Here's the problem and the diagram that goes with it
Fill in each empty space of the grid in the image below with a number from 1 to 8 so that every row & column contains each of these digits only once. Some diagonally adjacent spaces have been joined together. For these pairs of joined spaces the same number must be written in both.
We had a sub the other day and he gave us this question offering extra credit to the first person who solved it. Someone already did it and got the points but we never got an explanation for the answer which I want to hear. Yes, I already asked my teacher and she said she didn't know.
I shall use chess-like notation for fields.
We make definite conclusions in the order of green letters below. Non-final results are shown as blue or orange (where same colour means that the corresponding entries are correlated).
To begin with, field b8c7 cannot be 1,2,3,4,5,6 by the other row/column entries, hence must be 7 or 8 as shown. We draw the same conclusion for a6b5, but of course this must differ from b8c7. So much for (most of) the blue group; we will get to a3 later.
(a) g4h3 must be 6 because 1,2,3,4,5,7,8 are already in their rows/columns.
(b) Now g1f3 must be 7
(c) Here, 2,3,4,5,6,7,8 are already excluded, hence 1
(d) 1,2,3,4,6,7,8 forbidden, hence 5.
Meanwhile, we also know enough to note the blue info for a3
(e) In column b, only 1,2,6 are left, but 1 and 6 are already used in row 4. Hence 2
(f) The only field in column a where a 6 can still be, is in a1
(j) The only field in column b where a 6 can be, is in b2
(g) The only place for a 1 in column b is here
(h) c1d2 is not 1 and e2 is not 1, hence the 1 of row 2 must be here
(i) Every row and column needs a 1, this is the last of them
Meanwhile we note that the remaining fields in column 1 must be 3 and 4 in some order (orange). Actually, in row 4, only 3,4,8 are open, and h4 cannot be 8. hence the third orange entry.
Now the only place for 8 in row 4 is at c4, which resolves the blue group. From this point, we readily find that c3 must be 2, that e3 must be 8, that the 7 of row 5 must be at h5, that the 2 of row 5 must be at f5, and so we get a fourth orange note "43" at g5. But g3 is 4, which resolves the orange group!
The rest should not be too hard any more, now that a large part of the board is filled: