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dukuso
Joined: 25 Jun 2005
Posts: 378
Location: Germany
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 Posted: Fri Nov 04, 2005 1:54 am Post subject: |
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OK, I created 100000 random minimal magic sudokus.
average number of clues:8.02
minimum number of clues:7
maximum number of clues:11
here are the first 4:
| Code: |
.........
........1
2........
......9..
.........
...75....
.4..6....
.........
........6
.........
.....8...
........9
2......8.
.........
.......6.
....4....
..9......
3........
....3...4
.........
..2......
2.......5
.........
......4..
..1......
.....5...
.........
1.....6..
.........
....5....
.........
..5......
.....2...
...1.....
........7
.........
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a program to generate them is at :
http://magictour.free.fr/suex8g.exe
sourcecode attached to the executable, as usual
-Guenter.
Last edited by dukuso on Fri Nov 04, 2005 10:10 am; edited 1 time in total |
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udosuk
Joined: 17 Jul 2005
Posts: 2835
Location: Sydney, Australia
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| Posted: Fri Nov 04, 2005 7:25 am Post subject: |
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| BlueSpark wrote: | Udosuk,
4 + 5 + 6 does equal 15, but this combo is unusable in a "magic" sudoku as it would preclude those numbers from being used in the rest of the row/column/square (and every combo adding up to 15 has to contain at least one of those numbers). If you look at the above "magic" sudokus, every three-number row and column contains only one of those three numbers.
But you probably already knew that!  |
Yes, I definitely already knew that... Just wanted to correct the statement that "there are 7 sets of 3 numbers that sum to 15" to "8 sets" instead without considering any additional restraints first. Because with your reasoning than "2+5+8" should also be excluded too, right? Bear in mind in the perfect 3x3 magic square both 2+5+8 and 4+5+6 are used in the diagonals. |
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BlueSpark
Joined: 04 Oct 2005
Posts: 18
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| Posted: Fri Nov 04, 2005 8:21 am Post subject: |
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udosuk wrote:
| Quote: | | Because with your reasoning than "2+5+8" should also be excluded too, right? |
I don't think so, although my reasoning was meant to apply only to the sudokus that are "magic" according to the original poster's definition (where the diagonals' content doesn't matter). 2/5/8 contains exactly one of the numbers 4, 5, and 6 and so would be included in the list of "usable" triplets. All bets are off with respect to my reasoning when applied to perfect magic sudokus! |
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r.e.s.
Joined: 31 Aug 2005
Posts: 339
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| Posted: Fri Nov 04, 2005 9:53 am Post subject: |
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| dukuso wrote: | | OK, I created 10000 random minimal magic sudokus.[...] here are the first 5:[...] |
By the definition I posted earlier in this thread, no magic sudoku puzzle has yet been posted. (To be a magic puzzle by this definition, both the array of clues and the solution must be magic.) My original question was whether any of these exist. |
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BlueSpark
Joined: 04 Oct 2005
Posts: 18
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| Posted: Fri Nov 04, 2005 12:36 pm Post subject: |
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R.E.S. wrote:
| Quote: | | What you've described are sudoku solutions that are "magic". A magic sudoku puzzle might be one with such a unique magic solution and be such that the clues in each row, column, and box add to 15. Do these exist? (Less restrictively on the clues, perhaps just have them add to 15 in each box.) |
I think that the term "magic sudoku puzzle" could be used to describe a puzzle the solution to which is magic but the clues to which are not--but I guess it's just a verbal quibble .
I will work on that problem and try to come up with a puzzle that is magic in the sense you describe. |
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udosuk
Joined: 17 Jul 2005
Posts: 2835
Location: Sydney, Australia
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| Posted: Sat Nov 05, 2005 7:12 am Post subject: |
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| BlueSpark wrote: | | I don't think so, although my reasoning was meant to apply only to the sudokus that are "magic" according to the original poster's definition (where the diagonals' content doesn't matter). 2/5/8 contains exactly one of the numbers 4, 5, and 6 and so would be included in the list of "usable" triplets. All bets are off with respect to my reasoning when applied to perfect magic sudokus! |
What I could only say is your reasoning was just based on 1 of the 2 possible dimensions. From the start you've made the more obvious observation that for a 3x3 magic square (diagonals not considered) each row & column must contain 1 of {4,5,6}. What you failed to observe was each row & column must also contain 1 of {2,5,8} (check carefully if you don't believe me!!). In fact, each row & column must contain 1 from each of the 6 sets: {1,2,3},{4,5,6},{7,8,9},{1,4,7},{2,5,8},{3,6,9}. Just that {4,5,6} & {2,5,8} happen to sum to 15. So after you consider both dimensions only 6 sets could be remained! |
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BlueSpark
Joined: 04 Oct 2005
Posts: 18
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| Posted: Sat Nov 05, 2005 9:20 am Post subject: |
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Right you are! I hadn't given the list of triplets much thought--it just struck me that {4,5,6} was obviously unusable. This led me to the conclusion that each triplet (of row or column) could contain (and must contain) only 1 of those three numbers. I never even thought of {2,5,8} and never made the connection between what you said ("wouldn't your reasoning exlude 2,5,8 as well?") with the list given above. I misunderstood and didn't take my reasoning any further. Of course, a triplet's having exacty one of 4,5,6 is a necessary condition of a usable triplet, but not a sufficient condition. I simply assumed the latter and thus assumed that 2,5,8 would be included--but it ain't!
I don't suppose that my reasoning with respect to {4,5,6} implies that {2,5,8} is excluded as well, but a similar sort of reasoning does exclude {2,5,8}. |
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