Therefore Γ′ is not 1-lit for lit-only σ-game.
4.8 Combinatorial statements of Theorems 4.5.7 and 4.7.7
In order to easily execute Theorems 4.5.7 and 4.7.7, the goal of this section is to state the combinatorial versions of those results.
By Proposition 4.2.1 we restate Theorem 4.5.7 as follows.
Theorem 4.8.1. Assume that Γ is a tree with a perfect matching. Then Γ is 1-lit.
Assume that Γ = (S, R) is a tree with a perfect matching P. By an alternating path in Γ (with respect to P), we mean a path in which the edges belong alternatively to P and not to P.
Definition 4.8.2. Assume Γ = (S, R) is a tree with a perfect matchingP. For each s ∈ S define As to be the set consisting of all t ∈ S \ {s} such that the path between s and t is an alternating path which starts from and ends on edges in P. For each s ∈ S we say that As has even parity whenever the cardinality of As is even and odd parity otherwise.
Lemma 4.8.3. Assume Γ = (S, R) is a tree with a perfect matching P. Let As (s ∈ S) for all s∈ S. The result follows.
By Corollary 4.5.5 and Lemma 4.8.3, we restate Theorem 4.7.7 as follows.
Theorem 4.8.4. Assume that Γ = (S, R) is a tree with a perfect matching. Let x, y∈ S such that xy ∈ R. Let Γ′ denote the tree obtained from Γ by inserting a new vertex on the edge xy of Γ. Assume that Ax, Ay, defined as Definition 4.8.2, have distinct parities.
Then Γ′ is 1-lit.
We now illustrate Theorems 4.8.1 and 4.8.4 with two examples.
Example 4.8.5. Assume that Γ = (S, R) is the tree shown in Figure 5.
Figure 5: a tree of order 12.
Since Γ contains the perfect matching {{1, 2}, {3, 4}, {5, 6}, {7, 8}, {9, 10}, {11, 12}} and by Theorem 4.8.1, Γ is 1-lit. We next show that any tree obtained from Γ by inserting a new vertex on an edge of Γ is 1-lit. To see this, it suffices to show the four trees shown in Figure 6 are 1-lit. Figure 6: four 1-lit trees of order 13.
Let D ={{1, 2}, {3, 4}, {5, 6}, {4, 7}}. Observe that
A1 ={2, 4, 6, 8, 10, 12}, A2 ={1}, A3 ={4, 8, 12},
A4 ={1, 3}, A5 ={6}, A6 ={1, 5}, A7 ={8}. (4.40) Pick any xy ∈ D. By (4.40) and by Theorem 4.8.4 the tree obtained from Γ by inserting a new vertex on the edge xy of Γ is 1-lit. Therefore the four trees in Figure 6 are 1-lit. Example 4.8.6. The aim of this example is to show that the trees shown in class IV of Figure 1 are 1-lit by using Theorem 4.8.1 and Theorem 4.8.4. Let k ≥ 3 be an integer.
Suppose that Γ = (S, R) is the tree of order 2k shown in Figure 7. Let P denote the path in Γ between the two vertices 2 and 2k. It suffices to show that Γ and the tree obtained from Γ by inserting a new vertex on some edge of P are 1-lit.
c c c c c
Figure 7: a 1-lit tree of order 2k.
Since Γ contains the perfect matching {{1, 2}, {3, 4}, . . . , {2k − 1, 2k}} and by Theo-rem 4.8.1 Γ is 1-lit. It is routine to check that A2 ={1} and A6 ={1, 5}. Therefore there exists x ∈ {2, 6} such that A5 and Ax have distinct parities. By Theorem 4.8.4 the tree obtained from Γ by inserting a new vertex on the edge {5, x} of P is 1-lit. The result
follows.
Chapter 5
The edge-version of lit-only sigma-game
In this chapter we consider the edge-version of lit-only σ-game, which is called e-lit-only σ-game. We now describe this game on Γ = (S, R). A configuration is an assignment of one of two states, on or off, to all edges of Γ. Given a configuration, a move allows the player to choose one on edge ϵ of Γ and change the states of all adjacent edges ϵ′ of ϵ; i.e.
|ϵ∩ϵ′| = 1. Let L(Γ) denote the line graph of Γ. We may view this variation as the lit-only σ-game on L(Γ). We denote the flipping group of L(Γ) by WR, and call this the edge-flipping group of Γ. LetZ denote the additive group of integers. Let n and m denote the numbers of vertices and edges of Γ respectively. Assume n≥ 3. The goal of this chapter is to show that WR is isomorphic to a semidirect product of (Z/2Z)k and the symmetric group Sn of degree n, where k = (n− 1)(m − n + 1) if n is odd; k = (n − 2)(m − n + 1) if n is even.
5.1 The edge space and the bond space
In this chapter let |S| = n and |R| = m. In this section we mention some properties about the edge space and the bond space of Γ that we will need. The reader may refer to [24, p.23–p.28] for details.
Let R denote the power set of R. For any F, F′ ∈ R define F + F′ :={ϵ ∈ R | ϵ ∈ F ∪ F′, ϵ /∈ F ∩ F′}; i.e. the symmetric difference of F and F′. Define 1· F := F and 0· F := ∅, the empty set. The set R forms a vector space over F2 and this is called the edge space of Γ. Note that the zero element of R is ∅ and −F = F for F ∈ R. Observe {{ϵ} | ϵ ∈ R} is a basis of R. Therefore the dimension dim R of R is m.
For a subset U of S let R(U ) denote the subset of R consisting of all edges of Γ that have exactly one element in U. In graph theory R(U ) is often called an edge cut of Γ if U is a nonempty and proper subset of S. Notice that R(ϵ) = R({x, y}) for ϵ = {x, y} ∈ R.
For convenience R(s) := R({s}) for s ∈ S.
Proposition 5.1.1. The following (i), (ii) hold.
(i) Each ϵ = {x, y} ∈ R lies in exactly two edge cuts R(x) and R(y) among R(s) for all s∈ S.
(ii) For U ⊆ S we have R(U) =∑
s∈UR(s).
Proof. (i) is immediate from the definition of R(s) for s ∈ S. (ii) is immediate from (i) and the definition of R(U ).
For the rest of this chapter let B denote the subspace of R spanned by R(s) for all s∈ S. This is called the bond space of Γ.
Proposition 5.1.2. The following (i)–(iv) hold.
(i) B = {R(U) | U ⊆ S}.
(ii) The dimension dim B of B is n − 1.
(iii) For each t∈ S, R(t) =∑
s∈S\{t}R(s).
(iv) For each t∈ S the set {R(s) | s ∈ S \ {t}} is a basis of B.
Proof. (i) follows immediately from Proposition 5.1.1(ii). Similar to the edge space of Γ, the power set S of S forms a vector space over F2. Clearly the dimension of S is n.
Observe that the map from the vertex spaceS onto the bond space B of Γ, defined by U 7→ R(U) for U ∈ S,
is a linear transformation with kernel {∅, S}. It follows that dim B = n − 1. This proves (ii). Let u∈ S. Since R(S) = ∅ we have R(t) = R(t)+R(S). By this and Proposition 5.1.1, (iii) follows. (iv) is immediate from (ii), (iii).
For the rest of this chapter let T denote a minimal subset of R such that (S, T ) is connected. We call T a spanning tree of Γ. Note that |T | = n − 1.
Proposition 5.1.3. The subset{F ∈ R | F ⊆ R\T } of R is a set of coset representatives of B in R.
Proof. There are 2m−n+1 cosets of B in R because of dim B = n − 1 and dim R = m. It is clear that |{F | F ⊆ R \ T }| = 2m−n+1. For any two distinct F, F′ ⊆ R \ T the graph (S, R− (F − F′)) is still connected since T ⊆ R − (F − F′), which implies that F− F′ is not an edge cut of Γ. By Proposition 5.1.2(i), F − F′ ̸∈ B. Therefore {F | F ⊆ R \ T } is a set of coset representatives of B in R.