The New Sudoku Players' Forum

[champagne]

The tridagon impossible pattern has been clearly shown by mith with 3 digits in these cells

Code: Select all

..x x..   
.x. .x.   
x.. ..x   

x.. x..       
.x. .x.   
..x ..x   

Heavy influenced by the “loki” puzzle, most examples that I have seen are focusing on the box “bottom right” to insert another digit.
2 possible pattern can be seen in the solution grid (thanks to blue)

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..x x..   ..x x..
.x. .x.   .x. .x.
x.. ..x   x.. ..x

x.. x--   x.. x--   
.x. -x-   .x. -.-
..x --.   ..x --x
 type 1     type 2

We know that most of the puzzles in mith’s file don’t have one of these patterns in the solution grid.
In fact, any cell in the 12 cells pattern can have an extra digit and we have no reason to give more weight to the box bottom right patterns.

I proposed this puzzle
........1.....234...5.1..62....3.....5.1..4..21.6....5..3....5..7.24....89.56.... ED=10.9/1.2/1.2
1n another thread [url]http://forum.enjoysudoku.com/tridagon-t45572.html
[/url]
In this puzzle, several cells of the 12 clues pattern have an extra digit in the solution grid, this is already too complex for a first analysis.
Working in block on all patterns with one extra digit in the other boxes does not seem too hard.
I’ll see if this is another way to find other “loki families”

[champagne]

but it seems that with a single extra digit, the magic square can always be morphed to "type 1" or "type2"

[eleven]

Isn't there a 3rd type with the 3rd cell in box 9 (can't see an equivalent) ?

[champagne]

Isn't there a 3rd type with the 3rd cell in box 9 (can't see an equivalent) ?

I am not sure to see your point, but among the six perms along the diagonals in boxes 5,6,8, 2 keep the 2 clues in the main diagonal in box 9, but with r7c7<->r9c9
and the last cell can be anywhere, generally speaking
Do you have an example to make it clear

[eleven]

No, i didn't see how to get from this pattern

Code: Select all

    ..x x..   
    .x. .x.   
    x.. ..x   

    x.. ?..       
    .x. .x.   
    ..x ..x   

to one of your 2 types with equivalence operations (but then an extra candidate in one of the other 9 cells should be equivalent to the 3 types)

[champagne]

No, i didn't see how to get from this pattern

Code: Select all

    ..x x..   
    .x. .x.   
    x.. ..x   

    x.. ?..       
    .x. .x.   
    ..x ..x   

to one of your 2 types with equivalence operations (but then an extra candidate in one of the other 9 cells should be equivalent to the 3 types)

"no" what ??

to be sure to answer for this ( what I thought could be the problem)

Make the perm on the diagonals

r4c6<->r6c4
r7c4<->r9c6
r4c7<->r4c9
(hope to have made no typo)
and you get a "type 1 in box 9.

[eleven]

Ah yes, thanks, did not think of diagonal mirror.