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5 Partial Cubes 5 Partial Cubes This is the central chapter of the book. Unlike general cubical graphs, iso- metric subgraphs of cubes—partial cubes—can be effectively characterized in many different ways and possess much finer structural properties. In this chapter, we present what can be called a “micro” theory of partial cubes. 5.1 Definitions and Examples As cubical graphs are subgraphs of cubes, partial cubes are isometric sub- graphs of cubes (cf. Section 1.4). Definition 5.1. A graph G is a partial cube if it can be isometrically embed- ded into a cube H(X) for some set X. We often identify G with its isometric image in H(X) and say that G is a partial cube on the set X. Note that partial cubes are connected graphs. Clearly, they are cubical graphs. The converse is not true. The smallest counterexample is the graph G on seven vertices shown in Figure 5.1a. z {a,b,c} x y {a,b} {b,c} a) b) a c u v w { } {b} { } G s Ø Figure 5.1. A cubical graph that is not a partial cube. S. Ovchinnikov, Graphs and Cubes, Universitext, DOI 10.1007/978-1-4614-0797-3_5, 127 © Springer Science+Business Media, LLC 2011 128 5 Partial Cubes Indeed, suppose that ϕ : G → H(X) is an embedding. We may assume that ϕ(s)= ∅. Then the images of vertices u, v, and w under the embedding ϕ are singletons in X, say, {a}, {b}, and {c}, respectively. It is easy to verify that we must have (cf. proof of Theorem 4.42) ϕ(y) = {b,c}, ϕ(z)= {a,b,c} and [ϕ(x)= {a, b} or ϕ(x)= {a,c}]. We assume that ϕ(x) = {a, b} (see Figure 5.1b); the other case is treated similarly. The graph distance between vertices v and x in G is 3, whereas the distance between their images {b} and {a, b} in H(X) is 1. Therefore, ϕ is an embedding but not an isometric embedding. In Section 4.1 we proved that paths, even cycles, and trees are cubical graphs (cf. Figure 4.1). In fact, they are partial cubes. It is not difficult to prove this assertion directly using constructions from Section 4.1 (cf. Exercise 5.1). It also follows from one of our characterization theorems (see Section 5.4). These theorems can be used to show that all cubical graphs in Figure 4.5 are partial cubes, proving that the graph in Figure 5.1a is indeed the smallest counterexample (cf. Exercise 5.2). 5.2 Well-Graded Families of Sets Let G be a partial cube on X. Because G is an isometric subgraph of H(X), for any two vertices P and Q of G there is a sequence of vertices of G, R0 = P, R1,...,Rn = Q such that d(P,Q) = n and d(Ri, Ri+1) = 1 for 0 ≤ i < n, where d is the Hamming distance on H(X). The sequence (Ri) is a shortest path connecting P to Q in G (and in H(X)). Definition 5.2. A family1 F of distinct subsets of a set X is called a well- graded family of sets (wg-family for short) if, for any two distinct subsets P,Q ∈ F, there is a sequence of sets in F R0 = P, R1,...,Rn = Q such that |R0 △ Rn| = n and |Ri △ Ri+1| = 1, for all 0 ≤ i < n. (5.1) According to this definition, the vertex set of a partial cube on X is a well-graded family of finite subsets of X. X Let P(X) = K2 be the Cartesian power of K2 (cf. Section 3.5). Any nonempty family F of subsets of X induces a subgraph GF = (F, EF) of P(X), where EF = {{P,Q}∈ F : |P △ Q| = 1}. 1 To avoid trivialities, we assume that families of sets under consideration contain more than one element. 5.2 Well-Graded Families of Sets 129 Theorem 5.3. (i) The graph GF is an isometric subgraph of P(X) if and only if the family F is well-graded. (ii) If F is well-graded, then GF is a partial cube. Proof. (i) (Necessity.) Suppose that GF is an isometric subgraph of P(X). Because the graph GF is connected, it belongs to a connected component of the graph P(X). Hence, by Lemma 3.5, P △Q is a finite set for any P,Q ∈ F. Because GF is an isometric subgraph, F is a wg-family. (Sufficiency.) Let F be a wg-family and P , Q be two sets in F. It is clear that a sequence (Ri) of vertices of GF is a shortest PQ-path in GF if and only if it satisfies condition (5.1). Therefore the graph GF is an isometric subgraph of P(X). (ii) Let F be a wg-family of subsets of X. Then, by part (i), GF is an iso- metric subgraph of P(X). Let A be a vertex of GF. Because GF is connected, it is a subgraph of the connected component H(A) of P(X). By Theorem 3.9, the graph H(A) is isomorphic to the cube H(X). It follows that GF is a partial cube. The converse of claim (ii) does not hold as the following example demon- strates. Example 5.4. Let GF be a subgraph of the cube Q3 = H({a,b,c}) induced by the family F = {∅, {a}, {a, b}, {a,b,c}, {b,c}} (see Figure 5.2). Clearly, GF is not an isometric subgraph of Q3. On the other hand, GF is the path P5 and therefore is a partial cube (Exercise 5.1). Thus, GF is a partial cube, but not a partial cube on X = {a,b,c}. {a,b,c} {a,c} {a,b} {b,c} a { } {b} {c} Ø Figure 5.2. An induced subgraph of Q3. Example 5.5. Let F be the family of all subsets of Z in the form (−∞, n]. It is not difficult to see that F is a wg-family of infinite subsets of Z. The graph GF is isomorphic to the double ray Z. Therefore, by Theorem 5.3(ii), Z is an infinite partial cube (cf. Exercise 5.3). 130 5 Partial Cubes Corollary 5.6. The graph GF is a partial cube on X if and only if F is a wg-family of finite subsets of X. Sequences of vertices (Ri) satisfying condition (5.1) are shortest paths in the graph P(X). If F is a wg-family, then they are also shortest paths in the graph GF. The next two theorems establish some useful properties of these sequences. Theorem 5.7. Let F be a wg-family of subsets of a set X and let P = R0, R1,...,Rn = Q be a shortest path in GF. Then (i) d(Ri, Rj)= |j − i|. (ii) Ri,...,Rj is a shortest RiRj-path in GF. (iii) d(Ri, Rj)= d(Ri, Rk)+ d(Rk, Rj) for i ≤ k ≤ j. (iv) Ri ∩ Rj ⊆ Rk ⊆ Ri ∪ Rj for i ≤ k ≤ j, where d is the Hamming distance on P(X). Proof. (i) and (ii) are the results of Lemma 2.5. It remains to note that (iii) follows immediately from (i), and that (iii) and (iv) are equivalent conditions by Theorem 3.22. Theorem 5.8. Let F be a wg-family of subsets of a set X and let P = R0, R1,...,Rn = Q be a shortest path in GF. Then (i) Either Ri+1 = Ri ∪ {x} for some x ∈ Q \ P , or Ri+1 = Ri \{x} for some x ∈ P \ Q. Equivalently, for any 0 ≤ i < n there is x ∈ P △ Q such that Ri △ Ri+1 = {x}. (ii) For any x ∈ P △ Q there is 0 ≤ i < n such that Ri △ Ri+1 = {x}. Proof. (i) Inasmuch as |Ri△Ri+1| = d(Ri, Ri+1) = 1, we have Ri△Ri+1 = {x} for some x ∈ X. Hence, either Ri+1 = Ri ∪{x} and x∈ / Ri, or Ri+1 = Ri \{x} and x ∈ Ri. Suppose that Ri+1 = Ri ∪ {x}, x∈ / Ri. By Theorem 5.7(ii), Ri+1 lies between Ri and Rn = Q. Therefore, by Theorem 3.22, Ri+1 ⊆ Ri ∪ Q. Hence, x ∈ Q. Suppose now that Ri+1 = Ri \{x}, x ∈ Ri. Then, by Theorem 5.7(ii), Ri lies between Ri+1 and R0 = P . Therefore, by Theorem 3.22, Ri ⊆ Ri+1 ∪ P . It follows that x ∈ P . (ii) Let x ∈ P △Q. By symmetry, we may assume that x ∈ Q\P . We prove claim (ii) by induction on n = d(P,Q). For n = 1 the statement is trivial. For the induction step, suppose that n > 1 and assume that (ii) holds for paths 5.2 Well-Graded Families of Sets 131 of smaller length. If x ∈ Q \ Rn−1, we are done. Otherwise, x ∈ Rn−1 and the result follows from the induction hypothesis. We say that two distinct sets P and Q in a family of sets F are adjacent in F if P ∩ Q ⊆ R ⊆ P ∪ Q and R ∈ F implies R = P or R = Q. In other words, P and Q are adjacent in F if they are the only sets in F that lie between P and Q. One should distinguish two concepts of adjacency for a given family of sets F: adjacency in GF and adjacency in F. Clearly, two vertices P and Q that are adjacent in GF are also adjacent in F.
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