- Code: Select all
`*--------------------------------------------------------------*`

| 67 17 2 | 5 467 1367 | 346 9 8 |

| 4 9 b68 |a18 2 1367 | 36 167 5 |

| 3 178 5 | 9 *467-8 167 | 2 1467 467 |

|--------------------+--------------------+--------------------|

| 8 6 4 | 3 1 2 | 7 5 9 |

| 1 5 3 | 7 9 8 | 46 2 46 |

| 9 2 7 | 6 5 4 | 8 3 1 |

|--------------------+--------------------+--------------------|

| 2 478 1 |*4-8 678 5 | 9 67 3 |

| 67 47 69 | 14 3 1679 | 5 8 2 |

| 5 3 c689 | 2 d678 679 | 1 467 467 |

*--------------------------------------------------------------*

Here is the way to get further in puzzle 3 without guessing. A couple of simple moves have removed the 4's from row 7 column 5 (shortened to r7c5), r9c5 and r7c8.

Hopefully you'll know how to do this and in any case it doesn't affect what follows.

Now here's the important part : Look at the cell's I've marked a, b, c and d. Notice that they all contain 8's and that a and b contain the only two 8's in Row 2 and c and d contain the only two 8's in Row 9.

Now suppose r2c4 (cell a) is not 8. Then, since there are two 8's in Row 2, r2c3 (cell b) must be 8. That means that r9c3 (cell c) is not 8. Since there two 8's in Row 9, r9c5 (cell d) must be 8.

You can reverse this argument, assume r9c5 is not 8 and follow the cells in the order dcba to show that if r9c5 is not 8 then r2c4 must be 8.

All of this means that at least one of r2c4 and r9c5 must be 8. (They might both be 8 but they can't both be not 8.)

Now look at r3c5 and r7c4 (I've marked them with a * in the diagram. They can both see (ie share a row, column or box with) cells a and d. Since at least one of a and d must be 8 then r3c5 and r7c4 can't be 8.

In particular this solves r7c4 to 4 and r1c4 to 8 (since there are only two 8's in Box 2 or Column 4) and the puzzle solves easily after that. As Jason said this solving technique is called a Skyscraper.

Hope this helps, Leren