Place an A, B, or C in each empty cell below so that no three consecutive cells (horizonally, vertically or diagonally) contain a set consisting either of all three different (A, B, and C) letters or all the same letter.
From Mensa Puzzle Calendar 2019 by Fraser Simpson, Workman Publishing, New York. Puzzle for December 20.
The unique answer is:
ACAC
BBCC
BBAB
CAAB
Steps were deductive, using the constraints:
First, we quickly get two blanks filled, with "?" and "*" being the remaining unknowns:
?**C
B???
BBA?
CAAB
The "?"s can each have two values:
A/C * * C
B A/B B/C B/C
B B A A/B
C A A B
while the two "*"s are allowed one of five choices:
(A,A) (A,C) (B,B) (B,C), (C,A)
I tried each of these and they collapsed the other choices to an impossibility, except the last pair, which allowed the one solution.
Edited on January 21, 2020, 11:53 am
soln
