OK Things are getting a bit tougher from your current PM, so I'll try and finish off the puzzle for you with a minimum amount of fuss.
Here's your current PM with cells marked a and b and 4 being eliminated in r1c1.
- Code: Select all
*-----------------------------------------------------------------------*
| 235-4 b245 1 | 9 23 7 |a56 a456 8 |
| 6 b249 b29 | 1 8 5 | 7 3 249 |
| 23589 2589 7 | 4 6 23 | 1 59 259 |
|-----------------------+-----------------------+-----------------------|
| 1258 3 4 | 267 9 268 | 568 1567 156 |
| 1289 26789 2689 | 5 23 4 | 368 1679 1369 |
| 589 56789 689 | 367 1 368 | 4 2 3569 |
|-----------------------+-----------------------+-----------------------|
| 248 2468 5 | 236 7 1 | 9 48 36 |
| 29 1 2369 | 8 4 236 | 356 56 7 |
| 7 468 368 | 36 5 9 | 2 148 14 |
*-----------------------------------------------------------------------*
This move is called ALS ZX Rule. Here Z = 4 and X = 5. The way it works is as follows.
Look at the cells marked a ie r1c78. They are an ALS (Almost Locked Set) because they contain one more different digits (3) than the number of cells (2).
Now suppose r1c8 was not 4. Then r1c78 would be a locked pair (56), so in particular r1c2 would not be a 5.
Now look at the cells marked b. They are also an ALS with 4 digits and 3 cells. Now suppose r1c2 was not 5. These cells would become a locked triple and in particular either r1c2 or r2c2 would have to be 4.
So summing up, we can conclude that at least one of r1c8, r1c2 or r2c2 would have to be 4.
Since r1c1 can see all of these 3 cells we can conclude that r1c1 can't be 4. You can remove the 4 from r1c1.
The way I write this is ALS XZ Rule: X = 5, Z = 4: (4=5) r1c78 - (5=4) r1c2, r2c23 => - 4 r1c1
Z is the elimination digit and X is the digit that links the 2 Almost Locked sets. In general all the X's in ALS1 have to see all of the X's in ALS 2 and you can eliminate all Z's from cells that can see all of the Z's in both ALSs.
As 4 is one of 2 4's in Column 1 there are a few elementary eliminations that should lead you to the following PM:
- Code: Select all
*-----------------------------------------------------------------------*
| 235 245 1 | 9 23 7 | 56 456 8 |
| 6 249 29 | 1 8 5 | 7 3 249 |
| 23589 a2589 7 | 4 6 3-2 | 1 a59 a259 |
|-----------------------+-----------------------+-----------------------|
| 1258 3 4 | 267 9 268 | 568 1567 156 |
| 1289 26789 2689 | 5 23 4 | 368 1679 1369 |
| 589 56789 689 | 367 1 368 | 4 2 3569 |
|-----------------------+-----------------------+-----------------------|
| 4 b26 5 | 236 7 1 | 9 8 36 |
|b29 1 b2369 | 8 4 c236 |c356 c56 7 |
| 7 b68 368 | 36 5 9 | 2 14 14 |
*-----------------------------------------------------------------------*
The next move is called ALS XY Wing and is basically an extension of the previous move, except that this time there are 3 ALS's (which I have indicated by a, b and c in the PM).
So the logic of this move is as follows:
Suppose r3c2 and r3c9 were both not 2. Then the cells marked a would be a locked triple and in particular r3c2 would be 8.
In that case r9c2 would not be 8 and the cells marked b would be a locked quad and in particular r8c3 would be 3.
In that case r8c67 would not be 3, r8c678 would be a locked triple and in particular r8c6 would be 2.
So summing up, at least 1 of r3c29 and r8c6 must be 2. Since r3c6 can see all of these cells, r3c6 can't be 2. You can eliminate 2 from r3c6.
I write this move as ALS XY Wing: (2=8) r3c289 - (8=3) r7c2, r8c13, r9c2 - (3=2) r8c678 => - 2 r3c6
Simple ! Well maybe not, but the good news is that you'll find that the puzzle can now be solved by a cascade of singles.
Leren
