Magic Sudoku

For fans of Killer Sudoku, Samurai Sudoku and other variants

Magic Sudoku

Postby Pyrrhon » Wed Aug 30, 2006 7:58 am

The following sudoku variant was invented by Alexandre Owen Muniz. Here is an own example of this theme.

Fill in the grid so that every row, every column, and every 3x3 box contains the digits 1 through 9. Each number inside the blue cells must be no larger than the number of blue cells in its 3x3 block - the same as the given within that block. Each major diagonal also contains the numbers 1-9. Note that the 9 givens make a magic square.

Image
Pyrrhon
 
Posts: 240
Joined: 26 April 2006

Postby Smythe Dakota » Sun Sep 03, 2006 3:13 pm

Nice one.

I'm a little surprised that, even with the two shades and the diagonals, you managed a unique solution with only nine clues.

Do you have software you keep modifying for all these variants, or is everything done by hand?

Bill Smythe
Smythe Dakota
 
Posts: 534
Joined: 11 February 2006

Postby Pyrrhon » Sun Sep 03, 2006 8:06 pm

There a 9 given digits and 9 cages, that is more than 9 clues. The magic sudoku is handmade. But I'd used killer sudoku software (SumuCue and JSudoku) as tools to find this problem. In other cases I make this only by hand or use other programs as tools (Simple Sudoku, Sudoku Susser and an own one). The way to find a puzzle is different for problems with a low number of solution grids and problems with a high number of grids. In both cases it is important to understand how a variant works, which solving techniques are typical. And then there are some guys which publish somewhere variants, help to verify my problems, develop solving methods for special variants or identify my invalid puzzles.

Pyrrhon
Pyrrhon
 
Posts: 240
Joined: 26 April 2006

Postby SPARTAN-117 » Mon Sep 04, 2006 11:01 pm

Has anybody actually managed to do this??

I've had a look at it for a few hours and have only added 7 numbers. I've spotted a few pairs/triples but none of them seem to help much.

Can it be done without T&E?
SPARTAN-117
 
Posts: 6
Joined: 08 July 2006

Postby motris » Mon Sep 04, 2006 11:52 pm

I was able to complete this nice puzzle without "T&E" in a little under a half hour. The two sticking points I found required some careful coloring-type logic on the numbers 1 and 9, but both times considering both the region constraints, the diagonal constraints, and the set of hidden pairs/triplets you've already placed was enough to uniquely place those digits and get several others to follow.

Thomas Snyder
motris
 
Posts: 71
Joined: 13 March 2006

Postby Pyrrhon » Tue Sep 05, 2006 6:30 am

And here is my walkthrough. (In tiny text of course, take the quote button or copy it into a wordpad to read it):

Hidden Single R2C8 = 9, R8C1=1, Naked Single R8C9=3
Naked Pair 1 and 2 in B6, B9, C8, C9
Hidden Triple 789 in B1, Hidden Pair 89 in B4, Hidden Quad in B5
Naked Triple 789 in R3, Naked Pair 89 in R4
Hidden Single 8 R7C8 = 8
Box-Line interaction with 7 between B5 and R6
Hidden Single in R3 => R3C7 = 1
Line-Box interaction with 2,3 between R3 and B2, with 2 between R4 and B5, with 6 between R4 and B6
Naked Pair 2 and 3 in R1
Box-Line interaction with candidate 7 between B3 and R1, with 4 between B3 and C9, Naked Single R5C9 = 9
Hidden Single in C4 and R6 => R1C4, R6C5 = 9
Box-Line interaction with 6 between B5 and C4, with 7 between C5 and B8
Diagonal-Cell Interaction of D/ and R9C4 => R9C4 <> 8
Generalized X-Wing with strong links in R8 D\ and weak links in C3 C7 => R4C3 <> 9, R4C3=8, R4C1=9
Hidden Singles R6C6=8 (in D\), R6C4=7 (in B5), R5C4=6 (in C4), R2C6 = 7 (in C6), R9C1=8 (in D/),
R5C7=8 (in R5), R8C4 = 8 (in C4)
Naked Single 7,9,8 in B1
Hidden Single R1C8 = 7 (in B3), R1C5=8 (in B2), R7C2 = 9 (in C2), R7C7=7 (in D\), R8C7=9 (in B9),
R4C9=7
Box-Line interaction with 5 between B3 and R3
Diagonal-Cell Interaction of D/ and R7C6/R7C9 => R7C6 <> 3, R7C9 <> 6
Hidden Singles R9C7 = 6, R7C9=5 (both in B9), R3C8=5 (in B3), R4C8=6 (in C8)
Generalized X-Wing with strong links in C2 D/ and weak links in R4 B7 => R4C45, R7C1, R9C3 <> 3
Hidden Single R1C1=3 (in D\), R1C7=2 (in R1), R2C7=3 (in C7)
Box-Line interaction with candidate 3 between B5 and C6
Turbot Fish with 1, strong links in B5 and B9 => R5C8 <> 1
and the rest are hidden and naked singles.
Pyrrhon
 
Posts: 240
Joined: 26 April 2006

Postby Smythe Dakota » Tue Sep 05, 2006 11:26 am

SPARTAN-117 wrote:Has anybody actually managed to do this?? .... Can it be done without T&E?

I had solved it when I made my earlier post. As a relatively unsophisticated solver, I am not averse to T&E (I don't even know what an X-wing is, for example). I used T&E twice. The first was to guess whether the 9 in r8 appeared in r8c3 or r8c7. One of these led (fairly quickly) to a contradiction, the other eventually to the second T&E, which was to guess whether the 1 in the center box appeared in r4c4 or r5c6. Again, one led to a quick contradiction, so bingo.

In both cases, it just so happens I tried the wrong one first, which is always satisfying since I thereby establish uniqueness as well. So, for anybody wondering, I can hereby certify that the solution to this one is indeed unique. (I always worry about this for variants created (partially) by hand.)

Bill Smythe
Smythe Dakota
 
Posts: 534
Joined: 11 February 2006


Return to Sudoku variants