2
$\begingroup$

I was doodling around today and thought of this fun game. Two players take alternate turns playing this game. A function from now on refers to real valued functions with domain $\mathbb R$, and Odd and Even functions have the usual meaning.

A function $F$ is said to be complimentary to a function $H$, if one of the following is true:-

  1. $F$ is Odd when $H$ is Even.
  2. $F$ is Even when $H$ is Odd.

The first player starts with a function $h_n$, which is either Odd or Even. The second player is said to make a move when he/she provides another function $g$ along with an operation $*$, (where $*$ is one of $+,-,\times,\circ$), so as to create a function $h_{n+1} = h_n*g$ or $h_{n+1}=g*h_n$ such that $g$ is complimentary to $h_n$ and in accordance with the rules (stated below). Moves are made alternately between the two players, and the player who is unable to make a move loses.

Rules

  1. $h_n$ is added to a set $V$ for each $n$.

  2. $\forall g_1,g_2,.... \in V$ and $\forall a_1,a_2,.... \in \mathbb R$, $(a_1g_1^n+a_2g_2^n+....)$ is added to $V$, where $n \in \mathbb N$ and $g_k^n = {g_k} \times {g_k} \times \cdots g_k$ $n$ times.

  3. If $g \in V,$ $g$ cannot be used in a move.

  4. At the end of each move, the function used by the player is added to $V$.

  5. The $*$ provided by the player cannot be used in consecutive moves.

Example

Player 1: $h_1(x)=x$, which is Odd.

Player 2: $g(x)=|x|$ and $* = \circ$ so that $h_2(x) = (h_1 \circ g)(x) = |x|$, which is Even.

Player 1: $g(x)=sin(x)$ and $*=\times$ so that $h_3(x)=h_2(x)\times g(x) = |x|sin(x),$ which is Odd.

Player 2: $g(x) = cos(x)$ and $* = \circ$ so that $h_4(x) = (g \circ h_3)(x) = cos(|x|sin(x))$, which is Even.

And it goes on...

I don't know much of higher mathematics to analyze this game, so I would appreciate it if someone could answer the following questions which I had in mind:-

  1. Are the rules well defined? I don't think my statement of Rule #2 is correctly written, since $V$ turns out to be an infinite set.

  2. Are the rules strong enough to ensure that both players don't have a "trick" to play for an indefinitely long time? If not, how do I make them stronger?

  3. Does the game terminate, in theory?

  4. What is $V$ eventually? Can it ever contain all Even and Odd functions?

$\endgroup$
  • $\begingroup$ I'm having conceptual issues with your rule #2. In your example, at the start of turn 2, $$V = \left\{ h_1\right\} = \left\{ x\right\}$$ but if we choose $a_1 = 2, a_{k > 1} = 0$, the function $2x$ clearly isn't in $V$ and thus condition 2 fails to hold. Also, you would need to define whether $g_k^n$ is ${g_k} \circ {g_k} \circ \cdots $ or ${g_k} \times {g_k} \times \cdots$. $\endgroup$ – COTO Sep 3 '14 at 15:21
  • $\begingroup$ Edited, thanks. I meant to say that whenever a function is added to $V$, all possible linear combinations of it with other functions in $V$ are also added to $V$. And $g_k^n$ is ${g_k} \times {g_k} \times \cdots g_k$, $n$ times. $\endgroup$ – Train Heartnet Sep 3 '14 at 15:43
  • $\begingroup$ In that case, adding the function $x$ in the first turn also adds all polynomials with zero constant term to $V$. Any function expressible by a Taylor series with an infinite radius of convergence ($sin$ and $cos - 1$ being two examples) would then necessarily be ruled out unless you imposed finiteness on the number of terms in the polynomial. As a secondary note, notwithstanding the above issue, a game could go on forever with each player alternating $cos$-wrapping or $sin$-multiplying $h_n$ and adding symmetric or antisymmetric delta functions at unique locations for each $n$. $\endgroup$ – COTO Sep 3 '14 at 16:51
0
$\begingroup$

I think Player 2 can at least avoid losing by always playing a subtraction: Let $x_n$ be some more or less random sequence of (nonzero) real numbers and then let, $g=h_{2n-1}-h_{2n}$, where

$$h_{2n}(x)= \begin{cases} 0\quad\text{if }|x|\not= x_n\\ 1\quad\text{if }x=x_n\\ \pm1\quad\text{if }x=-x_n \end{cases}$$

with the $\pm1$ sign chosen according to whether Player 1 started the game Odd or Even. Rule 5 (no consecutive *'s) means that Player 1 cannot play the same strategy. (The OP's example shows that it's OK for Player 2 to use the same operation twice in a row.) Since there are an uncountable number of choices for each new $x_n$, you should be able to guarantee that each new $g$ is independent of all the functions that preceed it.

$\endgroup$
  • $\begingroup$ I'm sorry, but I'm unable to understand your argument. If $x_n$ is a sequence, what is meant by $|x| \neq x_n$? Could you explain with an example? $\endgroup$ – Train Heartnet Sep 3 '14 at 16:05
  • $\begingroup$ @JobinIdiculla, the displayed equation defines a function that is identically $0$ except at two symmetrically positioned points, $x_n$ and $-x_n$, where it takes either equal or opposite values (according to whether Player 1 started the game with an odd or even function). The basic idea is that there are uncountably many such functions and they are all linearly independent. The vector space $V$, on the other hand, is always finite-dimensional (it's built up one basis function at a time), so no matter how large it gets, there will always be an $x_n$ that gives a $g$ not in $V$. $\endgroup$ – Barry Cipra Sep 3 '14 at 16:35
  • $\begingroup$ Oh, I should have said that the dimension of $V$ is always countable (not finite). I didn't notice that you were including all (positive integer) powers of each new $g$. But the uncountability of the basis functions of the type I described still trumps whatever dimension $V$ has. $\endgroup$ – Barry Cipra Sep 3 '14 at 16:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.