Comments  Talk
... by: Tom
I think it's crucial to add that the UR can be used when there are exactly two candidates in the cell in question. It never worked for me when I had 3 candidates there..
... by: Duncan
Good and detailed explanation. I use this "impossible ambiguity" strategy in killer and in kakuro as well where it is equally valid.
One thing to look out for as well is when the ambiguity extends over three vertical pairs of cells in a horizontal line (or vice versa), which I have seen a couple of times.
... by: Eric
Nice strategy if you are out of options, but there must be another way to solve the puzzle. There MUST be another way to remove a candidate, otherwise multiple solutions are possible.
Andrew Stuart writes: Indeed, but that is because I dont possess all logical strategies although I do try and incorporate new ideas as soon as possible. Lack of a logical solution does not imply multiple solutions, however. But a multisolution puzzle definitely can't be solved logically.
... by: Yossarian
@ Paolo and Malikov:
Your argumentation is right. You should consider the following argument instead: Imagine at the end you have a solution containing the 8 in that cell. Then you can find a second solution swapping those 4 cells, which is also a solution to the given sudoku. This is impossible as we know there is only one solution to the given sudoku. So the 8 cannot go in that cell.
Hope this helps understanding, philipp
... by: Quescaisje
This elegant tactic finds good use in AI Escargot, viz. the rectangle formed by E2/3, H2/3.
bill
... by: CS Vidyasagar
Excellent exposition. One does come across such situations and this weapon will enhance our fire power in demolishing Sudoku puzzles. If one reads your excellent explanations of DEADLY RECTANGLES, then this article becomes crystal clear. I shall try and locate such Avoidable Rectangles and send them to you. thanks and regards, Vidyasagar
... by: Leo Salmonsson
Really good explenation. I liked it a lot. Easy to follow and easy to understand. Ill be back to this site for more.
Thanks a lot!
... by: paolo
can't get your example, but maybe I'm missing something. In the last example IF YOU PUT 5 in B7C4 and 8 in C7 this means that you had some reason to put that numbers there. In other words that 5s and 8 cannt be swapped. Otherwise they should not have been placed. So it is arbitrary to deduce from them the impossibility of having an 8 in B4. I mean that (IMHO of course) you won't ever arrive in that position (last example) because you should have applied the UNIQUE RECTANGLE tecnique before, in case. Hope my explanation is understandable.....
paolo
... by: Pritt Galford
Regarding your webpage about Avoidable Rectangles, I have a problem understanding your explanation. I think perhaps the problem arises not with your logic, but with the example instead.
Look at filledin cells A3 & A9, and C3 & C9. They are an interrelated pair, thus allowing two solutions to the puzzle which is a no, no. There are others of the same problem in the example.
Also see the top 1/3rd of the puzzle of your last example on the same webpage. It has cell A5 with candidates 8 & 9 but nine is already filled in at J5.
Bottom line: I think your Avoidable Rectangle logic is correct and sound, but I'm finding there sure are a lot, in fact it's rather a common occurrence, especially in difficult Sudokus, to find multiple solutions which are violations of the unique solution rule. Your thoughts, please.
Andrew Stuart writes: A3 & A9, and C3 & C9 are not four cells in two boxes. Avoidable Rectangles only apply if all four cells are in exactly two boxes. You will find that A3 & A9, and C3 & C9 are fixed by other numbers elsewhere and are not swappable.
... by: Malikov
I agree with the above comments, this being a great site and all, but don't start charging admission. You'll find that 99% of people do Sudoku for fun, and won't be willing to pay when there are hundreds of free sites they can access.
... by: Ken Stephens
I understand the logic based on the information given, that "the three other cells are solved cells (not clues)". However I do not understand how they CAN be solved cells if they would have been interchangeable but for the contents of the fourth cell.
Thanks for a great site which explains things so clearly and on just the right level.
... by: Gary Maness
I have a question on this one. I recently used my own variation on this idea that involved 2 pairs that didn't quite form a rectangle. In fact two cells could see the forth but the third could only be seen by one of the others that I had solved. But using the same logic I thought that since there were no clues on any of those cells, if the target number, 8 were possible then there would be two solutions.
Do you have Sudoku Dragon? I can send you the partially solved puzzle. i am interested to know if this forms a rule or did I just get lucky! I mean, in your example, the two end cells, the 5's could see the forht cell and the first cell, the 8. In mine the end cells could the target, but only one could see the first cell.
I hope this makes sense. BTW, I LOVE this site. I have learned so much about the game. You should consider charging admission! I do plan on buying the Index.Dat program and the book when I can though.
Thanks a bunch.

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