Reconstructing Words from Right-Bounded-Block Words

Proof. If $`w`$ is such that $`w_i`$ is a scattered factor of $`w`$, for all $`i\in [k]`$, and $`|w|_\ta=k_\ta`$, for all $`\ta\in \Sigma`$, then we can mark the $`i^{th}`$ occurrence of $`\ta`$ as $`(\ta)_i`$, for all $`\ta\in \Sigma`$ and $`i\in [k_\ta]`$. This induces a $`K`$-valid marking $`\psi`$ of the words $`w_i`$, and, moreover, the linear ordering of the nodes of $`G_\psi`$ induced by the order in which the marked letters (i.e., nodes of $`G_\psi`$) occur in $`w`$ is a topological sorting of $`G_{\psi}`$.

Let us now assume that there exists a $`K`$-valid marking $`\psi`$ of the words $`w_1,\ldots,w_k`$, and there exists a valid topological sorting of the $`\psi`$-marked letters of the words $`w_1,\ldots,w_k`$. Let $`w'`$ be the word obtained by writing the nodes of $`G_{\psi}`$ in the order given by its topological sorting and removing their markings. It is clear that $`w'`$ has $`w_i`$ as a scattered factor, for all $`i\in [k]`$, and that $`|w'|_\ta\leq k_\ta`$, for all $`\ta\in \Sigma`$. Let now $`w=w' \prod_{\ta\in \Sigma}\ta^{k_\ta-|w'|_\ta}`$, where $`\prod_{\ta\in \Sigma}\ta^{k_\ta-|w'|_\ta}`$ is the concatenation of the factors $`\ta^{k_\ta-|w'|_\ta}`$, for $`\ta\in \Sigma`$ in some fixed order. Now $`w`$ has $`w_i`$ as a scattered factor, for all $`i\in [k]`$, and $`|w|_\ta= k_\ta`$, for all $`\ta\in \Sigma`$. 0◻ ◻

Proof. Let $`w`$ be the word obtained by writing in order the letters of the unique valid topological sorting of the $`\psi`$-marked letters of the words $`w_1,\ldots,w_k`$ and removing their markings. It is clear that $`w'`$ has $`w_i`$ as a scattered factor, for all $`i\in [k]`$, and that $`|w|_\ta= k_\ta`$, for all $`\ta\in \Sigma`$. The word $`w`$ is uniquely defined (as there is no other $`K`$-valid marking nor valid topological sorting of the $`\psi`$-marked letters), and $`|w|_\ta=k_\ta`$, for all $`\ta\in \Sigma`$. 0◻ ◻

Proof. Notice firstly $`|W|=\frac{q(q-1)}{2}`$. Let $`k_\ta=|w_{\ta,\tb}|_\ta`$, for $`\ta<\tb\in \Sigma`$. These numbers are clearly well defined, by the second item in our hypothesis. Let $`K=(k_\ta)_{\ta\in \Sigma}`$. It is immediate that there exists a unique $`K`$-valid marking $`\psi`$ of the words $`(w_{\ta,\tb})_{\ta<\tb\in \Sigma}`$. As each two marked letters $`(\ta)_i`$ and $`(\tb)_j`$ (i.e., each two nodes $`(\ta)_i`$ and $`(\tb)_j`$ of $`G_\psi`$) appear in the marked word $`w^\psi_{\ta,\tb}`$, we know the order in which these two nodes should occur in a topological sorting of $`G_{\psi}`$. This means that, if $`G_\psi`$ is acyclic, then it has a unique topological sorting. Our statement follows now from Corollary [reconst_topo]. 0◻ ◻

Proof. The first part is, once again, a consequence of Corollary [reconst_topo]. The second part can be shown by assuming that $`\Sigma=\{\ta_1,\ldots,\ta_q\}`$ and the pair $`\{\ta_1,\ta_2\}`$ is not contained in any of the sets $`S_1,\ldots,S_k`$. Then, for $`w=\ta_1\ta_3 \ta_4\ldots \ta_q`$ and $`w'=\ta_2\ta_3 \ta_4\ldots \ta_q`$, we have that $`\pi_{S_1}(w)=\pi_{S_1}(w'), \ldots,`$ $`\pi_{S_k}(w)=\pi_{S_k}(w')`$. 0◻ ◻

Proof. We begin with a series of preliminaries. Let us recall the Orthogonal Vectors problem: Given sets $`A, B`$ consisting of $`n`$ vectors in $`\{0,1\}^k`$, decide whether there are vectors $`a\in A`$ and $`b\in B`$ which are orthogonal (i.e., for any $`i\in [k]`$ we have $`a[i]b[i] = 0`$). This problem can be solved naïvely in $`O(n^2k)`$ time, but under the Strong Exponential Time Hypothesis there is no $`O(n^{2-d} k^c)`$ algorithm for solving it, for any $`d,c>0`$ (once more, see the survey and the references therein).

We show that our problem is equivalent to the Orthogonal Vectors problem.

Let us first assume that we are given the sets $`S_1,\ldots, S_k\subset \Sigma`$, and we want to decide whether we can reconstruct any word uniquely from its projections $`\pi_{S_1}(\cdot), \ldots, \pi_{S_k}(\cdot)`$. This is equivalent, according to Theorem [char_proj], to checking whether each pair $`\{\ta,\tb\}\subseteq \Sigma`$ is included in at least one of the sets $`S_i`$. For each letter $`\ta`$ of $`\Sigma`$ we define the $`k`$-dimensional vectors $`x_\ta`$ where $`x_\ta[i]=1`$ if $`\ta \in S_i`$ and $`x_\ta[i]=0`$ if $`\ta \not\in S_i`$. This can be clearly done in $`O(q k)`$ time. Now, there exists a pair $`\{\ta,\tb\}`$ that is not contained in any of the sets $`S_1,\ldots, S_k`$ if and only if there exists a pair of vectors $`\{x_\ta,x_\tb\}`$ such that $`x_\ta[i] x_\tb[i]=0`$ for all $`i\in [k]`$. We can check whether there exists a pair of vectors $`\{x_\ta,x_\tb\}`$ such that $`x_\ta[i] x_\tb[i]=0`$ for all $`i\in [k]`$ by solving the Orthogonal Vectors problem by using for both input sets of vectors the set $`\{x_\ta\mid \ta \in \Sigma\}`$. As such, we can check whether there exists a pair $`\{\ta,\tb\}`$ that is not contained in any of the sets $`S_1,\ldots, S_k`$ in $`O(q^2 k)`$ time.

Let us now assume that we are given two sets $`A, B`$ consisting of $`n`$ vectors in $`\{0,1\}^k`$, and we want to decide whether there are vectors $`a\in A`$ and $`b\in B`$ which are orthogonal (i.e., for any $`i\in [k]`$ we have $`a[i]b[i] = 0`$). We can compute the set of $`(k+2)`$-dimensional vectors $`A'`$ containing the vectors of $`A`$ extended with two new positions (position $`k+1`$ and position $`k+2`$) set to $`10`$ and the vectors of $`B`$ extended with two new positions (position $`k+1`$ and position $`k+2`$) set to $`01`$. To decide whether there are vectors $`a\in A`$ and $`b\in B`$ which are orthogonal is equivalent to decide whether there are vectors $`a,b\in A'`$ which are orthogonal (if two such vectors exist, they must be different on their last two positions, so one must come from $`A`$ and one from $`B`$). Assume that $`A'=\{x_1,x_2,\ldots,x_{2n}\}`$. Now we define an alphabet $`\Sigma=\{\ta_1,\ldots,\ta_{2n}\}`$ of size $`2n`$ and the sets $`S_1,\ldots, S_k`$, where $`\ta_j\in S_i`$ if and only if $`x_j[i]=1`$. Computing $`A'`$ and then the alphabet $`\Sigma`$ and the sets $`S_i`$, for $`i\in [k]`$, takes $`O(nk)`$ time. Now, to decide whether there are vectors $`x_i,x_j\in A'`$ which are orthogonal is equivalent to decide whether there exists a pair of letters $`\{\ta_i, \ta_j\}`$ of $`\Sigma`$ that is not contained in any of the sets $`S_1,\ldots, S_k`$. The conclusion of the theorem now follows. 0◻ ◻

Proof. The case $`k=0`$ is trivial. Consider the case $`n=k`$, i.e., $`\sum_{i=1}^{k+1}s_i=0`$. This implies immediately $`s_i=0`$ for all $`i\in[k+1]`$ and the equivalence holds. Assume for the rest of the proof $`k

c_{k+1}(n-k)\leq
\sum_{i=j}^{k+1}c_i s_i' =\left(\sum_{i=j}^{k}c_i s_i' \right)+c_{k+1}(n-k-\ell).

This implies $`\sum_{i=j}^{k}c_i s_i' \geq c_{k+1}\ell`$. Since the coefficients are strictly increasing we get $`\sum_{i=j}^{k}c_i s_i' \leq c_k\sum_{i=j}^{k} s_i'

Proof. It follows directly from Equation [alb] and Lemma [maxvalues].0◻ ◻

Proof. Consider firstly $`k_{\ta^{\ell}\tb}\in[\ell]_0`$. By Remark [ci] we have $`c_{\ell+1}=1`$ and $`c_{\ell+2}=\ell+1`$. By $`c_ir}`$. Thus there exists $`x\in\N`$ with $`\binom{k_\ta}{\ell}(n-k_\ta-r')+x=\frac{(k_\ta-1)!(k_\ta(n-k_\ta)-\ell r)}{\ell!(k_\ta-\ell)!}`$, i.e., $`x=\frac{(k_\ta-1)!(k_\ta r'-\ell r)}{\ell!(k_\ta -\ell)!}`$. By $`k_\tb=n-k_\ta`$ we have $`x\leq \binom{k_\ta-1}{\ell}r'=\frac{(k_\ta-1)!(k_\ta r'-\ell r')}{\ell!(k_\ta-\ell)!}`$ (we only have $`r'`$ occurrences of $`\tb`$ left to distribute). By $`r'>r`$ we have $`\frac{(k_\ta-1)!(k_\ta r'-\ell r)}{\ell!(k_\ta-\ell)!}=x<\frac{(k_\ta-1)!(k_\ta r'-\ell r)}{\ell!(k_\ta-\ell)!}`$ - a contradiction.0◻ ◻

Proof. Let us first prove that $`w`$ is uniquely determined in these cases. It is obvious if $`k_\ta = 0`$ or $`k_\ta = n`$ since the word is composed of the same letter repeated $`n`$ times. If $`k_\ta = 1`$, then $`w = \tb^{s_1} \ta \tb^{n-1-s_1}`$ and $`\binom{w}{\ta\tb} = n-1-s_1 = k_{\ta\tb}`$. Therefore $`w`$ is uniquely determined. If $`k_\ta = n-1`$, then $`w = \tb^{s_1} \ta \tb^{s_2} \cdots \ta \tb^{s_n}`$ with exactly one of the $`s_i`$ being non zero and, in fact, equal to one. We have $`\binom{w}{\ta\tb} = \sum_{i=2}^{n} (i-1) s_i`$ and, if $`k_{\ta\tb}`$ is given (between $`0`$ and $`n-1`$), then $`s_{k_{\ta\tb}+1} = 1`$ is the only non zero exponent. Consider now $`k_\ta\in[n-2]_{\geq 2}`$, i.e. $`w=\tb^{s_1}\ta\tb^{s_2}\dots \tb^{s_{k_\ta}}\ta\tb^{s_{k_\ta+1}}`$. Thus $`k_{\ta\tb} = 0`$ implies $`s_1 = n - k_\ta`$ and $`s_2 = 0, \ldots, s_{k_\ta + 1} = 0`$ while $`k_{\ta\tb} = 1`$ implies $`s_2 = 1`$, $`s_1 = n - k_\ta - 1`$ and $`s_3 =0, \ldots, s_{k_\ta + 1} = 0`$. By Lemma [maxvalues], we know that [alb] is maximal if and only if $`s_{k_\ta +1 } = n - k_{\ta}`$ and all the other $`s_i`$ are equal to zero. In that case, the value of the sum equals $`k_{\ta}(n - k_{\ta})`$. Therefore, if $`\binom{w}{\ta\tb} = k_{\ta}(n - k_{\ta})`$, the word $`w`$ is uniquely determined. Finally, if $`k_{\ta\tb} = k_{\ta}(n - k_{\ta}) -1`$, we must have $`s_{k_\ta + 1} \leq n - k_\ta - 1`$. If we choose $`s_{k_\ta + 1} = n - k_\ta - 1`$, it remains that $`\sum_{i=1}^{k_\ta} s_i = 1`$ and $`\sum_{i=2}^{k_\ta} (i-1)s_i = k_\ta -1`$. We must have $`s_{k_\ta} = 1`$ and the other ones equal to zero. In fact, choosing $`s_{k_\ta + 1} = n - k_\ta - 1`$ is the only possibility: if otherwise $`s_{k_\ta + 1} = n - k_\ta - \ell`$ with $`\ell > 1`$, we obtain that $`\sum_{i=2}^{k_\ta} (i-1) s_i \geq \ell k_\ta

• 1$ with $\sum_{i=1}^{k_\ta} s_i = \ell`$. It is easy to check with Lemma [maxvalues] that these conditions are incompatible.

We now need to prove that $`w`$ cannot be uniquely determined if $`k_\ta \in [n-2]_{\geq 2}`$ and $`k_{\ta\tb} \in [k_\ta (n-k_{\ta}) - 2]_{\geq 2}`$. To this aim we will give two different sets of values for the $`s_i`$. The first decomposition is the greedy one. Let us put $`s_{k_\ta + 1} = \lfloor \frac{k_{\ta\tb}}{k_\ta} \rfloor`$, $`s_{(k_{\ta\tb} \bmod k_\ta) + 1} = 1`$ and the other $`s_i`$ equal to $`0`$. Let us finally modify the value of $`s_1`$ (which is, at this stage, equal to $`0`$ or $`1`$) by adding the value needed. By $`\sum_{i=1}^{k_{\ta}+1} s_i=n-k_\ta`$ we get $`s_1 \leftarrow s_1 + (n-k_\ta) -\lfloor \frac{k_{\ta\tb}}{k_\ta} \rfloor - 1`$. This implies $`\sum_{i=1}^{k_\ta + 1} s_i = 1 + (n-k_\ta) - \left \lfloor \frac{k_{\ta\tb}}{k_\ta} \right \rfloor - 1 + \left \lfloor \frac{k_{\ta\tb}}{k_\ta} \right \rfloor = n-k_\ta`$ and $`s_i \geq 0`$ for all $`i`$. Moreover we have $`\sum_{i=2}^{k_\ta + 1} (i-1) s_i = (k_{\ta\tb} \bmod k_\ta) + k_\ta \left \lfloor \frac{k_{\ta\tb}}{k_\ta} \right \rfloor = k_{\ta\tb}`$.

Now we provide a second decomposition for the $`s_i`$. First, let us assume that $`2 \leq k_{\ta\tb} < k_{\ta}`$. In that case, the greedy algorithm sets $`s_{k_{\ta\tb} + 1} = 1`$, $`s_1 = n - k_\ta - 1`$ and the other $`s_i`$ to $`0`$. Let us now set $`s_1 = n - k_{\ta} - 2`$ and all the other $`s_i`$ to $`0`$. Then, update $`s_{k_{\ta\tb}} \leftarrow s_{k_{\ta\tb}} + 1`$ and $`s_2 \leftarrow s_2 + 1`$ (in the case where $`k_{\ta\tb} = 2`$, $`s_2`$ will be equal to $`2`$ after these manipulations). We have that the sum in [alb] is equal to $`1 + (k_{\ta\tb} - 1)`$ as needed. Finally, if $`k_{\ta\tb} \geq k_\ta`$, then $`s_{k_\ta + 1}`$ was non zero in the greedy decomposition, and the idea is to reduce it of a value $`1`$. Let us set $`s_{k_\ta + 1} =\lfloor \frac{k_{\ta\tb}}{k_\ta} \rfloor - 1`$ and the other $`s_i`$ to $`0`$. Then, let us update some values: $`s_{(k_{\ta\tb} \bmod k_\ta) + 2} \leftarrow s_{(k_{\ta\tb} \bmod k_\ta) + 2} + 1`$ and $`s_{k_\ta} \leftarrow s_{k_\ta} + 1`$ if $`(k_{\ta\tb} \bmod k_\ta) \neq k_\ta - 1`$, and $`s_{k_\ta} = 2`$, $`s_2 = 1`$ otherwise. Finally, set $`s_1`$ to the right value, i.e., $`n-k_\ta - \sum_{i=2}^{k_\ta +1} s_i`$. It can be easily checked that, in both cases, $`s_1 \geq 0`$ (notice that $`(k_{\ta\tb} \bmod k_\ta) = k_\ta - 1`$ implies that $`\lfloor \frac{k_{\ta\tb}}{k_\ta} \rfloor \leq n - k_\ta -2`$) and that all $`s_i`$ sum up to $`n - k_\ta`$. Similarly, we can check that $`\sum_{i=2}^{k_\ta + 1} (i-1) s_i`$ is equal to $`k_{\ta\tb}`$ in both cases.

To sum up, we gave two different decompositions for the $`s_i`$ in cases where $`k_\ta \in [n-2]_{\geq 2}`$ and $`k_{\ta\tb} \in [k_\ta (n-k_{\ta}) - 2]_{\geq 2}`$. That implies that $`w`$ cannot be uniquely determined in those cases.0◻ ◻

Proof. If $`k_{\ta^{j}\tb}`$ is unique, the coefficients $`s_{j+1},\dots,s_{k_{\ta}+1}`$ are uniquely determined. Substituting backwards the known values in the first $`j-1`$ equations ([alb]) (for $`\ell = 1, \ldots, j-1`$) we can now obtain successively the values for $`s_{j},\dots,s_1`$.0◻ ◻

Proof. It follows directly from Theorem [uniquerecon].0◻ ◻

Proof. Assume w.l.o.g. $`k_\ta\leq k_\tb`$. Then $`k_{\ta} \leq \frac{n}{2}`$ and Theorem [uniquerecon] shows that words in the set $`\{ \tb, \ta \tb, \ldots, \ta^{\lfloor \frac{n}{2} \rfloor} \tb \}`$ can reconstruct $`w`$ uniquely. If $`k_{\ta} > k_{\tb}`$, the set $`S`$ is obtained by replacing the letter $`\ta`$ by $`\tb`$ and vice-versa.

Set $`N_2(i):=\frac{1}{i}\sum_{d|i}\mu(d)2^{\frac{i}{d}}`$ for all $`i\in[\lfloor\frac{16}{7}\sqrt{n}\rfloor+5]`$, i.e., $`\sum_{i=1}^{\lfloor \frac{16}{7}\sqrt{n}\rfloor +5} N_2(i)`$, which is Equation [eqform], binomial coefficients suffice. By we have

\begin{align*}
N_2(i) & \geq 
\frac{1}{i}\left(2^i-\frac{2^{\frac{i}{2}}-1}{2-1}\right)
= \frac{1}{i}\left(2^i-2^{\frac{i}{2}}+1\right)
= \frac{1}{i}\left(2^{\frac{i}{2}} (2^{\frac{i}{2}}-1)+1\right)
\geq \frac{2^{\frac{i}{2}}}{i}.
\end{align*}

This results in

\begin{align*}
\sum_{i=1}^{\lfloor 
\frac{16}{7}\sqrt{n}\rfloor +5} N_2(i) &\geq \sum_{i=1}^{\lfloor 
\frac{16}{7}\sqrt{n}\rfloor +5} \frac{2^{\frac{i}{2}}}{i}
\geq 
\frac{1}{\frac{16}{7}\sqrt{n} +5} \, \frac{\sqrt{2}^{\frac{16}{7} \sqrt{n}+5}-1}{\sqrt{2}-1}.
\end{align*}

We want to show that this quantity is at least equal to $`\frac{n+1}{2}`$. Let us define

f(x) = \frac{1}{\frac{16}{7}\sqrt{x} +5} \, \frac{\sqrt{2}^{\frac{16}{7} \sqrt{x}+5}-1}{\sqrt{2}-1} 
- \frac{x+1}{2}

for all $`x > 0`$, which is the continuous extension on $`\mathbb{R}^{+}`$ of the quantity we are interested in. It is easy to verify by hand that $`f(1), f(2), f(3)`$ and $`f(4)`$ are positive. Let us formally show that $`f(x) > 0`$ for all $`x \geq 5`$. Since this function is differentiable, we get with $`y = \frac{16}{7} \sqrt{x} + 5`$

f'(x) = \frac{1}{y} \, \sqrt{2}^{y} \, 
\frac{\ln(\sqrt{2})}{\sqrt{2}-1} \, \frac{8}{7 \sqrt{x}} - \frac{1}{y^2} 
\, \frac{8}{7 \sqrt{x}} \, \frac{\sqrt{2}^{y}-1}{\sqrt{2}-1} - \frac{1}{2}.

We thus have $`\sqrt{x} = \frac{7y-35}{16}`$ and $`y \geq 7`$ for all $`x \geq 1`$. By injecting $`y`$ in the previous expression, and reducing to the common denominator, we have to show that

\begin{align*}
 & \, 2y \sqrt{2}^y 128 \ln(\sqrt{2}) - 256 (\sqrt{2}^y - 1) - 7(7y-35) y^2(\sqrt{2}-1)  \\
 = & \,\sqrt{2}^y (128 \ln(2) y - 256) + 256 - 49y^3 (\sqrt{2} - 1) + 245 y^2 (\sqrt{2} - 1) \\
 \geq & \, \sqrt{2}^y 365 + 256 - 49y^3 (\sqrt{2} - 1) + 245 y^2 (\sqrt{2} - 1)
\end{align*}

is strictly positive. Let us call the last quantity $`g(y)`$. We will show that it is positive for all $`y \geq 10.05`$, which means that $`f(x)`$ is positive for all $`x`$ such that $`\frac{16}{7} \sqrt{x}

• 5 \geq 10.05$, i.e., for all $x \geq 5`$. We have

\begin{align*}
g'(y) &= 365 \sqrt{2}^y \ln(\sqrt{2}) - 147(\sqrt{2}-1) y^2 + 490(\sqrt{2}-1)y, \\
g''(y) &= 365 \sqrt{2}^y (\ln(\sqrt{2}))^2 - 294(\sqrt{2}-1) y + 490(\sqrt{2}-1), \\
g'''(y) &= 365 \sqrt{2}^y (\ln(\sqrt{2}))^3 - 294(\sqrt{2}-1),
\end{align*}

and $`g'''(7) > 50`$, $`g''(8.5) > 2`$, $`g'(10.05) > 8`$ and finally $`g(10.05) > 1787`$. Since $`g'''(y)`$ is increasing and positive in $`7`$, $`g''(y)`$ is increasing for $`y \geq 7`$. Therefore $`g'(y)`$ is increasing for $`y \geq 8.5`$ and finally $`g(y)`$ is increasing for $`y \geq 10.05`$ and positive.0◻ ◻

Proof. If $`w`$ is such that $`w_i`$ is a scattered factor of $`w`$, for all $`i\in [k]`$, and $`|w|_\ta=k_\ta`$, for all $`\ta\in \Sigma`$, then we can mark the $`i^{th}`$ occurrence of $`\ta`$ as $`(\ta)_i`$, for all $`\ta\in \Sigma`$ and $`i\in [k_\ta]`$. This induces a $`K`$-valid marking $`\psi`$ of the words $`w_i`$, and, moreover, the linear ordering of the nodes of $`G_\psi`$ induced by the order in which the marked letters (i.e., nodes of $`G_\psi`$) occur in $`w`$ is a topological sorting of $`G_{\psi}`$.

Let us now assume that there exists a $`K`$-valid marking $`\psi`$ of the words $`w_1,\ldots,w_k`$, and there exists a valid topological sorting of the $`\psi`$-marked letters of the words $`w_1,\ldots,w_k`$. Let $`w'`$ be the word obtained by writing the nodes of $`G_{\psi}`$ in the order given by its topological sorting and removing their markings. It is clear that $`w'`$ has $`w_i`$ as a scattered factor, for all $`i\in [k]`$, and that $`|w'|_\ta\leq k_\ta`$, for all $`\ta\in \Sigma`$. Let now $`w=w' \prod_{\ta\in \Sigma}\ta^{k_\ta-|w'|_\ta}`$, where $`\prod_{\ta\in \Sigma}\ta^{k_\ta-|w'|_\ta}`$ is the concatenation of the factors $`\ta^{k_\ta-|w'|_\ta}`$, for $`\ta\in \Sigma`$ in some fixed order. Now $`w`$ has $`w_i`$ as a scattered factor, for all $`i\in [k]`$, and $`|w|_\ta= k_\ta`$, for all $`\ta\in \Sigma`$. 0◻ ◻

Proof. Let $`w`$ be the word obtained by writing in order the letters of the unique valid topological sorting of the $`\psi`$-marked letters of the words $`w_1,\ldots,w_k`$ and removing their markings. It is clear that $`w'`$ has $`w_i`$ as a scattered factor, for all $`i\in [k]`$, and that $`|w|_\ta= k_\ta`$, for all $`\ta\in \Sigma`$. The word $`w`$ is uniquely defined (as there is no other $`K`$-valid marking nor valid topological sorting of the $`\psi`$-marked letters), and $`|w|_\ta=k_\ta`$, for all $`\ta\in \Sigma`$. 0◻ ◻

Proof. Notice firstly $`|W|=\frac{q(q-1)}{2}`$. Let $`k_\ta=|w_{\ta,\tb}|_\ta`$, for $`\ta<\tb\in \Sigma`$. These numbers are clearly well defined, by the second item in our hypothesis. Let $`K=(k_\ta)_{\ta\in \Sigma}`$. It is immediate that there exists a unique $`K`$-valid marking $`\psi`$ of the words $`(w_{\ta,\tb})_{\ta<\tb\in \Sigma}`$. As each two marked letters $`(\ta)_i`$ and $`(\tb)_j`$ (i.e., each two nodes $`(\ta)_i`$ and $`(\tb)_j`$ of $`G_\psi`$) appear in the marked word $`w^\psi_{\ta,\tb}`$, we know the order in which these two nodes should occur in a topological sorting of $`G_{\psi}`$. This means that, if $`G_\psi`$ is acyclic, then it has a unique topological sorting. Our statement follows now from Corollary [reconst_topo]. 0◻ ◻

Proof. The first part is, once again, a consequence of Corollary [reconst_topo]. The second part can be shown by assuming that $`\Sigma=\{\ta_1,\ldots,\ta_q\}`$ and the pair $`\{\ta_1,\ta_2\}`$ is not contained in any of the sets $`S_1,\ldots,S_k`$. Then, for $`w=\ta_1\ta_3 \ta_4\ldots \ta_q`$ and $`w'=\ta_2\ta_3 \ta_4\ldots \ta_q`$, we have that $`\pi_{S_1}(w)=\pi_{S_1}(w'), \ldots,`$ $`\pi_{S_k}(w)=\pi_{S_k}(w')`$. 0◻ ◻

Proof. We begin with a series of preliminaries. Let us recall the Orthogonal Vectors problem: Given sets $`A, B`$ consisting of $`n`$ vectors in $`\{0,1\}^k`$, decide whether there are vectors $`a\in A`$ and $`b\in B`$ which are orthogonal (i.e., for any $`i\in [k]`$ we have $`a[i]b[i] = 0`$). This problem can be solved naïvely in $`O(n^2k)`$ time, but under the Strong Exponential Time Hypothesis there is no $`O(n^{2-d} k^c)`$ algorithm for solving it, for any $`d,c>0`$ (once more, see the survey and the references therein).

We show that our problem is equivalent to the Orthogonal Vectors problem.

Let us first assume that we are given the sets $`S_1,\ldots, S_k\subset \Sigma`$, and we want to decide whether we can reconstruct any word uniquely from its projections $`\pi_{S_1}(\cdot), \ldots, \pi_{S_k}(\cdot)`$. This is equivalent, according to Theorem [char_proj], to checking whether each pair $`\{\ta,\tb\}\subseteq \Sigma`$ is included in at least one of the sets $`S_i`$. For each letter $`\ta`$ of $`\Sigma`$ we define the $`k`$-dimensional vectors $`x_\ta`$ where $`x_\ta[i]=1`$ if $`\ta \in S_i`$ and $`x_\ta[i]=0`$ if $`\ta \not\in S_i`$. This can be clearly done in $`O(q k)`$ time. Now, there exists a pair $`\{\ta,\tb\}`$ that is not contained in any of the sets $`S_1,\ldots, S_k`$ if and only if there exists a pair of vectors $`\{x_\ta,x_\tb\}`$ such that $`x_\ta[i] x_\tb[i]=0`$ for all $`i\in [k]`$. We can check whether there exists a pair of vectors $`\{x_\ta,x_\tb\}`$ such that $`x_\ta[i] x_\tb[i]=0`$ for all $`i\in [k]`$ by solving the Orthogonal Vectors problem by using for both input sets of vectors the set $`\{x_\ta\mid \ta \in \Sigma\}`$. As such, we can check whether there exists a pair $`\{\ta,\tb\}`$ that is not contained in any of the sets $`S_1,\ldots, S_k`$ in $`O(q^2 k)`$ time.

Let us now assume that we are given two sets $`A, B`$ consisting of $`n`$ vectors in $`\{0,1\}^k`$, and we want to decide whether there are vectors $`a\in A`$ and $`b\in B`$ which are orthogonal (i.e., for any $`i\in [k]`$ we have $`a[i]b[i] = 0`$). We can compute the set of $`(k+2)`$-dimensional vectors $`A'`$ containing the vectors of $`A`$ extended with two new positions (position $`k+1`$ and position $`k+2`$) set to $`10`$ and the vectors of $`B`$ extended with two new positions (position $`k+1`$ and position $`k+2`$) set to $`01`$. To decide whether there are vectors $`a\in A`$ and $`b\in B`$ which are orthogonal is equivalent to decide whether there are vectors $`a,b\in A'`$ which are orthogonal (if two such vectors exist, they must be different on their last two positions, so one must come from $`A`$ and one from $`B`$). Assume that $`A'=\{x_1,x_2,\ldots,x_{2n}\}`$. Now we define an alphabet $`\Sigma=\{\ta_1,\ldots,\ta_{2n}\}`$ of size $`2n`$ and the sets $`S_1,\ldots, S_k`$, where $`\ta_j\in S_i`$ if and only if $`x_j[i]=1`$. Computing $`A'`$ and then the alphabet $`\Sigma`$ and the sets $`S_i`$, for $`i\in [k]`$, takes $`O(nk)`$ time. Now, to decide whether there are vectors $`x_i,x_j\in A'`$ which are orthogonal is equivalent to decide whether there exists a pair of letters $`\{\ta_i, \ta_j\}`$ of $`\Sigma`$ that is not contained in any of the sets $`S_1,\ldots, S_k`$. The conclusion of the theorem now follows. 0◻ ◻

Proof. The claim that $`\sum_{i\in[q]}|w|_i(q+1-i)`$ binomial coefficients suffice to reconstruct $`w`$ uniquely follows by Theorem [recon2]: for each pair of letters we apply the method of the binary case. We are thus going to reconstruct words $`w_{\ta,\tb}`$ for all pairs of letters $`\ta < \tb`$. If $`\ta`$ is the $`i^\text{th}`$ letter in the alphabet, there are $`q-i`$ such pairs. To determine $`w_{\ta,\tb}`$ uniquely, $`\min(k_{\ta}, k_{\tb}) + 1 \leq k_{\ta} + 1`$ binomial coefficients from the set $`\{k_{\tb}, k_{\ta \tb} ,\ldots, k_{\ta^{|w|_\ta} \tb} \}`$ suffice. In total, we thus need the binomial coefficients of the set

\{ k_{\ta^j \tb} : \ta < \tb, j \in [|w|_{\ta}] \} \cup \{k_\tb : \tb \in \Sigma \backslash \{1\} \}.

There are $`\sum_{i \in [q]} |w|_i (q-i) + (q-1)`$ such coefficients. This quantity is less than or equal to $`\sum_{i \in [q]} |w|_i(q+1-i)`$ for every $`w`$ of length at least $`q-1`$.

We show the second claim about the bound by induction on $`q`$ where the binary case in Theorem [binarybound] serves as induction basis. This implies

\begin{align*}
\sum_{i\in[q]}|w|_i(q+1-i)
&=\left(\sum_{i\in[q-1]}|w|_i(q+1-i)\right)+|w|_q(q+1-q)\\
&=\left(\sum_{i\in[q-1]}|w|_i(q-i)\right)+\sum_{i\in[q-1]}|w|_i+|w|_q
\end{align*}

and therefore

\begin{align*}
\sum_{i\in[q]}|w|_i(q+1-i) &\leq \left(\sum_{i\in[\alpha]}\frac{1}{i}\frac{q^{\frac{i}{2}}-1}{q-1}\right)+n\\
&=\left(\sum_{i\in[\alpha]}\frac{1}{i}\frac{q(q^{\frac{i}{2}}-1)}{q(q-1)}\right)+n.
\end{align*}

On the other hand, we have to compare this quantity with [generalnumber] which can be rewritten as

\sum_{i\in[\alpha]}\frac{1}{i}\frac{(q+1)^{\frac{i}{2}}-1}{q}
=\sum_{i\in[\alpha]}\frac{1}{i}\frac{(q-1)((q+1)^{\frac{i}{2}}-1)}{q(q-1)}.

Thus the claim is proven, if the substraction of the latter one and the previous one is greater than zero, i.e., we show that

\begin{align}
\left(\sum_{i\in[\alpha]}\frac{1}{i} \frac{(q-1)((q+1)^{\frac{i}{2}}-1)-q(q^{\frac{i}{2}}-1)}{q(q-1)}\right)-n &>0,\mbox{ i.e.} \\
\left(\sum_{i\in[\alpha]}\frac{1}{i} 
\frac{(q-1)(q+1)^{\frac{i}{2}}-qq^{\frac{i}{2}}+1}{q(q-1)}\right)-n &> 0.\label{eqfinal}
\end{align}

With $`f(i)=\frac{(q-1)(q+1)^{\frac{i}{2}}-qq^{\frac{i}{2}}+1}{iq(q-1)}`$ for all $`i\in[\alpha]`$, the proof of ([eqfinal]) contains the following steps

1. For all $`i\geq 2`$ we have $`f(i)\geq 0`$,

2. $`f(5)+f(1)\geq 0`$,

3. $`f(\alpha)-n > 0`$.

Step [item1].: For $`i=2`$ we have

\begin{align*}
f(2)
=\frac{1}{2}\frac{(q-1)(q+1)-q^2+1}{q(q-1)}
=\frac{q^2-1-q^2+1}{2q(q-1)}
=0.
\end{align*}

For $`i=3`$ we have

\begin{align*}
f(3)
&=\frac{1}{3}\frac{(q-1)(q+1)\sqrt{q+1}-q^2\sqrt{q}+1}{q(q-1)}.
\end{align*}

Consider the function $`g:\R\rightarrow\R;q\mapsto q^4-2q^3-2q^2+q+1`$. This function has two minima (between $`-0.75`$ and $`-0.5`$ as well as between $`1.75`$ and $`2`$) and one maximum (between $`0.125`$ and $`0.25`$). Since $`g`$ has only two inflexion points and $`g`$ is strictly greater than zero at the first minima, $`g`$ has only two roots. The first root is between $`0.7`$ and $`0.8`$ and the second root is between $`2.5`$ and $`2.75`$. Thus for all $`q\geq 2.75`$ we have $`g(q)>0`$. This implies $`q^5+q^4-2q^3-2q^2+q+1>q^5`$. Hence equivalently we get $`(q+1)(q^4-2q^2+1)>q^5`$, i.e., $`(q+1)(q^2-1)^2>qq^4`$. This implies $`\sqrt{q+1}(q^2-1)>\sqrt{q}q^2`$ which proves that the numerator of $`f(3)`$ is positive and hence $`f(3)>0`$. Before we prove the claim for $`i\geq 4`$, we will prove that $`(q-1)(q+1)^j\geq q^{j+1}`$ for $`j\geq 2`$. Firstly we get

(q-1)(q+1)^j
=\left(\sum_{k\in [j]}\left (\binom{j}{k-1}-\binom{j}{k}\right) q^k\right) +q^{j+1}-1.

Due to the central symmetry of each row of the Pascal triangle and since the distribution of the binomial coefficient is unimodal, for $`k\le \lfloor j/2\rfloor`$, we have

\binom{j}{j-k}-\binom{j}{j-k+1}=-\left(\binom{j}{k-1}-\binom{j}{k}\right)>0

and thus

(q-1)(q+1)^j=\left(\sum_{k\in [\lfloor j/2\rfloor]} \left (\binom{j}{k}-\binom{j}{k-1}\right) (q^{j-k+1}-q^k)\right) +q^{j+1}-1.

Since $`k\le \lfloor j/2\rfloor`$, we have $`j-k+1>k`$ and each term of the above sum is thus positive. This shows that $`(q-1)(q+1)^j\ge q^{j+1}`$. This leads to the following estimations for $`f(i)`$. For $`i=2j`$ and $`j\geq 2`$ we get

\begin{align*}
f(i)
&= \frac{(q-1)(q+1)^j-qq^j+1}{iq(q-1)}\geq \frac{q^{j+1}-q^{j+1}+1}{iq(q-1)}>0.
\end{align*}

Finally for $`i=2j+1`$ and $`j\geq 2`$ we get

\begin{align*}
f(i)
&=\frac{(q-1)(q+1)^j\sqrt{q+1}-qq^j\sqrt{q}+1}{iq(q-1)}
&\geq \frac{q^{j+1}(\sqrt{q+1}-\sqrt{q})+1}{iq(q-1)}>0.
\end{align*}

Step [item2].: Notice that $`\alpha\geq 7`$ holds and thus $`f(5)`$ is always a summand. For $`f(5)+f(1)`$ we have to prove

\begin{align*}
\frac{(q-1)(q+1)^2\sqrt{q+1}-qq^2\sqrt{q}+1}{5q(q-1)} + \frac{(q-1)\sqrt{q+1}-q\sqrt{q}+1}{q(q-1)}\geq 0
\end{align*}

Thus we get for the numerator

\begin{align*}
&(q-1)(q+1)^2\sqrt{q+1}-qq^2\sqrt{q}+1+5(q-1)\sqrt{q+1}-5q\sqrt{q}+5\\
&= (q-1)\sqrt{q+1}((q+1)^2+5)-q\sqrt{q}(q^2+5)+6\\
&=(q-1)\sqrt{q+1}(q^2+2q+6)-q\sqrt{q}(q^2+5)+6\\
&=q^3\sqrt{q+1}+q^2\sqrt{q+1}+4q\sqrt{q+1}-6\sqrt{q+1}-q^3\sqrt{q}-5q\sqrt{q}+6.
\end{align*}

We have $`q^3 \sqrt{q+1} > q^3 \sqrt{q}`$ and, since $`q \geq 3`$,

q^2 \sqrt{q+1} \geq (6+q) \sqrt{q+1}.

Therefore $`q^2 \sqrt{q+1} + 4q \sqrt{q+1} \geq 6 \sqrt{q+1} + 5q \sqrt{q}`$ and the numerator is positive.

Step [item3].: Notice that for fixed $`i`$, $`f(i)`$ is monotonically increasing for increasing $`q`$. This implies

f(\alpha)\geq \frac{2\cdot 4^{\frac{\alpha}{2}}-3\cdot 3^{\frac{\alpha}{2}}+1}{6\alpha}
=\frac{2^{\alpha+1}-3^{\frac{\alpha}{2}+1}+1}{6\alpha}.

We are going to prove that

\begin{align}
\label{lastEquation}
2^{\alpha+1}-3^{\frac{\alpha}{2}+1}+1 > 6\alpha n.
\end{align}

Recall that $`\alpha`$ is a function of $`n`$, given by $`\alpha = \lfloor \frac{16}{7} \sqrt{n} \rfloor + 5`$.

First, we have

2^{\alpha+1}-3^{\frac{\alpha}{2}+1} > 2^{\alpha-1} - 2^{\frac{\alpha}{2}}.

Indeed, this inequality is equivalent to

\begin{align*}
2^{\frac{\alpha}{2}} \left(3 \cdot 2^{\frac{\alpha}{2}-1} + 1 \right) > 3^{\frac{\alpha}{2}+1} \quad \Leftrightarrow \quad
2^{\frac{\alpha}{2}-1} + \frac{1}{3} > \left(\frac{3}{2}\right)^{\frac{\alpha}{2}}.
\end{align*}

We can check that this last inequality is true by taking the logarithm of both sides, since $`\alpha > 5`$.

Therefore, it is sufficient for [lastEquation] to show that

2^{\alpha-1} - 2^{\frac{\alpha}{2}} = 2^{\frac{\alpha}{2}} (2^{\frac{\alpha}{2}-1} -1) > 6 \alpha n.

Note that $`2^{\frac{\alpha}{2}} > n`$ (indeed, $`\lfloor \frac{16}{7} \sqrt{n} \rfloor+5 > 2 \sqrt{n}+5`$, thus $`2^{\frac{\alpha}{2}} > 2^{\frac{5}{2}} \cdot 2^{ \sqrt{n}}`$). Once again, taking the logarithms, one can check that $`2^{\frac{5}{2}} \cdot 2^{ \sqrt{n}} > n`$ holds.

To verify [lastEquation], it remains to show that $`2^{\frac{\alpha}{2} - 1} -1 \geq 6 \alpha`$ or that $`2^{\frac{\alpha}{2} - 1} > 6 \alpha`$. Taking the logarithms, it is equivalent to

\begin{align*}
&\frac{\alpha}{2} - 1 > \log(6) + \log(\alpha) \\
\Leftrightarrow &\alpha - 2 \log(\alpha) > 2 \log(6) + 2,
\end{align*}

which is true for $`\alpha \geq 15`$, that is for $`n \geq 16`$. Equation [lastEquation] can be verified by a computer for $`q-1 \leq n < 16`$.

By [item1]., [item2]., and [item3]. Equation ([eqfinal]) is proven and this proves the claim.0◻ ◻