Happy Halloween From  The Muse Garden!

In the tradition of my Holiday Math Puzzles, I’m here with an appropriately themed puzzle for this time of year.

Candy Distribution

It’s that time of year all right. You’re out and about, trick or treating with your friends or family and when you come home, you decide to dump all the candy out on the floor to sort through it. But then, as siblings often do, you begin to bicker about who has “more” than the other. In fact, there are some candies that you really like, and some you don’t like. You’d rather have a bunch of chocolate than a bunch of peppermints, for instance.  But wait a minute! Of course you can’t like the same thing! Your sibling actually likes peppermints!

Here’s a table of the different candies you have. Each candy has a “value” to it; that is, how much you “want” it. Try to split up the candies such that you and your sibling both have as equal value as possible at the end. And no fighting!

 Candy Quantity Your Value (per piece) Sibling’s Value (per piece) Candy Corn 150 25 50 Peppermints 50 5 50 Peanut Butter Cups 10 100 75 Hershey Bars 25 50 10 Kit Kat 20 75 30

Chopsticks Game – A Combinatorial Challenge

So I don’t know if anyone else is familiar with this game, but I just remember playing it with friends in middle school and it occurred to me the other day that it would be an interesting game to analyze combinatorially, and perhaps write a game playing algorithm for. This game can be found in more detail here: http://en.wikipedia.org/wiki/Chopsticks_(hand_game)

Players: 2+.

Rules of Play: Players each begin with two “piles” of points, and each pile has 1 point to begin with. We used fingers to represent this, one finger on each hand.

On each turn, a player can choose to do one of two things:

1. Send points from one of the player’s pile to one of the opponents piles. So if Player 1 wanted to send 1 point to Player 2’s left pile, then Player 2 would have 2 points in the left pile and Player 1 would still have 1 point in his left pile. Player 1 does not lose points, they are simply “cloned” over to the opponents pile.
2. If the player has an even number of points in one pile and zero points in the other pile, the player may elect to split his points evenly between the two piles. This consumes the player’s turn. Example: if Player 1 has (0     4) then he can use his turn to split his points, giving him (2     2).

If a player gets exactly 5 points in either pile, that pile loses all of its points and reverts to 0. However, if points applied goes over 5 (such as adding 2 points to a 4 point pile), then the remainder of points are added. (meaning that 4 + 2 = 1). The opponent simply gets points mod 5.

If a player gets to 0 points in both their piles, then they lose. The last person that has points remaining wins.
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Okay, so let’s break this down. Here’s an example game for those of you that are more visually oriented (follow the turns by reading left to right, moves are marked with red arrows):

• On turn 7, player 2 adds 3 points to 3 points. $3+3=6$ as we all know, but $6 \equiv 1 \mod 5$, so player 1 now has one point in his pile.