Gurth's Puzzles

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

Emerald Challenge

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.

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...
gurth

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Joined: 11 February 2006
Location: Cape Town, South Africa

Re: Emerald Challenge

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...
udosuk

Posts: 2698
Joined: 17 July 2005

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........

tarek

tarek

Posts: 2793
Joined: 05 January 2006

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.

RW
RW
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re: GA16 and GA20

...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.
gurth

Posts: 358
Joined: 11 February 2006
Location: Cape Town, South Africa

GA22Emerald

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...
gurth

Posts: 358
Joined: 11 February 2006
Location: Cape Town, South Africa

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!

RW
RW
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GA25Emerald7

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.

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.
gurth

Posts: 358
Joined: 11 February 2006
Location: Cape Town, South Africa

Re: GA25Emerald7

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.
ronk
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GA25Emerald7

Nice fast reply, ronk.... but the small print looks wrong to me!
gurth

Posts: 358
Joined: 11 February 2006
Location: Cape Town, South Africa

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!

RW
RW
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RW wrote:Ronk, I'm afraid your solution isn't correct.

No wonder. I find myself doing single diagonal symmetry part of the time instead of 180-degree rotational all the time. After a 2nd .. and 3rd incorrect answer, I think I'll walk the dogs instead.
ronk
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Location: Southeastern USA

GA26Emerald

GA26Emerald

Meant to be a difficult one, so no redundant clues.

Code: Select all
`..64......5...2.6.4...873.23.....61.9.....8......................................`

With only 15 clues, it might rival GA22 for difficulty.
I hope to post an easier one soon, to fill more levels of choice.
gurth

Posts: 358
Joined: 11 February 2006
Location: Cape Town, South Africa

I must be misunderstanding a basic of the "emerald rules" ... so I'd appreciate a sanity check here.

GA25 Emerald: ..49...2..5..8.9..6....2...4.........6...94....2.4..6..4..6..9.7....46...........

My steps:
hs: r9c3=6
hs: r2c1=2
Eliminate unavoidable digits 1 and 3 from all locations 180-degree symmetric to clues (and placements eventually)
ns: r2c3=7
np: r2c49= 46 for
Code: Select all
` 138   138   4     | 9        1357    13567   | 13578  2      135678 2     5     7     |#46       8       136     | 9      134    46 6     89    1389  | 13457    57      2       | 13578  4578   134578-------------------+--------------------------+---------------------- 4     789   13589 | 1235678 *257     135678  | 2578   13578  1235789 1358  6     58    | 2578     12357   9       | 4      578    123578 13589 13789 2     | 13578   *4       13578   | 13578  6      5789-------------------+--------------------------+---------------------- 1358  4     1358  | 2578     6       13578   | 123578 9      2578 7     12389 589   | 12358    259    #4       | 6      58     12358 13589 289   6     | 123578   123579  578     | 2578   134578 1234578`

Now I already see a problem. r2c4=46 and r8c6=4 imply either 4=4 or 4=6 .... BUT r4c5 and r6c5 imply 4=2 or 4=5 or 4=7. No common ground. What am I missing?
ronk
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ronk wrote:I must be misunderstanding a basic of the "emerald rules" ... so I'd appreciate a sanity check here.

...

Eliminate unavoidable digits 1 and 3 from all locations 180-degree symmetric to clues (and placements eventually)

The rules don't say that digits 1 and 3 should be opposite to each other.

RW
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