Here is a 2-player game.
(I do not know how fun this would be to actually play, but I should
mention it.)
Players decide on an even positive integer n, where n is in the range
of
14 to 100 or is higher or is lower.
:)
You make 3 lists of primes.
(You only need list C, actually, which is constructed during play.
Lists A and B, which both have primes crossed off from them, help you
remember which primes have been used, and may be unnecessary for many
players.)
All of the first n primes are written down in list A, all of the first
(n-1) primes are written down in list B.
Players take turns adding primes to third list, list C, one prime per
turn.
Each prime is an uncrossed-out prime from list A.
When it is added to list C, it is crossed out of list A .
From the 2nd move on, after a player adds a prime to list C, he/she
finds
the sum of
c(m)+c(m-1), where c(m) is the m_th term of list C.
She/he then takes the highest prime divisor of (c(m)+c(m-1)), and then
crosses out this highest-prime-divisor from list B, if not already
crossed out.
Now, there are two variations of the game.
In variation 1,
Players cannot add a prime to list C if the resulting
highest-prime-divisor has already been crossed out of list B.
So, the winner of the game is the player who is last able to place
down
an integer in list C.
In variation 2,
Players play until all n primes have been crossed out of list A and
written down in list C.
A prime can be written down in C
no matter if the highest-prime-divisor of (c(m)+c(m-1)) has already
been
crossed out of list B or not.
But if the highest-prime-divisor has been not crossed out of B yet,
the
player making the move gets a point.
And the winner is the player with the highest score.
Any winning strategies for either variation?
thanks,
Leroy Quet
Thursday, September 18, 2008
Subscribe to:
Post Comments (Atom)
No comments:
Post a Comment