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+arXiv:2301.13456v1  [cs.FL]  31 Jan 2023
+ONE DETERMINISTIC-COUNTER AUTOMATA
+PRINCE MATHEW, VINCENT PENELLE, PRAKASH SAIVASAN, AND A. V. SREEJITH
+Abstract. We introduce one deterministic-counter automata (odca), which
+are one-counter automata where all runs labelled by a given word have the same
+counter effect, a property we call counter-determinacy. odcas are an extension
+of visibly one-counter automata - one-counter automata (oca) where the input
+alphabet determines the actions on the counter. They are a natural way to
+introduce non-determinism/weights to ocas while maintaining the decidabil-
+ity of crucial problems, that are undecidable on general ocas. For example,
+the equivalence problem is decidable for deterministic ocas whereas it is un-
+decidable for non-deterministic ocas. We consider both non-deterministic and
+weighted odcas. This work shows that the equivalence problem is decidable
+in polynomial time for weighted odcas over a field and polynomial space for
+non-deterministic odcas. As a corollary, we get that the regularity problem,
+i.e., the problem of checking whether an input weighted odca is equivalent
+to some weighted automaton, is also in polynomial time.
+Furthermore, we
+show that the covering and coverable equivalence problems for uninitialised
+weighted odcas are decidable in polynomial time.
+We also introduce a few reachability problems that are of independent
+interest and show that they are in P. These reachability problems later help
+in solving the equivalence problem.
+Introduction
+Visibly pushdown automata (vpda) was introduced by Alur and Madhusudan
+in 2004 [2]. They have received a lot of attention as they are a strict subclass of
+pushdown automata, suitable for program analysis. vpdas enjoy tractable decid-
+able properties, which are undecidable in the general case. The visibly restriction,
+in essence, is that the stack operations are input-driven, i.e., only depends on the
+letter read.
+In this paper, we investigate a relaxation in the visibly constraint on one-counter
+automata (oca): the counter actions are no longer input-driven, but are determin-
+istic. We could summarise this as: “any run on a given word has a fixed counter
+effect”. We give a model satisfying this new restriction, which includes all visibly
+oca.
+Syntactically, one deterministic-counter automata (odca) contain two parts:
+(1) Counter structure: This is a deterministic oca without epsilon transitions.
+The transitions are deterministic, and the state transitions depend only on
+the current state, the alphabet, and whether the counter is zero.
+(2) Finite state machine: This machine has finite states and no counters. It
+can be deterministic, non-deterministic, or weighted. The transitions of
+Key
+words
+and
+phrases. One
+counter
+automata,
+Equivalence,
+Reachability,
+Weighted
+automata.
+1
+
+2
+P. MATHEW, V. PENELLE, P. SAIVASAN, AND A.V. SREEJITH
+this machine depend on its current state, current counter structure state,
+input alphabet, and whether the counter value is zero.
+An odca will be called deterministic, non-deterministic, or weighted depending
+on the type of the finite state machine. One can observe that the class of deter-
+ministic oca and the class of visibly oca are specific cases of odcas. In a visibly
+oca, the input alphabet determines the counter structure.
+An odca represents a function that maps words (over a finite alphabet) to a
+weight. The run of a word over an odca determines its accepting weight. In the case
+of weighted odca, the weights come from a field, and in the case of deterministic
+and non-deterministic odca, the weights come from the boolean semiring. Hence
+a deterministic or non-deterministic odca represents a language, which is the set
+of all words whose weight is 1.
+A non-deterministic odca can have a succinct representation compared to the
+deterministic odca recognising the same language. For example, for any k ∈ N,
+let Lk denote the language {an(b + c)mb(b + c)k | m, n ∈ N and m > n}. The non-
+deterministic odca recognising the language Lk guesses whether a b encountered
+after reading the string an(b + c)n+1 for some n ∈ N is at the kth position from the
+end of the string. An example of a non-deterministic odca that recognises L2 is
+shown in Figure 1. The deterministic odca that recognises the same language will
+have to check whether every b encountered after reading the string an(b + c)n+1 is
+at the kth position from the end. This will require an additional 2k states.
+Our results. Two odcas are equivalent if the functions they represent are equal.
+Observe that deterministic real-time ocas are deterministic odcas. We also note
+that deterministic odcas are deterministic real-time ocas. B¨ohm et al. [5] proved
+that the equivalence of deterministic oca is in non-deterministic log space. We
+show that a non-deterministic odca is equivalent to an exponentially sized deter-
+ministic odca. Therefore, unlike non-deterministic ocas, the equivalence of two
+non-deterministic odcas is decidable and can be determined by a PSPACE machine.
+This paper also presents a polynomial time algorithm for deciding the equivalence
+problem of two weighted odcas. If the two odcas are non-equivalent, we output
+a word (whose length is polynomial in the size of the two odcas) that the two
+weighted odcas accept with different weights.
+We dedicate Section 4 to prove
+Theorem 1.
+Theorem 1. There exists a polynomial time algorithm that decides if two weighted
+odcas are equivalent and outputs a word that distinguishes them, if they are non-
+equivalent.
+To solve the equivalence problem for weighted odca, we introduce a few reacha-
+bility problems. These problems are also of independent interest. The complement
+to vector space (co-VS) reachability problem takes a weighted odca, a vector space,
+and an initial configuration as input. It asks whether it is possible, starting from
+the initial configuration, to reach a configuration in the complement of the given
+vector space. We develop novel ideas to show that the unary (resp. binary) co-VS
+reachability problem is in P (resp. NP). Let us call a word a witness if the run of
+the word ‘reaches’ a configuration desired by the reachability problem. Through a
+series of lemmas, we identify two interesting properties of witnesses.
+(1) The witnesses satisfy a small model property - a witness that is longer than
+a polynomial can be ‘cut’ to get a shorter witness. We remove parts of a
+
+ONE DETERMINISTIC-COUNTER AUTOMATA
+3
+q0
+q1
+q2
+q3
+q4
+q5
+({a}, {0, 1})
+({b, c}, {0})
+({b, c}, {1})
+({b, c}, {0})
+({b}, {0})
+({b, c}, {1})
+({b, c}, {0})
+({b, c}, {0})
+({b, c}, {0})
+Finite state machine
+Counter structure
+p0
+p1
+({b, c}, {1}, −1)
+({a}, {0, 1}, +1)
+({b}, {0}, 0)
+({b, c}, {0, 1}, 0)
+Figure 1. The figure shows a non-deterministic odca recognising
+the language L = {an(b + c)mb(b + c)2 | m, n ∈ N and m > n}.
+Let S ⊆ Σ, R ⊆ {0, 1} and T ∈ {−1, 0, +1} are non-empty sets.
+For i, j ∈ N, if a transition from qi to qj is labelled (S, R) and
+(s, r) ∈ S × R, then there is a transition from qi to qj on reading
+the symbol s.
+The current counter value should be 0 if r = 0
+and greater than 0 if r = 1. Similarly, if a transition from pi to
+pj is labelled (S, R, T ) and (s, r, t) ∈ S × R × {T }, then there is
+a transition from pi to pj on reading the symbol s with counter
+action t. The current counter value should be 0 if r = 0 and greater
+than 0 if r = 1.
+long run and join the remaining portions. The challenge is identifying cuts
+that preserve the counter actions during the run.
+(2) The lexicographically smallest word that witnesses the reachability is of the
+form uyr1
+1 vyr2
+2 w where u, v, w, y1 and y2 are ‘small’ words and r1, r2 ∈ N.
+The reachability problems, along with the ideas developed in the context of real-
+time oca by B¨ohm et al. [3] (also see [4] [6]), and Valiant, Paterson [21] help us
+solve the equivalence problem for odcas.
+Next, we consider the regularity problem - the problem of deciding whether a
+weighted odca is equivalent to some weighted automata. In Theorem 36, we show
+that regularity of odca is decidable in polynomial time. This is done by showing
+the existence of infinitely many equivalence classes by “pumping up” some parts of
+a run.
+Next, we look at uninitialised odcas - an odca without initial finite state dis-
+tribution and initial counter state. We show that the “equivalence” problem for
+unitialised odcas are in polynomial time.
+
+4
+P. MATHEW, V. PENELLE, P. SAIVASAN, AND A.V. SREEJITH
+Related work. Extensive studies have been conducted on weighted automata with
+weights from semirings. Tzeng [20] gave a polynomial time algorithm to decide
+the equivalence of two probabilistic automata. The result has been extended to
+weighted automata with weights over a field. On the other hand, the problem is
+undecidable if the weights are over the semiring (N, min, +) [14]. Unlike the exten-
+sive literature on weighted automata, the study on weighted versions of pushdown
+or one-counter machines is limited [11] [12] [16]. One of the major bottlenecks is
+the undecidability of many interesting problems.
+Probabilistic Pushdown Automata (pPDA) is equivalent to probabilistic recur-
+sive state machines (RSMs) or recursive Markov chains [8] [15].
+These models
+have been studied extensively for the analysis and model checking of procedural
+programs [9]. pPDAs can model probabilistic sequential programs with recursive
+procedure calls. They are also a generalisation of stochastic context-free grammars
+[1] used in natural language processing, molecular biology, and many variants of
+one-dimensional random walks [7]. Kucera et al. [16] have looked at model-checking
+of probabilistic pushdown systems and Br´azdil et al. [8] studied temporal proper-
+ties of probabilistic pushdown automata. The equivalence problem of pPDA was
+examined by Forejt et al. [10] and they showed that it is equivalent to the multiplic-
+ity equivalence of context-free grammars. The decidability of the latter problem is
+open. Kiefer et al. [13] show that the equivalence of probabilistic vpda is logspace
+equivalent to polynomial identity testing. The later problem in known to be in
+coRP.
+The bisimilarity problem of probabilistic vpda (resp. probabilistic oca) was
+shown to be EXPTIME-complete (resp. PSPACE-complete) by Forejt et al. [11].
+They also proved the decidability of the bisimilarity problem of pPDA. Etessami et
+al. [9] show that probabilistic oca and Quasi-Birth-Death processes are equivalent.
+Moving on to the non-weighted models, for non-deterministic pushdown au-
+tomata the equivalence problem is known to be undecidable. On the other hand,
+from the seminal result by S´enizergues [17], we know that the equivalence prob-
+lem for deterministic pushdown automata is decidable. The lower bound, though,
+is primitive recursive [18]. The equivalence problem for deterministic one-counter
+automata (with and without ǫ transitions) is decidable in polynomial time. In fact,
+similar to that of deterministic finite automata, the problem is NL-complete [5].
+Outline of the paper. The rest of this paper is organised as follows. Section 1
+contains the basic definitions and some lemmas from linear algebra. We also give
+a formal definition of odca. In Section 2, we look at the special cases of non-
+deterministic and deterministic odcas and show the decidability of their equiva-
+lence problems. Section 3 analyses a few reachability problems of weighted odcas.
+In Section 4, we prove Theorem 1 and show that the equivalence of weighted odcas
+is in polynomial time. Section 5, gives a polynomial time algorithm for the regular-
+ity problem of weighted odca, and in Section 6, we prove that the covering problem
+for weighted odca is in polynomial time. Section 7 gives a short conclusion.
+1. Preliminaries
+1.1. Basic notations. An alphabet is a non-empty finite set of letters. In this
+paper, we denote the alphabet by Σ. We use Σ∗ to denote the set of finite length
+words over Σ, and for all l ∈ N, we use Σ≤l (resp. Σl) to denote the set of words
+over Σ having length less than or equal to l (resp. exactly equal to l). Given a
+
+ONE DETERMINISTIC-COUNTER AUTOMATA
+5
+word w ∈ Σ∗, we use |w| to denote the length of the word w. We use the notation
+[i, j] to denote the interval {i, i + 1, . . . , j}. We say that a word u = a1 · · · ak is
+a subword of a word w, if w = u0a1u1a2 · · · akuk, where ai ∈ Σ, uj ∈ Σ∗ for all
+i ∈ [1, k] and j ∈ [0, k]. We call u a proper subword of w if u ̸= w. We say that a
+word u is a prefix of a word w if there exists v ∈ Σ∗ such that w = uv. Given a
+word w = a0 · · · an, we write w[i · · · j] to denote the factor ai · · · aj. Given d ∈ N,
+sign(d) = 0 if d = 0 and is 1 otherwise.
+1.2. Linear algebra. A field F = (S, +, ·, 0, 1) is a set S with operations + and
+· and distinguished elements 0 and 1 such that (S, +, 0) and (S, ·, 1) are groups.
+In this paper, we use x, y, z to denote row vectors over a field F, s, t, r to denote
+elements in a field F and A, B, M to denote matrices over a field F. We use U, V
+to denote vector spaces. We recall the following facts.
+Lemma 2 ([19]). The following are true for a field F.
+(1) For any set X of n vectors in Fm with n > m, there exists a vector x ∈ X
+that is a linear combination of the other vectors in X.
+(2) Given a set B of n vectors in Fm and a vector x ∈ Fm, we can check if x
+is a linear combination of vectors in B in time polynomial in m and n.
+(3) Let k, r ∈ N and M ∈ Fk×k. The matrix Mr can be computed in time
+polynomial in k and log(r).
+□
+The following properties of vector spaces are important.
+Lemma 3. Let V be a vector space, k ∈ N and for all r ∈ [0, k] zr ∈ Fk and
+Mr ∈ Fk×k.
+Then, there exists an i ∈ [1, k] such that the following conditions are
+true:
+(1) zi is a linear combination of z0, . . . zi−1, and
+(2) if ziMi /∈ V, then there exists j < i such that zjMi /∈ V.
+Proof. Let k ∈ N, r ∈ [0, k], zr ∈ Fk, Mr ∈ Fk×k be matrices over F and V be a
+vector space.
+1. Consider the set {z0, z1, . . . , zk} of k +1 vectors of dimension k. It follows from
+Lemma 2 that there are at most k independent vectors of dimension k, and hence
+not all elements of the set can be independent.
+2. Let i ∈ [1, k] be such that zi is a linear combination of z0, . . . zi−1 and ziMi /∈ V.
+Let us assume for contradiction that zjMi ∈ V for all j ∈ [0, i − 1]. Since zi is a
+linear combination on z0, . . . zi−1, there exists s0, . . . si−1 ∈ F such that
+zi = s0 · z0 + s1 · z1 + · · · + si−1 · zi−1
+Since ziMi = �i−1
+j=0 sj · zjMi and V is closed under linear combinations, we get
+that ziMi ∈ V contradicting our initial assumption.
+□
+Lemma 4. Let V be a vector space, k ∈ N and for all r ∈ [0, k2] Ar, Mr, Br ∈ Fk×k.
+Then, there exists an i ∈ [1, k2] such that for all x ∈ Fk the following conditions
+are true:
+(1) Mi is a linear combination of M0, . . . , Mi−1, and
+(2) if xAiMiBi /∈ V, then there exists a j < i such that xAiMjBi /∈ V.
+
+6
+P. MATHEW, V. PENELLE, P. SAIVASAN, AND A.V. SREEJITH
+Proof. Let Ar, Mr, Br ∈ Fk×k for r ∈ [0, k2], be matrices over F and V a vector
+space.
+1. Consider the set {M0, M1, . . . , Mk2} of k2 + 1 matrices of dimension k2.
+It
+follows from Lemma 2 that there are at most k2 independent vectors of dimension
+k2, and hence not all elements of this set can be independent.
+2. Let i ∈ [1, k2] be such that Mi is a linear combination of M0, . . . , Mi−1 and
+xAiMiBi /∈ V. Since Mi is dependent on M0, . . . , Mi−1, we prove that there exists
+j < i such that xAiMjBi /∈ V. Let us assume for contradiction that this is not the
+case. Since Mi is a linear combination on M0, . . . Mi−1, there exists s0, . . . si−1 ∈ F
+such that
+Mi = s0 · M0 + s1 · M1 + · · · + si−1 · Mi−1
+Since xAiMjBi ∈ V for all j ∈ [0, i−1] we get that xAiMiBi = �i−1
+j=0 sj ·xAiMjBi ∈
+V, which is a contradiction.
+□
+Lemma 5. Let k ∈ N, A ∈ Fk×k and V ⊆ Fk be a vector space. Then the following
+set is a vector space,
+U = {y ∈ Fk | yA ∈ V}.
+Proof. To prove that U is a vector space, it suffices to show that it is closed under
+vector addition and scalar multiplication. First, we prove that U is closed under
+vector addition. Let z1, z2 ∈ U be two vectors, since z1A, z2A ∈ V, (z1 + z2)A =
+z1A+z2A ∈ V. Therefore, z1 +z2 ∈ U. Now we prove that U is closed under scalar
+multiplication. For any vector z1 ∈ U, we know that z1A ∈ V. Since V is a vector
+space, for any scalar r ∈ F, (r · z1)A ∈ V, and therefore r · z1 ∈ U. This concludes
+the proof.
+□
+In particular, the above lemma holds for the vector space {0 ∈ Fk}.
+1.3. One deterministic-counter automata. A one deterministic-counter au-
+tomata (odca) consists of two parts, a finite state machine, which is a weighted
+automaton over a semiring, and a counter structure, which is a deterministic oca.
+An odca is defined as follows:
+Definition 6. A one deterministic-counter automata ( odca), A over an alphabet
+Σ and a semiring S is as defined below:
+A = (C, δ, p0; Q, λ, ∆, η)
+• C is a non-empty finite set of counter states.
+• δ : C ×Σ×{0, 1} → C ×{−1, 0, +1} is the deterministic counter transition.
+• p0 ∈ C is the start state for counter transition.
+• Q is a non-empty finite set of states of the finite state machine. We assume
+|C| = |Q| and use K to denote |Q|.
+• λ ∈ SK is the initial distribution where the ith component of λ indicates
+the initial weight on state qi ∈ Q.
+• ∆ : C × Σ × {0, 1} → SK×K gives the transition matrix for all p ∈ C,
+a ∈ Σ and d ∈ {0, 1}. The component in the ith row and jth column of
+∆(p, a, d) denotes the weight on the transition from state qi ∈ Q to state
+qj ∈ Q on reading symbol a from counter state p and counter value n with
+sign(n) = d.
+
+ONE DETERMINISTIC-COUNTER AUTOMATA
+7
+• η ∈ SK is the final distribution, where the ith component of η indicates the
+output weight on state qi ∈ Q.
+A configuration c of an odca is of the form (xc, pc, nc) ∈ SK × C × N. The
+configuration (λ, p0, 0) is the initial configuration of A.
+A transition is a tuple
+τ = (ιτ, dτ, aτ, ceτ, Aτ, θτ) where ιτ, θτ ∈ C are counter states, dτ ∈ {0, 1} is to
+denote whether the current counter value is zero or not, aτ ∈ Σ, ceτ ∈ {−1, 0, 1} is
+the counter-effect, Aτ ∈ SK×K such that ∆(ιτ, aτ, dτ) = Aτ, and δ(ciτ, aτ, dτ) = θτ.
+Given a transition τ and a configuration c, we denote the application of τ to c
+as τ(c) = (xcAτ , θτ, nc + ceτ) if pc = ιτ and dτ = 0 if and only if nc = 0, and is
+undefined otherwise. Note that the counter values always stay positive, implying
+that we cannot perform a decrement operation on the counter from a configuration
+with a counter value of zero.
+Given a sequence of transitions T = τ0 · · · τℓ−1, we denote word(T ) = aτ0 · · · aτℓ−1
+the word labelling it, we(T ) = Aτ0 · · · Aτℓ−1 its weight-effect matrix, and ce(T ) =
+ceτ0 + · · · + ceτℓ−1 its counter-effect. For all 0 ≤ i < j ≤ |ℓ − 1|, we use Ti···j to
+denote the sequence of transitions τi · · · τj and |T | to denote ℓ.
+We call a sequence of transition T = τ0 · · · τℓ floating if for all i ∈ [0, ℓ−1] dτi = 1
+and non-floating otherwise. We denote mince(T ) = mini(ce(τ0 · · · τi)) the minimal
+effect of its prefixes and call it its decrease and maxce(T ) = maxi(ce(τ0 · · · τi)) is
+the maximal effect of its prefixes. We say that the sequence of transitions T is
+valid if for every i ∈ [0, ℓ − 2], θτi = ιτi+1. We will only consider valid sequences of
+transitions.
+A run π is an alternate sequence of configurations and transitions denoted as
+π = c0τ0c1 · · · τℓ−1cℓ such that for every i, ci+1 = τi(ci). Given a sequence of
+transition T and a configuration c, we denote T (c) the run obtained by applying
+T to c sequentially (if it is defined). The word labelling it, its length, weight effect,
+and counter-effect are those of its underlying sequence of transitions.
+Observe that, for a valid floating sequence of transitions, T (c) is defined if and
+only if nc > − mince(T ), and for a valid non-floating sequence of transitions, T (c)
+is defined if and only if nc = − mince(T ) and for every i, dτi = 0 if and only if
+ce(τ0 · · · τi−1) = mince(T ). In particular, observe that if a valid floating sequence
+of transition T is applicable to a configuration (xc, pc, nc), then for every n′ ≥ nc
+and vector x′ ∈ SK, it is applicable to (x′, pc, n′).
+For any word w, there is at most one run labelled by w starting from a given
+configuration c0. We denote this run π(w, c0). A run π(w, c0) = c0τ0c1 · · · τℓ−1cℓ is
+also represented as c0
+w
+−→ cℓ. We use the notation c0 →∗ cℓ to denote the existence
+of some word w such that c0
+w
+−→ cℓ. The counter effect of a word w on a floating
+run c0
+w
+−→ cℓ is ncℓ − nc0. The weight with which a word w is accepted by A along
+the run c0
+w
+−→ cℓ is denoted by fA(w, c0) = λwe(π(w, c0))η⊤. We use the notation
+fA(w) to denote fA(w, (λ, p0, 0)).
+Let A and B be two odcas. Consider the configurations c of A and d of B. We
+say that c ≡l d if and only if for all w ∈ Σ≤l, fA(w, c) = fB(w, d) otherwise c ̸≡l d.
+We say that the configurations c and d are equivalent if and only if c ≡l d for all
+l ∈ N and we denote this by c ≡ d. We say that A and B are equivalent if for all
+w ∈ Σ∗, fA(w) = fB(w).
+If the odca is defined over a semiring which is also a field, then we call the
+model a weighted odca, and if it is the boolean semiring then we call it a non-
+deterministic/deterministic odca.
+Note that the equivalence problem of odca
+
+8
+P. MATHEW, V. PENELLE, P. SAIVASAN, AND A.V. SREEJITH
+defined over an arbitrary semiring is undecidable because of the undecidability of
+equivalence of weighted automata over semirings.
+The class of weighted odcas
+includes deterministic oca, visibly weighted oca, and deterministic weighted oca.
+We have to bring in known examples from literature.
+Have others considered
+weighted oca?
+Does our model capture it?
+Also, note that the δ need not be
+a function and A is always a function. In that case, the control states are non-
+deterministic. Like in the previous section, we can determinise it (with an expo-
+nential blow-up). The question is, Is equivalence checking in this model (as well as
+the weighted case) in PTIME?
+Given a weighted odca A over the alphabet Σ and a field F, we define it’s
+M-unfolding weighted automata AM as a finite state weighted automaton that
+recognises the same function as A for all runs where the counter value does not
+exceed M. A formal definition in given below.
+Definition 7 (M-unfolding weighted automata). Let A = (C, δ, p0; Q, λ, ∆, η) be
+a weighted odca over the alphabet Σ and a field F, let K = |Q| = |C|. For a
+given M ∈ N, we define an M-unfolding weighted automata AM of A as follows,
+AM = (C′, δ′, p′
+0; Q′, λ′, ∆′, η′
+F ) where,
+• C′ = C × [0, M] is the finite set of counter states.
+• δ′ : C′×Σ → C′ is the deterministic counter transition. Let p, q ∈ C, m ∈ N,
+a ∈ Σ and d ∈ {−1, 0, +1}. δ′((p, m), a) = (q, m + d), if δ(p, a, sign(m)) =
+(q, d).
+• p′
+0 = (p0, 0) is the initial counter state.
+• Q′ = Q × [0, M] is the finite set of states.
+• λ′ ∈ F|Q′| is the initial distribution.
+λ′[i] =
+�
+λ[i], if i < K
+0, otherwise
+• ∆′ : C′ × Σ → F|Q|′×|Q′| gives the transition matrix. For i, j ∈ |Q′|, p ∈
+C, m ∈ N and a ∈ Σ,
+∆′((p, m), a)[i][j] =
+
+
+
+
+
+
+
+
+
+
+
+
+
+∆(p, a, 0)[i][j], if i, j < K
+∆(p, a, 1)[i mod K][j mod
+K],
+if
+i
+K =
+j
+K
+0, otherwise
+• η′
+F ∈ F|Q′| is the final distribution.
+η′
+F [i] = η[i mod K]
+An uninitialised weighted odca A is a weighted odca without an initial counter
+state and initial distribution. Formally, A = (C, δ; Q, ∆, η). Given an uninitialised
+weighted odca A and an initial configuration c0 = (x, p, 0), we define the weighted
+odca A⟨c0⟩ = (C, δ, p; Q, x, ∆, η).
+Weighted automata (WA) is a restricted form of an odca where the counter value
+is fixed at zero. The above notions of transitions, runs, acceptance, etc. are used
+for WA also. We also use the classical notion and represent weighted automata as
+A = (Q, λ, ∆, η), without counter states.
+
+ONE DETERMINISTIC-COUNTER AUTOMATA
+9
+2. Nondeterministic / deterministic odca
+A deterministic/non-deterministic odca A is an odca over the boolean semiring
+S = ({0, 1}, ∨, ∧).
+The language recognised by A is given by L(A) = {w |
+fA(w) = 1}.
+We say an odca A = (C, δ, p0; Q, λ, ∆, η) is deterministic if for
+every transition sequence T = τ0 · · · τℓ−1, the vector λwe(T ) contains exactly one 1
+and non-deterministic otherwise.
+The following theorem is ‘analogous’ to the case of finite automata. The idea is
+a simple subset construction.
+Theorem 8. For every language recognised by a non-deterministic odca, there is
+a deterministic odca of at most exponential size that recognise it.
+Proof. Let A = (C, δ, p0; Q, λ, ∆, η) be a non-deterministic odca.
+Given a vector
+x ∈ Sk for some k ∈ N, we define the function IsDet: Sk → {true, false} as follows:
+IsDet(x) =
+�
+true, if ∃i < k s.t x[i] = 1 and ∀j ̸= i, x[i] = 0
+false, otherwise.
+Given a transition matrix A corresponding to the states Q, we define its determin-
+isation det(A) as follows. There are rows and columns corresponding to each set
+in 2Q. For any qi ∈ Q, let M(qi, A) = {qj | A[i][j] = 1} be the set of all states in
+the row of qi whose entries are 1. With the notation that det(A)[s][s′] corresponds
+to the entry of the cell corresponding to the sets s, s′ ∈ 2Q, we let det(A)[s][s′] = 1
+if and only if s′ = �
+q∈s M(qi, A). We claim that Adet = (C, δ, p0; Q, λ, ∆′, η′),
+with η′ such that for any S ∈ 2Q,η′[S] = �
+s∈S η[s] and for all p ∈ C, a ∈ Σ
+and d ∈ {0, 1}, ∆′(p, a, d) = det(∆(p, a, d)) is such that it is deterministic and
+L(A) = L(Adet).
+For this, for any sequence of operations T = τ0 · · · τℓ−1, let vT , v′
+T be the vectors
+corresponding to λwe(T ) in A and Adet respectively. Then we have IsDet(v′
+T ) = 1
+and for any S ∈ 2Q, v′
+T [S] = 1 if and only if for all qi ∈ S, vT [i] = 1.
+□
+The equivalence of deterministic odcas can be decided in non-deterministic log
+space [3]. From the above theorem, it follows that a PSPACE machine can decide
+on the equivalence of non-deterministic odcas.
+Theorem 9. Equivalence of non-deterministic odca is in PSPACE.
+As a corollary, we get the following.
+Corollary 10. The emptiness and the universality problems of non-deterministic
+odca are in PSPACE.
+3. Reachability problems in weighted odca
+In this section, we examine two reachability problems of weighted odcas. In the
+subsequent section, we develop the techniques that play a key role in proving the
+equivalence of weighted odca.
+A weighted odca A = (C, δ, p0; Q, λ, ∆, η). Without loss of generality, assume
+|C| = |Q| and denote |Q| by K. We use V ⊆ FK to denote a vector space and
+V = FK \ V to denote the set complement of V. Let S ⊆ C be a subset of the
+set of counter states, X ⊆ N a set of counter values and w ∈ Σ∗. The notation
+c
+w
+−→ V × S × X denotes the run c
+w
+−→ d where d ∈ V × S × X. We call z ∈ Σ∗
+a reachability witness for (c, V, S, X) if c
+z−→ V × S × X. Moreover, we say z is a
+
+10
+P. MATHEW, V. PENELLE, P. SAIVASAN, AND A.V. SREEJITH
+minimal reachability witness for (c, V, S, X) if c
+z−→ V × S × X and for all u ∈ Σ∗
+with c
+u−→ V × S × X, |u| ≥ |z|. We use c
+∗−→ V × S × X to denote that there exists
+a word u ∈ Σ∗ such that c
+u−→ V × S × X.
+We assume that the vector space V ⊆ FK will be provided by giving a suitable
+basis for V.
+(1) co-VS reachability problem:
+Input: a weighted odca A, an initial configuration c, a vector space V,
+set of counter states S and counter value m.
+Output: Yes, if there exists a run c
+∗−→ V × S × {m} in A. No, otherwise.
+(2) co-VS coverability problem:
+Input: a weighted odca A, an initial configuration c, a vector space V,
+and set of counter states S.
+Output: Yes, if there exists a run c
+∗−→ V × S × N in A. No, otherwise.
+Note that in the second problem, the counter value of the final configuration is not
+part of the input. We consider the cases where the counter values of the initial
+configuration and the final counter value, if part of the input, are given in unary
+or in binary notation separately. Note that the size of the unary representation is
+exponentially larger than the binary representation for the same value.
+First, we look at the particular case of co-VS reachability problem for weighted
+automata.
+Note that for weighted automata, the counter value is always zero.
+Given a weighted automata B, with k states, an initial configuration ¯c, a vector
+space U ⊆ Fk and a set of counter states S, the co-VS reachability problem asks
+whether there exists a run ¯c
+∗−→ U × S × {0}.
+Theorem 11. There is a polynomial time algorithm that decides the co-VS reach-
+ability problem for weighted automata and outputs a minimal reachability witness
+if it exists.
+Proof. Tzeng [20] gives a polynomial time algorithm for the equivalence of two
+probabilistic automata by reducing the problem to the co-VS reachability problem
+where V = {0}. The same algorithm can be modified to solve the general co-VS
+reachability problem.
+□
+The following lemma will help us break down both the reachability problems
+into smaller sub-problems.
+Lemma 12. Let z ∈ Σ∗ be a minimal reachability witness for (c, V, S, X). Consider
+arbitrary z1z2 ∈ Σ∗ such that z = z1z2. Let d, e be configurations such that c
+z1
+−→
+d
+z2
+−→ e and A ∈ FK×K be such that xdA = xe. Then z1 is a minimal reachability
+witness for (c, U, {pd}, {nd}), where U = {y ∈ FK | yA ∈ V}.
+Proof. Let z ∈ Σ∗ be a minimal reachability witness for (c, V, S, X), d, e be config-
+urations such that c
+z1
+−→ d
+z2
+−→ e where z1, z2 ∈ Σ∗ with z = z1z2 and A ∈ FK×K
+be such that xdA = xe. Let U = {y ∈ FK | yA ∈ V}. Assume for contradic-
+tion that there exists z′
+1 ∈ Σ∗ smaller than z1 and c
+z′
+1
+−→ f for some configuration
+f ∈ U × {pd} × {nd}. Note that for all y ∈ U, the vector yA ∈ V. Since nf = nd
+and pf = pd, the run c
+z′
+1
+−→ f
+z2
+−→ V × {pe} × {ne} is a valid run and the word z′
+1z2
+contradicts the minimality of z.
+□
+
+ONE DETERMINISTIC-COUNTER AUTOMATA
+11
+The following subsection shows that the unary version of co-VS reachability and
+coverability are in P. In the subsection after, we show that binary version of both
+problems are in NP.
+3.1. Unary reachability in P. In this subsection, we show that both the reacha-
+bility problems of weighted odca are solvable in polynomial time when the counter
+values are given in unary representation.
+Theorem 13. Unary co-VS reachability and co-VS coverability problems are de-
+cidable in polynomial time.
+The theorem is proved by showing a small model property. i.e., the length of a
+minimal witness of reachability is bounded by a polynomial in the number of states
+K and the input counter value(s). This is proved by showing that the maximum
+and minimum counter values encountered during the run of a minimal reachability
+witness do not exceed some bound. Assume this is not true. In this case, there are
+two sub-runs of the run which satisfy the following conditions. In the first part,
+the counter values increases and reaches a maximum counter value. In the second
+part, the counter values decreases. We show that in such a scenario, we can cut
+parts from both the sub-runs by maintaining the reachability conditions. This is
+proved in Lemma 15.
+Now, we prove that if the number of distinct counter values encountered during
+the run of a minimal reachability witness is polynomially bounded, then we can
+bound the length of that witness.
+Lemma 14. Let z ∈ Σ∗ be a minimal reachability witness for (c, V, S, X). If the
+number of distinct counter values encountered during the run c
+z−→ V × S × X is t,
+then |z| ≤ K2 · t.
+Proof. Let c = c1 and T (c1) = c1τ1c2 · · · τh−1ch be the run on word z from c1 and
+T the corresponding sequence of transitions. Let t be the number of distinct counter
+values encountered during this run. Now assume for contradiction that h > K2 · t,
+then by Pigeon-hole principle, there are K + 1 many configurations ci0, ci1, . . . , ciK
+with the same counter state and counter value during this run. Let Aj denote the
+matrix such that xcij Aj = xch for all j ∈ [0, K]. From Lemma 4 we get that there
+exists r ≤ K, and t ∈ [0, r − 1] such that xcit Ar ∈ V. Consider the sequence of
+transitions T ′ = τ1···itτr···ℓ−1 and v = word(T ′). The run π(v, c1) = T ′(c1) is a
+valid run since nct = ncr and pct = pcr. This is a shorter run than π(z, c1) and
+c1
+v−→ V × S × X. This is a contradiction since z was assumed to be minimal.
+□
+It now suffices to show that the number of distinct counter values encountered
+during the run of a minimal witness is polynomially bounded. We first show that if
+the run of a minimal reachability witness of (c, V, S, {m}) is a floating run, then the
+maximum and minimum counter values encountered during this run are bounded
+by a polynomial in K and the initial and final counter values.
+Lemma 15. Let z ∈ Σ∗ be a minimal reachability witness for (c, V, S, {m}). If
+c
+z−→ V × S × {m} is a floating run, then the maximum counter value during this
+run is less than max(nc, m) + K4.
+Proof. Let z ∈ Σ∗ be a reachability witness for (c, V, S, {m}) and c
+z−→ V ×S ×{m}
+be a floating run. Let f ∈ V × S × {m}, such that c
+z−→ f. We prove that the
+
+12
+P. MATHEW, V. PENELLE, P. SAIVASAN, AND A.V. SREEJITH
+vj
+d
+c
+e
+gi
+ui
+g′
+i
+wi
+gj
+g′
+j
+word length
+nc1 + M
+nc1
+ngi
+ngj
+counter value
+Figure 2. The figure shows the floating run from a configuration
+c to e such that xe ∈ U. The configurations gi and gj (resp. g′
+i
+and g′
+j) are where the counter values ngi and ngj are encountered
+for the last (resp. first) time before (resp. after) reaching counter
+value nc + M.
+Also, pgi = pg′
+i = pgj = pg′
+j.
+The dashed line
+denotes the part of the run which can be removed to get a shorter
+reachability witness for (c, U, {pe}, {ne}).
+maximum counter value encountered during this run are bounded. Let us assume
+that max(nc, m) = nc. The case where max(nc, m) = m can be proven analogously.
+Assume for contradiction that the maximum counter value encountered during this
+run is greater than nc + K4. There exists z1, z2, z3 ∈ Σ∗ such that z = z1z2z3 and
+configurations d, e such that the run on z from c can be written as follows:
+c
+z1
+−→ d
+z2
+−→ e
+z3
+−→ f
+where ne = nc and nd = nc + maxce(π(z, c)) (see Figure 2). Let M ∈ FK×K such
+that xf = xeM. From Lemma 5 we know that the set U = {y ∈ FK | yM ∈ V} is
+a vector space and hence the vector xe ∈ U. From Lemma 12 we know that z1z2 is
+a minimal reachability witness for (c, U, {pe}, {ne}). We contradict the minimality
+of z1z2.
+Let c1 = c and T (c1) = c1τ1c2 · · · τℓ−1cℓ denote the run on word z1z2 from
+the configuration c1 and T the corresponding sequence of transitions. Let M =
+maxce(π(z, c)). Note that M = nd − nc. For any i ∈ [0, M], we denote by li and
+di the indices such that the counter value nc1 + i is encountered for the last (resp.
+first) time before (resp. after) reaching counter value nc1 + M in T (c1). That is,
+ce(T1···li−1) = ce(T1···di−1) = i, and for any j ∈ [li, di − 2], ceT1···j > i. We call
+gi = T1···li−1(c1) and g′
+i = T1···di−1(c1).
+Let r = K2 + 1. Since M > K4, by Pigeonhole principle, there exists set of
+indices X = {i1, i2, · · · , ir} ⊆ [0, M] such that for any k < r, we have ik < ir and
+for all h, j ∈ X pgh = pg′
+h = pgj = pg′
+j. For all j ∈ X, let uj, vj, wj be words
+such that c1
+uj
+−→ gj
+vj
+−→ g′
+j
+wj
+−−→ g′
+1 as depicted in Figure 2. For all j ∈ X, let
+matrix Aj and Bj be such that xg′
+j = xgjAj and xg′
+1 = xg′
+jBj. We know that
+for all j ∈ X, xgjAjBj ∈ U. Now we list the matrices in the following sequence
+
+ONE DETERMINISTIC-COUNTER AUTOMATA
+13
+Air, Air−1, . . . , Ai1. From Lemma 4, it follows that, there exists h, j ∈ X with h < j
+such that xghAjBh ∈ U.
+Consider the sequence of transitions T ′ = τ1···lh−1τlj···dj−1τdh···ℓ.
+The word
+uhvjwh = word(T ′) is a proper subword of z1z2 and the run π(uhvjwh, c1) =
+T ′(c1) is a valid floating run shorter than π(w, c1) and c1
+uhvjwh
+−−−−−→ e′ such that
+e′ ∈ U × {pe} × {ne}. This contradicts the minimality of z1z2.
+□
+Now, we prove that for any run (need not necessarily be a floating run) of a mini-
+mal reachability witness z for (c, V, S, m), the maximum counter value encountered
+during the run c
+z−→ V ×S×{m} is bounded by a polynomial in the number of states
+of the machine and the initial and final counter values. This is achieved by applying
+Lemma 15 multiple times on the run of the minimal witness (refer Figure 3).
+e1
+e2 e3
+c
+e4
+d
+word length
+counter value
+Figure 3. The figure shows a run from configuration c to d such
+that xd ∈ V. Configurations e1, . . . , e4 denote the configurations
+where counter value zero is encountered during the run.
+The
+dashed lines denote the portions that can be removed to get a
+shorter reachability witness for (c, V, {pd}, {nd}).
+Lemma 16. If z ∈ Σ∗ is a minimal reachability witness for (c, V, S, {m}) then the
+maximum counter value encountered during the run c
+z−→ V × S × {m} is less than
+max(nc, m) + K4.
+Proof. Let z ∈ Σ∗ be a minimal reachability witness for (c, V, S, {m}). Consider
+the run of word z from c. Let d ∈ V × S × {m} such that c
+z−→ d. Assume for
+contradiction that the maximum counter value encountered during the run c
+z−→ d
+is greater than max(nc, m) + K4. Let e1, e2, · · · , et be all the configurations in this
+run such that nei = 0 for all i ∈ [1, t]. There exists words u1, u2, · · · , ut+1 ∈ Σ∗
+such that z = u1u2 · · · ut+1 and
+c
+u1
+−→ e1
+u2
+−→ e2
+u3
+−→ · · ·
+ut
+−→ et
+ut+1
+−−−→ d
+Note that c
+u1
+−→ e1, et
+ut+1
+−−−→ d and ei
+ui+1
+−−−→ ei+1 for all i ∈ [1, t − 1] are floating
+runs (refer Figure 3).
+
+14
+P. MATHEW, V. PENELLE, P. SAIVASAN, AND A.V. SREEJITH
+We show that the counter values are bounded during each of these floating runs.
+First, we consider the floating run c
+u1
+−→ e1. Let A ∈ FK×K be such that xd = xe1A.
+From Lemma 5 we know that the set U = {y ∈ FK | yA ∈ V} is a vector space
+and hence the vector xe1 ∈ U. From Lemma 12, we know that u1 is a minimal
+reachability witness for (c, U, {pe1}, {0}) and therefore by Lemma 15 we know that
+the maximum counter value encountered during the run π(u1, c) is less than nc+K4.
+Similarly for the floating run et
+ut+1
+−−−→ d, the maximum counter value is bounded
+by nd + K4. Now consider the floating runs ei
+ui+1
+−−−→ ei+1 for all i ∈ [1, t − 1]. Again
+by applying Lemma 15 we get that the maximum counter value encountered during
+each of these sub-runs is less than K4.
+Therefore, the maximum counter value
+encountered during the run c
+z−→ V × S × {m} is less than max(nc, m) + K4.
+□
+We have shown that the counter values are polynomially bounded during the
+run of a minimal reachability witness for the co-VS reachability problem. Our next
+objective is to prove an analogous result for the co-VS coverability problem. The
+problem is similar to co-VS reachability, except that now we are not given a final
+counter value. A crucial ingredient in proving this is Lemma 17 where we prove that
+if the run of a minimal reachability witness z for (c, V, S, N) is a floating run, then
+the number of distinct counter values encountered during the run c
+z−→ V × S × N is
+polynomially bounded in K and nc. Using this and the ideas presented earlier for
+co-VS reachability, we can prove the existence of a polynomial length witness for
+the co-VS coverability problem.
+Lemma 17. If z ∈ Σ∗ is a minimal reachability witness for (c, V, S, N) and c
+z−→
+V × S × {m} is a floating run then |m − nc| ≤ K2.
+word length
+counter value
+c1
+ci0
+ci1
+cil
+cik
+ciK
+cℓ
+Ad
+Figure 4. The figure shows a run from c1 to cℓ such that xcℓ ∈ V.
+The configurations cil and cik are where the counter values ncil
+and ncik are encountered for the last time. Also pcil = pcik . The
+dashed lines denote a part that can be removed to get a shorter
+reachability witness for (c, V, {pcℓ}, N).
+Proof. Let z ∈ Σ∗ be a minimal reachability witness for (c, V, S, N) and c
+z−→
+V × S × {m} is a floating run. Assume for contradiction that m > nc + K2. The
+
+ONE DETERMINISTIC-COUNTER AUTOMATA
+15
+case where nc > m + K2 can be proven analogously. Let c1 = c and π(z, c1) =
+c1τ1c2 · · · τℓ−1cℓ be such that m = ncℓ > nc1 + K2.
+Consider the sequence of
+transitions T = τ0τ1 · · · τℓ−1 in π(z, c1). Since there are only K counter states, by
+Pigeon-hole principle, there exists a strictly increasing sequence I = 0 < i0 < i1 <
+· · · < iK ≤ ℓ such that for all j, j′ ∈ I (Condition 1) pcj = pcj′ and (Condition 2) if
+j < j′ then ncj < ncj′ and for all d ∈ [j + 1, j′ − 1], ncj < ncd < ncj′ .
+Consider the set of configurations ci0, ci1, . . . , ciK. For any j ∈ [0, K], let Aj
+denote the matrix such that xcij Aj = xcℓ. Since xcidAd ∈ V for all d ∈ [0, K], from
+Lemma 3 we get that there exists l, k ∈ [0, K] with l < k such that xcilAk ∈ V.
+Consider a configuration e = (x, p, n). If π(u, e) is a valid floating run with
+mince(π(u, e)) > 0, then for all m ∈ N and y ∈ FK, π(u, (y, p, m)) is a valid
+run.
+Consider the sequence of transitions T ′ = τik···ℓ−1 and let u = word(T ′).
+Because of Condition 2, mince(π(u, cik)) > 0. Therefore the run T ′′(c1) where
+T ′′ = τ1···il−1τik···ℓ−1 is a valid run shorter than π(z, c1).
+This contradicts the
+minimality of z.
+□
+Now we show that for any run (need not be floating) of a minimal reachability
+witness z for (c, V, S, N), the maximum counter value encountered during the run
+c
+z−→ V × S × N is bounded by a polynomial in K and the initial counter value.
+Lemma 18. If z ∈ Σ∗ is a minimal reachability witness for (c, V, S, N) then the
+maximum counter value encountered during the run c
+z−→ V × S × N is less than
+max(nc, K2) + K4.
+Proof. Let z ∈ Σ∗ be a minimal reachability witness for (c, V, S, N). Consider the
+run of word z from c. Let d ∈ V × S × N such that c
+z−→ d. If c
+z−→ d is a floating
+run, then by Lemma 17 the maximum counter value encountered during this run
+will be less than nc + K2. Now if c
+z−→ d is not a floating run, then there exists
+u1, u2 ∈ Σ∗ such that z = u1u2 and c
+u1
+−→ e
+u2
+−→ d where, nei = 0 and e
+u2
+−→ d is a
+floating run.
+Let A ∈ FK×K be such that xd = xeA. From Lemma 5, we know that the the
+set U = {y ∈ FK | yA ∈ V} is a vector space and hence the vector xe ∈ U. Note
+that for all y ∈ U, the vector yA ∈ V. From Lemma 12, we know that u1 is a
+minimal reachability witness for (c, U, {pe}, {0}) and therefore by Lemma 16, we
+know that the maximum counter value encountered during the run π(u1, c) is less
+than nc + K4. Now since e
+u2
+−→ d is a floating run and u2 is the minimal such
+word, from Lemma 17, we get that nd ≤ K2, and by Lemma 15, we know that the
+maximum counter value encountered during this run is less than K2+K4. Therefore,
+we get that the maximum counter value encountered during the run c
+z−→ d is less
+than max(nc, K2) + K4.
+□
+Proof of Theorem 13. For solving the co-VS reachability problem when the weighted
+odca A = (C, δ, p0; Q, λ, ∆, η) with K = |Q| = |C| states, initial configura-
+tion c, vector space V, set of counter states S and counter value m are given
+as inputs, we first consider the max(nc, m) + K4-unfolding weighted automata
+Amax(nc,m)+K4 = (C′, δ′, p′
+0; Q′, λ′, ∆′, η′
+F ) of A as described in Definition 7. From
+Lemma 16, we know that the maximum counter value encountered during the run
+of the minimal reachability witness z for (c, V, S, {m}) is less than max(nc, m)+K4.
+We define a vector space U ⊆ F|Q′| as follows: A vector x ∈ F|Q′| is in U if there
+
+16
+P. MATHEW, V. PENELLE, P. SAIVASAN, AND A.V. SREEJITH
+exists y ∈ V such that for all i ∈ [0, K − 1], x[K · m + i] = y[i] and for all n ̸= m
+and i ∈ [0, K − 1], x[K · n + i] = 0.
+Given a configuration c of a weighted odca, we define the vector zc ∈ F|Q′|.
+zc[i] =
+�
+xc[i mod K], if
+i
+K = nc
+0, otherwise
+Now, consider the configuration ¯c = (zc, (pc, nc)) of Amax(nc,m)+K4 and check
+whether ¯c
+∗−→ U × S × {0}.
+This is a co-VS reachability problem of weighted
+automata. Using Theorem 11, this can be solved in polynomial time.
+For solving the co-VS coverability problem when the weighted odca A with K
+states, an initial configuration c, a vector space V and a set of counter states S are
+given as inputs, we consider the max(nc, K2) + K4-unfolding weighted automata
+Amax(nc,K2)+K4 = (C′, δ′, p′
+0; Q′, λ′, ∆′, η′
+F ) of A. From Lemma 18, we know that
+the maximum counter value encountered during the run of a minimal reachability
+witness z for (c, V, S, N) is less than max(nc, K2) + K4. We define a vector space
+U ⊆ F|Q′| as follows: A vector x ∈ F|Q′| is in U if there exists y ∈ V and m ∈ N
+such that for all i ∈ [0, K−1], x[K·m+i] = y[i] and for all n ̸= m and i ∈ [0, K−1],
+x[K·n+i] = 0. Now, consider the configuration ¯c = (zc, (pc, nc)) of Amax(nc,K2)+K4
+and check whether ¯c
+∗−→ U × S × {0}. This is a co-VS reachability problem of
+a weighted automaton. From Theorem 11, we know that this can be solved in
+polynomial time.
+□
+3.2. Binary reachability in NP. Consider the case where the counter values are
+specified in binary. Theorem 13 can still be applied to get an algorithm whose
+running time is polynomial in the input counter values. Since the counter values
+are represented in binary, their values can be exponentially large compared to their
+size. Therefore, we only get an exponential time algorithm for reachability from
+Theorem 13. This section shows that co-VS reachability can be tested in NP. The
+technically challenging part of the proof is proved in Lemma 22. It shows that the
+“lexicographically minimal” reachability witness z is of the form uyr1
+1 vyr2
+2 w, where
+the length of the words u, y1, y2, v and w are polynomially bounded in K and r1, r2
+are polynomial values dependent on K and the input counter values.
+This is a
+polynomial sized representation of the witness (r1, r2 in binary) whose reachability
+can be verified in polynomial time. A non-deterministic machine guesses the words
+u, y1, y2, v, and w and verifies reachability in polynomial time.
+Theorem 19. Binary co-VS reachability and co-VS coverability problems are in
+NP.
+We aim to show that there is an “encoding” of a minimal reachability witness of
+polynomial size with respect to the input size. The following lemma shows that the
+length of a minimal reachability witness is bounded by a polynomial in the input
+counter values. Note that this can be exponential in size with respect to the input
+size when the counter values are represented in binary.
+Lemma 20.
+(1) If z is a minimal reachability witness for (c, V, S, {m}) then
+|z| ≤ K2 · (max(nc, m) + K4).
+(2) If z is a minimal reachability witness for (c, V, S, N) then |z| ≤ K2·(max(nc, K2)+
+K4).
+
+ONE DETERMINISTIC-COUNTER AUTOMATA
+17
+Proof. 1. Let z ∈ Σ∗ be a minimal reachability witness for (c, V, S, {m}). From
+Lemma 16, we know that the maximum counter value encountered during the run
+c
+z−→ V × S × {m} is less than max(nc, m) + K4. Therefore, there are at most
+max(nc, m) + K4 many distinct counter values encountered during this run. Now
+from Lemma 14 we get that |z| ≤ K2 · (max(nc, m) + K4).
+2. Let z ∈ Σ∗ be a minimal reachability witness for (c, V, S, N). From Lemma 18,
+we know that the maximum counter value encountered during the run c
+z−→ V×S×N
+is less than max(nc, K2)+K4. Therefore, there are at most max(nc, K2)+K4 many
+distinct counter values encountered during this run. Now from Lemma 14 we get
+that |z| ≤ K2 · (max(nc, K2) + K4).
+□
+We define the counter effect of a word w with respect to a counter state q ∈ C as
+ce(π(w, c)) where c is any configuration with nc = |w| and pc = q. Note that for
+any two configuration c, d, ce(π(u, c)) = ce(π(u, d) if nc = nd = |w| and pc = pd.
+First, we consider the case of the run of a minimal reachability witness from c,
+which is a floating run. The following lemma is required for the special case where
+|nc − m| is bounded by a polynomial in K.
+Lemma 21. If z ∈ Σ∗ is a minimal reachability witness for (c, V, S, {m}) and
+c
+z−→ V × S × {m} is a floating run, then
+(1) the minimum counter value during this run is greater than min(nc, m)−K4,
+and
+(2) |z| ≤ K2 · (|nc − m| + 2K4).
+Proof. Let z ∈ Σ∗ be a minimal reachability witness for (c, V, S, {m}) and c
+z−→
+V × S × {m} is a floating run. We prove the claims in the lemma one by one.
+1. This case is symmetric to that of Lemma 15 and can be proven analogously.
+2. From Lemma 15 and Point 1, we get that the counter values encountered during
+the run c
+z−→ V × S × {m} lies between max(nc, m) + K4 and min(nc, m) − K4. Let
+t = |nc − m|. There are at most t + 2 · K4 distinct counter values during this run.
+Now from Lemma 14 we get that |z| ≤ K2 · (t + 2 · K4).
+□
+We assume a total order on the symbols in Σ. Given two words u, v ∈ Σ∗, we
+say that u precedes v in the lexicographical ordering if |u| < |v| or if |u| = |v| and
+there exists an i ∈ [0, |u| − 1] such that u[0, i − 1] = v[0, i − 1] and u[i] precedes v[i]
+in the total ordering assumed on Σ. A word z ∈ Σ∗ is called the lexicographically
+minimal reachability witness for (c, V, S, {m}), if c
+z−→ V × S × X and for all u ∈ Σ∗
+with c
+u−→ V × S × X, z precedes u in the lexicographical ordering. We show that
+the lexicographically minimal reachability witness z for (c, V, S, {m}) has a special
+form. First, we consider the case of floating runs.
+Lemma 22. If z ∈ Σ∗ is the lexicographically minimal reachability witness for
+(c, V, S, {m}) and c
+z−→ V × S × {m} is a floating run, then there exists u, y, w ∈ Σ∗
+and r ∈ N such that z = uyrw and for all configurations dk, k ∈ [0, r], where
+c
+uyk
+−−→ dk the following conditions hold:
+(1) |u|, |y| ≤ 3K7 and |w| < 6K7,
+(2) either ndi > ndj for all i, j such that 0 ≤ i < j ≤ r or
+ndi < ndj for all i, j such that 0 ≤ i < j ≤ r,
+(3) for all i, j ∈ [0, r], pdi = pdj, and
+
+18
+P. MATHEW, V. PENELLE, P. SAIVASAN, AND A.V. SREEJITH
+(4) r ∈ [0, K2 · |nc − m| + K6].
+Proof. Let z be the lexicographically minimal reachability witness for (c, V, S, {m})
+such that c
+z−→ V × S × {m} is a floating run. Let t = |nc − m|. If t ≤ K4, then from
+Lemma 21, Item 2, we get that |z| ≤ 3K6 and the claim is trivially true. Consider
+the case where nc > m. The case where m > nc can be proven analogously. Let us
+assume t > K4 and let d ∈ Z be such that d = −t + K4 + 1.
+Let c = c1 and T (c1) = c1τ1c2 · · · τℓ−1cℓ denote the run on word z from the
+configuration c1. For any i ∈ [0, K4], we denote by li the index such that the counter
+value nc1 −i is encountered for the first time and ri the index such that the counter
+value nc1 − i + d is encountered for the last time in T (c1) (see Figure 5). Since
+t > K4, there are at least K4 + 1 pairs of positions (li, ri), i ∈ [0, K4] such that for
+all i ∈ [0, K4] the factor z[li, ri] has counter effect d with respect to counter state
+pcli . Note that these factors need not be all distinct. Let X = {(li, ri)}i∈[0,K4] be
+the set containing these pairs of positions and W = {z[l, r] | (l, r) ∈ X} be the set
+containing the corresponding factors. Note that |X| > K4.
+z[li, ri]
+c
+g
+ei
+fi
+word length
+nc
+m
+nc − i + d
+nc − i
+counter value
+Figure 5. The figure shows the floating run from a configuration
+c to g such that xg ∈ V. The configurations cli and cri are where
+the counter values nc − i and nc − i + d are encountered for the
+first (resp. last) time during this run. The dashed line denotes the
+part of the run due to the factor z[li, ri] and has a counter effect
+d.
+Claim 1. |W| ≤ K4.
+Proof: Assume for contradiction that |W| > K4. Let g ∈ V × S × {m} be such
+that c
+z−→ g. Since number of counter states is K, by Pigeon-hole principle there
+exists Y ⊆ X with |Y | = K2 + 1 such that for all (l, r), (l′, r′) ∈ Y , pcl = pcl′,
+pcr = pcr′ , and z[l, r] ̸= z[l′, r′]. We say (l, r) < (l′, r′) if z[l, r] precedes z[l′, r′] in
+the lexicographical order. Therefore, the elements in Y have an ordering as follows:
+(l0, r0) < (l1, r1) < · · · < (lK2, rK2).
+For all i ∈ [0, K2], let ui = z[1, li], xi =
+z[li, ri], wi = z[ri, ℓ], configurations ei, fi be such that c
+ui
+−→ ei
+xi
+−→ fi
+wi
+−→ g and
+matrices Ai, Mi, Bi be such that xei = xcAi , xfi = xeiMi , xg = xfiBi.
+We know that for all i ∈ [0, K2], xcAiMiBi ∈ V.
+Consider the sequence of
+matrices M0, M1, · · · , MK2. From Lemma 4 we know that there exists an i ∈ [0, K2−
+
+ONE DETERMINISTIC-COUNTER AUTOMATA
+19
+1] and j < i such that xcAiMjBi ∈ V. Note that the word uixjwi precedes z in
+the lexicographical ordering. Therefore the run c
+uixjwi
+−−−−→ V × S × {m} contradicts
+minimality of z.
+□Claim:1
+Since |W| ≤ K4 and |X| > K4, there exists i, j ∈ [0, K4], with i < j and x ∈ Σ∗
+such that (li, ri) ∈ X, (lj, rj) ∈ X and x = w[li, ri] = w[lj, rj]. Let u1, w1, u2, w2 ∈
+Σ∗ such that z = u1xw1 = u2xw2. Since u1 ̸= u2, either u1 is a prefix of u2 or u2 a
+prefix of u1. Without loss of generality, let us assume u1 is prefix of u2. Therefore,
+there exists v ∈ Σ∗ such that u2 = u1v. Let e be a configuration such that c
+u1
+−→ e.
+Claim 2. |u1|, |v|, |w1| ≤ 3K6.
+Proof: Consider the set X. For any i, j ∈ [0, K4], with i < j, ncli − nclj ≤ K4 + 1
+and ncrj − ncri ≤ K4 + 1. Therefore the counter-effect of u2 and w2 can be at
+most K4. So the counter-effect of v with respect to counter state pe can be at most
+K4. Since it is a minimal floating run from Lemma 21, we get that |v| ≤ 3K6. By
+similar arguments, the counter-effect of u1 and w1 can be at most K4, and again
+by Lemma 21, we get that their lengths are at most 3K6.
+□Claim:2
+u1
+x
+w1
+u2
+x
+w2
+v
+v
+v
+v
+v
+v
+v′
+v
+v
+v
+v
+v
+v
+v′
+Figure 6. The figure shows the factorisation of a word z =
+u1xw1 = u2xw2, where x is an overlapping factor.
+The factor
+v is a prefix of x such that u2 = u1v. The word z can be written
+as u1viv′w2 for some i ∈ N and v′ prefix of v.
+Claim 3. There exists v′ ∈ Σ∗ and r ∈ [0, K2 · |nc − m| + K6] such that x = vrv′
+with |v′| ≤ |v|.
+Proof: Let r ∈ N be the largest number such that x is of the form vrv′ for some
+v′ ∈ Σ∗ (see Figure 6).
+We know that z = u2xw2 and u2 = u1v.
+Therefore,
+z = u1vxw2 = u1vvrv′w2 = u1vrvv′w2. Furthermore, z = u1xw1 = u1vrv′w1. Now
+since u1vrvv′w2 = u1vrv′w1, we get that vv′w2 = v′w1. Hence, if |v′| ≥ |v|, then v
+is a prefix of v′. This is a contradiction since r was chosen to be the largest number
+such that x is of the form vrv′.
+In order to show the bound on the value r, we observe the following. We know
+that the counter effect of the run π(x, e) is d. Therefore from Lemma 21 Point 2,
+we get that |x| ≤ K2 · (|d| + 2K4). Therefore, the value of r is less than or equal to
+K2 · (|d| + 2K4).
+□Claim:3
+From Claim 3 and Claim 2, we get that |u1v′w1| ≤ 9K6 and z = u1vrv′w1
+for some r ∈ [0, K2 · (|d| + 2K4)].
+Note that the factor v might start and end
+in different counter states during the run and, therefore need not always have a
+negative counter effect. However, we also know that the word vr has a negative
+counter effect. For i ∈ [1, 2K], let gi be the configuration such that e
+vi
+−→ gi. By
+
+20
+P. MATHEW, V. PENELLE, P. SAIVASAN, AND A.V. SREEJITH
+Pigeon-hole principle there exists j, k ∈ [1, 2K] with j < K and k − j ≤ K such that
+pgj = pgk. Also, note that the word y = vk−j has a negative counter-effect from
+the counter state pgj. Let r′ = r−j
+k−j and j′ = (r − j) (mod (k − j)). Now consider
+the word z = u1vjyr′vj′v′w1. Since |u1|, |w1|, |v| ≤ 3K6, j < K and k − j ≤ K, we
+get that |u1vj| ≤ 3K7, |vj′v′w1| < 6K7, |y| ≤ 3K7 and r′ ∈ [0, K2 · (|d| + 2K4)].
+□
+Lemma 23. If z ∈ Σ∗ is the lexicographically minimal reachability witness for
+(c, V, S, {m}), then there exists u, y1, v1, v2, v3, y2, w ∈ Σ∗ and r1, r2 ∈ N such that
+(1) z = uyr1
+1 v1v2v3yr2
+2 w,
+(2) |uy1v1v2v3y2w| ≤ 25K7,
+(3) r1, r2 ≤ max{m, nc} · K2 + K6,
+(4) π(uyr1
+1 v1, c) and π(v3yr2
+2 w, d) are floating runs for configuration d where
+c
+uyr1
+1 v1v2
+−−−−−−→ d. , and
+(5) ce(π(uyr1
+1 v1, c)) = ce(π(uyr1
+1 v1v2, c)) = −nc.
+Proof. Let z ∈ Σ∗ be the lexicographically minimal reachability witness for (c, V, S,
+{m}). Consider the run of word z from c. Let d ∈ V × S × {m} such that c
+z−→ d.
+Let c = c1 and T (c1) = c1τ1c2 · · · τℓ−1cℓ denote the run on word z from the
+configuration c1 and T the corresponding sequence of transitions. Let e1 be the first
+configuration with counter value zero and e2 be the last configuration with counter
+value zero during this run. Let z1, z2, z3 ∈ Σ∗ be such that c
+z1
+−→ e1
+z2
+−→ e2
+z3
+−→ cℓ
+and z = z1z2z3. Observe that c
+z1
+−→ e1 and e2
+z3
+−→ cℓ are floating runs.
+From Lemma 22, we know that there exists u1, u3, v1, v3, y1, y3 ∈ Σ∗ and r1, r3 ∈
+N such that z1 = u1yr1
+1 v1, z3 = u3yr3
+3 v3, |u1|, |u3| ≤ 3K7, |v1|, |v3| ≤ 6K7, |y1|, |y3| ≤
+3K7, r1 ∈ [0, nc · K2 + K6] and r3 ∈ [0, m · K2 + K6]. Also, from Lemma 20 we get
+that |z2| ≤ K6.
+□
+We now prove that the binary co-VS reachability and co-VS coverability prob-
+lems are in NP. From Lemma 23 we observe there is a polynomial-size encoding
+of the lexicographically minimal word (where r1 and r2 are in binary). A non-
+deterministic machine can guess this encoding and verify the reachability in poly-
+nomial time since Mr can be computed in log(r) time (see Lemma 2). A detailed
+proof is given below.
+Proof of Theorem 19. Let us first look at the binary co-VS reachability problem.
+Let z ∈ Σ∗ be the lexicographically minimal reachability witness for (c, V, S, {m}).
+Consider the run of word z from c. Let d ∈ V×S×{m} such that c
+z−→ d. Let c = c1
+and T (c1) = c1τ1c2 · · · τℓ−1cℓ denote the run on word z from the configuration c1
+and T the corresponding sequence of transitions. Let e1 be the first configuration
+with counter value zero and e2 be the last configuration with counter value zero
+during this run.
+Let z1, z2, z3 ∈ Σ∗ be such that c
+z1
+−→ e1
+z2
+−→ e2
+z3
+−→ cℓ and
+z = z1z2z3. Observe that c
+z1
+−→ e1 and e2
+z3
+−→ cℓ are floating runs.
+From Lemma 22, we know that there exists u1, u3, v1, v3, y1, y3 ∈ Σ∗ and r1, r3 ∈
+N such that z1 = u1yr1
+1 v1, z3 = u3yr3
+3 v3, |u1|, |u3| ≤ 3K7, |v1|, |v3| ≤ 6K7, |y1|, |y3| ≤
+3K7, r1 ∈ [0, nc · K2 + K6] and r3 ∈ [0, m · K2 + K6]. Also, from Lemma 20 we get
+that |z2| ≤ K6.
+Our NP algorithm starts by guessing the words u1, y1, v1, z2, u3, y3, v3, the values
+r1, r2, and the configurations e1 and e2. We first show how to verify if c
+u1yr1
+1 v1
+−−−−−→ e1.
+
+ONE DETERMINISTIC-COUNTER AUTOMATA
+21
+The algorithm computes configuration f0 such that c
+u1
+−→ f0. Now it constructs
+the matrix My1 and computes the configuration f1 such that f0
+y1
+−→ f1 and xf1 =
+xf0My1.
+From Lemma 2, we know that (My1)r1 can be computed by repeated
+powering in time polynomial in log(r1) and K. Let fr1 be a configuration such
+that f0
+yr1
+−−→ fr1. From Lemma 22, we know that pf0 = pfr1 and nfr1 = pf0 −
+r1 · (nf0 − nf1). Since xfr1 = xf0(My1)r1, we can construct the configuration fr1
+in polynomial time. We now verify in polynomial time whether fr1
+v1
+−→ e1 or not.
+We can verify if e2
+u3yr3
+3 v3
+−−−−−→ d in a similar manner. The algorithm can also check
+whether e1
+z2
+−→ e2 in polynomial time since |z2| ≤ K6. It finally checks whether
+d ∈ V × S × {m}. Hence the binary co-VS reachability problem is decidable in NP.
+As for the binary co-VS coverability problem, either the run of a minimal witness
+is a floating run or is not.
+In the former case where the run is floating, from
+Lemma 17, we know that the difference between the final and initial counter values
+is at most K2. In the latter case where the run is grounded, by Lemma 17, we get
+that the final value is at most K2. In both the cases, the algorithm guesses the final
+counter value, and the problem is reduced to the co-VS reachability problem, which
+is in NP. Hence the binary co-VS coverability problem is decidable in NP.
+□
+4. Equivalence of weighted odca
+In this section, we give a polynomial time algorithm to check the equivalence of
+two weighted odcas (Theorem 1). The algorithm returns a minimal distinguishing
+word if the odcas are non-equivalent. We use the reachability results presented
+in the previous section to show that the length of a minimal distinguishing word
+is short. The idea here is to prove that the maximum counter value encountered
+during the run of a minimal witness is polynomially bounded. We use this to reduce
+the equivalence problem to that of weighted automata.
+In the remainder of this section, we fix two weighted odcas A1 and A2 over an
+alphabet Σ and a field F. For i ∈ {1, 2},
+Ai = (Ci, δi, p0i; Qi, λi, ∆i, ηi).
+Without loss of generality assume K = |C1| = |Q1| = |C2| = |Q2|. We will reason
+on the synchronised runs on pairs of configurations. Given two weighted odcas,
+A1 and A2 and i ∈ N, we denote a configuration pair as hi = ⟨ci, di⟩ where ci is a
+configuration of A1 and di is a configuration of A2. We similarly consider transition
+pairs of A1 and A2, and consider synchronised runs as the application of a sequence
+of transition pairs to a configuration pair. We fix a minimal word z (also called
+witness) that distinguishes A1 and A2 and ℓ = |z|. Henceforth we will denote by
+Π = h0τ0h1 · · · τℓ−1hℓ
+the run pair of z from the initial configuration pair. We denote by T = τ0 · · · τℓ−1
+the sequence of transition pairs of this run pair.
+To prove Theorem 1, we use
+the following lemma, which states that the counter values in Π are bounded by a
+polynomial poly0(K).
+Lemma 24. There is a polynomial poly0 : N → N such that if two weighted odcas
+A1 and A2 are not equivalent, then there exists a witness z such that the counter
+values encountered during the run of z are less than poly0(K).
+
+22
+P. MATHEW, V. PENELLE, P. SAIVASAN, AND A.V. SREEJITH
+We use Lemma 24 to show that the length of the witness z is bounded by a
+polynomial poly1(K) = 2K5poly0(K).
+Lemma 25. There is a polynomial poly1 : N → N such that if two weighted odcas
+A1 and A2 are not equivalent, then there exists a witness z such that |z| is less than
+or equal to poly1(K).
+Proof. Assume for contradiction that the length of a minimal witness z is greater
+than poly1(K). From Lemma 24, we know that the counter values encountered
+during the run Π in less than poly0(K). Since |z| > poly1(K), by the Pigeonhole
+principle, we get that there exist indices 0 ≤ i0 < i2 < · · · < i2K ≤ ℓ such that for
+all configuration pairs hij, j ∈ [1, 2K], ncij = ncij−1 , ndij = ndij−1 , pcij = pcij−1
+and pdij = ndij−1 .
+For all j ∈ [0, 2K] we define the vector xj ∈ F2K such that xj[r] = xcij [r], if r <
+K and xdij [r − K], otherwise. We also define the vector η ∈ F2K such that η[r] =
+η1[r], if r < K and η2[r − K], otherwise. For all j ∈ [0, 2K], let Aj denote the
+matrix such that xjAj = xℓ. Since z is a minimal witness, we know that for all
+j ∈ [0, 2K], xjAjη⊤ ̸= 0. From Lemma 4, we get that there exists r, r′ ∈ [0, 2K],
+with r′ < r such that xr′Arη⊤ ̸= 0. The sequence of transitions τir+1 · · · τℓ can be
+taken from hi′r since the counter values and counter states are the same for both
+configurations. Consider the sequence of transitions T ′ = τ0 · · · τi′
+rτir+1 · · · τℓ and
+let w = word(T ′).
+The word w is a shorter witness than z and contradicts its
+minimality.
+□
+Lemma 25 helps us to reduce the equivalence problem of weighted odca to
+that of weighted automata by “simulating” the runs of weighted odcas up to
+length poly1(K) by two weighted automata. The naive algorithm will only give us
+a PSPACE procedure, but there is a polynomial time procedure to do this, and the
+proof is given below.
+Proof of Theorem 1. We consider the two weighted odcas A1 and A2.
+From
+Lemma 25, we know that the length of the minimal witness z is less than poly1(K).
+Let M = poly1(K). We construct the M-unfolding weighted automata AM
+1 and AM
+2
+as described in Definition 7. It follows that, A1 is non-equivalent to A2 if and only
+if there exists a word w ∈ Σ≤M such that fAM
+1 (w) ̸= fAM
+2 (w). Tzeng [20, Lemma
+3.4] gives a polynomial time algorithm to output a minimal word that distinguishes
+two probabilistic automata. We conclude the proof by noting that the algorithm
+can be extended to the case of weighted automata.
+□
+The rest of this section is dedicated to proving Lemma 24. We adapt techniques
+developed by B¨ohm et al. [3] for ocas. We start by labelling some configuration
+pairs as background points (see Figure 7). Consider the case where there is no
+background point in Π. By reducing the problem to co-VS reachability/coverability
+we show that the counter values in Π are polynomially bounded. Now consider the
+case where there is a background point hj in Π. We show that the counter values
+encountered during the run of Π till hj is polynomially bounded. This is shown by
+Lemma 29 and Lemma 35. We conclude by arguing that the length of the run from
+
+ONE DETERMINISTIC-COUNTER AUTOMATA
+23
+hj is polynomially bounded.
+counters poly-bounded
+�
+��
+�
+h0τ0h1τ1h2 · · · hj−1τj−1
+hj = ⟨cj, dj⟩
+�
+��
+�
+1st configuration pair in
+background space
+counters
+poly-bounded
+�
+��
+�
+τj · · · τℓ−1hℓ
+4.1. Configuration Space. Each pair of configuration h = ⟨c, d⟩ is mapped to a
+point in the space N × N × (C1 × C2) × FK × FK, henceforth referred to as the
+configuration space. Here, the first two dimensions represent the two counter val-
+ues, the third dimension C1 × C2 corresponds to the pair of counter states, and the
+remaining dimensions represent the weight vector. The projection of the configu-
+ration space onto the first two dimensions is depicted in Figure 7. We partition
+the configuration space into three: initial space, belt space, and background space.
+The size of the initial space and, thickness and number of belts will be polynomially
+bounded in K. This partition is indexed on two polynomials, poly2 : N → N and
+poly3 : N → N chosen so that all belts are disjoint outside the initial space. We use
+some properties of these partitions to show that the length of a minimal witness
+is bounded. We assume poly2(K) = 516K21 and poly3(K) = 42K14. The precise
+polynomials are required in the proofs of Lemma 26 and Lemma 32.
+• initial space: All configuration pairs ⟨c, d⟩ such that nc, nd < poly2(K).
+• belt space: Let α, β ∈ [1, 3K7] be co-prime.
+A belt of slope
+α
+β consists
+of those configuration pairs ⟨c, d⟩ outside the initial space that satisfies
+|α.nc − β.nd| ≤ poly3(K). The belt space contains all configuration pairs
+⟨c, d⟩ that is inside belts with slope α
+β .
+• background space: All remaining configuration pairs.
+N
+N
+initial space
+background space
+belt space
+belt space
+belt space
+poly2(K)
+poly3(K)
+Figure 7. Projection of configuration space
+The proof of the following lemma is similar to that of the non-weighted case
+presented in [3].
+Lemma 26. If ⟨c, d⟩ and ⟨e, f⟩ are configuration pairs inside two distinct belts and
+lie outside the initial space, then there is no a ∈ Σ such that ⟨c, d⟩
+a−→ ⟨e, f⟩.
+
+24
+P. MATHEW, V. PENELLE, P. SAIVASAN, AND A.V. SREEJITH
+Proof. Recall poly2(K) = 516K21 and poly3(K) = 42K14. Let B and B′ be two
+distinct belts with µ being the slope of the belt B and µ′ the slope of the belt B′.
+Hence µ ̸= µ′. Without loss of generality, let us assume that µ′ > µ. It suffices to
+show that for all x > poly2(K), we have
+µx + poly3(K) + 1 < µ′x − poly3(K) − 1.
+We know that µ′ − µ ≥
+1
+3K7 and x > 516K21.
+Therefore, 516K21
+6K7
+< (µ′ − µ) · x.
+=⇒ µx + 86K14
+2
+< µ′x − 86K14
+2
+=⇒ µx + 42K14 + K14 < µ′x − 42K14 − K14
+=⇒ µx + 42K14 + 1 < µ′x − 42K14 − 1
+□
+Lemma 26 ensures that the belts are disjoint outside the initial space and that
+no run can go from one belt to another without passing through the initial space
+or background space.
+4.2. Belt space. We look at two scenarios in this section. First, we show that if
+a sub-run of the run of a minimal witness enters and exists a belt from the initial
+space, then the counter values encountered during this sub-run are polynomially
+bounded in K. Secondly, we show that if the run of a minimal witness never enters
+the background space, then the counter values encountered during this run are
+polynomially bounded in K. This is shown by reducing to co-VS reachability of an
+odca.
+Let Πb = hiτihi+1 · · · τj−1hj be a sub-run of the run of z inside a belt with slope
+α
+β . Similar to the technique mentioned in [5], each configuration pair hr, where
+r ∈ [i, j] can alternatively be presented as ((xcr, xdr), pcr, pdr, lr) where lr denotes
+a line with slope α
+β inside the given belt that contains the point (ncr, ndr). Let L be
+the set of all lines with slope α
+β inside the given belt. Note that |L| = poly3(K). The
+run Πb is similar to the run of a weighted odca D that has the tuple (pcr, pdr, lr)
+as the state of the finite state machine and xr ∈ F2K as its weight vector where
+xr[i] = xcr[i], if i < K and xr[i] = xdr[i − K], otherwise. A formal definition of the
+odca D is given below.
+Definition 27. Let Ai = (Ci, δi, p0i; Qi, λi, ∆i, ηi) for i ∈ {1, 2}, be the two
+odcas given. Let L be the set of all lines with slope α
+β inside the given belt. We
+define the odca D = (C, δ, p0; Q, λ, ∆, η), where the initial state p0 and the initial
+distribution λ are arbitrarily chosen.
+• C = C1 × C2 × L is a non-empty finite set of states.
+• δ : C × Σ → C × {−1, 0, +1} is the deterministic counter transition. Let
+p1, q1 ∈ C1, p2, q2 ∈ C2, a ∈ Σ and d1, d2 ∈ {−1, 0, +1}. Let l1, l2 ∈ L and
+m1, m2 ∈ N, such that the point (m1, m2) lies on the line l1. δ((p1, p2, l1), a) =
+((q1, q2, l2), d1), if δ1(p1, a, 1) = (q1, d1) and δ2(p2, a, 1) = (q2, d2) and the
+point (m1 + d1, m2 + d2) lies on the line l2. It is undefined otherwise.
+• Q = Q1 ∪ Q2 is a non-empty finite set of states of the finite state machine.
+
+ONE DETERMINISTIC-COUNTER AUTOMATA
+25
+• ∆ : C × Σ × {0, 1} → F2K×2K gives the transition matrix for all p ∈ C,
+a ∈ Σ and d ∈ {0, 1}. For p1 ∈ C1, p2 ∈ C2, l ∈ L, m ∈ N and a ∈ Σ,
+∆((p1, p2, l), a)[i][j] =
+
+
+
+
+
+
+
+
+
+
+
+∆(p1, a, 1)[i][j], if i, j < K
+∆(p2, a, 1)[i
+−
+K][j
+−
+K],
+if i, j > K
+0, otherwise
+• η ∈ F2K is the final distribution.
+η[i] =
+�
+η1[i], if i < K
+η2[i − K], otherwise
+The sub-run Πb can now be seen as a floating run of a weighted odca D. If
+the run Π ends inside a belt, then Πb = hiτi · · · τℓ−1hℓ. In this case, we show that
+the difference between the counter values of the first and last configuration pairs is
+smaller than a polynomial in K.
+Lemma 28. There is a polynomial poly : N → N, such that if Πb = hiτi · · · τℓ−1hℓ
+lies inside a belt, then |ncℓ − nci| ≤ poly(K) and |ndℓ − ndi| ≤ poly(K).
+Proof. Let Πb = hiτihi+1 · · · τℓ−1hℓ be a sub-run of the run of a minimal wit-
+ness inside a belt and ends in the belt. As mentioned in Definition 27, we con-
+sider this as the run of the weighted odca D. Since it is the run of a witness,
+xjη⊤ ̸= 0. Consider the vector space U = {y ∈ F2K | yη⊤ = 0}. Our problem
+now reduces to the co-VS coverability problem in machine D and asks whether
+(xi, (pci, pdi, li), nci)
+∗−→ U × {(pcℓ, pdℓ, lℓ)} × N. From Lemma 20, we know that the
+length of a minimal reachability witness for ((xi, (pci, pdi, li), nci), U, (pcℓ, pdℓ, lℓ), N)
+is polynomially bounded in nci and K. Hence proved.
+□
+In the following lemma, we show that if Πb = hiτihi+1 · · · τj−1hj is a sub-run of
+Π inside a belt and either nci = ncj or ndi = ndj, then the counter values in Πb
+cannot increase more than a polynomial in K from nci and ndi.
+Lemma 29. There is a polynomial poly : N → N such that, if Πb = hiτihi+1 · · · τj−1hj
+is a run inside a belt with nci = ncj or ndi = ndj, then the counter effect of any
+sub-run of Πb is less than or equal to poly(K).
+Proof. Let Πb = hiτihi+1 · · · τj−1hj be a sub-run of the run of a minimal witness
+inside a belt such that nci = ncj. We consider this as the run of the weighted
+odca D as mentioned in Definition 27. Since it is the run of a witness, we know
+that there exists A ∈ F2K×2K such that xjAη⊤ ̸= 0. Consider the vector space
+U = {y ∈ F2K | yAη⊤ = 0}.
+Our problem now reduces to the co-VS reachability problem in machine D and
+asks whether (xi, (pci, pdi, li), nci)
+∗−→ U × {(pcj, pdj, lj)} × {nci}. From Lemma 20,
+length of a minimal reachability witness for ((xi, (pci, pdi, li), nci), U, (pcj, pdj, lj),
+{nci}) is bounded by a polynomial in nci and K. Hence proved.
+□
+We have now shown that if the run of a minimal witness does not enter the
+background space, then the counter values in this run are polynomially bounded in
+K. Now we look at the case where the run enters the background space.
+
+26
+P. MATHEW, V. PENELLE, P. SAIVASAN, AND A.V. SREEJITH
+4.3. Background space. In this subsection, we consider the case where the run of
+a minimal witness enters the background space. We show that during the run of the
+minimal witness the counter values of the first configuration pair in the background
+space and the remaining length of the run is polynomially bounded in K.
+Floating runs of a weighted ODCA are isomorphic to runs of a weighted au-
+tomaton obtained by ignoring counter values. In order to bound the length of the
+run of a minimal witness in the background space, we introduce the notion of an
+underlying uninitialised weighted automaton.
+Definition 30. For l ∈ {1, 2}, the underlying uninitialised weighted automaton of
+A is the uninitialised weighted automaton U(Al) = (Q′
+l, ∆′
+l, η′
+l), where Q′
+l = Cl ×Ql
+and η′
+l ∈ FK2 is the final distribution.
+For i < K2, η′
+l[i] = ηl[i mod K].
+The
+transition matrix is given by ∆′
+l : Σ → FK2×K2. Let a ∈ Σ, d ∈ {−1, 0, +1}, i, j <
+K2,
+∆′
+l(a)[i][j] =
+
+
+
+
+
+∆l(p i
+K , a, 1)[i mod K][j mod K],
+if δl(p i
+K , a, 1) = (p j
+K , d)
+0 otherwise
+Note that a configuration of U(Al) is a vector of dimension K2. A configuration
+c of a weighted odca A is said to be k-equivalent to a configuration ¯c of an
+uninitialised weighted automata B, denoted c ∼k ¯c, if for all w ∈ Σ≤k, fA(w, c) =
+fB(w, ¯c). We say that c is not k-equivalent to ¯c otherwise and denote this as c ̸∼k ¯c.
+As we need to test the equivalence of configurations from A1 and A2, we con-
+sider the uninitialised weighted automata B, which is a disjoint union of U(A1)
+and U(A2). This gives us a single automaton with which we can compare their
+configurations.
+Let i ∈ {1, 2} and c be a configuration of Ai. For all p ∈ Ci and
+m < 2K2, we define the sets Wp,m
+i
+. The set Wp,m
+i
+contains vectors x ∈ FK such
+that the configuration (x, p, m) is 2K2-equivalent to some configuration of B. The
+set W
+p,m
+i
+is the set FK \ Wp,m
+i
+. Formally,
+Wp,m
+i
+= {x ∈ FK|∃¯c ∈ F2K2, c = (x, p, m) ∼2K2 ¯c}
+Lemma 31. For any i ∈ {1, 2}, p ∈ Ci and m < 2K2, the set Wp,m
+i
+is a vector
+space.
+Proof. To prove this, it suffices to show that it is closed under vector addition and
+scalar multiplication. We fix a set Wp,m
+i
+. First, we prove that it is closed under
+scalar multiplication.
+For any vector z1 ∈ Wp,m
+i
+, we know that there exists a
+configuration c = (z1, p, m) and ¯c ∈ F2K2 such that c ∼2K2 ¯c. Now, for any scalar
+r ∈ F, the configuration (r.z1, p, m) ∼2K2 r · ¯c. Therefore r · z1 ∈ Wp,m
+i
+. Now, we
+show that it is closed under vector addition. Let z1, z2 ∈ Wp,m
+i
+be two vectors.
+Therefore, there exists configurations c1 = (z1, p, m), c2 = (z2, p, m), ¯c1 ∈ F2K2
+and ¯c2 ∈ F2K2, such that c1 ∼2K2 ¯c1 and c2 ∼2K2 ¯c2. Consider the configuration
+c3 = (z1 + z2, p, m), c3 ∼2K2 ¯c1 + ¯c2. Therefore, z1 + z2 ∈ Wp,m
+i
+.
+□
+The distance of a configuration c of Ai is the length of a minimal word that
+takes you from c to a configuration (x, p, m) for some m < 2K2 and p ∈ Ci such
+that x ∈ W
+p,m
+i
+. We define distAi(c) as:
+min{|w| | c
+w
+−→ (x, p, m) ∃p ∈ Ci, m < 2K2, x ∈ W
+p,m
+i
+}
+
+ONE DETERMINISTIC-COUNTER AUTOMATA
+27
+The notion of distance play a key role in determining which parts of the run
+of a witness can be pumped out if it is not minimal. The following lemma is a
+crucial component in proving the equivalence of weighted odca. By Lemma 22,
+the lexicographically minimal reachability witness has a special form and this plays
+a crucial role in proving Lemma 32.
+Lemma 32. Let c be a configuration of weighted odca A. If distA(c) < ∞ then,
+distA(c) = a
+b nc + d where a, b ∈ [0, 3K7] and |d| < 42K14.
+Proof. Without loss of generality, let us consider the weighted odca A1 and a
+configuration c of A1. Let us assume that distA1(c) < ∞. This means that c →∗ d
+with xd ∈ W
+p,m
+1
+for some p ∈ C1 and m < 2K2. Since nd = m, by Lemma 22, we
+know that there is a word u = u1ur
+2u3 (with r ≥ 0) such that that c
+u−→ d where
+|u| = distA1(c), |u1u3| ≤ 9K7, |u2| ≤ 3K7 and u2 has a negative counter effect ℓ.
+Let g be the combined counter effect of u1, u3 and α = |u2|
+ℓ . Since |u1u3| ≤ 9K7,
+we have |g| ≤ 9K7.
+distA1(c) = nc − nd − g
+ℓ
+|u2| + |u1u3|
+= αnc − α(nd + g) + |u1u3|
+�
+��
+�
+d
+Since 1 ≤ α ≤ 3K7 it follows that −42K14 < d < 42K14. Hence proved.
+□
+The polynomials poly1 and poly2 were picked so that the configuration pairs
+with equal distance always lie in the belt space. Therefore, the background space
+points either have unequal or infinite distances.
+Lemma 33. For any configuration pair ⟨c, d⟩, in the background space, either
+distA1(c) ̸= distA2(d) or distA1(c) = distA2(d) = ∞.
+Proof. Assume for contradiction that there is a configuration pair ⟨c, d⟩, in the back-
+ground space such that distA1(c) = distA2(d) < ∞. Since distA1(c) = distA2(d).
+From Lemma 32, there exists a1, b1, a2, b2 ∈ [0, 3K7] and d1, d2 < 42K14 such that
+a1
+b1
+nc + d1 = distA1(c) = distA1(d) = a2
+b2
+nd + d2
+Therefore | a1
+b1 nc − a2
+b2 nd| ≤ |d2 −d1| < 42K14. This satisfies the belt condition and is
+a configuration pair in the belt space. This contradicts our initial assumptions.
+□
+The following lemma shows that the length of the run Π in the background space
+is polynomially bounded in K and the counter values of the first background point
+in Π. The proof is similar to that in [3] and is given below.
+Lemma 34. If hj = ⟨cj, dj⟩ is the first configuration pair in the background space
+during Π, then ℓ − j is bounded by a polynomial in ncj, ndj and K.
+Proof. Let hj = ⟨cj, dj⟩ be the first configuration pair in the background space
+during the run Π, then from Lemma 33, either distA1(cj) = distA2(dj) = ∞ or
+distA1(cj) ̸= distA2(dj). We separately consider the two cases.
+Case-1, distA1(cj) = distA2(dj) = ∞: then we prove that the remaining length
+of the witness from ⟨cj, dj⟩ is bounded by 2K2. Assume for contradiction that this
+is not the case and cj ∼2K2 dj but cj ̸≡ dj. Let v ∈ Σ>2K2 be the word which
+
+28
+P. MATHEW, V. PENELLE, P. SAIVASAN, AND A.V. SREEJITH
+N
+N
+⟨c, d⟩
+⟨e, f⟩
+Figure 8. The figure depicts an α-β repetition inside a belt with
+slope α
+β . The configuration pairs ⟨c, d⟩ and ⟨e, f⟩ are α-β-related.
+i.e., they line on a line with slope α
+β , pc = pe and pd = pf.
+distinguishes c and d.
+Therefore, there exists a prefix of v, u ∈ Σ|v|−2K2, and
+i = ℓ − 2K2 such that ⟨cj, dj⟩
+u−→ ⟨ci, di⟩ and ci ̸≡2K2 di.
+Since v is a minimal witness ci ≡2K2−1 di and ci ̸≡2K2 di. Since distA1(cj) =
+distA2(dj) = ∞, there exists configurations ¯ci and ¯di in the underlying automa-
+ton B such that ci ∼2K2 ¯ci and di ∼2K2 ¯di. Since ci ≡2K2−1 di, it follows that
+¯ci ∼2K2−1 ¯di. From the equivalence result of weighted automata, we know that if
+two configurations of a weighted automata with k states are non-equivalent, then
+there is a word of length less than k which distinguishes them. Therefore, this is
+sufficient to prove that the underlying weighted automata with ¯ci and ¯di as initial
+distributions are equivalent, and thus ¯ci ∼2K2 ¯di. This allows us to deduce that
+ci ≡2K2 di, which is a contradiction. Therefore, the remaining length of the witness
+from ⟨cj, dj⟩ is bounded by 2K2.
+Case-2, distA1(cj) ̸= distA2(dj): Without loss of generality, we suppose distA1(cj)
+> distA2(dj).
+By definition of distA2, there exists u ∈ ΣdistA2 (dj), i > j and
+a configuration ¯c of the underlying automaton B such that cj
+u−→ ci, dj
+u−→ di,
+ci ∼2K2 ¯ci and di ̸∼2K2 ¯ci.
+Therefore ci ̸≡2K2 di.
+By definition, there exists
+v ∈ Σ≤2K2 such that fA1(v, ci) ̸= fA2(v, di) and hence fA1(uv, cj) ̸= fA2(uv, dj).
+As uv ∈ ΣdistA2(dj)+2K2, we get that cj ̸≡distA2 (dj)+2K2 dj.
+Therefore, there is
+w ∈ Σ≤min{distA1 (cj),distA2 (dj)}+2K2 that distinguishes cj and dj.
+□
+Let α, β ∈ [1, 3K7] be co-prime. We say configuration pairs ⟨c, d⟩ and ⟨e, f⟩ are
+α-β related if pc = pe, pd = pf and α·nc −β ·nd = α·ne −β ·nf. Roughly speaking,
+two configuration pairs are α-β related if they have the same state pairs and lie on
+a line with slope α
+β . An α-β repetition is a run ¯π1 = ciτici+1τi+1 · · · τj−1cj that lies
+inside a belt with slope α
+β such that ci and cj are α-β related.
+The following
+lemma bounds the counter values of the first configuration in the background space,
+if it exists, during the run Π.
+Lemma 35. If hj is the first background point in Π then, counter values of hj are
+less than K5 · 42K14.
+Proof. Let hj be the first point in the background space during the run Π. Assume
+for contradiction that ncj is greater than K5·42K14. Let Π = h0τ0 · · · hj−1τj−1hj · · · hℓ
+be a run of a minimal witness. Since hj is the first point in the background space
+
+ONE DETERMINISTIC-COUNTER AUTOMATA
+29
+N
+N
+Figure 9. The figure shows the run of a word that enters the
+background space from the belt such that the counter values of
+the first configuration pair in the background space exceed a poly-
+nomial bound.
+in this run and ncj > K5 · 42K14, there exists 0 < i < j such that the sub-run
+Πb = hiτihi+1 · · · τj−2hj−1 lies inside a belt B with slope α
+β for some α, β ∈ [1, 3K7].
+Since we are looking at the run of a minimal witness, from Lemma 33 either
+cj ̸≡2K2 dj or dist(cj) ̸= dist(dj). We separately consider the two cases.
+Case-1:
+distA1(cj) ̸= distA2(dj): Without loss of generality, let us assume
+distA1(cj) < distA2(dj). Therefore there exists t ∈ N with j < t ≤ ℓ and con-
+figuration pair ht such that
+m = nct < 2K2, p = pct and xct ∈ W
+p,m
+1
+.
+We
+show that we can pump some portion out from Πb to reach a configuration in the
+background space with unequal distance and smaller counter values.
+Since ncj > K5 · 42K14, by Pigeonhole principle, there exists indices i0 < i1 <
+i2 < · · · , iK2 < i′
+0 < i′
+1 < i′
+2, · · · , < i′
+K2 such that for all r ∈ [1, K2], (1) hir−1 and
+hir are α-β related and lie in belt B, (2) ncir−1 < ncir = nci′r , (3) pci′r = pci′
+r−1 ,
+(4) for all t ∈ N with ir < t < j, nct > ncir , and (5) for all t ∈ N with j < t < i′
+r,
+nct < nci′r .
+For r ∈ [0, K2] let Ar ∈ FK×K denote the matrix such that xcir Ar = xci′r and
+Br ∈ FK×K denote the matrix such that xci′r Br = xct ∈ W
+p,m
+1
+. Therefore for all
+r ∈ [0, K2], we have xcir ArBr ∈ W
+p,m
+1
+. From Lemma 4, we have that there exits
+s, r ∈ [0, K2] with s < r such that xcisArBs ∈ W
+p,m
+1
+. Consider the sequence of
+transitions T ′ = τ0, · · · , τis−1τir, · · · , τj−1 and let w = word(T ′). Let hj′ be the
+configuration such that h0
+w
+−→ hj′. Since we have removed an α-β repetitions inside
+the belt, the configuration pair hj′ is a point in the background space (see Figure 9).
+Moreover, ncj′ < ncj and distA1(cj′) < ∞. Since it is a point in the background
+space, from Lemma 33, we get that distA1(cj′) ̸= distA2(dj′). Therefore, there is
+a shorter path to a configuration in background space with smaller counter values
+and unequal distance. This is a contradiction.
+Case-2: cj ̸≡2K2 dj: Consider the sub-run Πb. Since it is a run inside a belt,
+we can consider this as the run of the weighted odca D. Since ncj > K4 · 42K14,
+by Pigeon-hole principle, there exists indices i0, i1, i2, · · · , iK2 such that for all r ∈
+
+30
+P. MATHEW, V. PENELLE, P. SAIVASAN, AND A.V. SREEJITH
+[1, K2], hir−1 and hir are α-β related with ncir−1 < ncir and for all t ∈ N with
+ir < t < j, nct > ncir .
+Since it is the run of a minimal witness, we know that there exists A ∈ F2K×2K
+such that xj−1Aη⊤
+F ̸= 0. Consider the vector space U = {y ∈ F2K | yAη⊤
+F = 0}.
+For r ∈ [0, K2], let Ar denote the matrices such that xirAr = xj ∈ U.
+Since
+xirAr ∈ U for all r ∈ [0, K2], from Lemma 4, we get that there exists r′ ∈ [0, r − 1]
+such that xci′r Ar ∈ V. The sequence of transitions τir+1 · · · τℓ can be taken from hi′r
+since the counter values always stay positive. Consider the sequence of transitions
+T ′ = τ0 · · · τi′rτir+1 · · · τℓ and let w = word(T ′). The word w is a shorter witness
+than z and contradicts its minimality.
+□
+Finally, we prove that the counter values encountered during the run Π are
+polynomially bounded in K using above lemmas.
+Proof of Lemma 24. Consider the run Π. From Lemma 28, Lemma 29 and Lemma 35,
+we get that the counter values of configuration pairs inside the belt space during
+this run in polynomially bounded in K. Therefore, if it exists, the first background
+point in Π has polynomially bounded counter values. From Lemma 34, the length
+of Π after the first background point is polynomially bounded in K. Since initial
+space is already bounded by a polynomial in K, the maximum counter value in Π
+is polynomially bounded in K.
+□
+5. Regularity of ODCA is in P
+We say that a weighted odca A = (C, δ, p0; Q, λ, ∆, η) is regular if there is a
+weighted automaton B that is equivalent to it. We look at the regularity problem
+- the problem of deciding whether a weighted odca is regular. We fix a weighted
+odca A = (C, δ, p0; Q, λ, ∆, η) and use N to denote |C| · |Q|.
+The proof technique is adapted from the ideas developed by B¨ohm et al. [6] in
+the context of real-time oca. The crucial idea in proving regularity is to check
+for the existence of infinitely many equivalence classes. The proof relies on the
+notion of distance of configurations. Distance of a configuration is the length of a
+minimal word to be read to reach a configuration that does not have an N equivalent
+configuration in the underlying automata. The challenge is to find infinitely many
+configurations reachable from the initial configuration, so that no two of them have
+same distance. Our main contribution is in designing a “pumping” like argument
+to show this.
+Theorem 36. There is a polynomial time algorithm to decide whether a weighted
+odca is equivalent to some weighted automata.
+Recall the definition of U(A) from Definition 30.
+We use c to denote some
+configuration of A and ¯c to denote some configuration of U(A). For a p ∈ C and
+m ∈ N, we define
+Wp,m = {x ∈ F|Q||∃¯c ∈ FN, c = (x, p, m) ∼N ¯c} .
+The set W
+p,m is F|Q|\Wp,m. The distance of a configuration c (denoted by dist(c))
+is
+min{|w| | c
+w
+−→ (x, p, m) ∃p ∈ C, m < N, and x ∈ W
+p,m} .
+The following lemma shows when A is not regular. Clean up, move to appendix,
+proof sketch.
+
+ONE DETERMINISTIC-COUNTER AUTOMATA
+31
+Lemma 37. Let c be an initial configuration of an odca A. Then the following
+are equivalent.
+(1) A is not regular.
+(2) for all t ∈ N, there exists configurations d, e s.t. ne < N,c
+∗−→ d
+∗−→ e,
+xe ��� W
+pe,ne and t < dist(d) < ∞.
+(3) there exists configurations d, e and a run c
+∗−→ d
+∗−→ e s.t. N2 + N ≤ nd ≤
+2N2 + N, xe ∈ W
+pe,ne with ne < N.
+Proof. 3 → 2: Consider an arbitrary q ∈ C, m < N and vector space V = Wq,m.
+Let us assume for contradiction the complement of Point 2. That is, there exists a
+t ∈ N such that for all configurations d′ where c
+∗−→ d′
+∗−→ V ×{q}×{m}, dist(d′) ≤ t.
+Note that for all d′ where nd′ > N, dist(d′) ≥ nd′ −N. Hence there exists an M ∈ N
+such that for all d′ where c
+∗−→ d′
+∗−→ V × {q} × {m}, nd′ ≤ M.
+Consider a configuration d where nd > N2+N and a run c
+∗−→ d
+∗−→ V ×{q}×{m}.
+Point 3 shows the existence of such a run. For contradiction, it suffices to show
+there exists a d′ such that c
+∗−→ d′
+∗−→ V × {q} × {m} and nd′ > nd.
+Let m = |Q|2 + 1. Since nd > N2 + N, by Pigeonhole principle, there exists set of
+indices X = {i1, i2, · · · , im} such that for k, l ∈ [1, m] with k < l, we have ik < il,
+and for all h, j ∈ X pch = pc′
+h = pcj = pc′
+j. Let uj, vj, wj be words such that for
+all j ∈ X, c
+uj
+−→ cj
+vj
+−→ c′
+j
+wj
+−−→ e. For all j ∈ X, let matrix Aj and Bj be such
+that xc′
+j = xcjAj and xe = xc′
+jBj. We know that for all j ∈ X, xcjAjBj ∈ V. List
+the matrices Ai1, Ai2, . . . , Aim in sequence. From Lemma 4, it follows that, there
+exists i, j ∈ X with i < j such that xcjAiBj ∈ V. Consider the run π(ujviwj, c1).
+It contains a configuration d′ where nd′ > nd.
+2 → 1: Assume for contradiction that for all t ∈ N, there exists configurations
+d, e such that c
+∗−→ d
+∗−→ e, xe ∈ W
+pe,ne, ne < N and t < dist(d) < ∞ but A is
+regular. Let B be the weighted automaton equivalent to A. We use |B| to represent
+the number of states of B.
+Let t1, t2, . . . t|B|+1 ∈ N such that for i ∈ [1, |B|], ti < ti+1, and dti be such
+that c
+∗−→ dti
+∗−→ (xi, pe, ne), xi ∈ W
+pe,ne and ti < dist(dti) < ti+1 < ∞. Clearly
+dti ̸≡ dtj for i ̸= j and hence corresponds to two different states of B. Since we
+can find more than |B| pairwise non-equivalent configurations, it contradicts the
+assumption that B is equivalent to A.
+1 → 3: We prove the contrapositive of the statement. Let us assume that there
+is no configurations d, e and a run c
+∗−→ d
+∗−→ e such that N2 + N ≤ nd ≤ 2N2 + N,
+xe ∈ W
+pe,ne with ne < N. This implies that there does not exists a configuration
+d′ such that nd′ > 2N2, c
+∗−→ d′
+∗−→ (y, pe, ne) for some y ∈ W
+pe,ne. Assume for
+contradiction that there is such a run, then there exists a portion in this run that
+can be “pumped down” to get a run c
+∗−→ d′′
+∗−→ (y′, pe, ne) for some configuration
+d′′ such that N2 + N ≤ nd′′ ≤ 2N2 + N and y′ ∈ W
+pe,ne. This is a contradiction.
+Therefore all runs starting from configuration with counter value greater than or
+equal to N2 + N “looks” similar to a run on a weighted automaton. This allows us
+to simulate the runs of A using a weighted automaton.
+□We now prove that the
+regularity problem for weighted odca is decidable in polynomial time.
+Proof of Theorem 36. Let A be a weighted odca. Lemma 37 shows that if A is
+not regular, then there are words u, v ∈ Σ∗ and configurations d, e such that there
+
+32
+P. MATHEW, V. PENELLE, P. SAIVASAN, AND A.V. SREEJITH
+is a run of the form c
+u−→ d
+v−→ e such that N2 + N ≤ nd ≤ 2N2 + N, xe ∈ W
+pe,ne
+with ne < N. The existence of such words u and v can be decided in polynomial
+time since the minimal length of such a path if it exists, is polynomially bounded
+in the number of states of the weighted odca by Lemma 20. This concludes the
+proof.
+□
+6. Covering
+Let A1 and A2 be two uninitialised weighted odcas.
+We say A2 covers A1
+if for all initial configurations c0 of A1 there exists an initial configuration d0 of
+A2 such that A1⟨c0⟩ and A2⟨d0⟩ are equivalent. We say A1 and A2 are coverable
+equivalent if A1 covers A2 and A2 covers A1.
+We show that the covering and
+coverable equivalence problems for uninitialised weighted odcas are decidable in
+polynomial time. The proof relies on the algorithm to check the equivalence of two
+weighted odcas and is given below.
+Theorem 38. Covering and coverable equivalence problems of uninitialised weighted
+odcas are in P.
+Proof. We fix two uninitialised weighted odcas A1 = (C1, δ1; Q1, ∆1, η1) and
+A2 = (C2, δ2; Q2, ∆2, η2) for this section.
+Without loss of generality, assume
+K = |C1| = |Q1| = |C2| = |Q2|. For i ∈ [1, K] we define the vector ei ∈ FK as
+follows:
+ei[j] =
+�
+1, if i = j
+0, otherwise
+For j ∈ [1, K], q ∈ C1, we use hj,q to denote the configuration (ej, q, 0) of A1 and
+for i ∈ [1, K], p ∈ C2, we use gi,p to denote the configuration (ei, p, 0) of A2.
+Claim 1. There is a polynomial time algorithm to decide whether A2 covers A1⟨hj,q⟩
+for any j ∈ [1, K] and q ∈ C1.
+Proof: First, we check, in polynomial time (equivalence with a zero machine),
+whether A1⟨hj,q⟩ accepts all words with weight 0 ∈ F. If that is the case, then
+A1⟨hj,q⟩ and A2⟨g0⟩ are equivalent for the configuration g0 = ({0}K, p, 0), for any
+p ∈ C2. Otherwise, there is some word w0 accepted by A1⟨hj,q⟩ with non-zero
+weight s (returned by the previous equivalence check). Without loss of generality,
+we consider the smallest one, whose size is polynomial in K.
+We pick a p ∈ C2 and check whether there exists an initial distribution from
+counter state p that makes the two machines equivalent.
+Assume that such an
+initial distribution exists and for all i ∈ [1, K], let αi denote the initial weight on
+state qi ∈ Q2. We use α to denote the resultant initial distribution. We initialise
+an empty set B to store a system of linear equations.
+The following steps will be repeated at most K times to check the existence of an
+initial distribution with initial state p ∈ C2. Let w be the counter-example returned
+by the equivalence query in the previous step.
+For all i ∈ [1, K], we compute
+fA2⟨gi,p⟩(w). We add the linear equation �K
+i=1 αi · fA2⟨gi,p⟩(w) = fA1⟨hj,q⟩(w) to
+B and compute values for αi, i ∈ [1, K], such that it satisfies the system of linear
+equations in B. We check whether A1⟨hj,q⟩ and A2⟨(α, p, 0)⟩ are equivalent or not.
+If they are not equivalent, we get a new counter example that distinguishes them.
+Now we repeat the procedure to compute a new initial distribution.
+
+ONE DETERMINISTIC-COUNTER AUTOMATA
+33
+Note that the above procedure is executed at most K times to find an initial dis-
+tribution if it exists. This is because we can find only K many linearly independent
+linear equations in K variables. Suppose the above procedure fails to find an initial
+distribution for which the machines are equivalent. In that case, there is an initial
+distribution of A1 with initial counter state q, for which A2 with initial counter
+state p does not have an equivalent initial distribution. We now pick a different
+counter state of C2 and repeat the process until we find a p ∈ C2 for which the
+algorithm finds an equivalent initial distribution. If for all p ∈ C2, the algorithm
+returns false, then A2 does not cover A1⟨hj,q⟩.
+□Claim:1
+First, we show the existence of a polynomial time procedure to check whether A2
+covers A1. For all j ∈ [1, K], we check whether there exists an initial state p ∈ C2
+such that A2 with initial counter state p has an initial distribution that makes it
+equivalent to A1⟨hj,q⟩ using Claim 1. If we fail to find such a state in C2 then
+we return false. We repeat this procedure for all q ∈ C1. If for all q ∈ C1 there
+exists a p ∈ C2 such that A2 with initial counter state p has an initial distribution
+that makes it equivalent to A1⟨hj,q⟩ for all j ∈ [1, K], then we say that A2 covers
+A1 otherwise we say that A2 does not cover A1. Let us see why this is true. For
+simplifying the arguments we fix a q ∈ C1. Assume that for all j ∈ [1, K], there
+exits p ∈ C2 such that A1⟨hj,q⟩ is equivalent to the configuration (xj,q, p, 0) for
+some xj,q ∈ FK. Now, any initial configuration (λ, q, 0) of A1 is equivalent to the
+configuration (�K
+j=1 λ[j] · xj,q, p, 0) of A2.
+The coverable equivalence problem can now be solved by checking whether A1
+covers A2 and A2 covers A1, which can be done in time polynomial in K.
+□
+7. Conclusion
+We introduced a new model called odca. The equivalence problem for non-
+deterministic odcas is in PSPACE. This is in contrast to non-deterministic oca,
+where the equivalence problem is undecidable.
+We observe that undecidability
+is a consequence of non-determinism occurring in counter actions. In the case of
+weighted odcas, we show that the reachability, equivalence, regularity, and covering
+problems are in P.
+The natural way to extend the work is to consider epsilon transitions in the
+odca. We conjecture that the equivalence, regularity, and covering problems will
+be polynomial time decidable. Another possible direction is to look at pushdown
+systems partitioned into a deterministic stack and a finite state machine. In this
+case, a non-deterministic model can be determinized (similar to Theorem 8). Even
+though all our algorithms are in polynomial time, they may not be ‘practical’. Con-
+siderable effort is required to find faster algorithms. We also leave open questions
+on learning and approximate equivalence of odcas.
+Acknowledgment
+The authors would like to thank Dr. Rahul C S, School of Mathematics and
+Computer Science, IIT Goa, for his valuable and intuitive suggestions which helped
+us in solving the binary Co-VS reachability problem. Sreejith would like to thank
+DST Matrics grant for the project “Probabilistic Pushdown Automata”.
+
+34
+P. MATHEW, V. PENELLE, P. SAIVASAN, AND A.V. SREEJITH
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+Indian Institute of Technology Goa
+Email address: prince@iitgoa.ac.in
+Universit´e de Bordeaux
+Email address: vincent.penelle@u-bordeaux.fr
+The Institute of Mathematical Sciences, HBNI
+Email address: prakashs@imsc.res.in
+Indian Institute of Technology Goa
+Email address: sreejithav@iitgoa.ac.in
+