How to maximize the number of operations in process In my research project I have encountered the following problem, concerning a tuple of words in the formal language $L=\{0,1\}^*$, with $\epsilon$ denoting the empty word.
If we are given an ordered triple of words  $(a,b,c)$, an operation on that tuple consists of replacing one of the $x\in\{a,b,c\}$ by $x0,\ x1,$ or $y$ such that $x\neq y\in\{a,b,c\}$, e.g:
$$(\epsilon,\epsilon,\epsilon)\xrightarrow{b:=b1}(\epsilon,1,\epsilon)\xrightarrow{b:=b0}(\epsilon,10,\epsilon)\xrightarrow{c:=c1}(\epsilon,10,1)\xrightarrow{a:=b}(10,10,1) $$
Given a non-negative integer $h$, we say that $(a,b,c)$ is $h$-complete if the length of each of $a,b,c$ is $h$, that is $|a|=|b|=|c|=h$. We say that a series of operations is healthy if and only any tuple $(a,b,c)$ appears at most once. For example, the sequence
$$(\epsilon,\epsilon,\epsilon)\xrightarrow{b:=b0}(\epsilon,1,\epsilon)\xrightarrow{ b:=c}(\epsilon,\epsilon,\epsilon)$$
is not healthy because $(\epsilon,\epsilon,\epsilon)$ appears twice.
The length of a tuple $(a,b,c)$ is defined as $|(a,b,c)|=\max(|a|,|b|,|c|)$
My question:

What is the maximum number of healthy operations in a sequence which can transform $(\epsilon,\epsilon,\epsilon)$ into an $h$-complete tuple with all the intermediate tuples have length less than or equal to $h$

The minimum number of operations is $h+2$, but I'm interested in the worst case, and my problem consists of a generalization for $n$ words and not just $3$.
If you know any references for such problems, or have any idea about this problem, it will be appreciated!

Edit: the answer for couples is $f_2(h)=\frac{(h+2)(h+1)}{2}-1$ when we are taking pairs $(x,y)$ instead of $3$-tuples $(a,b,c)$


*

*If $h=1$ there is only two operations which can transform $(\epsilon,\epsilon)$ to $(0,0),(0,1),(1,0),(1,1)$ so:
$$f_2(1)=2 $$

*Suppose that there is only that the result is true for $h$ given two words of lenght $|x|=|y|=h+1$ my idea is described by the following process:
$$(\epsilon,\epsilon)\underbrace{-------\rightarrow}_{m(h) \text{ operations}} \begin{Bmatrix}
(0,0)\\
(0,1)\\ 
(1,0)\\ 
(0,0) \\
(w,w') \big/ |w|\text{or }|w'|>1
\end{Bmatrix}\underbrace{-------\rightarrow}_{x(h) \text{ operations}}  (x,y)$$
the added intermediate step is necessary (we can not complete the process without passing throught it) and one can prove that : $m(h)\leq h+2$ and clearly $x(h)\leq f_2(h)$ so:
$$f_2(h+1)\leq f_2(h)+h+2$$
to complete the proof we have to construct a process with $f_2(h+1)=h+2+f_2(h)$ which can be done using the schema 

 A: (I wouldn't claim my answer to be complete, but it is too long for a comment.)
There can be exponentially many steps, as I show below. And an obvious upper bound is $2^{3(h + 1)}$.
Now let consider following series of steps ($x \leftarrow y$ means that $x$ takes value $y$).
$$(\epsilon, \epsilon, \epsilon) \xrightarrow{a \leftarrow a0} (0, \epsilon, \epsilon) \xrightarrow{a \leftarrow a0} \cdots \xrightarrow{a \leftarrow a0} (\underbrace{0\ldots0}_{h}, \epsilon, \epsilon) \\
= (\underbrace{0\ldots0}_{h}, \epsilon, \epsilon) \xrightarrow{b \leftarrow b0} (\underbrace{0\ldots0}_{h}, 0, \epsilon) \xrightarrow{b \leftarrow b0} \cdots \xrightarrow{b \leftarrow b0} (\underbrace{0\ldots0}_{h}, \underbrace{0\ldots0}_{h}, \epsilon) \\
\xrightarrow{a \leftarrow c}  (\epsilon, \underbrace{0\ldots0}_{h}, \epsilon) \xrightarrow{a \leftarrow a0} (0, \underbrace{0\ldots0}_{h}, \epsilon) \xrightarrow{a \leftarrow a0} \cdots \xrightarrow{a \leftarrow a0} (\underbrace{0\ldots0}_{h - 1}, \underbrace{0\ldots0}_{h}, \epsilon) \xrightarrow{a \leftarrow a1} (\underbrace{0\ldots0}_{h - 1}1, \underbrace{0\ldots0}_{h}, \epsilon) \\
\xrightarrow{b \leftarrow c} (\underbrace{0\ldots0}_{h - 1}1, \epsilon, \epsilon) \xrightarrow{b \leftarrow b0} (\underbrace{0\ldots0}_{h - 1}1, 0, \epsilon) \xrightarrow{b \leftarrow b0} \cdots \xrightarrow{b \leftarrow b0} (\underbrace{0\ldots0}_{h - 1}1, \underbrace{0\ldots0}_{h - 1}, \epsilon) \xrightarrow{a \leftarrow a1} (\underbrace{0\ldots0}_{h - 1}1, \underbrace{0\ldots0}_{h - 1}1, \epsilon) \xrightarrow{a \leftarrow c}\\
\vdots\\
\rightarrow(\underbrace{1\ldots1}_{h}, \underbrace{1\ldots1}_{h}, \epsilon)\rightarrow\ldots$$
In other words after some preparation $(a, b, c)$ runs through $2^{h + 1} \cdot (h + 1)$ possible values of form $(x, \mathrm{prefix}(x), \epsilon)$ or $(\mathrm{prefix}(x + 1), x, \epsilon)$ for $2h + 2$ steps between $(x, x,  \epsilon)$ and $(x + 1, x + 1, \epsilon)$, where $x$ is a number written in binary with exactly $h$ digits (leading zeros are allowed), for all $2^h$ possible values of $x$. So all polynomial upper bounds are wrong.
