Minimum Maximal Matching

Minimum Maximal Matching#

We show how to solve the minimum maximal matching problem using JijZept and JijModeling. This problem is also mentioned in 4.5. Minimal Maximal Matching on Lucas, 2014, “Ising formulations of many NP problems”.

What is Minimum Maximal Matching?#

The minimum maximal matching is a matching problem in graphs, where the matching is a set of edges in a graph that do not share any vertices with each other.

We assume that we have an undirected graph $G = (V,E)$ and that $C \subseteq E$ is a “coloring” of vertices. The constraints on $C$ are given as follows. For each edge in $C$, we color the two vertices it connects to be the same. We then promise that no two edges in $C$ share a same vertex and that $(u,v) \notin E$ if $u, v \notin D$, where $D$ is the set of vertices that are connected to any edge in $C$. This is maximal in the sense that we cannot add any more edge to $C$ without violating the first constraint while we must include all edges between uncolored vertices as long as they do not violate the first constraint; i.e. the empty set is not allowed. Such a coloring is the minimum when and only when the number of edges in $C$ is the smallest possible.

Example#

For example, consider that we have a graph $G = (V, E)$, where $V = {1,2,3,4,5}$ and where $E = {(1,2),(1,3),(2,3),(3,4),(4,5)}$. To find a maximal matching, we first have the empty set as a matching. Then, we add edges to the matching until we cannot add any more edge without violating the constraints. In such a small problem, one may be able to find the minimum maximal matching with some trials and errors. Indeed the minimum maximal matching for this graph is ${(1,2), (3,4)}$.

Mathematical model#

A binary variable $x_e$ denotes whether or not the edge $e$ is colored; the edge is colored if $x_e$ is 1, and vice versa. We also introduce a binary variable $y_v$ that denotes whether or not the vertex $v$ is included in the matching ($D$ in the above statements); the vertex is connected to a edge in $C$ if $y_v = 1$, and vice versa.

Constraint 1: relationship between $x$ and $y$

By definition, the condition below obviously holds. $$ \quad y_v = \sum_{e \in E_v} x_e \quad \forall v, $$ where $E_v$ is the set of edges that are connected to the vertex $v$.

Constraint 2: every unselected edge is connected to at least one vertex connected to a selected edge

To formulate this we use a basic observation: $1-y_{v}$ is 0 if $u$ is connected to a selected edge, and vice versa. Then, by counting $(1-y_{u})(1-y_{v})$ for all $(u,v) \in E$, we can check the violation of the constraint; i.e. if $(1-y_{u})(1-y_{v}) > 0$ for any $(u,v) \in E$, the edge $(u,v)$ can be selected without violating the constraints. $$ \quad \sum_{(u,v)\in E} (1-y_u)(1-y_v) = 0. $$

Objective function: minimize the size of the matching

The size here means the number of edges selected. $$ \quad \min \sum_{e \in E} x_e. $$

Modeling by JijModeling#

Next, we show how to implement above equations using JijModeling. We first define the variables in the mathematical model described above.

import jijmodeling as jm

# define variables
V = jm.Placeholder('V')
E = jm.Placeholder('E', ndim=2)
num_E = E.shape[0]
x = jm.BinaryVar('x', shape=(num_E,))
y = jm.BinaryVar('y', shape=(V,))
v = jm.Element('v', V)
e = jm.Element('e', num_E)

Constraint#

The constraints and the objective function are written as:

problem = jm.Problem('Minimum Maximal Matching')
problem += jm.Constraint('y_x_relation', y[v] - jm.sum((e, (E[e][0]==v)|(E[e][1]==v)), x[e])  == 0 ,forall=v)
problem += jm.Constraint('unselected_edge', jm.sum(e, (1-y[E[e][0]])*(1-y[E[e][1]]))==0)

problem += x[:].sum()

On Jupyter Notebook, one can check the problem statement in a human-readable way by hitting

problem

\[\begin{split}\begin{array}{cccc}\text{Problem:} & \text{Minimum Maximal Matching} & & \\& & \min \quad \displaystyle \sum_{\ast_{0} = 0}^{\mathrm{len}\left(E, 0\right) - 1} x_{\ast_{0}} & \\\text{{s.t.}} & & & \\ & \text{unselected\_edge} & \displaystyle \sum_{e = 0}^{\mathrm{len}\left(E, 0\right) - 1} \left(- y_{E_{e, 0}} + 1\right) \cdot \left(- y_{E_{e, 1}} + 1\right) = 0 & \\ & \text{y\_x\_relation} & \displaystyle y_{v} - \sum_{\substack{e = 0\\E_{e, 0} = v \lor E_{e, 1} = v}}^{\mathrm{len}\left(E, 0\right) - 1} x_{e} = 0 & \forall v \in \left\{0,\ldots,V - 1\right\} \\\text{{where}} & & & \\& x & 1\text{-dim binary variable}\\& y & 1\text{-dim binary variable}\\\end{array}\end{split}\]

Prepare an instance#

Here we use the same instance with an example described above.

import networkx as nx

# set empty graph
inst_G = nx.Graph()
V = [0, 1, 2, 3, 4]
# add edges
inst_E = [[0, 1], [0, 2], [1, 2], [2, 3], [3, 4]] 
inst_G.add_edges_from(inst_E)
# get the number of nodes
inst_V = list(inst_G.nodes)
num_V = len(inst_V)
instance_data = {'V': num_V, 'E': inst_E}

This graph is shown below.

import matplotlib.pyplot as plt

pos = nx.spring_layout(inst_G)
nx.draw_networkx(inst_G, pos=pos, with_labels=True)
plt.show()

../_images/c4031c02610eabd1f754f264f1afbca2cc46227571d4adab0fd1e69a3d980a03.png

Solve by JijZept’s SA#

We solve this problem using JijSASampler. We also turn on a parameter search function by setting search=True.

import jijzept as jz

# set sampler
sampler = jz.JijSASampler(config="../../../config.toml")
# solve problem
multipliers = {"y_x_relation": 0.5, "unselected_edge": 0.5}
response = sampler.sample_model(problem, instance_data, multipliers=multipliers, search=True)

---------------------------------------------------------------------------
ModuleNotFoundError                       Traceback (most recent call last)
Cell In[6], line 1
----> 1 import jijzept as jz
      3 # set sampler
      4 sampler = jz.JijSASampler(config="../../../config.toml")

ModuleNotFoundError: No module named 'jijzept'

Visualize the solution#

The optimized solution can be seen as below.

import numpy as np

# get sampleset
sampleset = response.get_sampleset()
# extract feasible samples
feasible_samples = sampleset.feasibles()
# get the values of feasible objectives
feasible_objectives = [sample.eval.objective for sample in feasible_samples]
if len(feasible_objectives) == 0:
    print("No feasible sample found ...")
else:
    # get the index of lowest value of feasible objectives
    lowest_index = np.argmin(feasible_objectives)
    # get the lowest solution
    lowest_solution = feasible_samples[lowest_index].var_values
    # get the indices of x == 1
    x_indices = [key[0] for key in lowest_solution["x"].values.keys()]
    # get the indices of y == 1
    y_indices = [key[0] for key in lowest_solution["y"].values.keys()]
    # set color list for visualization
    cmap = plt.get_cmap("tab10")
    # set vertex color
    vertex_colors = [cmap(1) if i in y_indices else cmap(0) for i in inst_V]
    # set edge color list
    edge_colors = ["red" if j in x_indices else "black" for j, _ in enumerate(instance_data["E"])]
    # dwaw the figure with two subplots
    fig, (ax1, ax2) = plt.subplots(1, 2, figsize=(10, 5))
    # plot the initial graph
    nx.draw_networkx(inst_G, pos, with_labels=True, ax=ax1)
    ax1.set_title('Initial Graph')
    plt.axis('off')
    ax1.set_frame_on(False) # Remove the frame from the first subplot
    # plot the minimum maximal matching graph
    nx.draw_networkx(inst_G, pos=pos, node_color=vertex_colors, edge_color=edge_colors, with_labels=True)
    plt.axis('off')
    ax2.set_title('Minimum Maximal Matching Graph')
    # Show the plot
    plt.show()

../_images/7194fd366ca7402ae2d24a38a8cc3cf39860ae39b044e5fe3e8d7c888b7a2e43.png

As expected, we have obtained the minimum maximal matching graph.