . . . . . .


This is an Emerald Challenge with a twist. There are no threes or ones among the clues! And SS says there are too many solutions to count.

But then what does SS know about Emeralds? If you have concluded that there cannot be a unique solution with only 7 clue-digits, even for an Emerald, then you are correct! But only in one way: in another way you are wrong.

Because, not having given you any 3s or 1s to work with, it is only fair that I don't require you to include them in your solution! You are only required to place the other 7 digits in their correct cells, all 9 of each.

And there is only one way to do that, observing the usual Emerald rules of rotational symmetry. So the REQUIRED solution IS unique.

RW,
I have no idea how difficult you will find this puzzle. I confess to leaving 2 redundant clues in, so as not to spoil my favourite shape, and another 2 cells are easily established, so you have as many as 26 clues to work with. that's quite a lot, for an Emerald. On the other hand, I don't know how much difficulty is going to be caused by the absence of all 3s and 1s.

Your comments will be highly appreciated, as usual, in due course.

At the same time I also now plan to offer easier Emeralds as well, so as not to limit the delights of Emerald solving to the exceptional few.
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Re: GA25Emerald7

Postby ronk » Thu Oct 26, 2006 10:57 am

gurth wrote:This is an Emerald Challenge with a twist. There are no threes or ones among the clues!

Nice puzzle. That's the first 2-digit unavoidable set of 18 cells I've actually seen ... and it has 180-degree rotational symmetry. [edit: But that's because my solution was incorrect.]

2=2, 4=8, 5=9, 6=7
Last edited by ronk on Fri Oct 27, 2006 8:01 am, edited 1 time in total.
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GA25Emerald7

Postby gurth » Thu Oct 26, 2006 11:30 am

Nice fast reply, ronk.... but the small print looks wrong to me!
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Postby RW » Thu Oct 26, 2006 11:31 am

Ronk, I'm afraid your solution isn't correct. How do you place a 9 opposite to the initial clue r2c2=5? I'm not going to give away the real solution though, it's not so hard so just keep trying!:)

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