Another view: it is solvable but you have to be very careful.

You should be able to get this far without difficulties:

- Code: Select all
`--- -9- -7-`

--7 --- --2

198 --2 5--

9-5 6-3 ---

3-- 2-7 --9

--- 9-4 3--

--2 --9 486

6-- -2- 1--

-1- -36 ---

But then you are stuck.

Now look at box 8

-?9

-2-

-36

Well, what could go in r7c5? (The place with the question mark)

Not 2369 (Numbers already in box 8)

Not 24689 (Numbers already in row 7)

Not 239 or 158 (Numbers already in column 5)

Hunhh, you say? What 158, you say?

True, we don't know the order, but 158 has to go into this box

6-3

2-7

9-4

Ah, so! OK, back to the question: what could go in r7c5?

Not 12345689 because of box, row, column analysis. Has to be a 7. Put the 7 there. Then you should get here without problems:

- Code: Select all
`2-3 -9- -7-`

4-7 3-- --2

198 7-2 5--

9-5 6-3 ---

3-- 2-7 --9

--- 9-4 3--

532 179 486

6-- -2- 1--

-1- -36 ---

Now you are in trouble. There is nothing else easy that I can see.

Staring at columns you might eventually think like this: What is missing in column 7? Answer: 26789.

Let's concentrate on cells that could be 6:

r1c7 must be 68 since 279 are eliminated by row/column

r2c7 must be 689 since 27 are eliminated by row/column

r5c7 must be 68 since 279 are eliminated by row/column

Now the interesting fact. Since there are only two candidates for r1c7 and r5c7 then one of those MUST be 6 and the other 8. We don't know which, but we dont care. The fact is that r2c7 must be a 9!!.

Fill it in.

Now if we analyze cell r2c5, we see that only 1568 "fit". But we already determined that 158 go somewhere in box 5. So r2c6 MUST be a 6. Fill that in.

Whew! Now the puzzle is easy to solve.

Mac