The New Sudoku Players' Forum

[Fortesque92]

https://postimg.cc/F1J6dDfM

[Leren]

Code: Select all

...2.3.6.7....53.16...7..4..53......2.......8......13..2..9...48.41....6.1.6.2...

Code: Select all

*---------------------------------------*
| 9-1 4   159 | 2  18  3    | 58  6  7  |
| 7   8   2   | 4  6   5    | 3   9  1  |
| 6   3  a15  | 89 7  b19   | 258 4  25 |
|-------------+-------------+-----------|
|d14  5   3   | 78 28 c146  | 46  27 9  |
| 2   67  9-1 | 3  15  1469 | 46  57 8  |
| 49  67  8   | 79 25  469  | 1   3  25 |
|-------------+-------------+-----------|
| 3   2   6   | 5  9   8    | 7   1  4  |
| 8   9   4   | 1  3   7    | 25  25 6  |
| 5   1   7   | 6  4   2    | 9   8  3  |
*---------------------------------------*

1. Skyscraper of 1's in cells a-b-c-d => - 1 r1c1, r5c3, basics get you to here.

Code: Select all

*-----------------------------------*
| 9 4  15 | 2  18  3    | 58  6  7  |
| 7 8  2  | 4  6   5    | 3   9  1  |
| 6 3  15 | 89 7   19   | 258 4  25 |
|---------+-------------+-----------|
| 1 5  3  | 78 28 *46   |*46  27 9  |
| 2 67 9  | 3  15 *1-46 |*46  57 8  |
| 4 67 8  | 79 25  69   | 1   3  25 |
|---------+-------------+-----------|
| 3 2  6  | 5  9   8    | 7   1  4  |
| 8 9  4  | 1  3   7    | 25  25 6  |
| 5 1  7  | 6  4   2    | 9   8  3  |
*-----------------------------------*

2. UR Type 1 : r56c67 => - 46 r5c6 which solves the puzzle.

There were a number of choices for the second move, but I thought that the UR was the easiest to spot. The UR is described here in Hodoku.

However, beware of their linguistically stupid reasoning, which confuses just about anyone coming across UR's for the first time. What they say is this:

A UR Type 1 exists, if one of the four cells of a possible UR has additional candidates. If those candidates were eliminated, the UR would exist, causing two solutions. It is therefore absolutely necessary, that one of the additional candidates has to be true.

The second sentence is misleading. What they should have said is that if either UR candidate is placed in the cell that has the extra candidate(s) then the puzzle would have either 0 or 2 solutions. Since it is generally accepted that no one is supposed to pose a puzzle with anything but one solution then in practice, the puzzle would always have no solutions, never two solutions.

There, I've had my UR rant for today. Leren

<edit> Toned down my UR rant a bit and hopefully made it clearer. Leren

[Fortesque92]

Haha OK, thanks.