To start with the simplest possible, I thought that this slightly modified template could be good:
- Code: Select all
. . .|. . .|. . .
. . .|A B C|. . .
. . .|. . .|. . .
-----+-----+-----
. . .|1 4 7|. . .
. . .|2 5 8|. . .
. . .|3 6 9|. . .
-----+-----+-----
. . .|. . .|. . .
. . .|. . .|. . .
. . .|. . .|. . .
Without any constraints, there are 162 possible ways to fill this grid, but only 6 unique ones! They (can) look like this:
- Code: Select all
............412...............147......258......369..............................
............421...............147......258......369..............................
............423...............147......258......369..............................
............471...............147......258......369..............................
............472...............147......258......369..............................
............483...............147......258......369..............................
Now from these it is already clear that we can fix A to a digit.
Now running again with the constraint:
A = 4, we get 27/6 (27 in total, 6 unique)
In other words fixing A to 4 does not loose generality. However, it still produces 27, so that constraint is obviously not enough if we want to only produce unique grids. (our goal!)
Trying to fix A as 5->9, B as 1->3,6->9 and C as 1->6 produces the same result: 27/6.
This is also well reflected in the fact that 162(our total amount) / 27 (total after fixing one digit) = 6 which is the number of of possibilities for each place:
A can be {4,5,6,7,8,9}
B can be {1,2,3,7,8,9}
C can be {1,2,3,4,5,6}
Now let's look at the other type of constraint:
As most people now know, two columns can be shifted inside a stack. If we take one configuration like this:
- Code: Select all
. . .|. . .|. . .
. . .|4 1 2|. . .
. . .|. . .|. . .
-----+-----+-----
. . .|1 4 7|. . .
. . .|2 5 8|. . .
. . .|3 6 9|. . .
-----+-----+-----
. . .|. . .|. . .
. . .|. . .|. . .
. . .|. . .|. . .
and swap column 4 and 6 we get:
- Code: Select all
. . .|. . .|. . .
. . .|2 1 4|. . .
. . .|. . .|. . .
-----+-----+-----
. . .|7 4 1|. . .
. . .|8 5 2|. . .
. . .|9 6 3|. . .
-----+-----+-----
. . .|. . .|. . .
. . .|. . .|. . .
. . .|. . .|. . .
now relabeling by saying that 7=1, 8=2 and 9=3, we get:
- Code: Select all
. . .|. . .|. . .
. . .|8 7 4|. . .
. . .|. . .|. . .
-----+-----+-----
. . .|1 4 7|. . .
. . .|2 5 8|. . .
. . .|3 6 9|. . .
-----+-----+-----
. . .|. . .|. . .
. . .|. . .|. . .
. . .|. . .|. . .
so we can then say that:
(A=8, B=7, C=4) is equivalent to (A=4, B=2, C=1)
In order to limit the number of the equivalent grids that arises from this swapping, Red Ed introduced a term "symmetry constraints", that says that one digit from one "swappable" sector must be smaller than the digit it can replace in the other "swappable" sector. (i like that word, "swappable"
)
This can be shown with the relationship between A and B in this example:
- Code: Select all
. . .|. . .|. . .
4 5 6|A B .| 7 8 9
. . .|x x .|. . .
-----+-----+-----
. . .|. . .|. . .
. . .|x x .|. . .
. . .|. . .|. . .
-----+-----+-----
. . .|x x .|. . .
. . .|x x .|. . .
. . .|. . .|. . .
the possible {A,B} configurations are:
{1,2}
{1,3}
{2,1}
{2,3}
{3,1}
{3,2}
since columns 4 and 5 can be swapped and relabled for yet another isomorph of this template, we have these equivalent combinations for A and B: (1-2 / 2-1), (1-3 / 3-1), (2-3 / 3-2)
What Red Ed did to prevent these from happening, was saying that A must be smaller than B -> (A < B)
Of the 6 possible combinations, this will limit us to:
{1,2}
{1,3}
{2,3}
and no isomorphs would arise from swapping of column 5 and 6, this constraint effectivly preventing it!
The simplest way of showing this in context would be with this example:
- Code: Select all
. . .|A . .|. . .
. . .|. . B|. . .
. . .|. . C|. . .
-----+-----+-----
. . .|1 4 7|. . .
. . .|2 5 8|. . .
. . .|3 6 9|. . .
-----+-----+-----
. . .|. . .|. . .
. . .|. . .|. . .
. . .|. . .|. . .
This gives 150 combinations / 15 unique.
Now by adding (B<C) as a constraint, preventing row 2 and row 3 freely swapping place, the results are: 75 / 15. In other words, we have exactly cut the number of combinations in two, without losing any generality!
This is what the output then looks like:
- Code: Select all
...4..........1........2......147......258......369..............................
...6..........1........2......147......258......369..............................
...7..........1........2......147......258......369..............................
...9..........1........2......147......258......369..............................
...5..........1........4......147......258......369..............................
...7..........1........4......147......258......369..............................
...8..........1........4......147......258......369..............................
...4..........1........5......147......258......369..............................
...6..........1........5......147......258......369..............................
...7..........1........5......147......258......369..............................
...8..........1........5......147......258......369..............................
...9..........1........5......147......258......369..............................
...6..........4........5......147......258......369..............................
...7..........4........5......147......258......369..............................
...9..........4........5......147......258......369..............................
We see that B is kept smaller than C, but notice how now A is moving around!
Does this mean that we can't fix A and have the symmetry constraint?
Fixing A=4 gives 10/6, so we are only getting 6 of the 15!
Interestingly: A=4,5,6 -> 10/6 but A= 7,8,9 -> 15/9.
In other words, by fixing A as 4,5,6, we are only giving B and C 5 numbers to play with! This is best illustrated by looking at the output:
- Code: Select all
...A..........B........C......147......258......369..............................
...4..........1........2......147......258......369..............................
...4..........1........5......147......258......369..............................
...4..........2........3......147......258......369..............................
...4..........2........5......147......258......369..............................
...4..........2........6......147......258......369..............................
...4..........5........6......147......258......369..............................
- Code: Select all
...A..........B........C......147......258......369..............................
...7..........1........2......147......258......369..............................
...7..........1........4......147......258......369..............................
...7..........1........5......147......258......369..............................
...7..........2........3......147......258......369..............................
...7..........2........4......147......258......369..............................
...7..........2........5......147......258......369..............................
...7..........2........6......147......258......369..............................
...7..........4........5......147......258......369..............................
...7..........5........6......147......258......369..............................
We see that combinations like B=1-C=4 and B=4-C=5 is effectively eliminated in the first one, since A is set to 4. Note that if A were 5, 1-5 and 5-6 would go, and if it were 6 2-6 and 5-6 would go!
So why does A=7,8,9 produce 15/9 when the full grid is 150/15? Why are we still 6 short!
Adding JUST A=(7,8,9) as a constraint gives the result: 30/9!
Aha! There is nothing wrong with the interaction of A=(7,8,9) and B<C, it is something wrong with fixing A!
Why? What has changed since our last example, where it was OK to fix the A! (pun intended!)
Let's look at the 15 unique:
- Code: Select all
...4..........1........2......147......258......369..............................
...4..........1........5......147......258......369..............................
...4..........2........3......147......258......369..............................
...4..........2........5......147......258......369..............................
...4..........2........6......147......258......369..............................
...4..........5........6......147......258......369..............................
...7..........1........2......147......258......369..............................
...7..........1........4......147......258......369..............................
...7..........1........5......147......258......369..............................
...7..........2........3......147......258......369..............................
...7..........2........4......147......258......369..............................
...7..........2........5......147......258......369..............................
...7..........2........6......147......258......369..............................
...7..........4........5......147......258......369..............................
...7..........5........6......147......258......369..............................
This clearly shows that there is no fixable clue with this template...
Why?!?
Well, the first observation is that:
A has the possibilities {4,5,6,7,8,9}
B AND C has the possibilities {1,2,3,4,5,6}
Second observation is that B and C share a sector, while A does not share anything.
Ok, I think I will wrap up this post now, and continue later. One more note for the last template. Since we got 75/15 for the last template with (B<C), cutting the possibilities down by two, it "should" mean that there are some sort of constraint out there that will reduce the grid with a factor 5... (75/15 = 5)
so, are you guys interested?
Havard
