Computing a Minimum Cost Matching with Half-Edges

In this section we describe the construction of a minimum-cost matching with half-edges (and some additional properties) M¹² in the extended instance (G⁰, c⁰), as defined in Sect. 2.

Recall that a square is a 4-cycle. We refer to a diagonal of cost i as an i-diagonal, for i = 1, 2.

Intuitively, we want to ensure that M¹² matches at least one vertex of each 1-square of Cminwith some vertex from outside the square or leaves at least one vertex of a 1-square unmatched. As we want to find such a matching M¹² efficiently, and we want the cost of M¹² to be at most c(opt)/2, we allow M¹² to contain half-edges. However, we will only allow half-edges in a controlled manner. The idea is to allow a half-edge {u, v_e} within M¹² only when the corresponding edge e is a native edge of a 1-square of Cmin. Also, we want to allow at most one half-edge per each 1-square of Cmin. For technical reasons (due to parity issues), we have to relax these simple conditions to a more complex set of conditions.

IDefinition 6. A matching with half-edges M¹² in (G⁰, c⁰) is good for Cmin if it satisfies the following properties.

(M¹².1) For each half-edge {u, ve} ∈ M¹², except at most one special half-edge, the edge e is a native edge of a 1-square of C_min. Also, each 1-square of C_min is incident to at most one half-edge of M¹².

(M¹².2) For every 1-square C ∈ C_min(i) there is a 1-edge e_C ∈ M¹² incident to C such that the other endpoint of eC is incident to a cycle of Cmin different from C, or (ii) at least one of the vertices of C is unmatched in M¹².

(M¹².3) M¹² may contain a special half-edge {u, ve} only if |V (G)| is odd and Cmindoes not contain a 2-edge. The following conditions are satisfied for the special half-edge {u, ve}:

a. e is an edge of a fixed odd-length 1-cycle C₀∈ C_mincalled a special cycle.

b. If Cmin contains a cycle of length at least 7, then C0 has length at least 7.

c. If C₀ is a triangle or pentagon and C_minconsists of at least two cycles, then at least two vertices of C0are incident to external edges of M¹², or some vertex of C0 is unmatched in M¹².

ILemma 7. Consider an instance (G, c) of (1, 2)-TSP with minimum-cost path-cycle cover C_min and extended instance (G⁰, c⁰). A matching M¹² of minimum cost among all matchings with half-edges M¹² which are good for Cmincan be computed in time O(n^2.5) where n = |V (G)|.

v1 v2

Proof. First, from the instance (G, c) we create an unweighted graph G⁰⁰₀ with a vertex capacity interval for each vertex of G⁰⁰₀. We do that by locally and independently modifying each 1-square in C_min. The modification introduces new vertices and edges, as well as vertex capacity intervals [`v, uv] for each vertex v ∈ V (G⁰⁰₀).

We start by setting V (G⁰⁰₀) = V (G), and assigning the capacity interval [0, 1] to each vertex. For each edge e ∈ E(G) which is not an internal 1-edge of a 1-square of Cmin, we add e to E(G⁰⁰₀). Then, for each 1-square C = (v1, . . . , v4) ∈ Cmin we proceed as follows.

For each internal 1-edge e_{i,j} = {vi, v_j} ∈ E(G) of C (note that we consider both the native 1-edges and the 1-diagonals of C here), we introduce two new vertices u_(i,j), u_(j,i) with capacity intervals [1, 1], and two new edges {v_i, u_(i,j)}, {u_(j,i), v_j}. We call these added vertices subdivision vertices.

The exact type of further modification depends on whether the number of 1-diagonals of C in (G, c) is two, one, or zero.

If C has two 1-diagonals: (See Fig. 1a) Introduce a vertex v_C of capacity interval [9, 12]; then connect vC to all 12 subdivision vertices u_(i,j), u_(j,i).

If C has exactly one 1-diagonal {v₂, v₄}: (See Fig. 1b) Introduce two vertices v_C¹, v_C² of capacity intervals [3, 4] and [5, 5], respectively, and one vertex vC of capacity interval [0, 1]. Connect v_C1 to each vertex u_(i,j), u_(j,i)that is a neighbour of v₁ or v₃, and connect vC₂ to each vertex u_(i,j), u_(j,i) that is a neighbour of v2or v4. Further, add two edges {vC, v¹_C}, {vC, v_C²}.

If C is a square with no 1-diagonal: (See Fig. 1c) Introduce two vertices v_C¹ and v²_C of capacity interval [3, 3] and one vertex v_C of capacity interval [0, 1]. Connect v_C1 to each vertex u(i,j), u(j,i) that is a neighbour of v1 or v3, and connect vC₂ to each vertex u_(i,j), u_(j,i) that is a neighbour of v2or v4. Further, add twoedges {vC, v_C¹}, {vC, v_C²}.

This completes the construction of the graph G⁰⁰₀ with vertex capacity intervals [`v, uv] for each v ∈ V (G⁰⁰₀).

The cost of a b-matching M₀⁰⁰ in G⁰⁰₀ is defined as c⁰(M₀⁰⁰) =¹₂|{v ∈ V (G) : v is matched in M₀⁰⁰}| + |{v ∈ V (G) : v is unmatched in M₀⁰⁰}|. For the graph G⁰⁰₀ and vertex capacity intervals [`_v, u_v] for each v ∈ V (G⁰⁰₀), we compute a minimum-cost b-matching M₀⁰⁰ that respects the vertex capacity intervals or, equivalently a b-matching that respects the vertex capacity intervals and minimizes the number of basic vertices unmatched in M₀⁰⁰.

IClaim 1. For the graph G⁰⁰₀ and vertex capacity intervals [`v, uv] for each v ∈ V (G⁰⁰₀), a minimum-cost b-matching M₀⁰⁰ that respects the vertex capacity intervals can be computed in O(n^2.5) time.

We defer the proof of Claim 1 to the full version of this paper.

IClaim 2. The b-matching M₀⁰⁰in G⁰⁰₀can be transformed into a matching with half-edges M¹² which is good for Cmin, has the same cost as M₀⁰⁰, and contains no special half-edge.

Proof of Claim 2. We construct the matching M¹² as follows. For any edge e = {u, v} ∈ E(G⁰⁰₀) such that both u, v are basic vertices of G⁰⁰₀ (i.e., they correspond to the vertices of G, and not to the vertices introduced during the gadget construction), we add the edge e to M¹². Then, consider each 1-square C = {v1, .., v4} ∈ Cmin and a gadget corresponding to it. If there are some two vertices vi, vj ∈ C such that both vi and vj are matched by the b-matching with subdivision vertices, and the edge e = {v_i, v_j} in G⁰ is a 1-edge, we add e to M¹². We construct such pairings greedily. For all vertices vi which are matched by the b-matching with subdivision vertices, and which have not been paired, we add a half-edge {vi, ve} to M¹², where e = {vi, v(i mod 4)+1}.

The degree of each basic vertex v ∈ V (G⁰⁰₀) in M¹² is the same as the degree of v in the b-matching. The degree of each extended vertex v ∈ V (G⁰⁰₀) in M¹² is either 0 or 1. Therefore, M¹² is a matching with half-edges. Also, is it easy to see that the cost of M¹² is the same as the cost of the b-matching. We now have to prove that M¹² is good for Cmin. As we did not denote any half-edge of M¹² as special, we only need to check properties 1 and 2 of Definition 6.

Consider a 1-square C = {v1, . . . , v₄} ∈ Cmin. From the gadgets construction we can see that the vertex capacities for vC, v_C¹, v_C² enforce that at most three of the vertices {v1, . . . , v4} are matched with a subdivision vertex. Therefore, at least one of the vertices {v₁, . . . , v₄} is matched by the b-matching via an edge not belonging to the gadget, i.e., an external 1-edge or at least one vertex of C is unmatched in M₀⁰⁰. Therefore, Property 2 holds.

From the construction of M¹², each half-edge of M¹² corresponds to a native edge e = {vi, v(i mod 4)+1} of a 1-square. To prove Property 1, we need to show that each 1-square C ∈ Cmin is incident to at most one half-edge. We already know that at most three of the vertices {v1, . . . , v4} of C are matched with a subdivision vertex. If there are three, some two of them are incident to neighboring vertices of C, and will be transformed into one native edge in M¹², which will result in only one half-edge of M¹² incident to C. If exactly two of the vertices {v₁, . . . , v₄} of C were matched with a subdivision vertex, they yield two half-edges within M¹² only if they are incident with the opposite corners of C, and the corresponding diagonal has cost 2. We show that the construction of the gadgets prevents this from happening.

First, consider the case when C has no 1-diagonal, see Fig. 1c. Assume, without loss of generality, that exactly the vertices v1, v3are matched by the b-matching with the subdivision vertices. Then, as the capacity interval of v¹_C is [3, 3], and the capacity interval of vC is [0, 1], v_C must be matched with v_C¹, and v¹_C must be matched with 2 of the subdivision vertices. Then, as the capacity interval of v_C² is [3, 3], v_C² must be matched with 3 subdivision vertices. But that leaves one subdivision vertex unmatched, and it therefore cannot happen.

Now, consider the case when C has exactly one 1-diagonal {v2, v4}, see Fig. 1b. Assume that exactly the vertices v₁, v₃ are matched by the b-matching with the subdivision vertices.

Then, by the capacity intervals of v_C¹ and vC, again vC must be matched with v_C¹, and v_C¹ must be matched with 2 of the subdivision vertices. Then, as the capacity interval of v_C² is [5, 5], v_C² must be matched with 5 subdivision vertices. But that leaves one subdivision vertex unmatched, and it therefore cannot happen.

Each 1-square of Cminis incident to at most one half-edge, and therefore Property 1 holds and the matching M¹² is good. This completes the proof of Claim 2. J IClaim 3. Any matching with half-edges M¹² which is good for Cmin and contains no special half-edge can be transformed into a b-matching M₀⁰⁰ in G⁰⁰₀ of the same cost.

Proof of Claim 3. Consider a matching with half-edges M¹² which is good for C_min and contains no special half-edge. We will construct a corresponding b-matching for M¹². For any edge e = {u, v} ∈ M¹² which is not a half-edge or a 1-edge of a 1-square, e is also present in the graph G⁰⁰₀, and we add e to the b-matching. Now, consider any half-edge {u, ve} ∈ M¹², where e = {u, u⁰}. From Property 1 of Definition 6, e is a native edge of a square C ∈ Cmin. In the b-matching, we connect u with any subdivision edge neighbouring with it. Last, for any edge {ui, uj} which is a 1-edge of a 1-square, we take the two edges {ui, u_(i,j)} and {u_(j,i), u_j} into the b-matching. From Property 2 of Definition 6, at most 3 subdivision edges corresponding to any 1-square C ∈ Cmin have been matched by this procedure. Moreover, if there were two or three, then some two of them must be incident to two endpoints of a 1-edge of C (either a native edge, or a diagonal).

We now show how to extend this matching to a b-matching. For any 1-square C ∈ C_min with two 1-diagonals, we match vC with the at least 9 unmatched subdivision vertices.

This completes the proof of Claim 3. J

If |V (G)| is odd and Cmin contains only 1-edges, we also build another unweighted graph G⁰⁰₁ from G⁰, in which we find a b-matching M₁⁰⁰. The graph G⁰⁰₁ is quite similar to G⁰⁰₀. The details of constructing G⁰⁰₁ and computing M₁⁰⁰ are given in the full version.

From M₁⁰⁰ we obtain a matching M

1 2

1 with half-edges good for Cmin. If |V (G)| is odd and Cmin does not contain any 2-edge, we set as M¹² that one of the matching M¹² and M

1 2

1

that has smaller cost. This completes the proof. J

ILemma 8. Any minimum-cost matching M¹² of (G⁰, c⁰) that is good for Cmin satisfies c⁰(M¹²) ≤ c(opt)/2.

Proof. Let (G⁰⁰, c⁰⁰) denote the extension of the graph G, in which we add two half-edges of each edge of G, also those of cost 2. Each edge e of G has cost c⁰⁰(e) = c(e) in G⁰⁰. Each half-edge of a 1-edge e ∈ G has cost ¹₂ and each half-edge of a 2-edge e ∈ G has cost 2. The cost of a matching M in G⁰⁰is defined in the usual way as c⁰⁰(M ) =P

e∈Mc⁰⁰(e). We notice that for any matching M in G⁰⁰it holds c⁰⁰(M ) = c⁰(M⁰), where M⁰= M ∩ E(G⁰).

To prove the lemma, we partition the edges of a fixed but arbitrary tour opt of minimum cost in (G, c) into two perfect matchings M1∪ M2 in (G⁰⁰, c⁰⁰) with half-edges, each of which constitutes in (G⁰, c⁰) a matching with half-edges, which is good for Cmin.

To this end, let S2 denote the set of squares in Cmin such that opt uses two of its internal 1-edges. Similarly, let S3 denote the set of squares in Cmin such that opt uses three of its internal 1-edges. Let us note that if S2∪ S3 6= ∅, then partitioning opt into two perfect matchings might yield a matching or matchings that are not good for C_min. Therefore, for each square C ∈ S2∪ S3, we take one of its internal 1-edges eCbelonging to opt, and split es

into two half-edges. For each such edge e_C, we place one of its half-edges into M₁and place its other half-edge into M2.

If the parities of |S₂∪ S₃| and |V (G)| are the same, then this way we have already decomposed opt into two perfect matchings with half-edges M1and M2. Assume, without loss of generality, that c⁰(M₁) ≤ c⁰(M₂). From M₁ we construct a matching M⁰ in G⁰. We first initialize M⁰ = M1. This way, the condition c⁰(M⁰) ≤ c⁰(opt)/2 is clearly satisfied.

However, M⁰ potentially is not a perfect matching with half-edges, as it might contain a half-edge of a diagonal of a square C ∈ S2. We can, however, replace such a half-edge with a half-edge of an edge of C, without increasing the cost of M⁰.

If the parities of |S2∪ S3| and |V (G)| differ and opt uses a 2-edge, then we choose any such 2-edge e ∈ opt and split it into two half-edges. Otherwise, if S₂∪ S₃ is empty, |V (G)| is odd. Then any path-cycle cover of G must contain at least one odd cycle C. We split any edges of opt which is incident to a vertex of C into two half-edges. We decompose opt into two perfect matchings M1 and M2. Since c⁰(M1) = c⁰(M2), we may choose that one which contains a half-edge incident to a vertex of C.

The remaining case is when the parities of |S2∪ S3| and |V (G)| differ, each edge of opt has cost 1 and S2∪ S3 is non-empty. Then we choose one square C ∈ S2∪ S3 and do not split any of its edges. At least one of the perfect matchings M₁, M₂ from the decomposition of opt is such that it does not use two internal edges of C. Since again c⁰(M1) = c⁰(M2), we may choose that one, which does not use two internal edges of C. J