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Theorem 5.2 (The “Fundamental Theorem of the OLPA Approach”) Let [MATH] be a uniform [MATH] -ary regular operation. Let [MATH] be regular languages, where [MATH] is recognized by a DFA [MATH] Let [MATH] and let [MATH] Then we have |
[EQUATION] where [MATH] , and [MATH] is defined to be [MATH] if [MATH] and [MATH] if [MATH] The state complexity of [MATH] is the function |
[EQUATION] where the maximum is taken over all possible values in the following ranges: [MATH] [MATH] and [MATH] To compute the worst-case state complexity of an [MATH] -ary operation for [MATH] input DFAs of sizes [MATH] through [MATH] , we use [MATH] different languages, each with an alphabet of size [MATH] . The num... |
Proof Define [MATH] We know from Lemma that [EQUATION] Let us prove the following: [EQUATION] where [MATH] for [MATH] are defined as in the statement of the theorem. |
By Lemma and the uniformity of [MATH] , it suffices to exhibit a morphism [MATH] such that [MATH] for [MATH] To define [MATH] , first we define bijections [MATH] for [MATH] As in the proof of Theorem 5.1 we take each [MATH] to be a bijection with the following properties: [MATH] and [MATH] The remaining elements of [MA... |
Now, for [MATH] , we show that [MATH] Let [MATH] , where each of these transformation tuples lies in [MATH] [EQUATION] Thus [MATH] , as required. ∎ |
Examples of Uniform and Non-Uniform Operations In this section, we prove that a number of common operations (as well as a few more esoteric ones) are uniform, and that the class of uniform operations is closed under composition. We also give some examples of non-uniform operations. |
6.1 Uniform Operations First we consider a class of operations called shuffles on trajectories Operations in this class include shuffle, literal shuffle, balanced literal shuffle, insertion, balanced insertion, concatenation, and anti-concatenation 17 , Remark 3.1] We denote the shuffle of languages [MATH] and [MATH] a... |
Proposition 5 For all regular languages [MATH] the shuffle on trajectories operation [MATH] is uniform. Proof Following , we define a DFA operation [MATH] equivalent to the shuffle on trajectories operation. Let [MATH] and [MATH] be arbitrary DFAs. Let [MATH] be a DFA for the set of trajectories [MATH] We set [MATH] to... |
[EQUATION] It was proved in 17 , Theorem 5.1] that [MATH] Let [MATH] and [MATH] be pairs of DFAs such that for [MATH] , the DFAs [MATH] and [MATH] have the same state configuration [MATH] , the DFA [MATH] has alphabet [MATH] and transition set [MATH] , and the DFA [MATH] has alphabet [MATH] and transition set [MATH] |
It is clear that the image DFAs [MATH] and [MATH] will have the same state configuration. Additionally, by inspecting the definitions of the transition sets [MATH] of [MATH] and [MATH] of [MATH] , it is clear that if [MATH] for [MATH] [MATH] and [MATH] , then [MATH] Indeed, let [MATH] . Then we have |
[EQUATION] Thus [MATH] is uniform, and it follows that the shuffle on trajectories operation is uniform. ∎ The above proof illustrates the fact that once one understands the definition of uniformity, it is often easy to determine whether an operation is uniform just by inspecting the DFA construction. There are no diff... |
One can also use the language-theoretic characterization of uniformity to prove that operations are uniform. Typically, proofs using the language-theoretic characterization require somewhat more thought to write and read, but are shorter and have less of the “boilerplate” needed for DFA-based proofs. The rest of our un... |
Proposition 6 The reversal operation [MATH] is uniform. Proof Fix a 1-uniform morphism [MATH] and suppose [MATH] Since [MATH] is 1-uniform, we have [MATH] for all [MATH] It follows that |
[EQUATION] Thus [MATH] Therefore, by Proposition , reversal is uniform. ∎ The cyclic shift operation is defined by [MATH] Proposition 7 |
The cyclic shift operation [MATH] is uniform. Proof Fix a 1-uniform morphism [MATH] and suppose [MATH] We want to show that [MATH] |
If [MATH] , we can write [MATH] for some [MATH] such that [MATH] Since [MATH] , we have [MATH] Thus [MATH] , and it follows that [MATH] |
If [MATH] , then [MATH] Thus we can write [MATH] for some [MATH] such that [MATH] Since [MATH] is 1-uniform, [MATH] has length [MATH] Write [MATH] where [MATH] and [MATH] Then [MATH] It follows that [MATH] , which implies [MATH] Hence [MATH] By Proposition , cyclic shift is uniform. ∎ |
Proposition 8 The star operation [MATH] is uniform. Proof Fix a 1-uniform morphism [MATH] and suppose [MATH] For a language [MATH] , let [MATH] be the set of all finite-length tuples of elements of [MATH] , and let [MATH] be the map [MATH] (the empty tuple is sent to [MATH] ). We claim that [MATH] Indeed, if [MATH] the... |
[EQUATION] Thus [MATH] Therefore, by Proposition , star is uniform. ∎ An [MATH] -ary boolean function is a function [MATH] Each [MATH] -ary boolean function defines a corresponding [MATH] -ary boolean operation on languages over [MATH] , as follows. For [MATH] , let [MATH] be the characteristic function of [MATH] : if ... |
[EQUATION] Examples of commonly used boolean operations on languages include union and intersection ( [MATH] -ary for [MATH] ), difference and symmetric difference (binary), and complement (unary). |
Proposition 9 Boolean operations on languages are uniform. Proof Let [MATH] be an [MATH] -ary boolean operation on languages. Fix a 1-uniform morphism [MATH] and suppose [MATH] for [MATH] We have |
[EQUATION] Therefore, by Proposition [MATH] is uniform. ∎ We have seen that binary concatenation is uniform, since concatenation belongs to the class of shuffles on trajectories. Next we give a direct proof that [MATH] -ary concatenation is uniform. |
Proposition 10 The [MATH] -ary concatenation operation [MATH] is uniform. Proof Fix a 1-uniform morphism [MATH] and suppose [MATH] for [MATH] Define [MATH] by [MATH] |
and similarly define [MATH] Using similar arguments to the proof of Proposition , we can show that [MATH] . Thus: [EQUATION] Therefore, by Proposition [MATH] -ary concatenation is uniform. ∎ |
Next, we show that the class of uniform operations is closed under composition. It is easy to see that this holds for unary uniform operations: if [MATH] and [MATH] are uniform and [MATH] for a 1-uniform morphism [MATH] , then [MATH] and subsequently [MATH] The general case is not much harder; the only difficulty is in... |
Proposition 11 Let [MATH] be an [MATH] -ary uniform operation, and let [MATH] be uniform operations where [MATH] has arity [MATH] . Set [MATH] for [MATH] , and consider the operation of arity [MATH] that maps [MATH] to |
[EQUATION] This operation, which we denote by [MATH] , is uniform. Proof Fix a 1-uniform morphism [MATH] and suppose [MATH] for [MATH] By Proposition , it suffices to show that |
[MATH] Let [MATH] ; then by the uniformity of [MATH] , for [MATH] we have [EQUATION] Set [MATH] and [MATH] Then [MATH] for [MATH] By the uniformity of [MATH] , we have |
[EQUATION] as required. ∎ This shows that all “combined operations” formed by compositions of the uniform operations we have seen so far are also uniform. |
The following “substitution lemma” can also be used to construct new uniform operations from known ones. Lemma 4 Let [MATH] be a [MATH] -ary operation. Fix [MATH] and [MATH] Then the operation [MATH] defined by [MATH] is uniform. |
Proof Fix a 1-uniform morphism [MATH] and suppose [MATH] for [MATH] Then by the uniformity of [MATH] , we have [EQUATION] Therefore, by Proposition , the operation [MATH] is uniform. ∎ |
As an example, we show that the power operation is uniform. Define [MATH] and for [MATH] , set [MATH] Proposition 12 For [MATH] , the power operation [MATH] is uniform. |
Proof For [MATH] , in Lemma , let [MATH] be the [MATH] -ary concatenation operation, set [MATH] and set [MATH] For [MATH] , it is immediate that [MATH] is uniform. For [MATH] , see Proposition 13 below. |
Another example is the anti-concatenation operation [MATH] . This belongs to the class of shuffles on trajectories, so we already know that it is uniform, but an alternate proof could be given using Lemma : let [MATH] be binary concatenation, set [MATH] , set [MATH] and set [MATH] |
Next we consider some operations which depend only on the alphabet of the input languages. These are not interesting from a state complexity perspective, but can be used to construct interesting combined operations. |
Proposition 13 Let [MATH] . The operation [MATH] , where [MATH] is the common alphabet of the inputs, is uniform. In particular, the following operations are uniform for all arities [MATH] |
1. [MATH] 2. [MATH] 3. [MATH] 4. [MATH] Proof Fix a 1-uniform morphism [MATH] and suppose [MATH] for [MATH] We claim that [MATH] for all [MATH] Indeed, take a word [MATH] ; then [MATH] is in [MATH] by 1-uniformity, and so [MATH] . Conversely, if [MATH] then certainly [MATH] . It follows that: |
[EQUATION] By Proposition , operations of this type are uniform. ∎ For a language [MATH] over [MATH] , the right ideal [MATH] is [MATH] , the left ideal |
[MATH] is [MATH] , the two-sided ideal [MATH] is [MATH] , and the all-sided ideal [MATH] is [MATH] , where [MATH] is (ordinary) shuffle. By combining our earlier results, we can show that the operations which map [MATH] to one of the ideals it generates are uniform. For example, let [MATH] be ternary concatenation, let... |
In summary, we have proved that the following operations are uniform: reversal, cyclic shift, star, power, [MATH] -ary concatenation, [MATH] -ary boolean operations (including union, intersection, difference, symmetric difference and complement), shuffles on trajectories (including shuffle, literal shuffle, balanced li... |
6.2 Non-Uniform Operations First we remark that constant operations , which output a fixed language regardless of the input, are not in general uniform. One issue is that our theoretical framework assumes that all regular operations are alphabet-preserving , so we cannot even define true “constant operations” that take... |
Our first example of an interesting non-uniform operation is the following: [EQUATION] This “half” operation is an example of a proportional removal ; the state complexity of proportional removals was studied by Domaratzki |
We could prove that this operation is not uniform directly from the definition, or using the language-theoretic characterization, but instead we will show something even stronger: the OLPA approach does not maximize the state complexity of this operation. |
If the OLPA approach worked for this operation, then by Lemma , the state complexity of the operation would be maximized by a language of the form [MATH] for some state configuration [MATH] . However, it is not hard to see that if [MATH] is non-empty, then [MATH] is either [MATH] or [MATH] , depending on whether [MATH]... |
shows that there are languages [MATH] of state complexity [MATH] such that [MATH] A similar argument shows that OLPA approach fails for many other proportional removal operations as well, although we have not tried to characterize the proportional removals for which the approach fails. |
Next we consider deletions along trajectories , a class of operations which includes left quotient, right quotient, deletion, scattered deletion, bi-polar deletion, and [MATH] -deletion , p. 296] . We will show that the left quotient operation and the deletion operation are not uniform. We have not investigated uniform... |
The case of left quotient is interesting, because the OLPA approach actually works for this operation despite its non-uniformity. The left quotient of [MATH] by [MATH] is |
[MATH] This operation satisfies a weak version of the language-theoretic characterization of uniformity: For all [MATH] -uniform morphisms [MATH] , if [MATH] and [MATH] for [MATH] , then ( [MATH] |
Because empty languages are excluded here, the OLPA approach would fail if maximizing the state complexity in certain cases required the use of empty languages. But this does not happen for left quotient. |
To see that left quotient is not uniform, let [MATH] and define [MATH] by [MATH] Then define [MATH] [MATH] [MATH] , and [MATH] If left quotient was uniform, we would have [MATH] But [MATH] , and so [MATH] Meanwhile, [MATH] |
The deletion of [MATH] from [MATH] is [MATH] We will show that the OLPA approach fails for this operation. If the OLPA approach worked, the state complexity would be maximized by some pair of OLPA witnesses. Consider the language |
[MATH] where [MATH] for some finite sets [MATH] and [MATH] We claim that [MATH] , which has state complexity one. Indeed, fix a word [MATH] Write [MATH] Let [MATH] set [MATH] , and choose a transformation [MATH] that sends [MATH] into [MATH] Next, choose a transformation [MATH] that sends [MATH] into [MATH] Then [MATH]... |
This shows that using OLPA witnesses for deletion only produces languages of state complexity one. However, Han, Ko and Salomaa proved that if [MATH] has state complexity [MATH] , then [MATH] is a tight upper bound on the state complexity of [MATH] . Hence the state complexity of deletion is not maximized by the OLPA a... |
It is interesting that our main examples of operations for which the OLPA approach fails involve the idea of “deletion” in some sense. |
Proofs using the OLPA Approach In the introduction, we used the OLPA approach to give a simple proof of the worst-case state complexity of reversal. We give two additional examples of proofs using the OLPA approach in this section. First we consider the star operation. |
Proposition 14 Let [MATH] be a regular language recognized by [MATH] , where [MATH] and [MATH] . Suppose [MATH] . If [MATH] then [MATH] , and otherwise we have the following tight upper bounds on [MATH] |
[EQUATION] Proof By Theorem 5.1 , it suffices to just compute the state complexity of [MATH] for [MATH] Given [MATH] , an FA for [MATH] is [MATH] where [MATH] It is easy to see that if [MATH] , then [MATH] recognizes [MATH] Henceforth assume [MATH] |
We show each non-empty set [MATH] is reachable by induction on [MATH] From [MATH] we reach [MATH] for each [MATH] by a transformation that sends [MATH] to [MATH] Now suppose [MATH] and smaller sets are reachable. Choose a set [MATH] of size [MATH] which contains a final state but does not contain [MATH] ; this is possi... |
We show that sets in the following collection are pairwise distinguishable: [EQUATION] If a non-empty set [MATH] is not in this collection, it contains a final state but does not contain [MATH] , and the set [MATH] is indistinguishable from [MATH] To distinguish distinct sets [MATH] and [MATH] in this collection, choos... |
For our other example, we consider boolean operations, defined in Section (see the discussion before Proposition ). In this case the proof is complicated, but the result is very general, and we believe it would be considerably more difficult to prove without the OLPA approach or a similar construction. |
It is a bit tricky to state a tight upper bound for the worst-case state complexity of an arbitrary [MATH] -ary boolean operation, because the operation’s result might not depend on all of its arguments. For example, if the inputs to a binary boolean operation have state complexity [MATH] and [MATH] respectively, the w... |
To state our upper bound, we introduce some notation. Given an [MATH] -ary boolean function [MATH] , we define functions [MATH] for [MATH] as follows. If there exist two binary [MATH] -tuples [MATH] and [MATH] which differ only in the [MATH] -th bit (that is, [MATH] and [MATH] for all [MATH] ) such that [MATH] , then w... |
depends on argument [MATH] , and if [MATH] is the constant function sending everything to [MATH] , we say that [MATH] does not depend on argument [MATH] |
Proposition 15 Let [MATH] be an [MATH] -ary boolean operation. Let [MATH] be regular languages, where [MATH] is recognized by [MATH] for [MATH] . Set [MATH] . Then [MATH] and this bound is tight. |
Proof Recall the usual direct product construction for boolean operations: [MATH] where the state set is [MATH] the transition set is [MATH] |
and the final state set is [MATH] This construction gives an upper bound of [MATH] . To get a tighter bound, we must consider distinguishability. Consider the following set of states: |
[EQUATION] There are precisely [MATH] states in this set, and we claim that every state lying outside this set is indistinguishable from a state within the set. To see this, fix a state [MATH] which is not in the above set. Then there exists [MATH] such that [MATH] and [MATH] does not depend on argument [MATH] . We cla... |
Now we show that the upper bound is tight. Our witnesses will be OLPA witnesses [MATH] where [MATH] and [MATH] for [MATH] (it does not really matter what we choose for [MATH] , as long as for [MATH] it is a proper non-empty subset of [MATH] |
The initial state of the direct product DFA is [MATH] . For each state [MATH] , let [MATH] be a transformation that sends [MATH] to [MATH] . Then the letter [MATH] sends the initial state to [MATH] ; thus all states are reachable. |
Next we show that all pairs of states in [MATH] which has size [MATH] , are distinguishable. Suppose we have two distinct states |
[MATH] and [MATH] Since they are distinct, they must differ in some component [MATH] , and for this [MATH] we must have [MATH] and [MATH] Hence there exist two binary [MATH] -tuples [MATH] and [MATH] which differ only in component [MATH] such that [MATH] Assume without loss of generality that [MATH] and [MATH] |
Now, choose a tuple of transformations [MATH] as follows: Choose [MATH] so that [MATH] and [MATH] For [MATH] if [MATH] depends on argument [MATH] , choose [MATH] so that [MATH] If [MATH] does not depend on argument [MATH] , let [MATH] be the identity map. |
Now, we claim that [MATH] is a final state. To determine whether the reached state [MATH] is final, we look at the binary [MATH] -tuple |
[MATH] If [MATH] depends on argument [MATH] (including the case [MATH] ) then we have [MATH] If [MATH] does not depend on argument [MATH] , then [MATH] , transformation [MATH] is the identity map, and [MATH] , so we have [MATH] Thus [MATH] , where [MATH] is [MATH] if [MATH] depends on argument [MATH] , and [MATH] is [M... |
[EQUATION] Hence [MATH] is a final state. On the other hand, consider the state [MATH] For this state we have [MATH] , where [MATH] is [MATH] if [MATH] depends on argument [MATH] , and [MATH] is [MATH] otherwise. Thus by the same bit-flipping argument, we have |
[EQUATION] Thus this state is not final, and we have distinguished the two states. ∎ Conclusions The “one letter per action” (OLPA) approach gives an easy way to find witnesses that maximize the state complexity of many regular operations, at the expense of requiring large alphabets. We defined a class of “uniform” reg... |
We list a few open questions that we find interesting. To what extent does the OLPA approach work in subclasses of the regular languages? It seems it will work for some “nice” subclasses but not for others. |
Can the OLPA approach be generalized to other state complexity measures, like incomplete state complexity or unrestricted state complexity |
Can we find a larger class than the class of uniform operations for which the OLPA approach provably works, without sacrificing the nice property of closure under composition? |
How do we maximize the state complexity of proportional removals like [MATH] ? Domaratzki’s work does not completely solve this problem. If we find a way to maximize their state complexity, can it be generalized to other operations for which the OLPA approach fails? |
Acknowledgements. I thank Jason Bell, Janusz Brzozowski, and the referees of the DLT 2018 version of this paper for their careful proofreading and helpful comments. I thank Lukas Fleischer for pointing me to some important references I overlooked, which allowed me to give a much more complete history of the ideas prese... |
# Source: arxiv 1806.08511 # Title: Search for CAC-integrable homogeneous quadratic triplets of quad equations and their classification by BT and Lax # Sections: all # Downloaded: 2026-03-02T08:54:09.773835+00:00 |
Search for CAC-integrable homogeneous quadratic triplets of quad equations and their classification by BT and Lax Abstract We consider two-dimensional lattice equations defined on an elementary square of the Cartesian lattice and depending on the variables at the corners of the quadrilateral. For such equations the pro... |
. We present here the results of a search and classification of homogeneous quadratic triplets of multidimensionally consistent lattice equations, allowing different equations on the three orthogonal planes (hence triplets) but using the same equation on parallel planes. No assumptions are made about symmetry or tetrah... |
Introduction 1.1 General setting When we discuss integrable equations, continuous or discrete, there is always the question of which definition of integrability should be used. Unfortunately at this time it is still not possible to give a universal definition of integrability. This is because although integrability can... |
In this paper we consider difference equations defined on an elementary square of the Cartesian lattice, with the dynamical variables located on the corners of the square, so called quadrilateral equations. The following basic assumptions are made: |
Definition 1 (Acceptable equations) 1. The equation depends on all corner variables of the quadrilateral. 2. The equation is affine linear in each corner variable. |
3. The equation is irreducible. In this paper we assume furthermore that the equations are homogeneous quadratic Within the present context of quadrilateral equations there are again several properties that are strongly associated with integrability. One is obtained using algebraic entropy |
, that is, the growth of complexity under evolution: If growth is linear the equation is linearizable, if the growth is polynomial the equation is integrable, if the growth is exponential the equation shows chaotic behavior. Other information is obtained from symmetry analysis |
The properties just mentioned involve analysis on the 2D lattice, but there is also a criterion that is based on a multidimensional lattice. It is well known that soliton equations come in hierarchies, with infinite number of different evolutionary times, and in the context of lattice equations this has been associated... |
In such an approach the original 2D quadrilateral equation is extended into [MATH] by introducing accompanying compatible equations on the other 2D-planes of the 3D-space. In the strictest version the accompanying equations are obtained from the original just by changing some lattice variables and parameters. This appr... |
, which has been important in resurrecting the study of integrable lattice equations. A more relaxed version is to allow completely different quad-equations on the three different [MATH] lattice planes, this approach was taken by Boll in his classification |
, where the tetrahedron condition was also used extensively. Even more generally, the accompanying equations could also live in bigger stencils (this is needed if the Lax matrix is bigger that [MATH] .) |
In any case the 3D system of equations must be compatible. If all equations are quad equations, the compatibility condition is called “Consistency-Around-a-Cube” (CAC) and is elaborated below. One statement relating 2D-integrability and CAC is that if an equation has been found to be integrable by a 2D condition, then ... |
In the present work we take completely free acceptable (by Definition ) quadratic equations on the three [MATH] lattices of [MATH] and classify those that have CAC. |
1.2 Detailed formulation For consistency analysis we consider a 3D cube and assign equations on each side of the cube. Furthermore we will extend the consideration from one cube to the full [MATH] lattice and embed the equation as follows: |
1. On a given [MATH] plane all elementary squares have equations of the same form (that is, only the corner variables change corresponding to the location). The coefficients of the equation may depend on the two lattice parameters associated with the plane but not on the location. |
2. When extended to the [MATH] lattice, the quadrilaterals on parallel planes all carry the same equation but intersecting planes may have different equations. |
In [MATH] the corner variables are usually notated as [MATH] where [MATH] gives the locations on the Cartesian plane, while in [MATH] the variables are indexed as [MATH] . When dealing with a specific quadrilateral or cube we use shorthand notation |
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