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I am trying to solve the following problem:

The Problem

Assume $S_n$ is the set of all ordered $n$-tuples of 0 and 1 and let $A_1, A_2, \ldots, A_{32}$ be a permutation of the elements of $S_5$. Also assume that $f(A_1) = 1$ and for every $1 \le i \le 32$ the value of $f(A_i)$ is equal to the smallest positive integer such that for every $j$ ($1 \le j < i$), where $A_i$ and $A_j$ differ in exactly one coordinate, it holds that $f(A_i) \neq f(A_j)$. (For instance, if $A_i = (0,1,1,0,1)$ and $A_j = (0,1,0,0,1)$ then $A_i$ and $A_j$ differ only in the third coordinate). What is the maximum possible value of $\max\{f(A_1), f(A_2), \ldots, f(A_{32})\}$?

Here's my progress till now:

Instead of 5-tuples, I took 3-tuples to understand the problem. So there are $2^3 = 8$ elements in $S_3$.

Suppose $A_1=(0,0,0)$, $A_2=(0,0,1)$, $A_3=(0,1,0)$. By definition, $f(A_1)=1$.

For $A_2$, since $A_2$ and $A_1$ differ by one coordinate, we must have $f(A_2) \neq f(A_1)$. The smallest positive integer for $f(A_2)$ is $2$.

For $A_3$, since $A_3$ and $A_1$ differ by one coordinate, we must have $f(A_3) \neq f(A_1)$. The smallest positive integer for $f(A_3)$ is also $2$.

I did this for the following permutation of the 8 elements: $$A_1 = (0,0,0)$$ $$A_2 = (0,0,1)$$ $$A_3 = (0,1,0)$$ $$A_4 = (0,1,1)$$ $$A_5 = (1,0,0)$$ $$A_6 = (1,0,1)$$ $$A_7 = (1,1,0)$$ $$A_8 = (1,1,1)$$

For this ordering, I calculated the following function values: $$f(A_1) = 1, f(A_2) = 2, f(A_3) = 2, f(A_4) = 1, f(A_5) = 2, f(A_6) = 1, f(A_7) = 3, f(A_8) = 2$$

So, for this specific permutation, $\max\{f(A_i)\} = 3$.

However, this result is only for one specific ordering of the $A_i$. Since any permutation is possible, maybe the maximum value could be 4 or 5? I don't know.

I related this to the Greedy algorithm, since the selection of the $A_i$'s would matter, but I don't know exactly what to do with that insight.

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    $\begingroup$ We have $f(A_7)=1$, because $A_7$ differ in exactly one coordinate with $A_3$ and $A_5$, and $f(A_3)=f(A_5)=2$. $\endgroup$ Commented Sep 12 at 2:34

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The $\max{\{f(A_i)\}} \leq 6$. This is true, since if $f(A_i)=k+1$ for some $i$, then for each $1\leq j\leq k$, there exists a $5$-tuple $A_{i_j}$ such that $f(A_{i_j})=j$ and such that $A_{i_j}$ differs in precisely one coordinate with $A_i$. Since $A_i$ has only $5$ coordinates to change, then $k\leq 5$ and the max must be less than or equal to $k+1=6$.

There is an ordering which yields a max of $6$. After the first $19$ vectors listed below, the remaining vectors can be concatenated at the end in an arbitrary order. \begin{array}{|c|c|c|} \hline Order & 5-tuple & f \\ \hline 1 & (0,1,0,1,1) & 1 \\ \hline 2 & (1,1,0,0,0) & 1 \\ \hline 3 & (1,0,0,1,1) & 1 \\ \hline 4 & (0,1,0,0,1) & 2 \\ \hline 5 & (1,1,0,1,0) & 2 \\ \hline 6 & (0,1,0,0,0) & 3 \\ \hline 7 & (1,0,0,1,0) & 3 \\ \hline 8 & (1,1,1,0,0) & 2 \\ \hline 9 & (0,0,0,0,0) & 1 \\ \hline 10 & (0,1,1,0,1) & 1 \\ \hline 11 & (1,1,1,1,0) & 1 \\ \hline 12 & (0,0,1,1,1) & 1 \\ \hline 13 & (0,0,1,1,0) & 2 \\ \hline 14 & (0,1,1,0,0) & 4 \\ \hline 15 & (0,0,0,1,0) & 4 \\ \hline 16 & (1,1,1,1,1) & 2 \\ \hline 17 & (0,1,1,1,1) & 3 \\ \hline 18 & (0,1,1,1,0) & 5 \\ \hline 19 & (0,1,0,1,0) & 6 \\ \hline 20 & \ldots & \ldots \\ \hline \ldots & \ldots & \ldots \\ \hline \end{array}

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  • $\begingroup$ I suppose you can reorder this so that all the uples with $f$ equal to $1$ come first, then the uples with $f$ equal to $2$, etc. $\endgroup$ Commented Sep 13 at 6:37
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    $\begingroup$ @EwanDelanoy It is easy to check that the enumeration of any permutation of $A_1,\dots,A_{32}$ reordered in such a way is still produced by the given algorithm. $\endgroup$ Commented Sep 14 at 9:36

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