Roee Anuar
Israel
Tel Aviv
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A riddle for the SET lovers. Use the following cards to build a 3x3 SET magic square so that each row, column and diagonal forms a SET:
Look here to see if you solved it correctly:
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Spoiler (mouseover to reveal):
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Lars Wagner Hansen
Denmark
Sorø
Any time, any place!
Fingers off, that's my car!
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There are many other solutions, for example this:
C#Ff: C=Color (Green, Purple, Red), #=number, Ff=Filling (Open, Solid, Striped)
P3St G2St R1St (all ovals)
P3So G2So R1So (All diamonds)
P3Bl G2Bl R1Bl (all squiggless)
Likewise you could turn the square 90 degrees, 180 degrees or 270 degrees (4 different soluitions), or you could swap any two columns (6 different solutions), or swap any two rows (6 different solutions), so there are 4*6*6=144 different solutions to your riddle.
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Julius Andrikonis
Lithuania
Vilnius
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Swapping of rows will not always work because of diagonals. You can swap first row with the last one though. The same could be said about columns.
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[wailing winter winds]
United States
Hillsboro
Oregon
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Juliusan wrote: Swapping of rows will not always work because of diagonals. You can swap first row with the last one though. The same could be said about columns.
You can always safely swap rows and columns in a set magic square.
For a given propety, fix the values of the top row, then try all combinations for one other cell. You can only generate the five patterns below. Note that in all cases you can safely switch columns and rows.
abc
abc
abc
abc
bca
cab
abc
cab
bca
aaa
aaa
aaa
aaa
bbb
ccc
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Julius Andrikonis
Lithuania
Vilnius
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I don't see how that happens, but yes, I am wrong... you can do this
Edit: Wait, now I see how that happens. Thanks for your explanation Dave, I finally understand what you said.
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Roee Anuar
Israel
Tel Aviv
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l-hansen wrote: There are many other solutions, for example this:
C#Ff: C=Color (Green, Purple, Red), #=number, Ff=Filling (Open, Solid, Striped)
P3St G2St R1St (all ovals)
P3So G2So R1So (All diamonds)
P3Bl G2Bl R1Bl (all squiggless)
Likewise you could turn the square 90 degrees, 180 degrees or 270 degrees (4 different soluitions), or you could swap any two columns (6 different solutions), or swap any two rows (6 different solutions), so there are 4*6*6=144 different solutions to your riddle.
Indeed
And it would happen with each magic square
However - what will be the case with a magic square when the diagonals can't form a SET but only the rows and columns (using different cards than those presented above).
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[wailing winter winds]
United States
Hillsboro
Oregon
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rouie_a wrote: However - what will be the case with a magic square when the diagonals can't form a SET but only the rows and columns (using different cards than those presented above).
Transformations are symmetric (if you can transform from A->B, you can transform to B->A), and given the statement I proved above, the new transformation would have to have non-SETs in its diagonals, as otherwise your transformation back to the original would have SET diagonals, which your claiming it doesn't.
For those wondering whether such a square is possible, consider aaa/bac/cab.
I suspect a more elegant proof than mine would employ modulo-3 arithmetic.
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