Your example of an 'invalid' puzzle actually has a very elegant move at the beginning.

- Code: Select all
` /6 /24 /2 /46`

+------------------------+------------------------+------------------------+

| 1 2 589 | 4 567 56789 | 3 678 679 |

| 3 789 489 | 2789 1 6789 | 24689 5 24679T | 246

| 489 5789 6 | 235789 2357 5789 | 1 2478 2479 |

+------------------------+------------------------+------------------------+

| 7 1568 1258 | 158 9 1458 | 24568 123468T 12346T | 246

| 289 4 12589 | 6 57 3 | 2589 1278 1279 |

| 689 15689 3 | 1578 457 2 | 45689 14678 14679 |

+------------------------+------------------------+------------------------+

| 5 1369 1249 | 1239 8 1469 | 7 12346 12346 | 246

| 24689 13689 7 | 1239 2346 1469 | 246B 12346 5 |

| 246 136 124 | 12357 234567 14567 | 246B 9 8 |

+------------------------+------------------------+------------------------+

Notice r89c7 (base cells). In r247 (cross-lines) each digit has to appear three times.

Each of the digits 246 can only appear in two columns (cover-houses) outside the right-hand stack as marked above the grid (other givens prevent them from appearing in r247c15).

Each of them then has to appear at least once in r247c789.

That means that the two digits in the base cells are forced in r2c9 r4c89 (target cells), where they form a triple with 3 (already forced in r4c89).

The target cells can therefore only contain the digits 2346.

This pattern is called Junior Exocet and it produces several inferences we can use.

Firstly no digit can appear twice in the target cells (as the other base digit could then only appear once in r247).

Secondly the digit in each target has to appear in the corresponding mirror node: r2c9 in r56c8, r4c89 in r13c89.

It's important to note that we know there is one base digit in r2c9 and r56c8 and one in r4c89 and r13c89.

If the 3 were forced in r24c9, instead of r4c89, we wouldn't be able to use the next step, as both base digits could just appear in r4c89 and r13c89.

Now look at the digits restricting 246 in c2346. We have 26b15 and 4b24.

There are several ways to prove that if 4 is a true base digit, we get a UR. The way I prefer starts in the mirror nodes:

Suppose 4r56c8 and (2 or 6)r13c89. Then in c1 both of them are forced into r89c1.

Suppose 4r13c89 and (2 or 6)r56c8. Then in c5 both of them are forced into r89c5.

If 4 is a true base digit, it and the other base digit will create a UR in r89c17 or r89c57.

Therefore 4 is not a true base digit.

What we ended up with is 26 in base cells and 236 in target cells.

Since 26 are now proven base digits, they can only appear once in r247c789, meaning they can be eliminated in their cover-houses (outside the cross-lines).

I have no doubt that Hodoku will be able to solve it from now on (I think the elimination of 4r89c7 was itself enough).

For a better explanation of Junior Exocet, see

this thread.