Place one of the four letters A, B, C or D in each of the blue-coloured cells. No letter can be horizontally or vertically adjacent to itself. The yellow-coloured tables above and to the left of the grid indicate how many times each letter appears in that column or row.

This is an extended version of ABCD.

				A	3	2	0	2	2	2	0	3	1
				B	2	1	3	0	0	1	3	0	2
				C	1	2	1	3	2	3	1	2	1
	A	B	C	D	0	1	2	1	2	0	2	1	2
	2	0	4	3
	4	4	1	0
	3	2	2	2
	3	3	2	1
	1	0	4	4
	2	3	3	1

(In reply to No Subject by ed bottemiller)

I got the same solution too.  Probably it is correct. :-)

My approach differs a bit from yours, which is the more interesting thing.  I started with the rows of zero 'B' and the colums of 3 'B.  these columns have only one possibility to place the B's.  Next was the row of 4 'A' and the two columns of zero 'A'.  That placed the middle two A's in that row.  After that, it was a pretty straightforward process of elimination to fill in the rest.

Cheers.

Posted by Steve on 2008-03-14 16:35:10