Place one of A, B, C or D into each of the 25 empty squares so that the number of letters in each row and column match the number indicated on that row and column for that letter.

Identical letters cannot be next to each other vertically or horizontally, but may be adjacent diagonally.

                 +---+---+---+---+---+
              A  | 2 | 1 | 1 | 0 | 1 |
                 +---+---+---+---+---+
              B  | 1 | 2 | 0 | 1 | 2 |
                 +---+---+---+---+---+
              C  | 1 | 0 | 3 | 2 | 2 |
                 +---+---+---+---+---+
  A   B   C   D  | 1 | 2 | 1 | 2 | 0 |
+---+---+---+---++===+===+===+===+===+
| 0 | 1 | 2 | 2 ||   |   |   |   |   |
+---+---+---+---++---+---+---+---+---+
| 2 | 1 | 1 | 1 ||   |   |   |   |   |
+---+---+---+---++---+---+---+---+---+
| 2 | 0 | 2 | 1 ||   |   |   |   |   |
+---+---+---+---++---+---+---+---+---+
| 1 | 2 | 1 | 1 ||   |   |   |   |   |
+---+---+---+---++---+---+---+---+---+
| 0 | 2 | 2 | 1 ||   |   |   |   |   |
+---+---+---+---++---+---+---+---+---+

---

From Mensa Puzzle Calendar 2016 by Mark Danna and Fraser Simpson, Workman Publishing, New York. ; puzzle for December 8.

To start, one can determine 5 C's and then 4 A's and then one B, giving

??C??
A??C?
?AC?A
A??CB
??C??

Then one can determine two more C's and then two more B's and then 2 D's and then one more A, giving
??CDC
A?ACB
?ACDA
A??CB
??CBC

Then one can complete column 3, and then rows 2-4 (in any order) and then column 2 and then rows 1 and 5 (in any order), giving

DBCDC
ADACB
CACDA
ABDCB
BDCBC

There is only one solution

Posted by Steve Herman on 2016-12-30 15:04:17