I am an engineer with limited mathematical proficiency, and I have encountered a quite challenging mathematical problem during my work. The problem as a whole is intricate, but I will try my best to provide a simplified version of it first.
I have 10 items to choose from, each of which has two attributes.
Attribute A has three states, and attribute B also has three states.
These states are represented as 0, 1, and 2. Therefore, the sample space for each item is { $(a, b)$ | $0\leq a \leq 2$, $0 \leq b \leq 2$ }.
The next step is to find a way to select 6 out of the 10 items such that both A state and B state distributions of the picked items are uniform. Here are the further details:
(1) For the uniform distribution of the A state, let's denote
$a_0$ = count of '0' in the first dimension of picks,
$a_1$ = count of '1' in the first dimension of picks,
$a_2$ = count of '2' in the first dimension of picks.
I define the "unevenness function" for A state as $u_A = (a_0 - \frac{6}{3})^2 + (a_1 - \frac{6}{3})^2 + (a_2 - \frac{6}{3})^2$.
This definition mimics the standard deviation, aiming for each state count to approach the expected value in a "perfectly uniform state", which equals the total picks divided by the number of states. Hence, a smaller $u_A$ indicates a more uniform distribution for the A state.
(2) For the uniform distribution of the B state, we can similarly denote
$b_0$ = count of '0' in the second dimension of picks,
$b_1$ = count of '1' in the second dimension of picks,
$b_2$ = count of '2' in the second dimension of picks.
I define the "unevenness function" for B state as $u_B = (b_0 - \frac{6}{3})^2 + (b_1 - \frac{6}{3})^2 + (b_2 - \frac{6}{3})^2$.
(3) The total unevenness function, denoted as $u$, is defined as $u = u_A + u_B$. The ultimate goal is to minimize $u$ to achieve the most uniform state.
Here's an example. Let's say the 10 items are as follows:
\begin{align*} \text{Item1} &= (0, 1) \\ \text{Item2} &= (0, 0) \\ \text{Item3} &= (0, 1) \\ \text{Item4} &= (1, 2) \\ \text{Item5} &= (1, 0) \\ \text{Item6} &= (1, 0) \\ \text{Item7} &= (1, 2) \\ \text{Item8} &= (2, 1) \\ \text{Item9} &= (2, 2) \\ \text{Item10} &= (2, 1) \\ \end{align*}
Assuming that I arbitrarily choose the first six items, then: $u_A = (a_0 - \frac{6}{3})^2 + (a_1 - \frac{6}{3})^2 + (a_2 - \frac{6}{3})^2 = (3 - 2)^2 + (3 - 2)^2 + (0 - 2)^2 = 2$, $u_B = (b_0 - \frac{6}{3})^2 + (b_1 - \frac{6}{3})^2 + (b_2 - \frac{6}{3})^2 = (3 - 2)^2 + (2 - 2)^2 + (1 - 2)^2 = 2$.
Therefore, the total unevenness for this picks set is $u = u_A + u_B = 4$.
On the other hand, when I make a "wiser" selection of items "1, 2, 5, 7, 8, 9", $u = u_A + u_B = 0$ because this choice ensures a very uniform distribution for both the A and B states, with each state appearing exactly twice.
My question is, is it possible to find the optimal "picks", i.e., the minimum value of $u$, without resorting to brute force?
Here is the complete version of the problem I am facing: Given $n$ items, each with $\alpha$ A states and $\beta$ B states, find the best $m$ items out of the $n$ to minimize $u = u_A + u_B = \left((a_0 - \frac{m}{\alpha})^2 + (a_1 - \frac{m}{\alpha})^2 + \dots + (a_\alpha - \frac{m}{\alpha})^2\right) + \left((b_0 - \frac{m}{\beta})^2 + (b_1 - \frac{m}{\beta})^2 + \dots + (b_\beta - \frac{m}{\beta})^2\right)$, and provide the selection method.
P.S. In practical situations, $n$ might be very small (might be less than 50)or exceed a thousand while $m$ is roughly 10, in most case making exhaustive enumeration almost impossible.
