## Tuesday, April 26, 2011

### Palindromes/Antipalindromes In Sequence

Game for 2 players.

Make a row of m squares drawn on paper, where m is decided by both players.
(I suggest that m be >= 16 for beginners.)

Players take turns. On a turn, a player places either a 0 or a 1 into any empty square.

After a total of m moves, and when each square has one digit in it, play is over.

Player 1 gets the sum of the lengths of all (not necessarily distinct but possibly overlapping*) palindromes
(a(1),a(2),a(3)...a(3),a(2),a(1))
of EVEN length (in the row of 0's and 1's) as her/his score.
Player 1's score =
sum{k=1 to [m/2]} 2k*(# of palindromes of length 2k).

Player 2 gets the sum of the lengths of all (not necessarily distinct but possibly overlapping*) antipalindromes
(a(1),a(2),a(3)...1-a(3),1-a(2),1-a(1))
(of even length) (in the row of 0's and 1's) as her/his score.
Player 2's score =
sum{k=1 to [m/2]} 2k*(# of antipalindromes of length 2k).

*(By "not necessarily distinct", I mean the patterns of 0's and 1's in different palindomes/antipalindromes may match. I of course don't mean the SAME palindrome/antipalindrome {in the same position and of the same length} can count more than once.)

The player with the largest score is the winner.

For example, let us say we have the following row (m=16):
1001001101000100

We have the following palindromes of even length:
00, 00, 11, 00, 00, 00
1001, 1001, 0110

So player 1 gets (2*6 + 4*3 =) 24 points.

We have the following antipalindromes:
10, 01, 10, 01, 10, 01, 10, 01, 10
0011, 1010
100110, 110100
01001101

So player 2 gets (2*9 + 4*2 + 6*2 + 1*8=) 46 points.

Player 2 wins.

(Did I make a mistake counting up the palindromes and antipalindromes in my example?)

I suggest, if this game isn't played on a computer, that players count their own (anti)palindromes, and confirm them with their opponent. (If a player misses any that would contribute to his/her own score, it would be his/her own fault.)

Thanks,
Leroy Quet