The Californian Garmo Lives Life

My Life as a gay Armenian-American living in California.

Same math quandry, different mathematical question September 18, 2008

Filed under: Math — thecaligarmo @ 4:24 pm
Tags: , , ,

So I was listening to the Math Factor Podcast [link](my favorite podcast, and the best podcast in the world!) and they proposed a mathematical question that is almost exactly similar to my question that I had asked just a few days ago.

First math problem
In my previous post [link] my final math question was:
Two people are playing a game. The rules are simple. Each person takes turns saying a number (or a 1-variable formula). The goal of the game is to say a number (or more likely a formula) that is so hi, that the other person cannot top it.

Ex:
Person A: 1
Person B: 2
Person A: Whatever you say +1 (aka n+1)
Person B: Whatever you say squared (aka n^2)
Person A: Whatever you say factorial (aka n!)
Person B: Whatever you say to the power of itself (aka n^n)

So… how do you win? Does there exist a formula that will allow any person to win?

Same math problem, different question:
So they ended up asking the following:
2 people are playing a game. One person is trying to get the game to go on forever, and the other one is trying to get it to end. Also, very simple rules. Forever puts down any number of balls with a certain number on it (any number). Then End takes away just one of the balls with a certain number n on it. For each ball End takes away, Forever gets to put down as many balls as she likes with the number n-1 on it. If End takes a ball with a number 0 on it, Forever can’t put anything down, so End gets to go again. For Example (Note, Ordered pairs are structured as [number of balls, number on balls]):

Forever: Puts down [2,3] (aka 2, #3 balls)
End: Picks up a #3 ball
Forever: Puts down [3,2] balls (Now on board: [1,3], [3,2])
End: Picks up a #2 ball
Forever: Puts down [1,1] (Now on board: [1,3], [2,2], [1,1])
End: Picks up a #2 ball
Forever: Puts down [2,1] (Now on board: [1,3], [1,2], [3,1])
End: Picks up #1 ball
Forever: Puts down [2, 0] (Now on board: [1,3], [1,2], [2,1], [2,0])
End: Picks up #0 ball (End gets to go again)
End: Picks up # 3 ball

So, this goes on for a while, but the game will eventually end. So the question with this problem is, is there a way to get the game to go on forever?

Very similar concept with mine where you need a formula to outdo the other person. In this case, only one person is supplying the formula, which is perfectly acceptable.

My assumption: There must be something out there that will allow forever to win. Although it may seem like it will always end, I feel it deep in my bones that there must be a way for forever to win.

Deadline: They will probably give out a solution to the problem next week, so I have until then to solve this conundrum! Fun times. Gives me something to do at the gym I guess!

Mathematically,
-The Cali Garmo