Noncontextual coloring of orthogonality hypergraphs

We discuss representations and colorings of orthogonality hypergraphs in terms of their two-valued states interpretable as classical truth assignments. Such hypergraphs, if they allow for a faithful orthogonal representation, have quantum mechanical realizations in terms of intertwined contexts or maximal observables that are widely discussed as empirically testable criteria for contextuality. Reconstruction is possible for the class of perfectly separable hypergraphs. Colorings can be constructed from a minimal set of two-valued states. Some examples from exempt categories are presented that either cannot be reconstructed by two-valued states or whose two-valued states cannot yield a chromatic number that is equal to the maximal clique number.

it still possible to characterize those (hyper)graphs by purely classical means?That is, given all two-valued states interpretable as classical truth assignments, is it possible to reconstruct the (hyper)graphs that, up to isomorphisms, characterize those observables?Inspired by and extending Theorem 0 of the 1967 paper of Kochen and Specker [1] we shall present demarcation criteria for (hyper)graph reconstructability.
We shall also explore the connection of two-valued states and coloring of (hyper)graphs, thereby presenting criteria for (re)construction of colorings and the chromatic number of a (hyper)graph from the set of its two-valued states.Thereby two-valued states may yield a systematic, constructive way to both colorings and/neither the reconstruction of (hyper)graphs.
The chromatic properties of such (hyper)graphs are directly related to their (non)classical aspects.For instance, in quantum logic [2] certain colorings-with the number of colors equal to the clique number that can be identified with the Hilbert space dimension-can be "reduced" to or "collapsed" into two-valued dispersionless states, which in turn are interpretable as noncontextual classical truth assignments of the elementary propositions represented by vertices of such graphs.If the chromatic number exceeds the Hilbert space dimension then no uniform, global two-valued state and also no corresponding uniform, global classical truth value assignment exist.
Thereby, the chromatic number signifies important properties and (non)existence of classical interpretations of the aforementioned configurations of observables; in particular, arrangements of observables realizable by quantum means.
One example of the usefulness of these reduction techniques is the construction of a dense yet discontinuous coloring of the rational sphere [3]-with the resulting states and propositions represented by unit vectors with rational coordinates-based on Pythagorean triples [4].Thereby, a two-valued dispersionless state can be straightforwardly obtained by identifying all but one colors of colorings of the sphere [5].
The usefulness of colorings is not restricted to reductions of the colors but can be extended to certain properties of operators.The chromaticity of observables within a given context-which in quantum mechanics can be essentially identified with an orthonormal basis of the associated finitedimensional Hilbert space-can be associated with particular types of spectral forms of maximal operators [6, § 84] formed by the elements of the contexts.Thereby, unit vectors are interpreted as the orthogonal projection operators formed by the respective dyadic products.Those orthogonal projection operators within any given context are mutually orthogonal and can be inserted into the spectral sum of a (normal, or, more specifically) self-adjoint operator.The respective (real) eigenvalues can be identified with some real-valued encoding of the color or chromaticity of the element of the context.Thereby a uniform way of defining (intertwining) context identifiable with maximal quantum operators representing quantum observables is obtained.
If the chromatic number equals the dimension of the Hilbert space-which is, at the same time, identical to the clique number of the graph-this yields a uniform way of defining (maximal) observables even among complementary observables and physical properties (associated with nonidentical contexts).However, if the chromatic number exceeds the dimension of Hilbert space, the construction yields entirely new potential features of observables: if one insists on the uniform global simultaneous (yet counterfactual [7]) existence of observables within such structures, then consistency dictates the abandonment of uniform eigenvalues associated with observables within such contexts.This holds even though the chromatic number of a (hyper)graph and its respective encoding by real values for the spectral sum within a single such context cannot exceed the dimension of the pertinent Hilbert space.As of today, neither such "omni-realistic" escape nor generalization of the Kochen-Specker theorem has been discussed or observed-so we might assume that it is an inapplicable option.In addition, yet it is feasible in principle.
In what follows, we shall first develop the necessary nomenclature and then present some criteria and results with regard to the reconstruction of (hyper)graphs representing logics-aka, collections of (intertwined) contexts containing a uniform number of elementary, atomistic propositions.
We shall also find criteria for the algorithmic (constructive) generation of colorings by the set of two-valued states on such (hyper)graphs or logics.

A. Hypergraphs
The following terms are used by authors synonymously: context, block, (maximal Boolean) subalgebra, (maximal) clique, complete subgraph and hyperedge.The same is applied to the terms atom, element and vertex.
Greechie has suggested [8]  In what follows, we shall refer to such a structure as Greechie diagram [9] (References [10][11][12][13][14][15] contain variants thereof).The Greechie diagrammatical representation of such, possibly intertwined, collection of blocks, is a hypergraph, which is a well-known structure in discrete mathematics.We shall briefly introduce the required terms here, but an interested reader might take a look at Ref. [14] for further theory of hypergraphs.
A hypergraph H is an ordered pair H = (V ; E) where V = V (H) is the set of vertices and E = E(H) is a family of subsets of V called the hyperedges.It depicts a collection of quantum contexts faithfully represented in an n-dimensional Hilbert space, whereby the vertices are identified with elementary quantum propositions whose label assignments are in terms of vectors (or with their respective one-dimensional orthogonal projection operators), and the hyperedges are identified with quantum contexts.An n-subset of atoms forms a hyperedge if its elements are mutually orthogonal.
Let H = (V ; E) be a hypergraph.The 2-section of H is a graph, denoted by [H] 2 , whose vertices are the same as V (H), and two distinct vertices form an edge if and only if they are on the same hyperedge of H.One may think of the 2-section of a hypergraph as the graph associated with the hypergraph.A hypergraph H is called conformal if any maximal clique (with respect to the inclusion) of its 2-section [H] 2 is on a hyperedge of H.If all the hyperedges of a hypergraph H consist of exactly n elements, then H is called n-uniform [14].
Quantum orthogonality hypergraphs are conformal n-uniform.Often these hypergraphs are referred to and depicted by their associated graphs; that is, by their 2-sections.Quantum contexts are represented by hyperedges of these quantum hypergraphs, that is, by the maximal cliques of their 2-sections.We also reserve the letter n for the clique number of those 2-sections, so we always mean n = ω([H] 2 ), which is a constant integer.
To represent orthogonality hypergraphs, we shall concentrate on Greechie diagrams which are pasting [10] constructions [16,Chapter 2] of a homogeneous single type of contexts K n where the clique number n is fixed.This means that every hyperedge is shown by a straight line segment or, more general, by a smooth curve which has exactly n elements as vertices on it; that is, the hypergraphs are conformal n-uniform.

B. Vertex labeling by vectors
A vertex labeling of an orthogonality (hyper)graph H is a function f : V (H) −→ L that assigns labels from a set L to vertices of H.Such a vertex labeling is called an n-dimensional faithful orthogonal representation [17][18][19] (FOR) for H. Any vertex labeling corresponds to a quantum realization in terms of the elementary propositions corresponding to its vector labels: every unit vector x x x spans a one-dimensional subspace of Hilbert space that is the orthogonal projection onto that subspace of the Hilbert space.

C. Coloring
A hypergraph coloring of H is a proper vertex coloring which associates colors to vertices of H so that every two vertices lying on a hyperedge receive different colors.That is, the n distinguished points of any single smooth curve in the hypergraph have n different colors.The coloring is noncontextual; that is, the coloring of atomic elements common to two or more contexts (intertwining there) is independent of the context.Finite examples for which the chromatic number exceeds the clique number, that is, m > n, are the logical structures involved in proofs of the Kochen-Specker theorem.Explicit constructions are, for instance, Γ 2 of Ref. [1], and the configurations enumerated in Figure 9 of [11], Figure 1-3 of [20], Ref. [21], and Table I and Figure 2 of Ref. [22], among numerous others which have a faithful orthogonal representation [17][18][19] in "small dimensions" greater than two.

D. Two-valued states
A state t on a conformal n-uniform orthogonality hypergraph H is a mapping t : V (H) → [0, 1] such that for any hyperedge h, we have ∑ v∈h t(v) = 1.A two-valued state is a state with values in {0, 1}.
A conformal n-uniform orthogonality hypergraph has a separable set of two-valued states if for any distinct pair of vertices u and v, there is at least one two-valued state, say t, such that t(u) = t(v) [11].A hypergraph H has a unital set of two-valued states if for each vertex v ∈ V (H) there is a state t for which we have t(v) = 1.A hypergraph H is said to be separable (respectively, unital) if it has a separable (respectively, unital) set of two-valued states on its vertices.
A "true implies false set" (a, b)-TIFS (gadget [23][24][25]) is a conformal n-uniform orthogonality hypergraph H containing two vertices a and b such that for all two-valued states of H, we have that b is true only if a is false.We call a the "head" and b the "tail" of H.
Similarly, a "true implies true set" (c, d)-TITS (gadget) is a conformal n-uniform orthogonality hypergraph H ′ containing two vertices c and d such that for all two-valued states of H ′ , we have that d is true whenever c is true [15] (in this case the converse need not be true as both c and d could be false, or d could be true and c false).That is, c true implies d true.

E. Completion criterion
The hypergraphs considered here are assumed to be on k vertices, and all their hyperedges (contexts) uniformly contain exactly n vertices.Equivalently, we can assume that our objects are connected graphs such as G on k vertices with clique number ω(G) = n, with the assumption that every vertex v ∈ V (G) lies on at least one maximal clique (context) of size n.This assumption is not always necessary, but it is not harmful to our argument, as we can always add vertices to those contexts that have less number of vertices.Therefore, we can state the completion criterion as follows: In other words, the completion criterion says that there must be no hyperedge with less than n vertices.The importance of this simple criterion becomes clear in Section III B. Note that for any two-valued state of the hypergraph H, exactly one of these n vertices is assigned true.

F. Separability, set representability and reconstructability
Another important criterion is about classical structure-preserving representability of quantum observables-that is, the classical representability of quantum logics-in terms of two-valued states.According to Theorem 0 of Kochen and Specker [1, p. 67] the possibility to "separate" and make a distinction between two arbitrary vertices of a (hyper)graph by at least one of its twovalued states is equivalent to homomorphic-that is, structure-preserving-embedability of the quantum observables into a "larger" classical Boolean algebra.
An early example of a nonseparable (hyper)graph that is not reconstructable from its set of two-valued states is Γ 3 introduced by Kochen and Specker [1, p. 70].There is another hypergraph, depicted in Figure 5 of Ref. [26] sharing the same Travis [27] matrix [16]-i.e., the same set of two-valued states-which is not isomorphic to the hypergraph corresponding to Γ 3 .However, separability does not imply graph theoretic reconstructibility.One reason for this is that end points of TIFS gadgets might not be separable from orthogonal vertices.We shall explicitly present an example (depicted in Figure 3) of such a configuration, although we are not able to present a hypergraph that has a faithful orthogonal representation in terms of vertex labeling by vectors.
In general a (hyper)graph is reconstructable if the Travis matrix-that is, the set of two-valued states-determines or encodes it completely.More explicitly, a reconstructable (hyper)graph is, up to isomorphism, determined by its Travis matrix-that is, effectively, by permutation of its vertices (the column vectors of the Travis matrix) or the ordering of two-valued states (the row vectors of the Travis matrix).Equivalently, a hypergraph H is reconstructable from its Travis matrix T H if, for any hypergraph G whose Travis matrix T G is equivalent to T H -that is, T G can be obtained from T H by permutations of rows and columns -the (hyper)graphs H and G are isomorphic.
The only connected conformal 1-uniform orthogonality (hyper)graph is the trivial singleton whose two-valued states consists of only one state that assigns true to the only vertex.In addition, the only connected conformal 2-uniform orthogonality hypergraph -which is actually a graph -is K 2 whose Travis matrix is equivalent to the 2 × 2 identity matrix.This is because all other connected graphs have a vertex, say a, adjacent to at least two other vertices, say b and c, and therefore any representation of the graph into a 2-dimensional Hilbert space has to assign collinear vectors to b and c.Hence, hereinafter, we suppose that the dimension of our space n, which is the uniform number of vertices on hyperedges, is greater than 2.

III. STRUCTURE RECONSTRUCTION FROM TWO-VALUED STATES
Suppose that we have a table of all two-valued states of a hypergraph H.Under what condition(s) can we reconstruct the graph theoretical structure of H from the set of its two-valued states?
This is what we are going to examine next.

A. Perfectly separable hypergraphs
Let H be a hypergraph with V (H) = {a 1 , a 2 , . .., a k }, and its set of two-valued states contains s elements [which can be shown by saying that nT S(H) = s].Its Travis matrix T (H) = [t i j ] s×k enumerates all two-valued states which are represented by row vectors on the vertices of H, arranged in the columns, such that each vertex corresponds to one column.Then, separability of H can be extended to the following statement: Definition 3. The hypergraph H is perfectly separable if and only if for all pairs a i and a j of vertices of H we obtain the following: 1.There is r 1 ∈ {1, . .., s} such that t r 1 i = 0 and t r 1 j = 1.
3. If a i and a j are not adjacent in H, then there is an r 3 ∈ {1, . .., s} such that t r 3 i = 1 and It might seem that an additional condition like "If a i and a j are not adjacent in H, then there is an r 4 ∈ {1, . . ., s} such that t r 4 i = 0 and t r 4 j = 0" would make a difference and be independent of conditions 1-3.However, this additional condition can be deduced from condition 1 or 3 because n ≥ 3 and so there is another vertex a k which is adjacent to a j , and if a i is also adjacent to a k , condition 1 implies it, and if a i and a k are not adjacent, condition 3 implies it.Thus, we did not include it in Definition 3.
Note that a hypergraph is separable if item 1 or 2 of Definition 3 holds.Therefore, every perfectly separable orthogonality hypergraph is also separable, but the converse is not always true.
Elementary counterexamples are true-implies-false gadgets such as the Specker bug depicted in Figure 8.

B. Reconstruction of perfectly separable hypergraphs
We know that whenever a i and a j are adjacent in H (or, equivalently, they are on the same context), then t ri = 1 implies t r j = 0 for r = 1, . . ., s.We want to know under what conditions the converse is true as well, i.e., the following adjacency criterion holds: Criterion 4. For all r = 1, . . ., s, if t ri = 1 implies t r j = 0 then a i and a j are adjacent.
The next theorem asserts that when H is a perfectly separable hypergraph, we can always reconstruct H from the information presented in its Travis matrix T (H) and the adjacency criterion.In addition, conversely, a reconstructable hypergraph using Criterion 4 has to be a perfectly separable hypergraph.In other words, Conjecture 6 is true for perfectly separable hypergraphs.Proof.First suppose that H is perfectly separable.Then, because of item 3 of Definition 3, for all a i and a j that are not adjacent in H there is a row r in T (H) such that t ri = 1 and t r j = 1.
Therefore, Criterion 4 in H ′ can only be satisfied for those vertices that are already adjacent in H.
Conversely, suppose that H = H ′ and a i and a j are two distinct non-adjacent vertices.Then the adjacency criterion does not meet for a i and a j .As a result, there is r, 1 ≤ r ≤ s such that t ri = 1 and t r j = 1, i.e., item 3 of Definition 3 already holds for a i and a j .Moreover, if item 1 (or 2) of Definition 3 does not hold for i and j, then the adjacency criterion makes a i adjacent to all vertices already adjacent to a j , which cannot happen unless a i = a j , or a i is colinear with a j , which is a contradiction.
Therefore, for every pair of non-adjacent vertices of H, items 1 to 3 of Definition 3 hold.If for any pair of adjacent vertices in H, namely a i and a j , both items 1 and 2 of Definition 3 are also true, then it can be inferred that H must be perfectly separable.To see this, first notice that if one of items 1 or 2 does not hold for adjacent vertices a i and a j , then one of them, say a j , is always assigned 0. Therefore, because H = H ′ , and H ′ is constructed via Criterion 4, a j must be adjacent to any vertex that is assigned 1 at least once.Hence, we can distinguish the following two cases: Case 1.The vertex a j belongs to all contexts of H. Then there is a two-valued state that assigns 1 to a j and 0 to all other vertices, a contradiction to our assumption that a j is always assigned 0. Case 2. There is at least one context, A , which does not contain a j .Since H = H ′ , only one vertex of A , say a p , can be assigned always 0. Then a j is adjacent to at least n − 1 vertices of A other than a p , and consequently, a j and a p are colinear, a contradiction.Therefore, items 1 and 2 of Definition 3 also hold for every pair of adjacent vertices of H. Now the conclusion is evident.
From the proof of Theorem 5 we see that, for a reconstructable hypergraph H, the most important factor is that at least item 1 or 2 is true for every pair of non-adjacent vertices.It should be noted that item 2 is actually the contraposition for item 1.Yet conversely, if we only want to imply Criterion 4, we see that for separable hypergraphs which are not perfectly separable, the reconstruction produces some extra hyperedges for H ′ and makes it different from H.For example, when we try to reconstruct the Specker bug from its two-valued states using the adjacency criterion, we produce an extra hyperedge on two vertices, as depicted in Figure 1.
However, one might think that we may be able to reconstruct H from H ′ by finding the extra hyperedge(s) that must be eliminated.To do this, suppose that ω([H] 2 ) = n ≥ 3 and every vertex is on a hyperedge of size n.Then the reconstructed H ′ must have the same property, i.e., we must have Criterion 2 for all hyperedges of H ′ , and if a hyperedge does not meet this criterion, then it has to be eliminated.This "unwanted" adjacency between a i and a j in H ′ appear only when these two are not on the same context but whenever one is assigned 1 the other has to be 0 and vice versa-a typical situation in a TIFS gadget-like what is happening when we try to reconstruct the Specker bug from its two-valued states.However, we show in Sec.III C that this method cannot always guarantee that we can identify these unwanted hyperedges from the table of two-valued states.

C. Examples of non-reconstructable hypergraphs using Criteria 2 and 4
Knowing that an orthogonality hypergraph is reconstructable from its two-valued states gives us the opportunity to find perpendicular pairs of propositions without actually measuring angles between them.Recall that, for two-valued states, whenever a vertex is assigned true, all other vertices adjacent to it have to be assigned false.This adjacency criterion (Criterion 4) is the most evident property that every pair of adjacent vertices has.
For a nonseparable hypergraph, however, this does not let us to reconstruct the hypergraph because for two distinct vertices a and b with t(a) = t(b) for every two-valued state t, it cannot be understood whether or not a is adjacent to all the neighbors of b.Therefore, it might be tempting to speculate that separability is a "good" criterion for reconstructability.Indeed, it might be tempting to claim the following general conjecture, against which we shall shortly present a counterexample.

Conjecture 6. A hypergraph H is reconstructable from the table of its two-valued states if and only if H is separable.
From Theorem 5 we know that for perfectly separable hypergraphs Conjecture 6 holds.Can we generalize this to separability?
In what follows, we present a counterexample: a hypergraph that is separable but not perfectly separable.For the sake of a contradiction with Conjecture 6 we introduce a configuration for which adjacency of vertices cannot be determined by the enumeration of all two-valued states (the Travis matrix) alone.The argument uses a maximal clique number n = 3, but can be generalized to higher clique numbers.
A counterexample to Conjecture 6 is an orthogonality hypergraph H that has some non-adjacent vertices but Criteria 2 and 4 detect and identify their adjacency "as if" they were on a hyperedge in the reconstructed hypergraph H ′ .Consequently these two hypergraphs H and H ′ are not the same.Note that, up to permutations of rows and columns, their Travis matrices are the same.
Since our dimension is 3, we must have at least three vertices a, b, and c, which are not adjacent in H, but whenever one is assigned true by a two-valued state, the other two have to be assigned false.Moreover, there must not be a two-valued state that simultaneously assigns a, b, and c false.
Suppose that G is an (a, b)-TIFS gadget such that whenever a is assigned true by a two-valued state, b has to be false.As mentioned earlier, the reverse is also true; that is, whenever b is assigned true by a two-valued state, a has to be false.This means that a and b cannot be assigned true at the same time (but they can both be false).Therefore, if we make a triangle, using a true implies false set (TIFS) as its edges, then we have the desired property that whenever one end is true, the other two are false.However, this hypergraph might have several two-valued states that assign false to all its three ends a, b and c.For an (a i , b i )-TIFS gadget G, Criteria 2 and 4 do not detect a hyperedge containing a 1 b 3 , a 2 b 1 and a 3 b 2 from the two-valued states of a hypergraph such as in Figure 2.This is because usually there are some two-valued states that assign false to all of these three vertices.Therefore, the layer hypergraph of Figure 2 must become a part of a larger hypergraph so that in every two-valued state, exactly one of the vertices a, b or c be assigned true.
To do so, one possibility is to use three copies of the layer hypergraph of Figure 2 and use three Again, since G 9 is a TIFS gadget, a ′′ has to be false.Now, a has to be assigned true by t r because a ′ and a ′′ are both false.
Therefore, for nonadjacent vertices a, b and c we have that exactly one of them has to be assigned true, and the other two are false.This is just like if these three vertices are on the same context.It means that if one tries to reconstruct B(G) from its table of two-valued states using criteria 2 and 4, extra hyperedges of {a, b, c}, {a ′ , b ′ , c ′ } and {a ′′ , b ′′ , c ′′ } are found.This implies that B(G) is not reconstructable using these criteria.
One question that arises in this regard is the following: suppose we know that G is separable, then does this imply that B(G) is also separable?This is not too hard to answer, and it is always "yes" if it does not contain a true implies true set (TITS) gadget.
To proceed, we need one more thing.A function f : S −→ T defined on a subset S X is said to be lifted to f : for each a ∈ S. It must be mentioned that when f possesses a property, like if f is a proper coloring [28] or a two-valued state, there might not always be a lift with the same property.
Lemma 7. Let G be a separable unitary TIFS gadget which does not contain a TITS gadget and Proof.Since G is unitary, there is no vertex that has to be assigned 0 by all the two-valued states of G. On the other hand, there is no vertex that is given 1 by all the two-valued states of G because else it must lie on at least one hyperedge with two other vertices, those that have to be always assigned 0, a contradiction to separability (and being unitary) of G. Thus, it can be inferred that for a vertex u of G, there are states ϕ and ψ such that ϕ(u) = 0 and ψ(u) = 1.
For an (a, b)-TIFS gadget G, the set of two-valued states are nonempty.Let s 1 be the set of those states in which a is true, s 2 be the set of those states that b is true and s 3 be the set of those states in which both a and b are false.Then these sets s 1 , s 2 and s 3 partition the set of twovalued states of G.For an (a, b)-TIFS gadget copy G i , these sets are shown here by s i1 , s i2 and s i3 .Therefore, if for example a two-valued state is in s 1 , it is a two-valued state of G so that its head, i. e. a, is assigned 1.
To show that B(G) is separable, we show that for any pair of distinct vertices x and y, there is a two-valued state of it that gives them different values.There are the following cases: Case 1. x and y belong to the same copy of G, say G i .Because of the symmetry, we can assume without loss of generality that i = 1.Since G is separable, there must be a two-valued state, say t on G, such that t(x) = t(y).We know that t ∈ s 1 ∪ s 2 ∪ s 3 .If t ∈ s l for l = 1, 2, 3, then there is a two-valued state for the underlying hypergraph of B(G) in Section A 4 such that it agrees with the values of t(a) and t(b).Now, using this two-valued state we define t for the end vertices a, b, c, a ′ , b ′ ,c ′ , a ′′ , b ′′ and c ′′ .Then using appropriate two-valued states of G, we can find a suitable two-valued state for internal vertices of G 2 , . .., G 9 .Therefore, there are two-valued states such as t of B(G) which is a lifting for t and t(x) = t(y).
Case 2. x and y lie on different copies of G, say G i and G j respectively.Then there are three other cases.
Case 2.1.G i and G j lie on the same layer of B(G).Without loss of generality, suppose that it is the layer consisting of the vertices a, b, and c.Then G i and G j have a common vertex that, again without loss of generality, we can assume it is a.Let G i 's head and tail be a and b, and G j 's head and tail be c and a, respectively [see Figure 4 (a)].Therefore, every two-valued state of the induced subhypergraph . Suppose on contrary that x and y receive the same value by all two-valued states of B(G).Then, since there is a two-valued state of B(G) that assigns 1 to x, at least one of the following statements holds: 1.If x is assigned 1 by a two-valued state of s i1 , then y has to be assigned 1 by all two-valued states of s j2 .This means that G j (and therefore G) is a (a, y)-TITS gadget, a contradiction to our assumption.

If
x is assigned 1 by a two-valued state of s i2 , then y has to be assigned 1 by all two-valued states of s j3 .Consequently, x cannot be assigned 0 by s i2 because else, it can be lifted to the required separation of x and y.Hence, G i (and therefore G) is a (b, x)-TITS gadget, a contradiction to our assumption.

If
x is assigned 1 by a two-valued state of s i3 , then y has to be assigned 1 by all two-valued states of s j1 .This means that G j (and therefore G) is a (c, y)-TITS gadget, again a contradiction to our assumption.

Case 2.2 G i and G j lie on different layers of B(G), and both ends of G i and G j lie on the same contexts of B(G).
Without loss of generality suppose that G i has a and b and G j has a ′ and b ′ as their heads and tails, respectively [see Figure 4 (b)].Therefore, every twovalued state of the induced subhypergraph G i ∪G j in B(G) is a member of s i1 ∪s j2 ∪s j3 , completely similar).Suppose on contrary that x and y receive the same value by all two-valued states of B(G).Then, since there is a two-valued state of B(G) that assigns 1 to x, at least one of the following statements holds: 1.If x is assigned 1 by a two-valued state of s i1 , then y has to be assigned 1 by all two-valued states of s j2 ∪ s j3 .This means that G j (and therefore G) is a (b ′ , y)-TITS gadget, a contradiction to our assumption.

If
x is assigned 1 by a two-valued state of s i2 , then y has to be assigned 1 by all two-valued states of s j1 ∪ s j3 .Consequently, x cannot be assigned 0 by a s i2 because else, it can be lifted to the required separation of x and y.Hence, G i (and therefore G) is a (b, x)-TITS gadget, a contradiction to our assumption.
3. If x is assigned 1 by a two-valued state of s i3 , then y has to be assigned 1 by all twovalued states of s j1 ∪ s j2 .This means that G j (and therefore G) is a (b ′ , y)-TITS gadget (and also a (c ′ , y)-TITS gadget), again a contradiction to our assumption.

Case 2.3 G i and G j lie on different layers of B(G), but only one end from G i and one end from
G j lie on the same context.Again without loss of generality, suppose that G i has a and b and G j has b ′ and c ′ as their heads and tails, respectively [see Figure 4 (c)].
Therefore, every two-valued state of the induced subhypergraph G i ∪ G j in B(G) is a member of s i1 ∪s j1 ∪s j2 ∪s j3 , s i2 ∪s j2 ∪s j3 , or s i3 ∪s j1 ∪s j2 ∪s j3 (or s j1 ∪s i1 ∪s i2 ∪s i3 , s j2 ∪s i2 ∪s i3 , or s j3 ∪s i1 ∪s i2 ∪s i3 which can be treated similarly).Suppose on contrary that x and y receive the same value by all two-valued states of B(G).Then, since there is a two-valued state of B(G) that assigns 1 to x, at least one of the following statements holds: 1.If x is assigned 1 by a two-valued state of s i1 , then y has to be assigned 1 by all two-valued states of s j1 ∪ s j2 ∪ s j3 , a contradiction to the fact that no vertex of G can be assigned 1 by all the two-valued states.
2. If x is assigned 1 by a two-valued state of s i2 , then y has to be assigned 1 by all two-valued states of s j2 ∪ s j3 .Consequently, G j (and therefore G) is a (c ′ , y)-TITS gadget, a contradiction to our assumption.
3. If x is assigned 1 by a two-valued state of s i3 , then y has to be assigned 1 by all two-valued states of s j1 ∪ s j2 ∪ s j3 , a contradiction to the fact that no vertex of G can be assigned 1 by all the two-valued states.
We showed that in any case, x and y can be separated by a two-valued state of B(G) (or else there is a contradiction) which concludes the proof.the elementary methods of counting, the number of two-valued states of For the sake of an example take the Specker bug G discussed in the Appendix Section A 3 of Appendix A. From its Travis matrix A3 we know that n a = 3, n b = 3 and n n = 8.Therefore, Formula 1 implies that B(G), which is a separable hypergraph on 108 vertices and 66 contexts, two-valued states, a number that can easily be checked via an ordinary computer.
However, it is not difficult to show that this hypergraph B(G), when G is the Specker bug, does not meet our requirements.This is because the two ends of the Specker bug cannot be orthogonal in the 3-space [29].It can also be discussed using graph theoretical terminology; one orthogonality hypergraph cannot have a cycle of length 4 because else any pair of antipodal vertices of the cycle of length 4 have to be colinear.Therefore, even if B(G) is an orthogonality hypergraph, the reconstructed hypergraph with an extra context of {a, b, c} is certainly not.
We have to find a TIFS gadget H, other than the Specker bug, in which the distance between its two ends is not 3 (so the two ends can be orthogonal in the reconstructed hypergraph).A candidate for such a hypergraph is shown in Figure 5, which is a TIFS on 43 vertices whose end points (say a and b) are far enough, so that it is not only separable, but also probably a FOR.This hypergraph If we construct B(H), then Lemma 7 implies that the resulting hypergraph on 378 vertices is separable and there would be enough space for the three end vertices to be perpendicular.
Another pertinent problem is to show that these hypergraphs-namely B(H) and also its counterpart, the reconstructed hypergraph B(H) ′ from B(H)'s table of two-valued states-have a faithful orthogonal representation; and to enumerate an explicit example of such a representation.
However, it seems that Criterion 1 is independent of Criteria 2 and 4, having an example such as B(H) on 378 vertices raises the possibility that Conjecture 6 is false for separable hypergraphs that are not perfectly-separable.
One challenge is to find either TIFS that allows for FORs-that is, vertex labellings by vectors-with end points that are orthogonal (that is, their relative angle is π/2) and at the same time have a separating set of two valued states; or a proof of nonexistence thereof.Note that since, unlike TIFS, TITS gadgets in general perform asymmetric, it is not possible to employ a serial composition strategy similar to the one of Kochen and Specker [1] for a construction of their Γ 2 : to concatenate a couple of TITS with a single TIFS at their respective end points and thereby to To prove the converse, suppose that we have a partition S = {S 1 , . . ., S n } of vertices of H which satisfies Properties 1 and 2 above.Define, σ : V (G) −→ {1, . . ., n} such that While Property 2 implies that every S i is non-empty, for i = 1 . . ., n, it also shows that every hyperedge contains a vertex v such that σ (v) = i.In other words, every color i = 1, . . ., k is used in each hyperedge.Furthermore, Property 1 implies that every S i is an independent set.Hence σ is a proper coloring of H and consequently, H is n-colorable.
As This does not exclude the existence of partition logics which are not semi-perfect.Indeed, in general, their chromatic number can exceed their 2-section's clique number.A concrete example is Greechie's G 32 [8,Figure 6,p. 121] mentioned in Appendix B, and depicted in Figure 11.

B. Reconstructing coloring from logical assignments
From Theorem 8 we know that when there is an n-coloring for an orthogonality hypergraph H, there are n two-valued states corresponding to it so that they induce a partition logic.In other words, there are n rows in the Travis matrix of H such that when one of them assigns 1 to a vertex one has exhausted all combinations of two-valued states one could still attempt to "complete" the coloring by identifying the missing colors with "suitable segments" of the remaining two-valued states (if there are any leftovers).Of course, in this way, the column sums of all the respective two valued states cannot be 1, and hence Formula (2) is no longer valid.In any case, Brooks' theorem [30,31]-stating that for any connected undirected graph G, the chromatic number of G is at most its maximum degree (the maximal number of edges that are incident to some vertex) ∆ unless G is a complete graph or an odd cycle, in which case the chromatic number is ∆ +1-and its generalization to hypergraphs [14, page 45, Theorem 3.2] yield an upper bound for the chromatic number of such (hyper)graphs.

V. SUMMARY AND CONCLUDING REMARKS
We have presented a constructive, algorithmic way to generate a coloring of a (hyper)graph from its set of two-valued states.The only criterion for the success of this approach is the assertion that the respective hypergraph is semi-perfect, that is, its chromatic number equals the clique number of its 2-section.We have been able to find a "compact" partition logic within the logical states of the hypergraph by showing that n-colorability is equivalent to finding a partition logic based on exactly n two-valued states.We also presented a detailed algorithm for constructively finding this partition logic and its associated coloring.
With regard to representing and reconstructing (hyper)graphs or logics in terms of their twovalued states, in particular, regarding separability of vertices or elementary propositions, we conjecture that there exist quantum logics with a separable set of two-valued states that cannot be reconstructed from these states.We have presented a hypergraph, namely B(G) depicted in Fig- ure 3 of Section III C with a TIFS gadget such as the one depicted in Figure 5, that has this characteristic but we could not find a faithful orthogonal representation in a Hilbert space.
Yet, stronger forms of separability, in particular, perfect separability, can be identified that allow (hyper)graphs or logics to be represented and reconstructed in terms of their two-valued states (that is, by their Travis matrices).In addition, while the conditions on perfectly separable (hyper)graphs are rather strong, one can be certain that such a reconstruction exists.
Indeed, such a reconstruction helps to directly identify mutually perpendicular elementary propositions, and thus the contexts corresponding to the maximal operators they form: if an orthogonality (hyper)graph is reconstructible from its set of two-valued states we can deduce the mutual orthogonality of the elementary quantum propositions by just looking at these two-valued states.This facilitates the construction of the (mutually perpendicular) orthogonal operators in the spectral sums associated with the contexts, and thus supports finding a global faithful orthogonal representation, i. e., the assignment of vectors to vertices, of (hyper)graphs.
Stated differently, we showed that there is a class of hypergraphs, namely perfectly separable ones, that are always reconstructable from their two-valued states.However, not all separable graphs are guaranteed to be reconstructible by these means.
Hence, while for perfectly separable (hyper)graphs we can be certain that they can be reconstructed; and for Kochen-Specker type (hyper)graphs that they cannot be reconstructed because there is no two-valued state associated with any classical value assignment, for the remaining (hyper)graphs reconstructability remains an open question.
Appendix A: Examples

Triangle logic
The coloring procedure of the triangle hypergraph is depicted in Figure 6.Consider the set of all four two-valued states on the six atoms which can be tabulated by a (compactified) Travis matrix T i j whose rows indicate the ith state s i and whose columns indicate the atoms a j , respectively; that is, T i j = s i (a j ): It is not too difficult to see that the first three measures, represented by the first three row vectors of the Travis matrix, add up to 1, 1, 1, 1, 1, 1 .They can thus be taken as the basis of a coloring.3 with the second color (green) and 2 with the third color (blue), thereby identifying a 2 and a 5 with green, and a 3 and a 6 with blue, respectively.Note that s 1 , s 2 , and s 3 "generate" a 3-partitioning of the set of atoms

House or pentagon or pentagram logic
The Travis matrix of the house or pentagon or pentagram logic depicted in Fig. 7 is a matrix representation of its 11 dispersion free states [32] T A coloring can be obtained from the earlier mentioned construction which results in three states partitioning all 10 atoms.The associated 1st, the 8th and the 11th row vectors of T i j are partitioning the 10 atoms.
Another way of seeing this is to associate a color to, say, the first state.As a consequence, all other states, namely states number 2, 3, 4, 5, and 6, need to be eliminated, leaving no state which can be associated with another color.
One possibility for finding a proper coloring is to drop "exclusivity", or rather, the unique asso- to (amendments are indicated by square brackets "[. ..]") [. ..] present [. ..] lattices as unions of [contexts] intertwined or pasted together in some fashion [. ..] by replacing, for example, the 2 n elements in the Hasse diagram of the power set of an n-element set with the [context aka] complete [sub]graph [K n ] onn elements.The reduction in numbers of elements is considerable but the number of remaining "links" or "lines" is still too cumbersome for our purposes.We replace the [context aka] complete [sub]graph on n elements by a single smooth curve (usually a straight line) containing n distinguished points.Thus we replace n(n + 1)/2 "links" with a single smooth curve.This representation is propitious and uncomplicated provided that the intersection of any pair of blocks contains at most one atom.

Criterion 1 .
There is a vertex labeling x x x : V (H) −→ L that assigns a set of k mutually non-colinear vectors in an n-dimensional Hilbert space L to vertices of H such that any pair of vertices a and b are adjacent if and only if x x x(a) is orthogonal to x x x(b).
For a hypergraph H, the chromatic number m = χ(H) is the minimum number of colors required for a proper coloring of vertices of H. Obviously the clique numbern = ω([H] 2 ) is a lower bound.If these numbers are the same, that is, if m = χ(H) = ω([H] 2 )= n, then one could obtain two-valued measures from colorings by "projecting" one of the colors into the value 1, and all the other n − 1 colors into the value 0[3][4][5].A hypergraph H, whose chromatic and clique numbers are equal, is called here semi-perfect.

Criterion 2 .
If a and b are adjacent in an orthogonality hypergraph H, then there is a hyperedge that contains a and b along with n − 2 other vertices.

Figure 1 .
Figure 1.Reconstruction of the Specker bug using Criterion 4 which gives the gray snake-shaped extra hyperedge that contains only two (the red and green) vertices.This extra hyperedge can be easily eliminated if we use Criterion 2.

Figure 2 .
Figure2.A layer hypergraph serving as a quasi-block for the construction of a counterexample to Conjecture 6.For i = 1, 2, 3, the snake-shaped edges are distinct copies of an arbitrary separable (a i , b i )-TIFS gadget G such that whenever a i is assigned true by a two-valued state, b i has to be false: their respective ends have been "cyclically folded" on each other, eg., a 1 from G 1 is identified with b 3 from G 3 .Note that vertices a 1 b 3 , a 2 b 1 and a 3 b 2 are not adjacent.

Figure 2
depicted this layer hypergraph, with a = a 1 b 3 , b = a 2 b 1 and c = a 3 b 2 .

Figure 3 .
Figure 3.A hypergraph B(G) depicting a counterexample to Conjecture 6.For i = 1, . . ., 9, the snakeshaped curves indicate different copies of an (a i , b i )-TIFS gadget G with their terminals suitably identified, that is, a 1 from G 1 is identified with b 3 from G 3 as the vertex a. Straight lines are ordinary hyperedges, i. e., vertices a, a ′ and a ′′ are on a context, drawn in brown.
We can go further by calculating the number of two-valued states of B(G) based on G's.Suppose that G is an (a, b)-TIFS and has, respectively, n a , n b , and n n two-valued states that give a true, b true, and none of a and b true.In other words, n a = |s 1 |, n b = |s 2 |, and n n = |s 3 |.Then, by

Figure 5 .
Figure 5.An (a, b)-TIFS gadget whose distance between its two terminal points a and b is at least five contexts.The snake-like decorated curves indicate Specker bugs, so this hypergraph has 43 vertices.

Figure 1 ,
p. 123], among others).It is a subgraph of G 32 introduced later in Figure 11.Its Travis matrix is

Figure 8 .Figure 9 .
Figure 8. Coloring scheme of the "Specker bug" gadget[33,35] from two-valued states.The set-theoretic representation is in terms of the canonical partition logic as an equipartitioning of the set {1, 2, . . ., 14} obtained from all 14 two-valued states on this gadget.

Figure 10 .
Figure 10.Coloring scheme of the "tight GHZ" logic[36] from two-valued states.The set-theoretic representation is in terms of the canonical partition logic as an equipartitioning of the set {1, 2, . . ., 8} obtained from all eight two-valued states on this gadget.
Theorem 5. Let H be a hypergraph on {a 1 , a 2 , . .., a k } whose Travis matrix T (H) is available and ω([H] 2 ) = n ≥ 3.Moreover, suppose that H ′ is the hypergraph on {a 1 , a 2 , . . ., a k } whose adjacency is defined by Criterion 4. Then H = H ′ if and only if H is perfectly separable.
Therefore, the following corollary (Corollary 9) is a consequence of Theorem 8.
a result, we can say that for each proper n-coloring of H we have n different two-valued states on vertices of H. Conversely we can construct exactly n! proper n-colorings for H from an available n-partition system on vertices of H.
Algorithm1 Finding an n-coloring for H from its set of two-valued states encoded by the Travis matrix Append to RemovedRows[i] all AvailableRows[s] for which there is a vertex u such that the state of rows AvailableRows[s] and AvailableRows[ j] both assign 1 to u Moreover, one could conjecture that Algorithm 1 could be modified to render a coloring even if the (hyper)graph is not n-partitionable, in which case Theorem 8 does not apply.Because even if Coloring scheme of the house or pentagon or pentagram logic from the set of two-valued states.