Comments - Talk
... by: RenStimpy
To bcorso,
I believe example 3 proves this theory to be incorrect. Unless I misunderstand what you're saying, the 38 pair on E2 and the 37 pairs on column 6 would imply that 7 & 8 are the only choices for the two cells in column 7. However, J7 eventually evaluates to be 7, meaning you really only had a choice between 3 and 8 for three of the cells in the extended UR.
To Andrew,
Great website! If I had an extra arm I'd give you three thumbs up!
... by: bcorso
Hey Andrew,
Regarding my previous post. I've come to the conclusion that the only deadly pattern available for a 3x2 extended UR is of the form
xy | xy
xz | xz
yz | yz
That is, to get a 3 x 2 deadly pattern each row must only contain 2 candidates, those 2 candidates must be the same for each row, and those 2 candidates can not be the same for any column. This leaves only permutations of the above patterns as possible deadly patterns.
Proof:
First off, each cell MUST only contain two candidates AND those two candidates MUST be the same for both deadly cells in that row. This is because a deadly pattern means that if ALL other cells are filled in on the board, there still exists multiple solutions. Thus, if all other cells in the rows are filled in, there are only two candidates that can be placed in the deadly cells of that row, and those candidates are the same for both of the cells.
Second, the pairs in the columns cannot be the same. This follows from the fact that if there were two pairs that matched, there would no solution (instead of multiple solutions). For example, if xz was repeated twice in the column:
xy | xy
xz | xz
xz | xz
Then there would be no solution because the second and third column would have to have the same solution, which is impossible.
... by: bcorso
Hey Andrew,
Please let me know if I am missing something important, but I see a flaw in the extended URs that you present here.
The way I understand it, a deadly solution only exists if the candidates in the "deadly" cells cannot see any other cell with those candidates (because the deadly cells cannot solve themselves, thus they need other cells to make further eliminations).
For example, in the 2x2 deadly URs it is deadly because the 4 cells contain a naked pair in the row, col, and box and the "naked pair" technique will remove those candidates from every cell those deadly cells can see. Thus, the naked pairs have multiple solutions because they cannot solve themselves.
Now, in the extended URs I image you would need the same circumstance, the deadly cells cannot see any other cells with their candidates, otherwise that cell can be used to break the deadly pattern. This brings me to your examples here. In both examples you only require a triple in the row OR col. but this obviously leaves cells (outside the deadly cells) with the same candidates which are in the same box and col OR row respectively.
I'm interested in hearing your solution, thanks Andrew.
... by: Dale Kloss
Help!!! I am having a problem with your first Extended Unique Rectangle example. Let me explain by tabulating the possible solutions of [C1,C3,E1,E3,G1,G3].
First the two Extended Deadly pattern solutions:
1 5 5 1
3 1 1 3
5 3 3 5
I don't see a problem with these as horizontally the 2 solutions are deadly, the patterns are deadly inside the 3 boxes, and they are deadly vertically. Since we are told the solution to the entire puzzle is unique they can be eliminated as possible solutions.
Now let's tabulate the other two possible solutions:
Pattern A: 1 5 Pattern B: 6 | (1,5)
5 3 (1,3,5) | (1,5)
3 1 (1,3,5) | (1,3,5)
This is where I have the problem. I don't see Pattern A as Deadly with respect to row E of the first two solutions nor Deadly with respect to row G of the first two solutions since the candidate sets are different. Nor is it Deadly with respect to Boxes 4 and 7 since the candidate sets are different.
Conclusion ? Either Pattern A or B destroys the Deadly pattern. We can't say that just the 6 destroys the pattern ?
One bright spot: By comparing Patterns A and B we see that we can eliminate the 5 in cell C1.
Andrew Stuart writes: I've had another good look at this example.
Let's call the first two patterns you described as C and D as they are unnamed.
Pattern B is not deadly because it's actually going to be the solution, ie 6 in C1 and some combination of the other candidates in the other cells which will come out in the wash later.
Pattern A is deadly with both C and D. Now, A with C doesn't look swappable in isolation because they both share 1 and 5 in C1 and C3. But they do swap candidates in cells E1,E3 and G1,G3. So in isolation just thinking about A and C we are effectively using four cells out of 6 to prove a point which seems insufficient because C1 and C3 are not being swapped. However, the super-set pattern AC can be swapped with pattern D. So we have three patterns which in combination can create swappable cells. This is the neat thing about triples - we don't need every candidate to be swappable in all cells, just as long as each cell and each candidate can be swapped in at least one combination.
Another way to think about it: It would be easier to show the total swapability of all patterns if we had a 1,3 and 5 in all six cells. The fact we are missing a 3 in C1 and C3 and a 5 in E3 merely restricts the number of swappable patterns available to patterns A, C and D. The minimum number of swappable patterns is two and we would get that in slightly more rare situations where all cells contained 2 candidates (apart from the extra one candidate) similar to the second example.
... by: Ehsan Vessal
Very cool Andrew! I just wanted to note that Peter Gordon in his Mensa Guide to solving Sudoku somewhat touches on what you are talking about here when he introduces what he calls "Gordonian Polygons". He has some examples here of not only rectangular shapes, but pentagonal and hexagonal shapes formed by possible candidates.
Andrew Stuart writes: Interesting, I'll check into that |
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