Menguji apakah surat dapat dijadwalkan untuk mencapai kata dalam bahasa biasa

Aku memperbaiki bahasa reguler $L$ pada alfabet $\Sigma$ , dan saya menganggap masalah berikut yang saya sebut penjadwalan surat untuk $L$ . Informal, input memberi saya $n$ huruf dan interval untuk setiap huruf (yaitu, posisi minimal dan maksimal), dan tujuan saya adalah menempatkan setiap huruf dalam intervalnya sehingga tidak ada dua huruf yang dipetakan ke posisi yang sama dan sehingga dihasilkan kata $n$ newsletter di$L$ . Secara formal:

• Input: $n$ tiga kali lipat $(a_i, l_i, r_i)$ mana $a_i \in \Sigma$ dan $1 \leq l_i \leq r_i \leq n$ adalah bilangan bulat
• Output: apakah ada bijih $f: \{1, \ldots, n\} \to \{1, \ldots, n\}$ sedemikian rupa sehingga $l_i \leq f(i) \leq r_i$ untuk semua $i$ , dan $a_{f^{-1}(1)} \cdots a_{f^{-1}(n)} \in L$ .

Jelas masalah ini adalah di NP, dengan menebak bijection $f$ dan memeriksa keanggotaan untuk $L$ di PTIME. Pertanyaan saya: Apakah ada bahasa $L$ yang umum sehingga masalah penjadwalan huruf untuk $L$ adalah NP-hard?

Beberapa pengamatan awal:

• Tampaknya masalah serupa telah dipelajari dalam penjadwalan: kita bisa melihat masalah sebagai penjadwalan tugas unit-biaya pada satu mesin sambil menghormati tanggal mulai dan selesai. Namun, masalah terakhir ini jelas di PTIME dengan pendekatan serakah, dan saya tidak melihat apa pun dalam literatur penjadwalan untuk kasus di mana tugas diberi label dan kami ingin mencapai kata dalam bahasa target biasa.
• Salah satu cara lain untuk melihat masalahnya adalah sebagai kasus khusus dari masalah pencocokan maksimum bipartit (antara huruf dan posisi), tetapi sekali lagi sulit untuk mengungkapkan kendala yang harus kita hadapi.$L$ .
• Dalam kasus tertentu di mana $L$ adalah bahasa dari bentuk $u^*$ untuk beberapa kata tetap $u$ (misalnya, $(ab)^*$ ), maka masalah surat penjadwalan untuk $L$ adalah PTIME dengan algoritma serakah mudah: membangun kata dari kiri ke kanan, dan letakkan di setiap posisi salah satu huruf yang tersedia yang benar relatif terhadap $L$ dan memiliki waktu terkecil $r_i$ . (Jika tidak ada huruf yang tersedia yang benar, gagal.) Namun, ini tidak digeneralisasikan ke bahasa biasa yang sewenang-wenang $L$ karena untuk bahasa-bahasa seperti itu kita mungkin memiliki pilihan jenis surat yang akan digunakan.
• Sepertinya algoritme dinamis seharusnya berfungsi, tetapi kenyataannya tidak sesederhana itu: tampaknya Anda perlu mengingat serangkaian huruf yang telah Anda ambil sejauh ini. Memang, ketika membangun sebuah kata dari kiri ke kanan, ketika Anda telah mencapai posisi $i$ , negara Anda tergantung pada huruf mana yang telah Anda konsumsi sejauh ini. Anda tidak dapat menghafal seluruh rangkaian karena dengan demikian akan ada banyak negara yang secara eksponensial. Tetapi tidak mudah untuk "meringkas" nya (misalnya, dengan berapa banyak salinan dari setiap huruf yang digunakan), karena untuk mengetahui salinan mana yang Anda gunakan, tampaknya Anda perlu mengingat kapan Anda telah mengkonsumsinya (yang kemudian Anda konsumsi mereka, semakin banyak surat yang tersedia). Bahkan dengan bahasa seperti $(ab|ba)^*$, mungkin sudah ada kendala rumit tentang kapan Anda harus memilih untuk mengambil $ab$ dan kapan Anda harus memilih untuk mengambil $ba$ tergantung pada huruf mana yang akan Anda perlukan nanti dan kapan surat-surat itu tersedia.
• Namun, karena bahasa reguler $L$ diperbaiki dan tidak dapat mengingat begitu banyak informasi, saya mengalami kesulitan menemukan masalah NP-hard yang dapat saya kurangi.

Masalahnya adalah NP-hard untuk L = A ∗ di mana A adalah bahasa yang terbatas yang mengandung kata-kata berikut:$L = A^*$$A$

• x 111 , x 000 ,$x111$$x000$
• y 100 , y 010 , y 001 ,$y100$$y010$$y001$
• 00 c 11 , 01 c 10 , 10 c 01 , dan 11 c $00c11$$01c10$$10c01$$11c00$

Pengurangan ini dari masalah Orientasi Grafik, yang dikenal sebagai NP-hard (lihat https://link.springer.com/article/10.1007/s00454-017-9884-9 ). Dalam masalah ini, kita diberi grafik tidak berarah 3-reguler di mana setiap titik diberi label " { 1 } " atau " { 0 , 3 } ". Tujuannya adalah untuk mengarahkan tepi grafik sehingga outdegree dari setiap vertex ada di set yang melabeli vertex tersebut.$\{1\}$$\{0,3\}$

Pengurangan perlu mengambil sebagai contoh instance Orientasi Grafik dan menghasilkan daftar tiga kali lipat sebagai output. Dalam pengurangan ini, tiga kali lipat yang kami hasilkan akan selalu memenuhi batasan tertentu. Kendala-kendala ini tercantum di bawah ini, dan kami akan merujuk pada daftar tiga kali lipat sebagai valid jika dan hanya jika mereka memenuhi batasan-batasan ini:

• Karakter x , y , dan c hanya diberi interval yang berisi tepat satu indeks. Dengan kata lain, setiap kali karakter ini ditempatkan, mereka ditempatkan di lokasi tertentu.$x$$y$$c$
• Untuk setiap triple ( i , l , r ) yang ada dalam instance dengan i ∈ { 0 , 1 } , triple ( 1 - i , l , r ) juga ada.$(i, l, r)$$i \in \{0,1\}$$(1-i, l, r)$
• Jika ( α , l , r ) dan ( α ′ , l ′ , r ′ ) keduanya ada tiga kali lipat dalam contoh maka baik l < l ′ ≤ r ′ < r , atau l ′ < l ≤ r < r ′ , atau { α , α ′ } = { 0 , 1 } dengan l = $(\alpha, l, r)$$(\alpha',l',r')$$l < l' \le r' < r$$l' < l \le r < r'$$\{\alpha,\alpha'\} = \{0,1\}$$l = l' < r = r'$.
• If $(\alpha, l, r)$ is a triple then the number of triples $(\alpha', l', r')$ with $l \le l' \le r' \le r$ is exactly $r-l+1$.

Note the following lemma, proved at the end of this post.

Lemma: untuk daftar tiga kali lipat yang valid, karakter x , y , dan c harus ditempatkan persis seperti yang ditunjukkan oleh tiga kali lipat, dan untuk setiap pasangan tiga kali lipat ( 0 , l , r ) dan ( 1 , l , r ) , dua karakter untuk triple harus ditempatkan pada indeks l dan r .$x$$y$$c$$(0, l, r)$$(1, l, r)$$l$$r$

Maka ide pengurangannya adalah sebagai berikut.

We use pairs of triples $(0, l, r)$, and $(1, l, r)$ to represent edges. The edge goes between endpoints at index $l$ and at index $r$. Assuming we produce a valid list of triples, the characters from these two triples must be placed at $l$ and $r$, so we can treat the order in which they are placed as indicating the direction of the edge. Here $1$ is the "head" of the edge and $0$ is the "tail". In other words, if the $1$ is placed at $r$ then the edge points from $l$ to $r$ and if the $1$ is placed at $l$ then the edge points from $r$ to $l$.

To represent vertices, we place an $x$ or $y$ character at an index and use the next three characters as the endpoints of the three edges which touch the vertex. Note that if we place an $x$, all three edges at the vertex must point in the same direction (all into the vertex or all out of the vertex) simply due to the strings that are in finite language $A$. Such vertices have outdegree $0$ or $3$, so we place an $x$ exactly for the vertices labeled $\{0,3\}$. If we place a $y$, exactly one of the three edges at the vertex must point in the same direction due to the strings in $A$. Such vertices have outdegree $1$, so we place a $y$ exactly for the vertices labeled $\{1\}$.

In some sense, we are done. In particular, the correspondence between solving this instance and solving the Graph Orientation instance should be clear. Unfortunately, the list of triples we produce may not be valid, and so the "edges" described may not work as intended. In particular, the list of triples might not be valid because the condition that the intervals from the triples must always contain each other might not hold: the intervals from two edges may overlap without one containing the other.

To combat this, we add some more infrastructure. In particular, we add "crossover vertices". A crossover vertex is a vertex of degree $4$ whose edges are paired such that within each pair one edge must point into the crossover vertex and one out. In other words, a crossover vertex will behave the same as just two "crossing" edges. We represent a crossover vertex by placing the character $c$ at some index $i$. Then note that the language $A$ constrains the characters at $i-1$ and $i+2$ to be opposite (one $0$ and one $1$) and the characters at $i-2$ and $i+1$ to be opposite. Thus, if we use these indices as the endpoints for the four edges at the crossover vertex, the behavior is exactly as described: the four edges are in pairs and among every pair one points in and one points out.

How do we actually place these crossovers? Well suppose we have two intervals $(l, r)$ and $(l', r')$ which overlap. WLOG, $l < l' < r < r'$. We add the crossover character into the middle (between $l'$ and $r$). (Let's say that all along we spaced everything out so far that there's always enough space, and at the end we will remove any unused space.) Let the index of the crossover character be $i$. Then we replace the four triples $(0, l, r)$, $(1, l, r)$, $(0, l', r')$, and $(1, l', r')$ with eight triples with two each (one with character $0$ and one with character $1$) for the following four intervals $(l, i-1)$, $(i+2, r)$, $(l', i-2)$, $(i+1, r')$. Notice that the intervals don't overlap in the bad way anymore! (After this change, if two intervals overlap, one is strictly inside the other.) Furthermore, the edge from $l$ to $r$ is replaced by an edge from $l$ to the crossover vertex followed by the edge from there to $r$; these two edges are paired at the crossover vertex in such a way that one is pointed in and one is pointed out; in other words, the two edges together behave just like the one edge they are replacing.

In some sense, putting in this crossover vertex "uncrossed" two edges (whose intervals were overlapping). It is easy to see that adding the crossover vertex can't cause any additional edges to become crossed. Thus, we can uncross every pair of crossing edges by inserting enough crossover vertices. The end result still corresponds to the Graph Orientation instance, but now the list of triples is valid (the properties are all easy to verify now that we have "uncrossed" any crossing edges), so the lemma applies, the edges must behave as described, and the correspondence is actually an equivalence. In other words, this reduction is correct.

proof of lemma

Lemma: for a valid list of triples, the characters $x$, $y$, and $c$ must be placed exactly as indicated by the triples, and for any pair of triples $(0, l, r)$ and $(1, l, r)$, the two characters for that triple must be placed at indices $l$ and $r$.

proof:

We proceed by induction on the triples by interval length. In particular, our statement is the following: for any $k$ if some triple has interval length $k$ then the character in that triple must be placed as described in the lemma.

Base case: for $k = 0$, the triple must be placing a character $x$, $y$, or $c$ at the single index inside the interval. This is exactly as described in the lemma.

Inductive case: assume the statement holds for any $k$ less than some $k'$. Now consider some triple with interval length $k'$. Then that triple must be of the form $(i, l, r)$ with $r = l+k'-1$ and $i \in \{0,1\}$. The triple $(1-i, l, r)$ must also be present. The number of triples $(\alpha',l',r')$ with $l \le l'\le r' \le r$ is exactly $r-l+1 = k'$. These triples include triples $(0, l, r)$ and $(1, l, r)$ but also $k'-2$ other triples of the form $(\alpha',l',r')$ with $l < l' \le r' < r$. These other triples all have interval length smaller than $k'$, so they all must place their characters as specified in the lemma. The only way for this to occur is if these triples place characters in every index starting at index $l+1$ and ending at index $r+1$. Thus, our two triples $(0, l, r)$ and $(1, l, r)$ must place their characters at indices $l$ and $r$, as described in the lemma, concluding the inductive case.

By induction, the lemma is correct.

@MikhailRudoy was the first to show NP-hardness, but Louis and I had a different idea, which I thought I could outline here since it works somewhat differently. We reduce directly from CNF-SAT, the Boolean satisfiability problem for CNFs. In exchange for this, the regular language $L$ that we use is more complicated.

The key to show hardness is to design a language $L'$ that allows us to guess a word and repeat it multiple times. Specifically, for any number $k$ of variables and number $m$ of clauses, we will build intervals that ensure that all words $w$ of $L'$ that we can form must start with an arbitrary word $u$ of length $k$ on alphabet $\{0, 1\}$ (intuitively encoding a guess of the valuation of variables), and then this word $u$ is repeated $m$ times (which we will later use to test that each clause is satisfied by the guessed valuation).

To achieve this, we will fix the alphabet $A = \{0, 1, \#, 0', 1'\}$ and the language: $L' := (0|1)^* (\# (00'|11')^*)^* \# (0|1)^*$. The formal claim is a bit more complicated:

Claim: For any numbers $k, m \in \mathbb{N}$, we can build in PTIME a set of intervals such that the words in $L'$ that can be formed with these intervals are precisely:

where $\tilde{u}$ denotes the result of reversing the order of $u$ and swapping $0$'s and $1$'s, where $u'$ denotes the result of adding a prime to all letters in $u$, and where $x \bowtie y$ for two words $x$ of $y$ of length $p$ is the word of length $2p$ formed by taking alternatively one letter from $x$ and one letter from $y$.

Here's an intuitive explanation of the construction that we use to prove this. We start with intervals that encode the initial guess of $u$. Here is the gadget for $n = 4$ (left), and a possible solution (right):

It's easy to show the following observation (ignoring $L'$ for now): the possible words that we can form with these intervals are exactly $u \# \tilde{u}$ for $u \in \{0, 1\}^k$. This is shown essentially like the Lemma in @MikhailRudoy's answer, by induction from the shortest intervals to the longest ones: the center position must contain $\#$, the two neighboring positions must contain one $0$ and one $1$, etc.

We have seen how to make a guess, now let's see how to duplicate it. For this, we will rely on $L'$, and add more intervals. Here's an illustration for $k = 3$:

For now take $L := (0|1)^* (\# (00'|11')^*)^* \# (0' | 1')^*$. Observe how, past the first $\#$, we must enumerate alternatively an unprimed and a primed letter. So, on the un-dashed triangle of intervals, our observation above still stands: even though it seems like these intervals have more space to the right of the first $\#$, only one position out of two can be used. The same claim holds for the dashed intervals. Now, $L$ further enforces that, when we enumerate an unprimed letter, the primed letter that follows must be the same. So it is easy to see that the possible words are exactly: $u \# (\tilde{u} \bowtie \tilde{u}') \# u'$ for $u \in \{0, 1\}^k$.

Now, to show the claim, we simply repeat this construction $m$ times. Here's an example for $k=3$ and $m=2$, using now the real definition of $L'$ above the statement of the claim:

As before, we could show (by induction on $m$) that the possible words are exactly the following: $u (\# \tilde{u} \bowtie \tilde{u}' \# u \bowtie u')^2 \# \tilde{u}$ for $u \in \{0, 1\}^k$. So this construction achieves what was promised by the claim.

Thanks to the claim we know that we can encode a guess of a valuation for the variables, and repeat the valuation multiple times. The only missing thing is to explain how to check that the valuation satisfies the formula. We will do this by checking one clause per occurrence of $u$. To do this, we observe that without loss of generality we can assume that each letter of the word is annotated by some symbol provided as input. (More formally: we could assume that in the problem we also provide as input a word $w$ of length $n$, and we ask whether the intervals can form a word $u$ such that $w \bowtie u$ is in $L$.) The reason why we can assume this is because we can double the size of each interval, and add unit intervals (at the bottom of the picture) at odd positions to carry the annotation of the corresponding even position:

Thanks to this observation, to check clauses, we will define our regular language $L$ to be the intersection of two languages. The first language enforces that the sub-word on even positions is a word in $L'$, i.e., if we ignore the annotations then the word must be in $L'$, so we can just use the construction of the claim and add some annotations. The second language $L''$ will check that the clauses are satisfied. To do this, we will add three letters in our alphabet, to be used as annotations: $+$, $-$, and $\epsilon$. At clause $1 \leq i \leq m$, we add unit intervals to annotate by $+$ the positions in the $i$-th repetition of $u$ corresponding to variables occurring positively in clause $i$, and annotate by~$-$ the positions corresponding to negatively occurring variables. We annotate everything else by~$\epsilon$. It is now clear that $L''$ can check that the guessed valuation satisfies the formula, by verifying that, between each pair of consecutive $\#$ symbols that contain an occurrence of $u$ (i.e., one pair out of two), there is some literal that satisfies the clause, i.e., there must be one occurrence of the subword $+1$ or of the subword $-0$.

This concludes the reduction from CNF-SAT and shows NP-hardness of the letter scheduling problem for the language $L$.