Constructing Suffixient Arrays Revisited

Recently, Navarro et al. [13] introduced an online construction algorithm that computes a minimum-size suffixient set – the set of indices forming a suffixient array – directly from the text. Their approach adapts Ukkonen’s algorithm [15], originally designed for the online construction of the suffix tree. Specifically, their method incrementally maintains a minimum-size suffixient set for the prefix $T[1..i]$ as the text is scanned left to right.

The algorithm runs in linear time over an alphabet of size $\sigma$ under the trans-dichotomous RAM model [8], but requires $𝒪(n\log \sigma )$ time in the standard RAM model [2]. Its working space is comparable to that of a suffix tree built over the text, which is known to have a large memory footprint in practice – up to 20 bytes per input character in the worst case [10]. Navarro et al. also demonstrated that a minimum-size suffixient set can be constructed incrementally while scanning the text from right to left.

Cenzato et al. [5] addressed the problem of constructing a minimum-size suffixient set given the suffix array $\mathrm{𝚂𝙰}[1..n]$, the Burrows–Wheeler transform (BWT) [4] array $\mathrm{𝙱𝚆𝚃}[1..n]$, and the longest common prefix array $\mathrm{𝙻𝙲𝙿}[1..n]$ of the reversed text. They presented two linear-time algorithms that require multiple passes over these arrays, resulting in a working space of $\mathrm{\Omega }(n)$ words. They also proposed a construction algorithm that avoids storing the full $\mathrm{𝚂𝙰}$, $\mathrm{𝙱𝚆𝚃}$, and $\mathrm{𝙻𝙲𝙿}$ arrays and processes them in a single pass. This algorithm runs in $𝒪(n+\overline{r}\log \sigma )$ time, where $\overline{r}$ denotes the number of runs in the BWT, and uses only $𝒪(\sigma )$ words of working space. When computing a suffixient array, the running time remains unchanged, while the working space increases from $𝒪(\sigma )$ to $𝒪(\sigma +\chi )\subseteq 𝒪(\chi )$ words, where $\chi$ denotes the size of the suffixient array and $\sigma \le \chi$ [5].

Using the technique of prefix-free parsing (PFP) [3, 9, 14], the entries of the $\mathrm{𝚂𝙰}$, $\mathrm{𝙱𝚆𝚃}$, and $\mathrm{𝙻𝙲𝙿}$ arrays can be generated in a left-to-right streaming fashion while using compressed space. By combining their one-pass algorithm with PFP, Cenzato et al. [5] were able to construct a suffixient array in compressed space.

In this paper, we revisit the problem of constructing a suffixient array in the standard RAM model, focusing on the specific setting introduced by Cenzato et al. [5]. Our main contribution is a linear-time streaming algorithm that constructs a suffixient array in $n$ iterations by processing the $\mathrm{𝚂𝙰}$, $\mathrm{𝙻𝙲𝙿}$, and $\mathrm{𝙱𝚆𝚃}$ arrays of the reversed text $T[1..n]$ in a single pass with one-step look-ahead. The efficiency of the algorithm relies on two key mechanisms.

Dynamic Candidate Management. At each iteration $i$, we maintain a set of candidate positions $j_{\text{text}}$ for the suffixient array in a doubly-linked list ($\mathrm{𝚛𝚘𝚠𝙻𝚒𝚜𝚝}$). The list has size at most $\sigma$ and is kept sorted in non-increasing order by the weight $𝚠(j)$. Here, $𝚠(j)$ denotes the maximum length of a right-maximal substring that is a suffix of $T[1..j_{\text{text}}-1]$, where $j={\mathrm{𝚂𝙰}}^{-1}[n-j_{\text{text}}+1]$. The array ${\mathrm{𝚂𝙰}}^{-1}$ denotes the inverse suffix array, defined by ${\mathrm{𝚂𝙰}}^{-1}[\mathrm{𝚂𝙰}[k]]=k$.

To decide at iteration $i$ whether the new position $i_{\text{text}}=n-\mathrm{𝚂𝙰}[i]+1$ should replace or join the existing candidates, we use an auxiliary array $\mathrm{𝚙𝚛𝚎𝚟𝚆}[1..\sigma ]$ that records, for each character, the most recent weighted occurrence (its $\mathrm{𝑖𝑛𝑑𝑒𝑥}$) together with the corresponding $\mathrm{𝑤𝑒𝑖𝑔ℎ𝑡}$. In addition, we employ a monotone stack to compute in amortized $𝒪(1)$ time $𝚋(i)$, the largest index such that . By comparing $𝚠(i)$ with $\mathrm{𝙻𝙲𝙿}[i]$, $\mathrm{𝚙𝚛𝚎𝚟𝚆}[c].\mathrm{𝑖𝑛𝑑𝑒𝑥}$ with $𝚋(i)$, and $\mathrm{𝚙𝚛𝚎𝚟𝚆}[c].\mathrm{𝑤𝑒𝑖𝑔ℎ𝑡}$ with $𝚠(i)$ (where $c=\mathrm{𝙱𝚆𝚃}[i]$), we can determine whether the position $i_{\text{text}}$ is a superior candidate to those currently stored. This strategy guarantees that, at any time, we retain only the most relevant candidate for each distinct character.  $\mathrm{⌟}$

The Ejection Mechanism. As the algorithm streams through the arrays, the current $\mathrm{𝙻𝙲𝙿}$ value, $\mathrm{𝙻𝙲𝙿}[i]$, acts as a threshold. When $\mathrm{𝙻𝙲𝙿}[i]$ drops below the weight of a candidate in $\mathrm{𝚛𝚘𝚠𝙻𝚒𝚜𝚝}$, this indicates that the corresponding right-maximal extension can no longer be extended. The candidate position is then ejected from the doubly-linked list and appended to the output list, which stores a sublist of the suffixient array. Since $\mathrm{𝚛𝚘𝚠𝙻𝚒𝚜𝚝}$ is sorted, each ejection takes constant time, resulting in overall linear running time.  $\mathrm{⌟}$

Our algorithm is asymptotically faster than the one-pass algorithm of Cenzato et al. [5]. Its working space is bounded by $𝒪(h+\chi )$, in addition to the $\mathrm{𝚂𝙰}$, $\mathrm{𝙻𝙲𝙿}$, and $\mathrm{𝙱𝚆𝚃}$ arrays, while computing a minimum-size suffixient set requires only $𝒪(h+\sigma )$ words of working space. Here, $h$ denotes the height of the suffix tree built over the reversed text, that is, the maximum number of branching nodes along any root-to-leaf path. Since the algorithm processes these arrays in a one-pass streaming fashion, it naturally integrates with prefix-free parsing [3].

Let $x\in [1..n]$ be an index. We say that $x$ is a start-run boundary if $x>1$ and $\mathrm{𝙱𝚆𝚃}[x]\ne \mathrm{𝙱𝚆𝚃}[x-1]$, and an end-run boundary if and $\mathrm{𝙱𝚆𝚃}[x]\ne \mathrm{𝙱𝚆𝚃}[x+1]$. Any index that is either a start- or end-run boundary is called a run boundary.

To every index $x\in [1..n]$ we assign a weight $𝚠(x)$, defined as the maximum length of a right-maximal substring that is a suffix of $T[1..n-\mathrm{𝚂𝙰}[x]]$, if $x$ is a run-boundary, and $-1$, otherwise. Note that if $x$ is a run-boundary, then a right-maximal substring ending at position $n-\mathrm{𝚂𝙰}[x]$ surely exists, hence $𝚠(x)$ is well-defined.

The value $𝚠(x)$ can then be computed as follows: if $x$ is not a run boundary, then we set $𝚠(x):=-1$, as per the definition. If $x$ is only a start-run (resp., only an end-run) boundary, then $𝚠(x)=\mathrm{𝙻𝙲𝙿}[x]$ (resp., $𝚠(x)=\mathrm{𝙻𝙲𝙿}[x+1]$). If $x$ is both a start- and end-run boundary, then $𝚠(x)=\max \{\mathrm{𝙻𝙲𝙿}[x],\mathrm{𝙻𝙲𝙿}[x+1]\}$. Each value can be computed in $𝒪(1)$ time using $𝒪(1)$ working space. The pseudocode is deferred to Appendix A.

For any $x\in [1..n]$, define $𝚋(x)$ as the largest index for which . If no such index exists, set $𝚋(x):=1$. Define $𝚎(x)$ as the smallest index $𝚎(x)>x$ such that . If no such index exists, set $𝚎(x):=n+1$. The value $𝚎(x)$ is only used conceptually throughout this paper, so we only show an algorithm that computes $𝚋(x)$.

In our setting, the entries in $\mathrm{𝙻𝙲𝙿}[1..n]$ are enumerated sequentially from $\mathrm{𝙻𝙲𝙿}[1],\mathrm{𝙻𝙲𝙿}[2],\mathrm{\dots }$ in a stream; immediately after reading the entry $\mathrm{𝙻𝙲𝙿}[i]$ for each $i>1$, our goal is to compute the index $𝚋(i)$ in amortized constant time.

Our data structure is a regular stack that stores tuples of form $(index,val,b\mathrm{\_}val)$, where $index\in [1..n]$, $val=\mathrm{𝙻𝙲𝙿}[index]$, and $b\mathrm{\_}val=𝚋(index)$. We call this stack monotone stack as the tuples in the stack are always sorted decreasingly by the $val$ entries from top to the bottom. Initially, we create an empty stack $S$ and push the triple $(1,-1,-1)$ onto $S$.

When $\mathrm{𝙻𝙲𝙿}[i]$ is available, we first check if or not the tuple at the top of $S$ has $index=i$. If so, then we return the $b\mathrm{\_}val$ value of the top tuple; this means that $𝚋(i)$ has already been computed before (note that in our setting, when $\mathrm{𝙻𝙲𝙿}[i]$ is available, we might call $𝚋(i)$ more than once). Otherwise, we pop every tuple in the stack with $val\ge \mathrm{𝙻𝙲𝙿}[i]$ out of $S$, until we find the tuple $t$ with . Then, we return $t.index$ right after we push the tuple $(i,\mathrm{𝙻𝙲𝙿}[i],t.index)$ onto the top of $S$. The pseudocode can be found in Appendix B.

Since the tuple associated with each $\mathrm{𝙻𝙲𝙿}[i]$ is pushed onto $S$ exactly once and popped from $S$ at most once, the overall running time is $𝒪(n)$. Consequently, $𝚋(i)$ can be computed in amortized constant time. The correctness is standard and omitted; a similar idea has appeared in [7, Theorem 2] and [1, Section 4] for a related purpose.

Let ${|S|}_i$ denote the size of the stack immediately after processing $\mathrm{𝙻𝙲𝙿}[i]$. By Proposition 4 (below), is upper bounded by the height of the suffix tree [16] built over $T^{rev}[1..n]$, plus one. In particular, if $T[1..n]=a^n$, then $|S|\subseteq \mathrm{\Omega }(n)$ , and if the text is uniform, then $|S|\subseteq 𝒪(\log n)$ with high probability [6]. However, as noted in [1, Section 5.1], the stack size in practice is much smaller.

Let $Trie$ denote the suffix tree constructed over $T^{rev}[1..n]$. For each $i$, let $A_i$ denote the list of all index entries, excluding $1$, of the triples stored in the monotone stack $S$ (in increasing order) upon completion of the execution of $𝚋(i)$. Thus, $|A_i|={|S|}_i-1$.

Consider any $j^{\prime },j^{\prime \prime }\in A_i$ with , and any $s\in (j^{\prime }..j^{\prime \prime }]$. Following the algorithm, we have ; otherwise, the triple indexed by $j^{\prime }$ would have been popped from the stack. Consequently, $T^{rev}[\mathrm{𝚂𝙰}[j^{\prime }]..\mathrm{𝚂𝙰}[j^{\prime }]+\mathrm{𝙻𝙲𝙿}[j^{\prime }]-1]$ is a prefix of $T^{rev}[\mathrm{𝚂𝙰}[j^{\prime \prime }]..\mathrm{𝚂𝙰}[j^{\prime \prime }]+\mathrm{𝙻𝙲𝙿}[j^{\prime \prime }]-1]$.

For each $j\in A_i$, let $u_j$ denote the node in $Trie$ that is the lowest common ancestor of the $\mathrm{𝚂𝙰}[j-1]$-leaf and the $\mathrm{𝚂𝙰}[j]$-leaf. Observe that $u_j$ is the locus of the string $T^{rev}[\mathrm{𝚂𝙰}[j]..\mathrm{𝚂𝙰}[j]+\mathrm{𝙻𝙲𝙿}[j]-1]$. Hence, $u_{j^{\prime }}$ is an ancestor of $u_{j^{\prime \prime }}$, and the nodes $\{u_j\mid j\in A_i\}$ all lie on a single root-to-leaf path in $Trie$.

It follows that $|A_i|\le h$. Since $|A_i|={|S|}_i-1$, we conclude that ${|S|}_i\le h+1$. $◀$

Since $\mathrm{𝙱𝚆𝚃}[x]\ne \mathrm{𝙱𝚆𝚃}[x-1]$ and $c\in \{\mathrm{𝙱𝚆𝚃}[x],\mathrm{𝙱𝚆𝚃}[x-1]\}$, Definition 5 implies that ${\mathrm{\ell }}_{x,c}={\mathrm{𝙲𝚊𝚗𝚍𝚝}}_c(𝚋(x),𝚎(x))>-1$ and that .

Let $\mathrm{\ell }\in \{x,x-1\}$ be the index with $\mathrm{𝙱𝚆𝚃}[\mathrm{\ell }]=c$. By the definition of $𝚠(\cdot )$, it follows that $𝚠(\mathrm{\ell })\ge \mathrm{𝙻𝙲𝙿}[x]$. Moreover, since $\mathrm{\ell }\in [𝚋(x),𝚎(x))$ by the definitions of $𝚋(\cdot )$ and $𝚎(\cdot )$, and $\mathrm{𝙱𝚆𝚃}[\mathrm{\ell }]=c$, the maximality of $𝚠({\mathrm{\ell }}_{x,c})$ implies $𝚠({\mathrm{\ell }}_{x,c})\ge 𝚠(\mathrm{\ell })\ge \mathrm{𝙻𝙲𝙿}[x]$. $◀$

Our goal is to show that $\mathrm{𝙵𝚞𝚕𝚕𝙻}$, defined in Definition 7, is a minimum-size suffixient set, thereby providing a new characterization of such sets in terms of $𝚠(\cdot )$, $𝚋(\cdot )$, and $𝚎(\cdot )$. To this end, one possible approach is to establish a bijection between $\mathrm{𝙵𝚞𝚕𝚕𝙻}$ and the set of all super-maximal extensions, as defined in [5, Definition 32]. This approach implies, as demonstrated in [5, Section 5], that $\mathrm{𝙵𝚞𝚕𝚕𝙻}$ is a minimum-size suffixient set. Instead, we present a new proof that does not rely on the notion of super-maximal extensions.

We first prove that $\mathrm{𝙵𝚞𝚕𝚕𝙻}$ is suffixient for the text $T[1..n]$, applying Definition 2.

By Definition 2, it suffices to show the following: for any right-maximal substring $str$ of $T[1..n]$ and for any one-character right-maximal extension $str\cdot c$ of $str$, there exists a position $x_{\text{text}}\in \mathrm{𝙵𝚞𝚕𝚕𝙻}$ such that $str\cdot c$ is a suffix of $T[1..x_{\text{text}}]$.

Since $str$ is right-maximal in $T[1..n]$, there exists another character $a\ne c$ such that $str\cdot a$ also occurs as a substring of $T[1..n]$. Consequently, there exists at least one interval $[\mathrm{\ell },z]$ such that both characters $a$ and $c$ occur in $\mathrm{𝙱𝚆𝚃}[\mathrm{\ell }..z]$, and the prefix $T[1..n-\mathrm{𝚂𝙰}[i]]$ has $str$ as a suffix for every index $i\in [\mathrm{\ell },z]$.

Let $[\mathrm{\ell },z]$ be the smallest such interval. Without loss of generality, assume that $\mathrm{𝙱𝚆𝚃}[\mathrm{\ell }]\ne c$ and $\mathrm{𝙱𝚆𝚃}[z]=c$; the case $\mathrm{𝙱𝚆𝚃}[\mathrm{\ell }]=c$ and $\mathrm{𝙱𝚆𝚃}[z]\ne c$ is symmetric. By the minimality of the interval $[\mathrm{\ell },z]$, we have $\mathrm{𝙱𝚆𝚃}[z]\ne \mathrm{𝙱𝚆𝚃}[z-1]$. Let $q:={\mathrm{𝙲𝚊𝚗𝚍𝚝}}_c(𝚋(z),𝚎(z))$. Since $\mathrm{𝙱𝚆𝚃}[z]=c=\mathrm{𝙱𝚆𝚃}[q]$ and $\mathrm{𝙱𝚆𝚃}[z]\ne \mathrm{𝙱𝚆𝚃}[z-1]$, it follows that $𝚠(q)\ge 𝚠(z)\ge \mathrm{𝙻𝙲𝙿}[z]\ge |str|$; moreover, as $T[n-\mathrm{𝚂𝙰}[q]+1]=\mathrm{𝙱𝚆𝚃}[q]=c$, we have $str\cdot c$ is a suffix of $T[1..n-\mathrm{𝚂𝙰}[q]+1]$. Setting $x_{\text{text}}:=n-\mathrm{𝚂𝙰}[q]+1$, we conclude that $x_{\text{text}}\in \mathrm{𝙵𝚞𝚕𝚕𝙻}$, completing the proof. $◀$

By Definition 5, we have ${\mathrm{\ell }}_{x,c}>-1$, $𝚠({\mathrm{\ell }}_{x,c})>-1$, ${\mathrm{\ell }}_{x^{\prime },c^{\prime }}>-1$, and $𝚠({\mathrm{\ell }}_{x^{\prime },c^{\prime }})>-1$.

Any index $\mathrm{\ell }$ with $𝚠(\mathrm{\ell })\ge 0$ induces the right-maximal substring $T[n-\mathrm{𝚂𝙰}[\mathrm{\ell }]+1-𝚠(\mathrm{\ell }),n-\mathrm{𝚂𝙰}[\mathrm{\ell }]]$, which cannot contain the symbol $\$$. Recall that the suffix array $\mathrm{𝚂𝙰}$ is constructed over $T^{rev}[1..n]$. Therefore, the reverse of a right-maximal substring induces an interval in the suffix array, called its SA-interval [11, Section 2.2.2]. Formally, for any index $\mathrm{\ell }$ with $𝚠(\mathrm{\ell })\ge 0$, let $[s(\mathrm{\ell }),t(\mathrm{\ell })]$ denote the SA-interval of the string $T[n-\mathrm{𝚂𝙰}[\mathrm{\ell }]+1-𝚠(\mathrm{\ell })..n-\mathrm{𝚂𝙰}[\mathrm{\ell }]]$, where $s(\mathrm{\ell })$ is the minimum index $i\in [1..n]$ such that the string is a suffix of $T[1..n-\mathrm{𝚂𝙰}[i]]$, and $t(\mathrm{\ell })$ is defined analogously as the maximum such index.

It suffices to show that $s({\mathrm{\ell }}_{x,c})\ge 𝚋(x)$ and . By the definition of SA-intervals, and .

Assume for contradiction that . Since $𝚋(x)\le {\mathrm{\ell }}_{x,c}$, this implies $\mathrm{𝙻𝙲𝙿}[𝚋(x)]\ge 𝚠({\mathrm{\ell }}_{x,c})$; otherwise, $s({\mathrm{\ell }}_{x,c})\ge 𝚋(x)$ by definition. By the definition of $𝚋(\cdot )$, , yielding $\mathrm{𝙻𝙲𝙿}[x]>𝚠({\mathrm{\ell }}_{x,c})$, which contradicts Observation 6.

A symmetric argument shows that . If $𝚎(x)\le t({\mathrm{\ell }}_{x,c})$, then since we obtain $\mathrm{𝙻𝙲𝙿}[𝚎(x)]\ge 𝚠({\mathrm{\ell }}_{x,c})$, which together with $\mathrm{𝙻𝙲𝙿}[x]>\mathrm{𝙻𝙲𝙿}[𝚎(x)]$ again contradicts Observation 6. This proves the claim. $\mathrm{⊲}$

We now prove the lemma by contradiction. Assume that $T[n-\mathrm{𝚂𝙰}[{\mathrm{\ell }}_{x,c}]+1-𝚠({\mathrm{\ell }}_{x,c})..n-\mathrm{𝚂𝙰}[{\mathrm{\ell }}_{x,c}]+1]$ is a suffix of $T[n-\mathrm{𝚂𝙰}[{\mathrm{\ell }}_{x^{\prime },c^{\prime }}]+1-𝚠({\mathrm{\ell }}_{x^{\prime },c^{\prime }})..n-\mathrm{𝚂𝙰}[{\mathrm{\ell }}_{x^{\prime },c^{\prime }}]+1]$, which implies $c=c^{\prime }$. Consequently, $T[n-\mathrm{𝚂𝙰}[{\mathrm{\ell }}_{x,c}]+1-𝚠({\mathrm{\ell }}_{x,c})..n-\mathrm{𝚂𝙰}[{\mathrm{\ell }}_{x,c}]]$ is a suffix of $T[n-\mathrm{𝚂𝙰}[{\mathrm{\ell }}_{x^{\prime },c^{\prime }}]+1-𝚠({\mathrm{\ell }}_{x^{\prime },c^{\prime }})..n-\mathrm{𝚂𝙰}[{\mathrm{\ell }}_{x^{\prime },c^{\prime }}]]$, and hence ${\mathrm{\ell }}_{x^{\prime },c^{\prime }}\in [s({\mathrm{\ell }}_{x,c}),t({\mathrm{\ell }}_{x,c})]$.

If , then by Claim 10 we have ${\mathrm{\ell }}_{x^{\prime },c^{\prime }}\in [𝚋(x),𝚎(x))$, which implies $𝚠({\mathrm{𝙲𝚊𝚗𝚍𝚝}}_c(𝚋(x),𝚎(x)))>𝚠({\mathrm{\ell }}_{x,c})$, contradicting the definition of ${\mathrm{\ell }}_{x,c}$. If instead $𝚠({\mathrm{\ell }}_{x,c})=𝚠({\mathrm{\ell }}_{x^{\prime },c^{\prime }})$, then the two substrings are equal, so $s({\mathrm{\ell }}_{x,c})=s({\mathrm{\ell }}_{x^{\prime },c^{\prime }})$ and $t({\mathrm{\ell }}_{x,c})=t({\mathrm{\ell }}_{x^{\prime },c^{\prime }})$. Assuming without loss of generality that , we obtain , which contradicts the choice of ${\mathrm{\ell }}_{x^{\prime },c^{\prime }}$ as ${\mathrm{𝙲𝚊𝚗𝚍𝚝}}_{c^{\prime }}(𝚋(x^{\prime }),𝚎(x^{\prime }))$. In both cases, we reach a contradiction, completing the proof. $◀$

Finally, we show that $\mathrm{𝙵𝚞𝚕𝚕𝙻}$ has minimum possible size, applying the pigeonhole principle.

Let $F$ be an arbitrary suffixient set. Our goal is to show that $|F|\ge |\mathrm{𝙵𝚞𝚕𝚕𝙻}|$.

Suppose, for the sake of contradiction, that . Recall that each $x_{\text{text}}\in \mathrm{𝙵𝚞𝚕𝚕𝙻}$ corresponds to a right-maximal extension $T[x_{\text{text}}-𝚠({\mathrm{𝚂𝙰}}^{-1}[n-x_{\text{text}}+1])..x_{\text{text}}]$. By the definition of suffixient sets, the string $T[x_{\text{text}}-𝚠({\mathrm{𝚂𝙰}}^{-1}[n-x_{\text{text}}+1])..x_{\text{text}}]$ must be a suffix of $T[1..f_{\text{text}}]$ for some $f_{\text{text}}\in F$.

Since , the pigeonhole principle implies that there exist two distinct indices $x_{\text{text}},y_{\text{text}}\in \mathrm{𝙵𝚞𝚕𝚕𝙻}$ such that both $T[x_{\text{text}}-𝚠({\mathrm{𝚂𝙰}}^{-1}[n-x_{\text{text}}+1])..x_{\text{text}}]$ and $T[y_{\text{text}}-𝚠({\mathrm{𝚂𝙰}}^{-1}[n-y_{\text{text}}+1])..y_{\text{text}}]$ are suffixes of the same prefix $T[1..f_{\text{text}}]$ for some $f_{\text{text}}\in F$. Consequently, one of these two strings must be a suffix of the other.

However, by Lemma 9, no such pair $x_{\text{text}}$ and $y_{\text{text}}$ can exist. This contradiction shows that our assumption was false, and therefore $|F|\ge |\mathrm{𝙵𝚞𝚕𝚕𝙻}|$. $◀$

Lemmas 11 and 8 together imply that $\mathrm{𝙵𝚞𝚕𝚕𝙻}$ is a minimum-size suffixient set.