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21.11.2020, 07:59
(Dieser Beitrag wurde zuletzt bearbeitet: 13.01.2022, 22:53 von DaleERKloss.
Bearbeitungsgrund: changed word from occurences to sets
)
After reading Phistomefel’s post about his new finding, I have been thinking about using red rectangles in place of Phistomefel’s red squares as the corner elements (I have been retired for a long time so I have lots of time on my hands and in my brains- a small joke).
I have found 4 other relationships (although the fourth one is obvious & simple but it illustrates the use of the technique)
Remember, for all puzzles, the red cells are counted twice the first sum, but once the second count.
SEE THE ATACHED SUDOKUPATTERNS.PDF FILE FOR PICTURES; I COULD NOT GET THE PICTURES TO POST CORRECTLY. BAH , pictures will not attach. I have written up a description of cell placements in each puzzle. Sorry
The donut hole is colored white, green cells are all unmentioned cells.
For all puzzles, R= red cells, G=green cells, B = blue cells, the donut hole is the box in the center of the puzzle
======================================
The first puzzle uses a 1-cell height by 2-cell width rectangle.
+----+-----+----+
| RRG | GGG | GRR |
| GGB | BBB | BGG |
| GGB | BBB | BGG |
+------+-----+------+
| GGB | | BGG |
| GGB | W | BGG |
| GGB | | BGG |
+------+----+------+
| GGB | BBB | BGG |
| GGB | BBB | BGG |
| RRG | GGG | GRR |
+-----+------+------+
1X2 pattern
Red cells = Row 1 & 9 Columns 1 & 2
Blue cells = rows 2,3,8,9 in columns 4 to 6
columns 3 & 7 in rows 2 to 8
Counting the cells along rows 1 and 9, then along columns 1, 2, 8, and 9 we get:
2R + G = 6 sets of all the numbers 1 to 9
Counting the cells in the 8 edge boxes we get:
R + G + Blue = 8 sets of all the numbers 1 to 9
Taking the difference, we get
R + 2 sets of all the numbers 1 to 9 = B
Or
The contents of the red cells + 2 sets of all the numbers 1 to 9 = the contents of blue cells
Including the rotations & interchanges, I get 13,122 possible versions of the puzzle as follows:
2 rotations
9 positions of the donut hole
3 positions of the line for red cells at the top
3 positions of the blue cell in the leftmost top box
3 positions of the blue cell in the rightmost top box
3 positions of line for red cells at the bottom
3 positions of the blue cell in the leftmost bottom box
3 positions of the blue cell in the rightmost bottom box
======================
The second puzzle uses just the 4 corner cells.
+-----+-----+----+
| RGG | GGG | GGR |
| GBB | BBB | BBG |
| GBB | BBB | BBG |
+-----+------+------+
| GBB | | BBG |
| GBB | W | BBG |
| GBB | | BBG |
+-----+------+------+
| GBB | BBB | BBG |
| GBB | BBB | BBG |
| RGG | GGG| GGR |
+-----+-----+------+
1X1 pattern
Red = r1c1, r1c9, r9c1, r1c9
Blue = Columns 2,3,7,8 in rows 2 to 8
Rows 2,3,7,8 in columns 4,5,6
Counting the cells along rows 1 and 9, then along columns 1 and 9 we get:
2R + G = 4 sets of all the numbers 1 to 9
Counting the cells in the 8 edge boxes we get:
R + G + Blue = 8 sets of all the numbers 1 to 9
Taking the difference, we get
R + 4 sets of all the numbers 1 to 9 = B
Or
The contents of the red cells + 4 sets of all the numbers 1 to 9 = the contents of blue cells
Since the rotation produces the same positions of red & blue cells I get (after interchanges) 6,561 different versions of the puzzle as follows:
9 positions of the donut hole
3 possible rows for the red cells at the top
3 possible positions of the of the leftmost red cell at the top
3 possible positions of the of the rightmost red cell at the top
3 possible rows for the red cells at the bottom
3 possible positions of the of the leftmost red cell at the bottom
3 possible positions of the of the rightmost red cell at the bottom
==============================
The third puzzle uses a 2-cell height by 3-cell width rectangle.
+-----+-----+----+
| RRR | GGG | RRR |
| RRR | GGG | RRR |
| GGG | BBB | GGG |
+------+------+-----+
| GGG | | GGG |
| GGG | W | GGG |
| GGG | | GGG |
+-----+ -----+------+
| GGG | BBB | GGG |
| RRR | GGG | RRR |
| RRR | GGG | RRR |
+-----+------+------+
2X3 pattern
Red = rows 1, 2, 8, & 9 in columns 1,2,3,7,8,9
Blue = rows 3 & 7 in columns 4, 5, & 6
Counting the cells along rows 1, 2, 8, and 9, then along columns 1, 2, 3, 7, 8, and 9 we get:
2R + G = 10 sets of all the numbers 1 to 9
Counting the cells in the 8 edge boxes we get:
R + G + Blue = 8 sets of all the numbers 1 to 9
Taking the difference, we get
R = B + 2 sets of all the numbers 1 to 9
Or
The contents of the red cells = the contents of blue cells + 2 sets of all the numbers 1 to 9
After rotations & interchanges, I get 162 different versions of the puzzle
2 rotations
9 positions of the donut hole
3 possible rows for the blue cells at the top
3 possible rows for the blue cells at the bottom
=============================================
The fourth puzzle uses a 1-cell height by 3-cell width rectangle
(A simple example of the technique just for completeness)
+------+------+------+
| RRR | GGG | RRR |
| GGG | BBB | GGG |
| GGG | BBB | GGG |
+------+------+------+
| GGG | | GGG |
| GGG | W | GGG |
| GGG | | GGG |
+------+------+-----+
| GGG | BBB | GGG |
| GGG | BBB | GGG |
| RRR | GGG | RRR |
+------+------+------+
1X3 pattern
Red = Row 1 in columns 1, 2, 3, 7, 8, & 9
Blue = Rows 3 & 4 in columns 4, 5, & 6
Counting the cells along row 1 , we get
R + G = 1 sets of all the numbers 1 to 9
Counting the cells in the top center box, we get:
G + Blue = 1 set of all the numbers 1 to 9
Taking the difference, we get
R = B
Or
The contents of the top red cells = the contents of top blue cells
==================================
For Phistomefel’s original post, I get 729 different versions as follows:
No difference between rotations.
9 donut hole positions
3 possible lines for the blue horizontal line at top
3 possible positions of the vertical blue line in the leftmost side of the puzzle
3 possible positions of the vertical blue line in the rightmost side of the puzzle
3 possible lines for the blue horizontal line at bottom
=================================
Obviously, of mine, the 2x3 is the least complicated to implement, has the least number of versions, and the least number of cells to be considered. Phistomefel’s puzzle or the 2x3 puzzle could be used in some solver.
It would seem to me that if one of these patterns were to be added to a solver, it would fastest to find eliminations by first searching for missing numbers that fell on the corner of a rectangle.
That would determine the donut hole position. That determines the positions of all the elements of the rest of the pattern.
Please let me know if I made any mistakes. First time poster.
Dale E. Kloss, Portland, Oregon, USA