SudokuSolver Forum

Will try harder

In passing I note the solution I gave was disjoint group and I assume the other 15 are as well.

I've been feeling remorse, this puzzle is meant to give you fun during the holiday season, not to torture your minds. I really think it is a very straight forward puzzle, all the information required to solve it is included in the image, but it seems the most obvious thing managed to escape from all of your eyes.

I planned to reveal the answer today, but for players like HATMAN (and hopefully enxio27) who are still trying, I will put off until the New Year's Eve. Meanwhile I will show one more hint, the lengths and first/last letters of the 4 words in the variant rule:



Next I will present two more similar puzzles which will also (hopefully) help on this one.

Mystery Symbols Sudoku #2


Note: use the information from #1 to get all symbols for this puzzle. Also the variant rule used in this puzzle IS NOT the same one used in #1.


Mystery Symbols Sudoku #3


Note: don't be fooled by optical illusion, there are really only 6 colours used for all symbols in this puzzle. Also the variant rule used in this puzzle IS the same one used in #1.

Revelation time.

The one and only variant rule that works, which none of you managed to spot is this:

Neighbours have different attributes.

Obviously each symbol has these 2 attributes:

1. Square structure: 1-sq/2-sq/3x3-sq

2. X structure: no-X/small-X/large-X

So for example, R2C5 is a 1-sq with small-X, therefore its 4 neighbours, R13C5 and R2C46, must not have the same attributes. Therefore they must be 2-sq/3x3-sq with no-X/large-X.

If one is observant enough, one can actually spot these relationships from the initial givens.

Therefore, the unique solution for the original puzzle is:



You can solve the puzzle using hidden singles and naked singles only, using my variant rule, which is similar to "non-consecutive", but more like "non-connective".

#2, with the identical initial givens structure, actually uses a completely opposite rule:



But it can be solved just as easily.

The following two puzzles combined the concepts of the previous 3 puzzles together:

Mystery Symbols Sudoku #4


If one actually solve this on paper in colours using crayons etc, one can see one of the most beautiful images associated with Sudoku. Too bad I don't think anyone here has the interest and motivation.


Mystery Symbols Sudoku #5


While I personally think the solution for this puzzle is less attractive than the previous, it does has a very nice property which is absent in the other one. Good luck finding it yourself.

As for the "2 Latin Squares" property, which both dyitto and HATMAN guessed, I will explain how it gets to appear in the solution using my correct rule.

First, I have to stress that the correct mathematical term for this is "Euler Square". Please google it if you have never heard of it before.

Now let's use the following notation for the symbols:

1=1-sq
2=2-sq
3=3x3-sq

N=no-X
S=small-X
L=large-X

So the original puzzle can be shown this way:



Now, without loss of generality, assume a 3x3 nonet (say N5) has 2S in the central cell (R5C5):



Using the "non-connective" rule, R46C5+R5C46={1N,1L,3N,3L}. Without loss of generality, assume R4C5=1N:



Now look at R5C46. They must not be 3L, or it will force one of R4C46 to be 2S, same as R5C5.
--> Hidden single: R6C5=3L



Now R5C46={1L,3N}. Without loss of generality, assume R5C4=1L:



Using the "non-connective" rule, the 4 other cells can be easily solved now:



As you can see, the nonet automatically becomes an "Euler Square".

Final wrap-up:

The solution grid to the original #1 puzzle has many other constraints:

• X (diagonals)
• Disjoint groups (e.g. R147C147, R258C369)
• 3 mini-rows in the same vertical stack (e.g. R147C123, R258C789)
• 3 mini-columns in the same horizontal band (e.g. R123C147, R789C258)
• Asterisk (R258C5+R37C37+R5C28)
• Small central X (\34567+/3467)
• Large central X (\13579+/1379)
• Small central + (R34567C5+R5C3467)
• Large central + (R13579C5+R5C1379)
• Various 3x3 squares with centre at R5C5 (e.g. R159C159, R357C357, R1C3+R2C6+R3C9+R4C2+R5C5+R6C8+R7C1+R8C4+R9C7)
• Each nonet is an Euler Square (semi magic square if applying a certain scheme of using 1..9 to represent the 9 symbols)

But to obtain a unique solution, there is only one simple rule which can achieve it, and that is the rule I used.

So, that is it from me. Wish all a Happy New Year and a great 2011!

Fair rule Simon

I had the other three words but could not get neighbours (even using a scrabble helper).

I played around with complex integers a couple of years ago -1-i, -i, 1-i, -1, 0, 1, -1+i, i, 1+i. I did not get much use out of them but they fit in with your formulation as two dimensional numbers.

Perhaps I'll try a kenken with them.

How did you produce the pretty diagrams?