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 JPF
 Joined: 05 Dec 2005  Posts: 29  :  Location: Paris  Items 

Posted: Sat Mar 29, 2008 5:49 pm Post subject: 


I hope I'm right*.
There are 36 non isomorphic minimal puzzles :
4 clues : 13 puzzles
Code:  0000000101023000
0000000101200300
0000000102000304
0000000102000320
0000000102003040
0000001200001300
0000010200000340
0000010200001030
0000010200002030
0000010200003040
0000010200304000
0001010000203000
0001020000304000

5 clues : 22 puzzles
Code:  0000000101020203
0000000101202003
0000000102003102
0000000102033010
0000001201033000
0000001201034000
0000001201300400
0000001201303000
0000010200000213
0000010200000231
0000010200000321
0000010200030240
0000010200201003
0000010200203001
0001002001002003
0001002001002300
0001002001003002
0001002001003200
0001002001003400
0001002003004002
0001002003004100
0001012000023000

6 clues : 1 puzzle
* Edit : see Red Ed's and sg's posts.
JPF 

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 tpgames
 Joined: 13 Feb 2007  Posts: 7  :   Items 

Posted: Sun Mar 30, 2008 11:53 pm Post subject: 


Fascinating! I'm going to look at your data and the other links in June, when I finally have the time and try once more to write that sudoku game. Currently, I'm using someone else's code with permission thats a 9x9 grid. The code does not give you authentically unique sudokus games, but it works good enough for me. I'm beginning to understand JS better. So, June should be doable. I'll post some of the code if I ever figure it out. It will be for the 2x2 sudoku game, with images. 

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 mathrec
 Joined: 15 Jul 2005  Posts: 10  :  Location: Carlsbad, CA  Items 

Posted: Tue Apr 01, 2008 11:09 pm Post subject: Minimal 2x2 


I got the same 36 minimal puzzles that you did. My results were generated separately by hand and by computer and they match with yours, so I'm pretty sure that 36 is the correct total.
Here's my list of all 36, sorted according to the arrangement of clues:
Four clues all different in four boxes:
4: 1.2.....3..4....
4: 1.2......3.4....
4: 1.2......3.....4
4: 1.....2..3.....4
Four clues all different in three boxes:
4: 12......3.4.....
4: 1.2..3.....4....
Four clues with one value clued twice in adjacent boxes:
4: 12......2.....3.
4: 1.2......1.3....
4: 1.....1..2.....3
Four clues with one value clued twice in diagonal boxes:
4: 1.2......2.3....
4: 12......3.....1.
4: 12........13....
4: 1.2..3....1.....
Five clues with one value clued twice in adjacent boxes:
5: 1.2..2....3....4
Five clues with one value clued twice in diagonal boxes:
5: 12..3.....1....4
5: 12..3.....2....4
5: 12......3.1...4.
5: 12......3.2....4
5: 1.2..3...24.....
Five clues with two values clued twice in the same adjacent boxes:
Five clues with two values clued twice in overlapping adjacent boxes:
5: 1.2..2.3.1......
5: 1.2..2...1.....3
Five clues with two values clued twice in nonoverlapping adjacent boxes:
5: 1.2..2..3.....3.
Five clues with two values clued twice in the same diagonal boxes:
5: 12..3.....1....2
5: 12..3.....2....3
5: 1.2..3....3....1
Five clues with two values clued twice in different diagonal boxes:
5: 123.....3.1.....
5: 123.....3.2.....
Five clues with two values clued twice in adjacent and diagonal boxes:
5: 123.....23......
5: 123.3.....1.....
5: 123.....2.1.....
5: 12......3.1....3
5: 12......3.2....3
5: 1.2..2.33.......
5: 1.2..2..3.....1.
5: 1.2..2....3....1
Six clues:
6: 12......2.3..31.
The same script that I wrote to determine the number of minimal puzzles also found the total number of distinct (nonisomorphic) puzzles: 4708 (not including underdetermined configurations or the two fullboard configurations). 

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 mrmarky2
 Joined: 13 Jan 2007  Posts: 64  :   Items 

Posted: Tue May 13, 2008 2:56 pm Post subject: 


Quote:  Let me try
Row1 : 4x3x2x1
Row2: 3x2x1x1
Row3: 2x1x1x1
Row4: 1x1x1x1
So 24x6x2 = 288. Correct? 
I think you might have gone slightly wrong somewhere:
Isn't it:
1st Row> 4x3x2x1
2nd Row > 2x1x2x1
3rd Row > 2x2x1x1
4th Row > 1x1x1x1
12x4x4 = 192
1st row: is 4!(4x3x2x1), dont need much explaining for that.
2nd row: 1st cell on the second row is 2 options, because 2 numbers are already filled in in that box. 2nd cell now has one option because 3 out of four numbers are used. 3rd cell 2 options because 2 numbers are already filled in, and because nothing else affects it. 4th cell, 1 option because 3 out of four numbers are filled in on that box.
3rd row: 1st cell there are 2 numbers already filled in, in that column, so 2 remaining options. 2nd cell, 2 numbers already filled in, in that column, and the value in the 1st cell is one of the two numbers in the column of the second cell's column. Then from there the puzzle has only 1 solution, so all remaining values are 1.
By my reconing theres only 192 valid solutions.
But frazer seems to have a good explanation, am i going wrong somewhere in my calculations? 

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 JPF
 Joined: 05 Dec 2005  Posts: 29  :  Location: Paris  Items 

Posted: Wed May 14, 2008 5:02 pm Post subject: 


Frazer's explanation is clear.
If it's useful, here's one considering the rows :
First row : 4! = 24 possibilities
Here's one :
Code:  1 2  3 4
. .  . .
+
. .  . .
. .  . . 
To fill the first box there are now 2 possibilities [34 or 43]
Here's one :
Code:  1 2  3 4
3 4  . .
+
. .  . .
. .  . . 
Now, to complete the grid there are 6 possibilities :
4 with these 2 rows :
Code:  1 2  3 4
3 4  1 2
+
. .  . .
. .  . . 
2 with these 2 rows :
Code:  1 2  3 4
3 4  2 1
+
. .  . .
. .  . . 
The total number of grids is therefore 4! x 2 x (4+2) = 288
JPF 

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 coloin
 Joined: 05 May 2005  Posts: 97  :   Items 

Posted: Thu May 15, 2008 6:02 pm Post subject: 


Well this was a good question
It was my first ever forum post, a long time ago  I thought there were 192......on paper.
There are 4+2 = 6 ways to compete the grid in JPFs example above.
tannedblondbloke here  he got 192, which bugged me a bit.
But.....
Code:  1st Row > 4x3x2x1
2nd Row > 2x1x2x1
3rd Row > 2x2x1x1
4th Row > 1x1x1x1 
should read
Code:  1st Row > 4x3x2x1
2nd Row > 2x1x2x1
3rd Row > 2x2x2x1
4th Row > 1x1x1x1 
But r3c3 only gives a valid grid three times out of four.
= 288
C 

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