#### 2.1. Nondeterministic Constraint Logic

Let

G be a 3-regular graph with edge weights among

$\{1,\phantom{\rule{3.33333pt}{0ex}}2\}$. A vertex of

G is an

AND vertex if exactly one incident edge has weight 2, and a vertex of

G is an

OR vertex if all the incident edges have weight 2. A graph

G is a

constraint graph if it consists of only AND vertices and OR vertices. An orientation

F of

G is

legal if for every vertex

v of

G, the sum of weights of in-coming edges of

v is at least 2. A

legal move from a legal orientation is the reversal of a single edge that results in another legal orientation.

Figure 1 illustrates all the possible orientations of edges incident to an AND vertex. We can also verify that all the possible legal move of an incident edge of the AND vertex are those depicted by the arrows in

Figure 1. Given a constraint graph

G and two legal orientation

${F}_{0}$ and

${F}_{t}$ of

G, the

nondeterministic constraint logic problem asks whether there is a sequence of legal orientations

${F}_{0},{F}_{1},\dots ,{F}_{t}$ such that

${F}_{i}$ is obtained from

${F}_{i-1}$ by a legal move for each

i with

$1\le i\le t$. Such a sequence of legal orientations is called a

reconfiguration sequence. The nondeterministic constraint logic problem is known to be PSPACE-complete even if the constraint graph is planar [

11]. See [

18] for more information on constraint logic.

For convenience of the reduction, we define a problem slightly different from the nondeterministic constraint logic problem. Let G be a bipartite graph with bipartition $(A,B)$ such that every vertex of A has degree 3 and every vertex of B has degree 2 or 3. The graph G has edge weights among $\{1,2\}$ such that for every vertex of A, exactly one incident edge has weight 2. An orientation F of G is legal if

for every vertex $v\in A$, the sum of weights of in-coming edges of v is at least 2, and

every vertex of B has one or two in-coming edges, but at most one vertex of B has two in-coming edges.

A

legal move from a legal orientation is the reversal of a single edge that results in another legal orientation. Notice that, in the legal moves, the vertices of

A behave in the same way as the AND vertices of the nondeterministic constraint logic problem, that is, as shown in

Figure 1. Given such a bipartite graph

G and two legal orientation

${F}_{0}$ and

${F}_{t}$ of

G, the problem

$\Pi $ asks whether there is a sequence of legal orientations

${F}_{0},{F}_{1},\dots ,{F}_{t}$ such that

${F}_{i}$ is obtained from

${F}_{i-1}$ by a legal move for each

i with

$1\le i\le t$. We further add a constraint to the instance of the problem

$\Pi $ so that every vertex of

B has exactly one in-coming edge in

${F}_{0}$ and

${F}_{t}$.

**Lemma** **1.** The problem Π is PSPACE-complete.

**Proof.** We can observe that the problem

$\Pi $ is in PSPACE ([

8], Theorem 1). We thus show a polynomial-time reduction from the nondeterministic constraint logic problem. Let

$(G,{F}_{0},{F}_{t})$ be an instance of the problem, that is,

G is a constraint graph, consisting of AND vertices and OR vertices, and

${F}_{0}$ and

${F}_{t}$ are two legal orientations of

G. We construct an instance

$({G}^{\prime},{F}_{0}^{\prime},{F}_{t}^{\prime})$ of the problem

$\Pi $ such that

$(G,{F}_{0},{F}_{t})$ is a yes-instance if and only if

$({G}^{\prime},{F}_{0}^{\prime},{F}_{t}^{\prime})$ is a yes-instance.

Let

${G}^{\u2033}$ be the bipartite graph obtained from

G by replacing each edge

$uv$ with two edges

$uw$ and

$wv$ so that

$uw$ and

$wv$ have the same weight as

$uv$, where

w is a newly added vertex. The bipartite graph

${G}^{\prime}$ with bipartition

$(A,B)$ is obtained from

${G}^{\u2033}$ by replacing each OR vertex with a subgraph shown in

Figure 2, where

A consists of the AND vertices of

G and the white points in the subgraphs (see

Figure 2) while

B consists of the newly added vertices of

${G}^{\u2033}$ and the gray points in the subgraphs. We can check that all the vertices of

A are incident to one weight-2 edge and two weight-1 edges.

Let

F be a legal orientation of

G. We define a legal orientation

${F}^{\prime}$ of

${G}^{\prime}$associated withF. Let

${F}^{\u2033}$ be the orientation of

${G}^{\u2033}$ obtained from

F by replacing each edge

$(u,v)\in F$ with two edges

$(u,w)$ and

$(w,v)$, where

w is the newly added vertex. Let

${F}^{\u2033\prime}$ be an orientation of

${G}^{\prime}$ obtained from

${F}^{\u2033}$ by replacing each OR vertex with the subgraph in

Figure 2 such that if

L is directed inward (resp. outward) in

${F}^{\u2033}$ then the edges

${L}_{0}$ and

${L}_{1}$ and the weight-1 edges between them are directed inward (resp. outward) in

${F}^{\u2033\prime}$ (and similarly for the edges

R and

D). The legal orientation

${F}^{\prime}$ is obtained from

${F}^{\u2033\prime}$ by reversing the direction of the edges incident to the OR vertices so that exactly one edge of

$\{{L}_{1},{R}_{1},{D}_{1}\}$ is directed inward for each OR vertex. Notice that at least one edge of

$\{{L}_{1},{R}_{1},{D}_{1}\}$ can be directed inward, since at least one edge of

$\{L,R,D\}$ is directed inward in

F. We can see that

${F}^{\prime}$ has no vertex of

B having two in-coming edges. The legal orientations

${F}_{0}^{\prime}$ and

${F}_{t}^{\prime}$ are the orientations associated with

${F}_{0}$ and

${F}_{t}$, respectively. This completes the construction of the instance

$({G}^{\prime},{F}_{0}^{\prime},{F}_{t}^{\prime})$ of the problem

$\Pi $.

Assume that there is a reconfiguration sequence

${F}_{0},{F}_{1},\dots ,{F}_{t}$ from

${F}_{0}$ to

${F}_{t}$. Let

${F}_{i}^{\prime}$ be a legal orientation of

${G}^{\prime}$ associated with

${F}_{i}$. If

${F}_{i+1}$ is obtained from

${F}_{i}$ by a legal move of an edge joining two AND vertices, we have a reconfiguration sequence from

${F}_{i}^{\prime}$ to

${F}_{i+1}^{\prime}$. Suppose that

${F}_{i+1}$ is obtained by a legal move of an edge incident to an OR vertex. Let

L,

R, and

D be the edges incident to the OR vertex. We assume without loss of generality that

${F}_{i+1}$ is obtained by a legal move of the edge

L. When

L is directed inward in

${F}_{i}$, the edge

L is directed outward in

${F}_{i+1}$, and thus the edges

R or

D are directed inward in

${F}_{i}$. Hence, in

${F}_{i}^{\prime}$ the edge

${R}_{1}$ or

${D}_{1}$ can be directed inward (see

Figure 2). Therefore, the edges

${L}_{0}$ and

${L}_{1}$ together with the weight-1 edges between them can be directed outward to obtain

${F}_{i+1}^{\prime}$. When

L is directed outward in

${F}_{i}$ and inward in

${F}_{i+1}$, in

${F}_{i}^{\prime}$ the edges

${L}_{0}$ and

${L}_{1}$ together with the weight-1 edges between them can be directed inward to obtain

${F}_{i+1}^{\prime}$. Since there is a reconfiguration sequence from

${F}_{i}^{\prime}$ to

${F}_{i+1}^{\prime}$ for any

i with

$0\le i<t$, the instance

$({G}^{\prime},{F}_{0}^{\prime},{F}_{t}^{\prime})$ is a yes-instance if

$(G,{F}_{0},{F}_{t})$ is a yes-instance. Notice that, in the subgraph shown in

Figure 2, if two edges of

$\{{L}_{0},{R}_{0},{D}_{0}\}$ are directed outward, then the remaining edge must be directed inward. Thus, a reconfiguration sequence from

${F}_{0}$ to

${F}_{t}$ can be obtained from a reconfiguration sequence from

${F}_{0}^{\prime}$ to

${F}_{t}^{\prime}$. It follows that the instance

$(G,{F}_{0},{F}_{t})$ is a yes-instance if

$({G}^{\prime},{F}_{0}^{\prime},{F}_{t}^{\prime})$ is a yes-instance.

Since the graph ${G}^{\prime}$ and the legal orientations ${F}_{0}^{\prime}$ and ${F}_{t}^{\prime}$ can be obtained in polynomial time, we have the claim. □

We can further see from the proof of Lemma 1 that the problem

$\Pi $ is PSPACE-complete for planar graphs, since the nondeterministic constraint logic problem is PSPACE-complete even if the constraint graph is planar [

11]. We can also see the following observation, which we will use in the proof of Lemma 2.

**Proposition** **1.** Let $(G,{F}_{0},{F}_{t})$ be an instance of the problem Π with a reconfiguration sequence ${F}_{0},{F}_{1},\dots ,{F}_{t}$ from ${F}_{0}$ to ${F}_{t}$. If i is even, then ${F}_{i}$ has no vertex of B having two in-coming edges, while ${F}_{i}$ has one vertex of B having two in-coming edges if otherwise. If a vertex ${b}_{i}\in B$ has two in-coming edges $({a}_{i},{b}_{i})$ and $({a}_{i}^{\prime},{b}_{i})$ in ${F}_{i}$, then we can assume without loss of generality that ${F}_{i}$ is obtained from ${F}_{i-1}$ by reversing the direction of the edge ${a}_{i}{b}_{i}$, while ${F}_{i+1}$ is obtained from ${F}_{i}$ by reversing the direction of the edge ${a}_{i}^{\prime}{b}_{i}$.

**Proof.** Let ${F}_{i}$ be a legal orientation such that every vertex of B has exactly one in-coming edge. Suppose that ${F}_{i+1}$ is obtained from ${F}_{i}$ by reversing the direction of an edge ${a}_{i}{b}_{i}$, where ${a}_{i}$ and ${b}_{i}$ are the vertices of A and B, respectively. Since all the vertices of B has one in-coming edge in ${F}_{i}$, we have $({b}_{i},{a}_{i})\in {F}_{i}$ and $({a}_{i},{b}_{i})\in {F}_{i+1}$. Now, ${b}_{i}$ has two in-coming edges in ${F}_{i+1}$. Let $({a}_{i}^{\prime},{b}_{i})\in {F}_{i}$ be the in-coming edge of ${b}_{i}$ in ${F}_{i}$. If we reverse the direction of an edge other than ${a}_{i}{b}_{i}$ or ${a}_{i}^{\prime}{b}_{i}$, then the orientation is no longer legal. Thus, we can reverse the direction of either ${a}_{i}{b}_{i}$ or ${a}_{i}^{\prime}{b}_{i}$ to obtain ${F}_{i+2}$, in which every vertex of B has exactly one in-coming edge. However, if we reverse the direction of ${a}_{i}{b}_{i}$, then we have the same orientation as ${F}_{i}$. Thus, we can assume without loss of generality that $({a}_{i}^{\prime},{b}_{i})\in {F}_{i+1}$ and $({b}_{i},{a}_{i}^{\prime})\in {F}_{i+2}$. Now, we have the claim. □

#### 2.2. Reduction

Let $(G,{F}_{0},{F}_{t})$ be an instance of the problem $\Pi $. In this section, we construct a reduction graph H together with two Hamiltonian cycles ${C}_{0}$ and ${C}_{t}$ such that there is a reconfiguration sequence from ${F}_{0}$ to ${F}_{t}$ if and only if there is a reconfiguration sequence from ${C}_{0}$ to ${C}_{t}$. That is, $(G,{F}_{0},{F}_{t})$ is a yes-instance if and only if $(H,{C}_{0},{C}_{t})$ is a yes-instance of the Hamiltonian cycle reconfiguration problem.

We use three types of gadgets corresponding to the vertices in

A, the vertices in

B, and the edges of

G. A gadget for a vertex in

A and a gadget for an edge of

G is shown in

Figure 3a,b respectively. Double lines in the figures denote

edges with ears, where an

ear of an edge

$uw$ is a path of length 3 joining

u and

w. Recall that, in the legal moves, the vertices in

A behave in the same way as the AND vertices. We thus refer to the gadgets for the vertices in

A as AND gadgets. Let

b be a vertex in

B of degree

k, and recall that

k is 2 or 3. A gadget for

b is a cycle

$({u}_{0},{w}_{0},{u}_{1},{w}_{1},\dots ,{u}_{k-1},{w}_{k-1})$ of length

$2k$ such that the edge

${w}_{i}{u}_{i+1}$ has a ear for each

i with

$0\le i<k$ (indices are modulo

k).

We construct the reduction graph H from G as follows: (1) Let a be a vertex in A, and let ${e}_{l},{e}_{r},{e}_{d}$ be the edges of G incident to a such that ${e}_{l}$ and ${e}_{r}$ have weight 1 and ${e}_{d}$ has weight 2. We identify the vertices ${l}_{u}^{\prime}$ and ${l}_{w}^{\prime}$ of the gadget for a with the vertices ${x}_{u}$ and ${x}_{w}$ of the gadget for ${e}_{l}$, respectively. Similarly, we identify the vertices ${r}_{u}^{\prime}$ and ${r}_{w}^{\prime}$ of the gadget for a with the vertices ${x}_{u}$ and ${x}_{w}$ of the gadget for ${e}_{r}$, respectively. Moreover, we identify the vertices ${d}_{u}^{\prime}$ and ${d}_{w}^{\prime}$ of the gadget for a with the vertices ${x}_{u}$ and ${x}_{w}$ of the gadget for ${e}_{d}$, respectively. (2) Let b be a vertex in B of degree k, and let ${e}_{0},{e}_{1},\dots ,{e}_{k-1}$ be the edges of G incident to b. We identify, for each i with $0\le i<k$, the vertices ${u}_{i}$ and ${w}_{i}$ of the gadget for b with the vertices ${y}_{u}$ and ${y}_{w}$ of the gadget for ${e}_{i}$, respectively. (3) We finally concatenate the gadgets for the vertices in A cyclically using edges with ears joining the vertices ${c}_{u}^{\prime}$ and ${c}_{w}^{\prime}$ of the gadgets.

Before describing the construction of the Hamiltonian cycles

${C}_{0}$ and

${C}_{t}$, we consider the possible configurations of a Hamiltonian cycle of the reduction graph

H passing through the gadgets. We will show that all the possible configurations in an AND gadget and an edge gadget are shown in

Figure 4a,b, respectively. We can also verify that all the possible transformations of Hamiltonian cycles by a single switch occurred in a gadget are those depicted by the arrows in the figures. Let

C be a Hamiltonian cycle. We first consider the configurations of

C in an AND gadget. The Hamiltonian cycle

C passes through all the edges on the ears, since interior vertices of an ear has degree 2. Thus,

C passes through any of the edges

${c}_{u}{d}_{w}$,

${c}_{u}{c}_{w}$,

${c}_{u}{r}_{w}$, or

${c}_{u}{l}_{w}$. We also have that

C does not pass through the edges

${l}_{u}^{\prime}{l}_{w}^{\prime}$,

${r}_{u}^{\prime}{r}_{w}^{\prime}$, or

${d}_{u}^{\prime}{d}_{w}^{\prime}$, since when we construct the reduction graph

H the vertices

${l}_{u}^{\prime}$,

${l}_{w}^{\prime}$,

${r}_{u}^{\prime}$,

${r}_{w}^{\prime}$,

${d}_{u}^{\prime}$ and

${d}_{w}^{\prime}$ are identified with the vertices of the edge gadgets incident to the edges with ears. Suppose that

C passes through

${c}_{u}{d}_{w}$. Since

C cannot pass through

${d}_{u}{d}_{w}$, it passes through

${d}_{u}{c}_{w}$. Since

C cannot pass through

${c}_{u}{l}_{w}$, it passes through

${l}_{u}{l}_{w}$. Since

C cannot pass through

${l}_{u}{r}_{w}$, it passes through

${r}_{u}{r}_{w}$, and we have the configuration

${S}_{0}$ in

Figure 4a. Suppose that

C passes through

${c}_{u}{c}_{w}$. Since

C cannot pass through

${c}_{u}{d}_{w}$, it passes through

${d}_{u}{d}_{w}$. Since

C cannot pass through

${c}_{u}{l}_{w}$, it passes through

${l}_{u}{l}_{w}$. Since

C cannot pass through

${l}_{u}{r}_{w}$, it passes through

${r}_{u}{r}_{w}$, and we have the configuration

${S}_{1}$ in

Figure 4a. Suppose that

C passes through

${c}_{u}{r}_{w}$. Since

C cannot pass through

${c}_{u}{d}_{w}$, it passes through

${d}_{u}{d}_{w}$. Since

C cannot pass through

${c}_{u}{l}_{w}$, it passes through

${l}_{u}{l}_{w}$. Since

C cannot pass through

${r}_{u}{r}_{w}$, it passes through

${r}_{u}{c}_{w}$, and we have the configuration

${S}_{3}$ in

Figure 4a. Suppose that

C passes through

${c}_{u}{l}_{w}$. Since

C cannot pass through

${c}_{u}{d}_{w}$, it passes through

${d}_{u}{d}_{w}$. Since

C cannot pass through

${l}_{u}{l}_{w}$, it passes through either

${l}_{u}{r}_{w}$ or

${l}_{u}{c}_{w}$. If

C passes through

${l}_{u}{r}_{w}$, then it passes through

${r}_{u}{c}_{w}$ since it cannot pass through

${r}_{u}{r}_{w}$, and we have the configuration

${S}_{4}$ in

Figure 4a. If

C passes through

${l}_{u}{c}_{w}$, then it passes through

${r}_{u}{r}_{w}$ since it cannot pass through

${l}_{u}{r}_{w}$, and we have the configuration

${S}_{2}$ in

Figure 4a. Therefore, all the possible configurations in an AND gadget are shown in

Figure 4a. We next consider the configurations of the Hamiltonian cycle

C in an edge gadget. Since

C passes through all the edges on the ears, it passes through either

$xy$ or

$x{y}^{\prime}$. If

C passes through

$xy$ then it passes through

${x}^{\prime}{y}^{\prime}$, while if

C passes through

$x{y}^{\prime}$, then it passes through

${x}^{\prime}y$. We also have that

C does not pass through the edges

${x}_{u}{x}_{w}$ or

${y}_{u}{y}_{w}$, since when we construct the reduction graph

H the vertices

${x}_{u}$,

${x}_{w}$,

${y}_{u}$, and

${y}_{w}$ are identified with the vertices of the AND gadgets incident to the edges with ears. Therefore, all the possible configurations in an edge gadget are shown in

Figure 4b.

Let

v be a vertex of

A. We next make a correspondence between the possible configurations of a Hamiltonian cycle in the gadget for

v and the possible orientations of the edges incident to

v such that the configuration

${S}_{i}$ in

Figure 4a corresponds to the orientation

${f}_{i}$ in

Figure 1 for each

$i\in \{0,1,\dots ,4\}$. We also make a correspondence between switches occurred in the gadget for

v and legal moves of the edges incident to

v such that switching the configuration from

${S}_{i}$ to

${S}_{j}$ in the gadget for

v corresponds to the legal move from

${f}_{i}$ to

${f}_{j}$ of the edges of

v, where

$i,j\in \{0,1,\dots ,4\}$.

We define a legal orientation

F of

G associated with a Hamiltonian cycle

C of

H so that for each vertex

$v\in A$, the edges incident to

v are oriented

according to the configuration of

C in the gadget for

v. That is, the edges of

v are oriented as

${f}_{i}$ in

F if the configuration of

C in the gadget for

v looks like

${S}_{i}$ (see

Figure 1 and

Figure 4a). Notice that a Hamiltonian cycle

C of

H has exactly one legal orientation of

G associated with

C, but a legal orientation

F may have some Hamiltonian cycles that are associated with

F, due to the two possible configurations in an edge gadget shown in

Figure 4b.

Now, we construct the Hamiltonian cycle

${C}_{0}$ from

${F}_{0}$ as follows, and

${C}_{t}$ is constructed similarly from

${F}_{t}$. (1) For each vertex

$v\in A$, we take the configuration in the gadget for

v according to the orientations of the edges incident to

v. That is, we take the configuration

${S}_{i}$ in

Figure 4a for the gadget for

v if the edges of

v are oriented as

${f}_{i}$ in

Figure 1. (2) We choose the configuration in each edge gadget arbitrarily among those in

Figure 4b. (3) The remaining parts are uniquely determined, since any Hamiltonian cycle pass through all the edges on the ears.

Figure 5b illustrates the Hamiltonian cycle constructed in this way from the legal orientation in

Figure 5a. Recall that every vertex of

B has exactly one in-coming edge in

${F}_{0}$ and

${F}_{t}$. This guarantees that

${C}_{0}$ and

${C}_{t}$ are Hamiltonian. This completes the construction of the instance

$(H,{C}_{0},{C}_{t})$ of the Hamiltonian cycle reconfiguration problem. We remark two facts, which we use in the proof of the following lemma. First, we can see that

${C}_{0}$ and

${C}_{t}$ are associated with

${F}_{0}$ and

${F}_{t}$, respectively. Second, if every vertex of

B has exactly one in-coming edge in a legal orientation

F, then for any two Hamiltonian cycles that are associated with

${F}_{t}$, there is a reconfiguration sequence from one to the other, in which the switches occur only in edge gadgets.

**Lemma** **2.** The instance $(G,{F}_{0},{F}_{t})$ of the problem Π is a yes-instance if and only if $(H,{C}_{0},{C}_{t})$ of the Hamiltonian cycle reconfiguration problem is a yes-instance.

**Proof.** We first prove the if direction. Assume that there is a reconfiguration sequence ${C}_{0},{C}_{1},\dots ,{C}_{t}$ from ${C}_{0}$ to ${C}_{t}$. Let ${F}_{i}$ be the legal orientation of G associated with ${C}_{i}$ (Recall that a Hamiltonian cycle C of H has exactly one legal orientation associated with C). Notice that ${F}_{i}={F}_{i+1}$ if and only if ${C}_{i+1}$ is obtained from ${C}_{i}$ by a switch occurred in an edge gadget. When ${F}_{i}={F}_{i+1}$ for some i with $0\le i<t$, we remove ${F}_{i+1}$ from the sequence ${F}_{0},{F}_{1},\dots ,{F}_{t}$ to obtain the reconfiguration sequence from ${F}_{0}$ to ${F}_{t}$.

We next prove the only-if direction. Assume that there is a reconfiguration sequence ${F}_{0},{F}_{1},\dots ,{F}_{t}$ from ${F}_{0}$ to ${F}_{t}$. Recall that, for any two Hamiltonian cycles that are associated with ${F}_{t}$, there is a reconfiguration sequence from one to the other, since every vertex of B has exactly one in-coming edge in ${F}_{t}$. Thus, it suffices to show that for each Hamiltonian cycle ${C}_{i}$ with $0\le i<t$, there is a Hamiltonian cycle ${C}_{i+1}$ together with a reconfiguration sequence from ${C}_{i}$ to ${C}_{i+1}$, where ${C}_{i}$ and ${C}_{i+1}$ are Hamiltonian cycles associated with ${F}_{i}$ and ${F}_{i+1}$, respectively. Suppose that ${F}_{i+1}$ is obtained from ${F}_{i}$ by reversing the direction of an edge ${a}_{i}{b}_{i}$, where ${a}_{i}$ and ${b}_{i}$ are the vertices of A and B, respectively.

We first consider the case when

$({b}_{i},{a}_{i})\in {F}_{i}$ and

$({a}_{i},{b}_{i})\in {F}_{i+1}$. We have from Proposition 1 that

${F}_{i}$ has no vertex of

B having two in-coming edges. Let

C be a graph obtained from

${C}_{i}$ by switching the configuration in the gadget for

${a}_{i}$ according to the legal move. If

C is a Hamiltonian cycle, the claim holds. However, there is some possibility that

C is disconnected. (In

Figure 5b, for example, when we replace the configuration in the gadget for

${a}_{2}$ from

${S}_{3}$ to

${S}_{4}$, we have two cycles, while, in

Figure 5a, this replacement corresponds to the reversal of the edge

$({b}_{2},{a}_{2})$ that results in another legal orientation). In this case, we use two steps as follows: Let

${C}^{\prime}$ be a graph obtained from

${C}_{i}$ by switching the configuration in the edge gadget for

${a}_{i}{b}_{i}$ as shown in

Figure 4b. Let

${C}^{\u2033}$ be a graph obtained from

${C}^{\prime}$ by switching the configuration in the gadget for

${a}_{i}$ according to the legal move. We show that

${C}^{\prime}$ and

${C}^{\u2033}$ are Hamiltonian cycles. Suppose that

C is obtained from

${C}_{i}$ by switching edges

${v}_{1}{v}_{2}$ and

${v}_{3}{v}_{4}$ with edges

${v}_{1}{v}_{3}$ and

${v}_{2}{v}_{4}$. Suppose also that

${C}^{\prime}$ is obtained from

C by switching edges

${v}_{5}{v}_{6}$ and

${v}_{7}{v}_{8}$ with edges

${v}_{5}{v}_{7}$ and

${v}_{6}{v}_{8}$. Since

C is disconnected while

${C}_{i}$ is Hamiltonian, the vertices

${v}_{1}$,

${v}_{2}$,

${v}_{3}$, and

${v}_{4}$ appear on

${C}_{i}$ as

${C}_{i}=({v}_{1},{v}_{2},\dots ,{v}_{4},{v}_{3},\dots )$. Since

$({b}_{i},{a}_{i})\in {F}_{i}$ and the switch occurs in the edge gadget, we can assume without loss of generality that the vertices

${v}_{5}$,

${v}_{6}$,

${v}_{7}$, and

${v}_{8}$ appear on

${C}_{i}$ as

Thus,

${C}^{\prime}$ and

${C}^{\u2033}$ are the following Hamiltonian cycles.

We can see that ${C}^{\prime}$ is also associated with ${F}_{i}$ since the switch occurs in an edge gadget. Hence, ${C}^{\u2033}$ is associated with ${F}_{i+1}$, and the claim holds.

We then consider the case when

$({a}_{i},{b}_{i})\in {F}_{i}$ and

$({b}_{i},{a}_{i})\in {F}_{i+1}$. Let

C be a graph obtained from

${C}_{i}$ by switching the configuration in the gadget for

${a}_{i}$ according to the legal move. We show that

C is the Hamiltonian cycle. We have from Proposition 1 that there is the vertex

${a}_{i}^{\prime}\in A$ with

${a}_{i}^{\prime}\ne {a}_{i}$ such that

$({a}_{i}^{\prime},{b}_{i})\in {F}_{i}$ while

$({b}_{i},{a}_{i}^{\prime})\in {F}_{i-1}$. Let

${C}^{\prime}$ be the Hamiltonian cycle associated with

${F}_{i-1}$ from which

${C}_{i}$ is obtained by a single switch. We can see that this switch occurs in the gadget for

${a}_{i}^{\prime}$. Suppose that

C is obtained from

${C}_{i}$ by switching edges

${v}_{1}{v}_{2}$ and

${v}_{3}{v}_{4}$ with edges

${v}_{1}{v}_{3}$ and

${v}_{2}{v}_{4}$. Suppose also that

${C}_{i}$ is obtained from

${C}^{\prime}$ by switching edges

${v}_{5}{v}_{6}$ and

${v}_{7}{v}_{8}$ with edges

${v}_{5}{v}_{7}$ and

${v}_{6}{v}_{8}$. Since

$({a}_{i},{b}_{i})$ is the only in-coming edge of

${b}_{i}$ in

${F}_{i-1}$, the vertices

${v}_{1}$,

${v}_{2}$,

${v}_{3}$, and

${v}_{4}$ appear on

${C}^{\prime}$ as

${C}^{\prime}=({v}_{1},{v}_{2},\dots ,{v}_{4},{v}_{3},\dots )$. Since

$({b}_{i},{a}_{i}^{\prime})\in {F}_{i-1}$, we can assume without loss of generality that the vertices

${v}_{5}$ and

${v}_{6}$ appear on

${C}^{\prime}$ as

${C}^{\prime}=({v}_{1},{v}_{2},\dots ,{v}_{5},{v}_{6},\dots ,{v}_{4},{v}_{3},\dots )$. Since

${C}_{i}$ is also a Hamiltonian cycle, the vertices

${v}_{7}$ and

${v}_{8}$ appear on

${C}^{\prime}$ as

Thus,

${C}_{i}$ and

C are the following Hamiltonian cycles.

Since C is associated with ${F}_{i+1}$, the claim holds. □

Obviously, the reduction graph H is bipartite. We can easily check that H has maximum degree 6 (The vertices ${c}_{v}$ and ${c}_{w}$ of each AND gadget have degree 6). Since the instance $(H,{C}_{0},{C}_{t})$ can be constructed from $(G,{F}_{0},{F}_{t})$ in polynomial time, we have the following.

**Theorem** **1.** The Hamiltonian cycle reconfiguration problem is PSPACE-complete for bipartite graphs with maximum degree 6.

A bipartite graph is chordal bipartite if each cycle in the graph of length greater than 4 has a chord, that is, an edge joining two vertices that are not consecutive on the cycle. Let D be the vertices of the reduction graph H incident with two edges having ears. We construct a graph ${H}^{\prime}$ from H by adding edges $uv$ for all vertices $u\in D$ and all vertices v of H that is in the color class different from u and is not an interior vertex of any ear. It is obvious that ${H}^{\prime}$ is bipartite. Suppose that ${H}^{\prime}$ has a chordless cycle Z of length greater than 4. Clearly, Z has no interior vertices of any ear. We also have that Z has no vertices in D, for otherwise Z would have a chord. Thus, Z is a cycle in a single AND gadget or a single edge gadget, but these gadgets contains no chordless cycle of length greater than 4. Therefore, ${H}^{\prime}$ is a chordal bipartite graph.

Since every added edges in ${H}^{\prime}$ is incident to a vertex in D, any Hamiltonian cycle does not pass through the added edges. Thus, there is a reconfiguration sequence from ${C}_{0}$ to ${C}_{t}$ in H if and only if there is a reconfiguration sequence from ${C}_{0}$ to ${C}_{t}$ in ${H}^{\prime}$. Now, we have the following.

**Theorem** **2.** The Hamiltonian cycle reconfiguration problem is PSPACE-complete for chordal bipartite graphs.