Lobachevskii Journal of Mathematics http://ljm.ksu.ru ISSN 1818-9962 Vol. 22, 2006, 35–46

Abstract. Let

P

be a closed triangulated manifold,

dim P = n

. We consider the group of simplicial 1-chains

C_{1} (P) = C_{1} (P, ℤ_{2})

and the homology group

H_{1} (P) = H_{1} (P, ℤ_{2})

. We also use some nonnegative weighting function

L : C_{1} (P) \to ℝ

. For any homological class

[x] \in H_{1} (P)

the method proposed in article builds a cycle

z \in [x]

with minimal weight

L (z)

. The main idea is in using a simplicial scheme of space of the regular covering

p : \hat{P} \to P

with automorphism group

G ≅ H_{1} (P)

. We construct this covering applying the index vector-function

J : C_{1} (P) \to ℤ_{2}^{r}

relative to any basis of group

H_{n - 1} (P)

r = rank H_{n - 1} (P)

1. Index Vector-Function

Consider a triangulated closed manifold

P

dim P = n

, and a basis

[z_{1}^{n - 1}], \dots, [z_{r}^{n - 1}]

of homology group

H_{n - 1} (P) = H_{n - 1} (P, ℤ_{2})

. Let

Ind : H_{1} (P) \times H_{n - 1} (P) \to ℤ_{2}

be the intersection index.

Proof. Let $\{[z_{1}^{1}], \dots, [z_{r}^{1}]\}$ be a basis of group $H_{1} (P) = H_{1} (P, ℤ_{2})$ , that is dual to the given basis $\{[z_{1}^{n - 1}], \dots, [z_{r}^{n - 1}]\}$ . Assume now that $z = x + y$ . Then $z \in Z_{1} (P)$ and $[z] = \sum_{i = 1}^{r} l^{i} [z_{i}^{1}]$ , where $l_{i} \in ℤ_{2}$ . This implies that $J^{k} (z) = Ind ([z], [z_{k}^{n - 1}]) = l^{k}$ for all $k = 1, \dots, r$ . So $J (x) = J (y)$ if and only if $l^{1} = \dots = l^{r} = 0$ . And this latter expression is equivalent to the equality $[z] = 0$ . □

ALGORITHM 1. Construction of index vector-function relative to the basis of group $H_{n - 1} (P)$ .

Input:

1) simple basis cycles $z_{1}^{n - 1}, z_{2}^{n - 1}, \dots, z_{r}^{n - 1}$ which are lists of $(n - 1)$ -dimensional simplices;

2) list $K^{1} (P)$ of edges for polyhedron $P$ ;

3) lists $K_{1}^{n} (P, z_{1}^{n - 1}), \dots, K_{r}^{n} (P, z_{r}^{n - 1})$ consisting of $n$ -dimensional simplices from neighborhoods of cycles $z_{1}^{n - 1}, \dots, z_{r}^{n - 1}$ respectively;

Output:

1) vectors $J (a) = (J^{1} (a), \dots, J^{r} (a)) \in ℤ_{2}^{r}$ for all edges $a \in K^{1} (P)$ ;

2) chains $M_{1}, \dots, M_{r}$ of edges indexed relative to cycles $z_{1}^{n - 1}, \dots, z_{r}^{n - 1}$ respectively;

3) lists $M_{k} (u)$ , $k = 1, \dots, r$ of edges, that we add to $M_{k}$ when considering vertex $u$ of cycle $z_{k}^{n - 1}$ ;

4) sets $Σ_{k} (u)$ , $k = 1, \dots, r$ of $n$ -simplices incident to edges from $M_{k} (u)$ .

Algorithm Description.

Step 0. For all $k = 1, \dots, r$ execute steps 1 – 3.

Step 1. Start operations. Assume $M_{k} = \emptyset$ , $J^{k} (a) : = 0$ for all $a \in K^{1} (P)$ . We denote $z_{k}^{n - 1}$ by $X$ and $K_{k}^{n} (P, z_{k}^{n - 1})$ by $K^{n} (P, X)$ . We create then lists of vertices and edges for all simplices of cycle $X$ , $K^{0} (X)$ and $K^{1} (X)$ respectively.

Step 2. Indexing edges that do not belong to the cycle. For each vertex $u \in K^{0} (X)$ execute steps 2.1 – 2.4.

Step 2.1. Initializing vertex neighborhood. Let us create a list $K^{n} (P, u) \subset K^{n} (P, X)$ of $n$ -dimensional simplices of the polyhedron $P$ , that contain $u$ , and a list $K^{n - 1} (P, u)$ of all $(n - 1)$ -dimensional faces of simplices from $K^{n} (P, u)$ . At the same time, for each simplex $σ^{n - 1} \in K^{n - 1} (P, u)$ we get a list $\partial^{n - 1} (σ^{n - 1}, u)$ of $n$ -dimensional simplices from $K^{n} (P, u)$ those are incident to $σ^{n - 1}$ , and assume $μ (σ^{n - 1}) : = 0$ . Then we create empty lists $M_{k} (u) : = \emptyset$ and $Σ_{k} (u) : = \emptyset$ .

Step 2.2. Creating the queue to keep $n$ -simplices. We chose a simplex $σ_{0}^{n} \in K^{n} (P, u)$ , create a queue $R : = {σ_{0}^{n}}$ and remove $σ_{0}^{n}$ from $K^{n} (P, u)$ .

Step 2.3. Main procedure of Algorithm. While the queue $R$ is not empty we will do the following actions. Take the ﬁrst simplex $σ^{n} \in R$ and remove it from the queue $R$ . For each $(n - 1)$ -dimensional face $σ^{n - 1}$ of the simplex $σ^{n}$ we check the following: whether it belongs to the cycle $X$ , whether $μ (σ^{n - 1})$ is equal to zero, whether the list $\partial^{n - 1} (σ^{n - 1}, u)$ contains any simplices different from $σ^{n}$ . If all above conditions are satisﬁed we will execute steps 2.3.1 – 2.3.2.

Step 2.3.1. Take the simplex $σ_{*}^{n} \in \partial^{n - 1} (σ^{n - 1}, u) ∖ {σ^{n}}$ , remove it from $K^{n} (P, u)$ and enqueue to $R$ ; set $μ (σ^{n - 1}) : = 1$ and $Σ_{k} (u) = Σ_{k} (u) \cup {σ_{*}^{n}}$ .

Step 2.3.2. For all vertices $w \neq u$ of the simplex $σ^{n - 1}$ we check whether the edge $a = [u w]$ is in the list $K^{1} (X)$ ; having $a \notin K^{1} (X)$ set $J^{k} (a) : = J^{k} (a) + 1 mod 2$ , $M_{k} (u) = M_{k} (u) \cup {a}$ , $M_{k} : = M_{k} + a mod 2$ .

Step 2.4. Main procedure repeated. If the list $M_{k} (u)$ is empty then go back to step 2.2.

Step 3. Indexing the edges of cycle. For each edge $a = [u v] \in K^{1} (X)$ we search any edges $b \in M_{k} (u)$ and $c \in M_{k} (v)$ such that $b \cap c \neq \emptyset$ and that $a$ , $b$ and $c$ are sides of some triangle of polyhedron $P$ . If the edges $b$ and $c$ do not exist then we set $J^{k} (a) : = 1$ and $M_{k} = M_{k} + a mod 2$ .

End of algorithm.

Proof. Let $x \in Z_{1} (P)$ . We will prove that $J^{k} (x) = Ind ([x], [z_{k}^{n - 1}])$ for all $k = 1, \dots, r$ . Set $z_{0}^{*} = z_{k}^{n - 1}$ . For all $p = 1, \dots, N$ we will make the following constructions; here $N$ is the power of the set $K^{0} (z_{k}^{n - 1})$ . Consider a vertex $u_{p} \in K^{0} (z_{k}^{n - 1})$ and its barycentric star $bst (u_{p}, P)$ .

Let $Σ_{k}^{*} (u_{p})$ be the set of all $n$ -simplices from the barycentric subdivision of $Σ_{k} (u_{p})$ . Construct the chain $c (u_{p})$ of simplices $σ_{1} \in bst (u_{p}, P) \cap Σ_{k}^{*} (u_{p})$ .

Then we write the chain boundary $c (u_{p})$ as a sum $Y_{1} + Y_{2}$ , where $Y_{1}$ is the sum of all its $(n - 1)$ -dimensional simplices, that belong to the cycle $z_{k}^{n - 1}$ and $Y_{2}$ is the sum of all remaining simplices from the chain $\partial c (u_{p})$ . Set $z_{p}^{*} = z_{p - 1}^{*} + Y_{1} + Y_{2} mod 2$ .

By construction $z_{p}^{*} \sim z_{p - 1}^{*}$ for all $p = 1, \dots, N$ . Hence, the cycle $z^{*} = z_{N}^{*}$ is homologous to the cycle $z_{k}^{n - 1} = z_{0}^{*}$ .

Let now prove that for any edge $a = [u v] \in K^{1} (P)$ and $σ_{b} \in bst (a)$ the simplex $σ_{b}$ belongs to $z^{*}$ if and only if $a \in M_{k}$ .

Let view all possible positions of the edge $a$ . At the same time we also agree to think that $M_{k} (u) = \emptyset$ and that $Σ_{k} (u) = \emptyset$ for all $u \notin K^{0} (z_{k}^{n - 1})$ .

0. If $a \notin M_{k} (u) \cup M_{k} (v)$ and $a \notin K^{1} (z_{k}^{n - 1})$ , then, according to the algorithm, $a \notin M_{k}$ . On the other hand, the edge $a$ can not be incident to simplices from the lists $Σ_{k} (u)$ and $Σ_{k} (v)$ and hence $σ_{b} \notin z^{*}$ .

1. Let $u \in K^{0} (z_{k}^{n - 1})$ , $a \in M_{k} (u)$ and $v \notin K^{0} (z_{k}^{n - 1})$ . Then the edge $a$ will be still in the chain $M_{k}$ when algorithm 1 is completed. At the same time the barycentric star $bst (a)$ belongs to the boundary of the chain $c (u)$ and does not belong to the cycle $z_{k}^{n - 1}$ . Thus in this case $a \in M_{k}$ and the chain $bst (a)$ belongs to the cycle $z^{*}$ .

2. Further, assume that $u, v \in K^{0} (z_{k}^{n - 1})$ and $a \in M_{k} (u)$ . At that, $a \notin K^{1} (z_{k}^{n - 1})$ .

2.1. If $a \in M_{k} (v)$ , then $a \notin M_{k}$ , and simplices of its barycentric star will be added twice to the initial cycle $z_{k}^{n - 1}$ and will not be in the resulting cycle $z^{*}$ .

2.2. If $a \notin M_{k} (v)$ , then $a \in M_{k}$ and any simplex $σ_{b} \in bst (a)$ is added to the cycle $z^{*}$ exactly once. So $σ_{b} \in z^{*}$ .

3. Finally, let $a \in K^{1} (z_{k}^{n - 1})$ .

3.1. Let assume that the condition from step 3 of algorithm 1 is satisﬁed, i.e.:

there exist edges $b \in M_{k} (u)$ and $c \in M_{k} (v)$ such that $b \cap c \neq \emptyset$ and that $a$ , $b$ and $c$ are sides of some triangle $σ^{'} \in K^{2} (P)$ .

In this case, according to the algorithm $a \notin M_{k}$ .

Let view all triangles $σ^{'}$ from ( $*$ ), and all $n$ -dimensional simplices incident to them. The such $n$ -simplices belong both to $Σ_{k} (u)$ and $Σ_{k} (v)$ . Consider an $n$ -dimensional simplex $σ$ , $σ_{b} \in σ$ . If $σ_{b} \in bst (a)$ , then $σ$ either belongs to the both sets $Σ_{k} (u)$ and $Σ_{k} (v)$ or does not belong to them. Hence, the simplex $σ_{b}$ either is not added to the cycle $z^{*}$ or is added twice. Therefore $σ_{b} \notin z^{*}$ .

3.2. Assume now that condition ( $*$ ) is not satisﬁed. Then according to step 3 of algorithm 1, $a \in M_{k}$ .

Barycentric star $bst (a)$ of the edge $a = [u v]$ belongs to the union $D (a)$ of all $n$ -simplices that contain the edge $a$ . We will prove that the sub-polyhedron $D (a)$ belongs to the union of simplices from the sets $Σ_{k} (u)$ and $Σ_{k} (v)$ .

The cycle $z_{k}^{n - 1}$ divides $D (a)$ into two components of strong connectivity $D^{+} (a)$ and $D^{-} (a)$ .

By construction the set $Σ_{k} (u)$ can not be empty. Moreover, if the simplex $σ \in z_{k}^{n - 1}$ is incident to the vertex $u$ , then $σ$ is a face of some $n$ -simplex from $Σ_{k} (u)$ . So there exists a simplex $σ^{n} \in Σ_{k} (u)$ that contains the edge $a$ .

Let the simplex $σ^{n}$ belong to $D^{+} (a)$ . Then under the strong connectivity $D^{+} (a)$ and according to algorithm 1, all $n$ -simplices from $D^{+} (a)$ also belong to $Σ_{k} (u)$ .

This implies, in accordance with our assumption, that no $n$ -simplex from $D^{+} (a)$ can belong to the set $Σ_{k} (v)$ .

The set $Σ_{k} (v)$ cannot be empty also. Since each simplex of $z_{k}^{n - 1}$ incident to the vertex $v$ is a face of some $n$ -simplex from $Σ_{k} (v)$ , it follows that there exists a simplex $σ_{*}^{n} \in Σ_{k} (v)$ that contains the edge $a$ . By the above proof, $σ_{*}^{n}$ belongs to $D^{-} (a)$ . Then all $n$ -simplices from $D^{-} (a)$ belong to the set $Σ_{k} (v)$ , too. Consequently all $n$ -simplices of the polyhedron $D (a) = D^{+} (a) \cup D^{-} (a)$ belong either to the set $Σ_{k} (u)$ or to $Σ_{k} (v)$ .

Consider $σ_{b} \in bst (a)$ . If there exists a simplex $σ \in Σ_{k} (u)$ containing $σ_{b}$ , then $σ \notin Σ_{k} (v)$ . Otherwise, in accordance to the above proof, there is a simplex $\tilde{σ} \in Σ_{k} (v)$ such that $σ_{b} \subset \tilde{σ}$ . It follows that $σ_{b}$ is involved in the cycle $z^{*}$ exactly once, so $σ_{b} \in z^{*}$ .

Thus we have proved that the cycle $z^{*} \sim z_{k}^{n - 1}$ consists of barycentric stars of the edges from chain $M_{k}$ . That means that this cycle intersects transversally only the edges of the cycle $x$ , that are in the list $M_{k}$ . According to algorithm 1 $J^{k} (a) = 1$ for all $a \in M_{k}$ and $J^{k} (b) = 0$ for all edges $b \notin M_{k}$ . So

Ind ([x], [z_{k}^{n - 1}]) = Ind ([x], [z^{*}]) = \sum_{a \in x} J^{k} (a) mod 2 = J^{k} (x) .

□

2. Regular Covering with the Automorphism Group

H_{1} (P)

Let

P

be an

n

-dimensional triangulated closed manifold and

S = (V, K)

be its simplicial scheme. We will construct an abstract simplicial scheme

\hat{S} = (\hat{V}, \hat{K})

as follows.

Set

\hat{V} = V \times G

, where

G = ℤ_{2}^{r}

. Let

{\hat{v}}_{0}, {\hat{v}}_{1}, \dots, {\hat{v}}_{m} \in \hat{V}

, where

{\hat{v}}_{i} = (v_{i}, b_{i})

for all

i = 0, 1, \dots, m

. We will think that

{{\hat{v}}_{0}, {\hat{v}}_{1}, \dots, {\hat{v}}_{m}} \in \hat{K}

if the below conditions are satisﬁed:

Let deﬁne now a mapping

p^{0} : \hat{V} \to V

and a left action

λ^{0} : G \times \hat{V} \to \hat{V}

of group

G

\hat{V}

, assuming

Let

\hat{P}

deﬁne some realization of the scheme

\hat{S} = (\hat{V}, \hat{K})

. At that we identify the set of vertices of the polyhedron

\hat{P}

with

\hat{V}

Proof. Simplicial and surjective properties of the mapping $p^{0}$ follow directly from its deﬁnition and from the construction of the complex $\hat{K}$ . If $\hat{s} = {(v_{0}, g_{0}), (v_{1}, g_{1}), \dots, (v_{m}, g_{m})} \in \hat{K}$ , then ${v_{0}, v_{1}, \dots, v_{m}} \in K$ and $g_{0} + g_{i} = J ([v_{0} v_{i}])$ for all $i = 1, \dots, m$ . On the other hand, $g \cdot \hat{s} = {(v_{0}, g + g_{0}), (v_{1}, g + g_{1}), \dots, (v_{m}, g + g_{m})}$ for an arbitrary $g \in G$ . Since $g + g_{0} + g + g_{i} = g_{0} + g_{i} = J ([v_{0} v_{i}])$ , then $g \cdot \hat{s} \in \hat{K}$ . So the action $λ^{0}$ is also simplicial.

Let $s = {v_{0}, v_{1}, \dots, v_{m}} \in K$ and ${\hat{v}}_{0} \in {(p^{0})}^{- 1} (v_{0})$ . Then ${\hat{v}}_{0} = (v_{0}, g_{0})$ , where $g_{0} \in G$ . Set $g_{i} = g_{0} + J ([v_{0} v_{i}])$ and ${\hat{v}}_{i} = (v_{i}, g_{i})$ for all $i = 1, \dots, m$ . At that $\hat{s} = {{\hat{v}}_{0}, {\hat{v}}_{1}, \dots, {\hat{v}}_{m}} \in \hat{K}$ , ${\hat{v}}_{0} \in \hat{s}$ and $p^{0} (\hat{s}) = s$ . Hence, the mapping $p^{0}$ has the following property:

for each abstract simplex $s \in K$ and for any vertex $\hat{v} \in {(p^{0})}^{- 1} (s)$ there is the unique abstract simplex $\hat{s} \in \hat{K}$ containing the vertex $\hat{v}$ and satisfying the equality $p^{0} (\hat{s}) = s$ .

Let choose an abstract simplex $\hat{s} = {{\hat{v}}_{0}, {\hat{v}}_{1}, \dots, {\hat{v}}_{m}} \in \hat{K}$ , and an element $g$ of group $G$ and assume that $g \cdot \hat{s} = \hat{s}$ . Then ${\hat{v}}_{i} = (v_{i}, g_{i})$ and $g \cdot {\hat{v}}_{i} = (v_{i}, g + g_{i})$ for all $i = 1, \dots, m$ . At the same time it follows from the equality $g \cdot \hat{s} = \hat{s}$ that $(v_{0}, g + g_{0}) = (v_{k}, g_{k})$ for some $k \in {0, 1, \dots, m}$ . The latter is possible only if $k = 0$ and $g = 0$ . Thus the action $λ^{0}$ has the following property:

if $g \cdot \hat{s} = \hat{s}$ for at least one non-empty simplex $\hat{s} \in \hat{K}$ , then $g$ is the neutral element of the group $G$ .

Let now consider the simplices $\hat{s} = {(v_{0}, g_{0}), (v_{1}, g_{1}), \dots, (v_{m}, g_{m})}$ and ${\hat{s}}^{'}$ of the complex $\hat{K}$ .

First, if $g \in G$ and ${\hat{s}}^{'} = g \cdot \hat{s}$ , then ${\hat{s}}^{'} = {(v_{0}, g + g_{0}), (v_{1}, g + g_{1}), \dots, (v_{m}, g + g_{m})}$ . At that $p^{0} ({\hat{s}}^{'}) = {v_{0}, v_{1}, \dots, v_{m}} = p^{0} (\hat{s})$ .

Further, assume that $p^{0} ({\hat{s}}^{'}) = p^{0} (\hat{s}) = {v_{0}, v_{1}, \dots, v_{m}}$ . Then according to (1), ${\hat{s}}^{'} = {(v_{0}, g_{0}^{'}), (v_{1}, g_{1}^{'}), \dots, (v_{m}, g_{m}^{'})}$ , where $g_{0}^{'}, g_{1}^{'}, \dots, g_{m}^{'}$ are some elements of group $G$ , and $g_{0} + g_{i} = J ([v_{0} v_{i}]) = g_{0}^{'} + g_{i}^{'}$ for $i = 1, \dots, m$ . Set $g = g_{0}^{'} + g_{0}$ . Then according to the above equalities $g_{i}^{'} = g + g_{i}$ for all $i = 0, 1, \dots, m$ and hence ${\hat{s}}^{'} = g \cdot \hat{s}$ .

This proves that $p^{0}$ and $λ^{0}$ have the following property:

for arbitrary abstract simplices $\hat{s}, {\hat{s}}^{'} \in \hat{K}$ the equality $p^{0} (\hat{s}) = p^{0} (\hat{s}^{'})$ is equivalent to the existence of an element $g \in G$ such that $g \cdot \hat{s} = \hat{s}^{'}$ .

It is known that $p^{0}$ and $λ^{0}$ may have the unique continuation to the simplicial mapping $p : \hat{P} \to P$ and the simplicial action $λ : G \times \hat{P} \to \hat{P}$ of group $G$ on $\hat{P}$ . It also follows from (C1) – (C3) that $p$ is a regular covering, and $G$ is a corresponding group of covering transformations (see, for example, [4], chapter 2, section 6, theorem 7). □

Proof. Let $z = [w_{0} w_{1}] + [w_{1} w_{2}] + \dots + [w_{s - 1} w_{s}]$ be a path in the polyhedron $P$ and $g_{0} \in G = ℤ_{2}^{r}$ . Then the unique path $\hat{z}$ of the polyhedron $\hat{P}$ , starting in the vertex ${\hat{w}}_{0} = (w_{0}, g_{0})$ and covering the path $z$ , is deﬁned by the formulas

{\hat{w}}_{i} = (w_{i}, g_{0} + J (z_{i})), i = 1, \dots, s,

(2)

where $z_{i} = [w_{0} w_{1}] + [w_{1} w_{2}] + \dots + [w_{i - 1} w_{i}]$ , and

\hat{z} = [{\hat{w}}_{0} {\hat{w}}_{1}] + [{\hat{w}}_{1} {\hat{w}}_{2}] + \dots + [{\hat{w}}_{s - 1} {\hat{w}}_{s}] .

(3)

Set $g_{i} = g_{0} + J (z_{i})$ for $i = 1, \dots, s$ and $z_{0} = 0$ . Then $J (z_{i}) = J (z_{i - 1}) + J ([w_{i - 1} w_{i}])$ for all $i = 1, \dots, s$ . At the same time $g_{i} = g_{i - 1} + J ([w_{i - 1} w_{i}])$ and the vertices ${\hat{w}}_{i - 1}$ and ${\hat{w}}_{i}$ from $\hat{V}$ , deﬁned by the formula (2), are connected by the edge $[{\hat{w}}_{i - 1} {\hat{w}}_{i}] \in \hat{K}$ . Then in the polyhedron $\hat{P}$ there is deﬁned a path (3) starting at the vertex ${\hat{w}}_{0} = (w_{0}, b_{0})$ . As $p ({\hat{w}}_{i}) = p ((w_{i}, g_{0} + J (z_{i}))) = w_{i}$ for all $i = 0, 1, \dots, s$ , then $\hat{z}$ covers the path $z$ . Since $p$ is a covering then the path $\hat{z}$ is unique.

Assume now that ${\hat{v}}_{0} = (v_{0}, g_{0})$ , where $g_{0} \in G$ . By the above proof, the equalities $p (\hat{x}) = x$ , $p (\hat{y}) = y$ and ${\hat{v}}_{0} = {\hat{u}}_{0}$ imply that ${\hat{v}}_{s} = (v_{s}, g_{0} + J (x))$ and ${\hat{u}}_{t} = (u_{t}, g_{0} + J (y))$ . So ${\hat{v}}_{s} = {\hat{u}}_{t}$ if and only if $J (x) = J (y)$ . According to proposition 1, the last equality is equivalent to the homology of the chains $x$ and $y$ . □

3. Minimal Cycles Searching

Let

E (P) = K^{1} (P)

be the set of edges of the polyhedron

P

, and

L : E (P) \to ℝ

be a non-negative function. Using the formulas

we can extend

L

to the function

L : C_{1} (R) \to (R)

. This function is often called weight function. And for an arbitrary chain

x \in C_{1} (P)

the value

L (x)

is called its weight (see, for example, [5], chapter 25).

ALGORITHM 2. Searching for the minimal path homologous to the given 1-chain.

Input:

1) list $V (P)$ of vertices for polyhedron $P$ ;

2) lists $U (v, P)$ of vertices incident to $v$ for all vertices $v \in V (P)$ ;

3) index vector-function $J : C_{1} (P) \to ℤ_{2}^{r}$ relative to some basis of group $H_{n - 1} (P)$ ;

4) weight function $L : C_{1} (P) \to ℝ$ ;

5) vertices $u_{1} \in V (P)$ and $u_{2} \in V (P)$ ;

6) 1-chain $x \in C_{1} (P)$ , $\partial x = {u_{1}, u_{2}}$ .

Output:

1-chain $z \in C_{1} (P)$ .

Algorithm Description.

Step 1. Determination of index of chain $x$ . Calculate the vector $i = J (x)$ .

Step 2. Initializing chain $z$ . Set $z : = \emptyset$ .

Step 3. Initializing sets $\hat{T} \subset V (P) \times G$ , ${\hat{P}}^{*} \subset V (P) \times G$ and a mapping $\hat{D} : V (P) \times G \to ℝ$ . Let $\hat{T} : = {(u_{1}, 0)}$ , where $0$ – null vector of space $G = ℤ_{2}^{r}$ , ${\hat{P}}^{*} : = \emptyset$ and $\hat{D} (u_{1}, 0) : = 0$ .

Step 4. First extension of ${\hat{P}}^{*}$ and $\hat{D}$ . For each vertex $v \in U (u_{1}, P)$ set $j : = J ([u_{1} v])$ and add the pair $(v, j)$ into the list ${\hat{P}}^{*}$ . At the same time set $\hat{D} (v, j) : = L ([u_{1} v])$ , $F (v, j) : = (u_{1}, 0)$ .

Step 5. Choosing a next element to add to $\hat{T}$ . Find the pair $(w, k) \in ({\hat{P}}^{*} ∖ \hat{T})$ such that $\hat{D} (w, k) = min_{(v, j) \in ({\hat{P}}^{*} ∖ \hat{T})} \hat{D} (v, j)$ .

Step 6. Stop criterion of $\hat{T}$ , ${\hat{P}}^{*}$ , $\hat{D}$ construction. If $w = u_{2}$ $k = i$ , then go to step 10.

Step 7. Extension of the set $\hat{T}$ . Add the pair $(w, k)$ into the list $\hat{T}$ .

Step 8. Next extension of ${\hat{P}}^{*}$ and $\hat{D}$ . For each vertex $v \in U (w, P)$ set $j : = k + J ([w v])$ . If the pair $(v, j) \notin {\hat{P}}^{*}$ , then set $\hat{D} (v, j) : = \hat{D} (w, k) + L ([w v])$ , $F (v, j) : = (w, k)$ and add the pair $(v, j)$ into ${\hat{P}}^{*}$ . If $(v, j) \in ({\hat{P}}^{*} ∖ \hat{T})$ and $\hat{D} (w, k) + L ([w v]) < \hat{D} (v, j)$ , then set $\hat{D} (v, j) = \hat{D} (w, k) + L ([w v])$ and $F (v, j) : = (w, k)$ .

Step 9. Continuation of $\hat{T}$ , ${\hat{P}}^{*}$ , $\hat{D}$ construction. Go back to step 5.

Step 10. Construction of chain $z$ .

Step 10.1. Take a pair $(v, j) = F (w, k)$ and set $z : = z + [v w]$ .

Step 10.2. If $(v, j) \neq (u_{1}, 0)$ , then set $(w, k)$ equal to $(v, j)$ and go back to step 10.1.

End of algorithm.

Proof. By construction $J (z) = i = J (x)$ , so, according to statement 1, $z \sim x$ .

Let $T^{*}$ be the result set of Dijkstra’s algorithm for a one-dimensional skeleton ${\hat{P}}^{1}$ of the polyhedron $\hat{P}$ if we choose the pair $(u_{1}, 0)$ as the start point, and the pair $(u_{2}, i)$ as the end point (see, for example, [6], chapter 6, section 6.3).

According to the deﬁnition of the complex $\hat{K}$ , the pairs $\hat{v} = (v, j)$ and $\hat{u_{1}} = (u_{1}, 0)$ in step 4, as well as the pairs $\hat{v} = (v, j)$ and $\hat{w} = (w, k)$ in step 8 are connected by the edges $[\hat{v} \hat{u_{1}}] \in E (\hat{P})$ and $[\hat{v} \hat{w}] \in E (\hat{P})$ respectively. Also, according to (5), we have the equalities $\hat{L} ([\hat{v} \hat{u_{1}}]) = L ([v u_{1}])$ in step 4 and $\hat{L} ([\hat{v} \hat{w}]) = L ([v w])$ in step 8. This implies that the set $\hat{T}$ constructed by step 10 is the same that $T^{*}$ .

Let note that step 10 is not limited to compute $z = [v_{0} v_{1}] + \dots + [v_{q - 1} v_{q}]$ starting at $v_{0} = u_{1}$ and ending at $v_{q} = u_{2}$ , but it also gives us the possibility to construct the vector sequence $j_{0}, j_{1}, \dots, j_{q} \in ℤ_{2}^{r}$ , that will satisfy the equalities $j_{0} = 0$ , $j_{q} = i$ and $j_{s} = j_{s - 1} + J ([v_{s - 1} v_{s}])$ .

Set ${\hat{v}}_{s} = (v_{s}, j_{s})$ for all $s = 0, 1, \dots, q$ . Then $[{\hat{v}}_{s - 1} {\hat{v}}_{s}] \in E (\hat{P})$ for the same $s$ and $\hat{z} = [{\hat{v}}_{0} {\hat{v}}_{1}] + [{\hat{v}}_{1} {\hat{v}}_{2}] + \dots + [{\hat{v}}_{q - 1} {\hat{v}}_{q}]$ is a path in the skeleton ${\hat{P}}^{1}$ , starting at $(u_{1}, 0)$ and ending at $(u_{2}, i)$ . Since it can be computed by Dijkstra’s algorithm, $\hat{L} (\hat{z})$ is not over than weight of any other path in ${\hat{P}}^{1}$ , running from $(u_{1}, 0)$ to $(u_{2}, i)$ .

By the construction of the path $\hat{z}$ and according to (5), $L (z) = \hat{L} (\hat{z})$ . Now, in the polyhedron $P$ , let consider another path $z^{'}$ connecting the vertices $u_{1}$ , $u_{2}$ and homologous to $x$ . Since $p : \hat{P} \to P$ is a covering, there exists the unique path ${\hat{z}}^{'}$ in $\hat{P}$ , that covers $z^{'}$ and starts at the vertex $(u_{1}, 0)$ . At the same time, by statement 3 the end points of these paths $\hat{z}$ and ${\hat{z}}^{'}$ coincide. But then according to the above proof, $L (z) = \hat{L} (\hat{z}) \leq \hat{L} ({\hat{z}}^{'}) = L (z^{'})$ . □

ALGORITHM 3. Searching for the minimal cycle from ﬁxed gomology class.

Input:

1) list $V (P)$ of vertices for polyhedron $P$ ;

2) lists $U (v, P)$ of vertices incident to $v$ for all vertices $v \in V (P)$ ;

3) simple basis cycles $z_{1}^{n - 1}, z_{2}^{n - 1}, \dots, z_{r}^{n - 1}$ of homology group $H_{n - 1} (P)$ ;

4) lists $V (z_{1}^{n - 1}), \dots, V (z_{r}^{n - 1})$ of vertices from cycles $z_{1}^{n - 1}, \dots, z_{r}^{n - 1}$ respectively;

5) index vector-function $J : C_{1} (P) \to ℤ_{2}^{r}$ relative to basis $[z_{1}^{n - 1}]$ , …, $[z_{r}^{n - 1}]$ of $H_{n - 1} (P)$ ;

6) weight function $L : C_{1} (P) \to ℝ$ ;

7) cycle $x \in Z_{1} (P)$ .

Output:

1-cycle $z \in Z_{1} (P)$ .

Algorithm Description.

Step 1. Set $Z : = \emptyset$ .

Step 2. Determine the vector $i = J (x)$ .

Step 3. If $i = 0$ , then set $z = 0$ and go to step 7.

Step 4. Find a number $k \in {1, \dots, r}$ such that coordinate $i^{k}$ of the vector $i$ is equal 1.

Step 5. For each vertex $v \in V (z_{k}^{n - 1})$ execute steps 5.1 – 5.3.

Step 5.1. Using algorithm 2 we ﬁnd containing $v$ cycle $z_{v} \in Z_{1} (P)$ homologous to the cycle $x$ and having minimal weight $L (z_{v})$ in set of all cycles with the same properties.

Step 5.2. Add the cycle $z_{v}$ into the list $Z$ .

Step 5.3. Take the next vertex $v \in V (z_{k}^{n - 1})$ .

Step 6. Choose the cycle $z \in Z$ such that $L (z) = min_{z^{'} \in Z} L (z^{'})$ .

Step 7. Quit.

End of algorithm.

Proof. First, if $i = 0$ , then according to proposition 1, cycle $x$ is homologous to zero. At the same time we assume in step 3 that $z = 0$ . According to (4), $L (0) = 0$ . Thus, in this case $z \sim x$ and $L (z) = min_{y \in [x]} L (y)$ .

Further, let $i \neq 0$ . Then according to step 4 $i^{k} = 1$ for $k \in {1, \dots, r}$ .

Let now consider an arbitrary element $z_{v}$ in the list $Z$ . It is chosen in step 5.1, and according to this step $z_{v} \sim x$ . Since $z = z_{v}$ for some $v \in V (z_{s}^{n - 1})$ then $z \sim x$ too.

Let assume that some one-dimensional cycle $y$ of the polyhedron $P$ belongs to the class $[x]$ . Then $J (y) = J (x) = i$ . Hence, $Ind ([y], [z_{k}^{n - 1}]) = J^{k} (y) = 1$ , and therefore the cycles $y$ and $z_{k}^{n - 1}$ have at least one common vertex $u \in V (z_{k}^{n - 1})$ . In this case, according to the selection of cycle $z_{u}$ in step 5.1 of algorithm 3, $L (z_{u}) \leq L (y)$ . This implies according to step 6, that $L (z) \leq L (z_{u}) \leq L (y)$ . □

For all algorithms we created program realization of C++ classes. Then we found cycles of minimal length on several computer models of closed 2-dimensional manifolds.

References

Nizhni Novgorod State University, Gagarina pr., 23, Nizhny Novgorod: 603950, Russia

1. Index Vector-Function

2. Regular Covering with the Automorphism Group H1 (P)

3. Minimal Cycles Searching

References

2. Regular Covering with the Automorphism Group $H_{1} (P)$