Posle dužeg vremena, najzad jedan originalni sudoku. Autor ovog zadatka Amerikanac Vei Va Huang, nazvao ga je sferoidni sudoku. Zamolio sam ga da mi objasni pravila ovog zadatka i evo šta je odgovorio:
"The rules are pretty much same as a normal sudoku. Fill in the cells with numbers 1-8 so that every region contains all numbers 1-8, every row contains all numbers 1-8, and all columns contain all numbers 1-8.The only catch is that "rows" and "columns" zigzag a bit using 5-way symmetry.A good way to think about it goes like this. Imagine trying to figure out what a "row" is in a normal sudoku. You start with a cell on an edge. Imaging "entering" that edge and exiting out the other edge parallel to that; you'll get the next cell in the row. Now continue across and exit across the next edge parallel to that, and you get the next cell in the row. Eventually, after crossing a series of cell-edge-cell-edge (where all the edges you cross are parallel), you'll exit out the other side. That's what a "row" is."
(edit)
Milovan Kovačević se potrudio da nam što bolje prikaže kako izgledaju ti "redovi" i "kolone".
3 коментара:
Rešio sam!
Dr A. Vasiljević
Nije pretežak, samo treba dobro gledati!
Pogledajte i ovako zadatak sa postavljenim strelicama http://www.stanford.edu/~tsnyder/barepenrose.jpg
Postavljeno je i dodatno objasnjenje uz tu sliku:
There are 10 bands (what would be rows/columns in a 2d sudoku) labeled A to I. All cells can be described as some pair of the letters A through I with no AB, CD, EF, GH, IJ for a total of 40 cells.
Look at HJ (I choose this cell in part because it interacts with all the empty cells but one in the region on the right). Let's say HJ is not 4 or 5, but some other X. Then DF is also X. Checking these shows IB is also X. As are EG and AC. You'll see that all of these cells are in the same relative position in their groupings (and it turns out if any two of those positions are a single number, I'm pretty sure they all must be). Now, none of the X's overlap with a number yet, so X must be 1. Once you've filled in those places with ones, it is MUCH easier to make progress on this sphere. For example, an 8 in AE or EJ now forces an 8 in BD (which can be worked around the diagram to force all other 8s). The placement of an X in DF happens to force a 7 in AH which is also valuable to work from. Working forward, you should be able to find an answer once all those 1's go in.
So, to know this is THE answer, we must show HJ is not a 4 or 5. Well, the easy part is the 5. If HJ is a 5, then BI is a 5, but then none of the cells in the lower-left region can be 5. So HJ cannot be 5.
Proving HJ is not 4 is nowhere near as simple and I can eventually do it but only with a couple chains of eliminations from a bifurcated guess. I welcome any good descriptions of how to eliminate 4 from HJ, but I really liked HJ as the key cell to work in on this puzzle.
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