# Solving Robust Variants of the Maximum Weighted Independent Set Problem on Trees

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## Abstract

**:**

## 1. Introduction

## 2. Definitions and Preliminaries

- Let $G=(V,E)$ be an undirected graph, where V is the set of vertices, and E the set of edges. An independent set of G is a subset X of V such that no two vertices in X are adjacent (connected by an edge from E).
- Let $G=(V,E)$ be an undirected graph where its vertices have weights. Suppose that all weights are nonnegative integers. A maximum weighted independent set of G is an independent set of G in which the sum of vertex weights is as large as possible.
- The problem of finding a maximum weighted independent set in a given weighted graph is called the (conventional) maximum weighted independent set problem (the MWIS problem).

- An absolute robust solution is the one in which the worst (conventional) objective-function value, measured over all scenarios, is the best among all feasible solutions.
- A robust deviation solution is the one in which the maximum deviation from the conventional optimum, measured over all scenarios, is the smallest among all feasible solutions.
- A relative robust deviation solution is the one in which the maximum relative deviation from the conventional optimum, measured over all scenarios, is the smallest among all feasible solutions.

- An absolute robust solution for the MWIS problem is an independent set ${X}_{A}$ that maximizes the function ${min}_{s\in S}F(X,s)$ over the whole collection of possible independent sets X.
- A robust deviation solution for the MWIS problem is an independent set ${X}_{D}$ that minimizes the function ${max}_{s\in S}({F}_{s}^{*}-F(X,s))$ over the whole collection of possible independent sets X.
- A relative robust deviation solution for the MWIS problem is an independent set ${X}_{R}$ that minimizes the function ${max}_{s\in S}(({F}_{s}^{*}-F(X,s))/{F}_{s}^{*})$ over the whole collection of possible independent sets X.

- An undirected graph G is a tree if it is connected and acyclic. Or, equivalently, an undirected graph G with n vertices is a tree if it is connected and its number of edges is $n-1$.A tree is usually organized in a hierarchy, so that one vertex is chosen to be the root, its neighbors become its children, remaining neighbors of children become children’s children, etc. Then, any vertex can have 0, 1 or more children. The root has no parent, and any other vertex has exactly one parent. Vertices with no children are called leaves.
- Let ${I}_{i}=\left[{a}_{i},{b}_{i}\right]$, denote a closed interval, where ${a}_{i},{b}_{i}\in \mathbb{R}$ and ${a}_{i}<{b}_{i}$. An undirected graph $G=(V,E)$ with $\left|V\right|=n$ vertices is called interval graph for a finite family $\mathcal{I}=\{{I}_{1},{I}_{2},...,{I}_{n}\}$ of intervals on the real line if there is a one-to-one correspondence between $\mathcal{I}$ and V such that two intervals of $\mathcal{I}$ have non-empty intersection if and only if their corresponding vertices in V are adjacent to each other.

- Consider first ${I}_{2}$, ${I}_{3}$, and ${I}_{4}$; since their vertices are not adjacent, they should be disjunct and placed on the real line in some sequence. Without any loss of generality, we can assume that ${I}_{2}$ is on the left, ${I}_{3}$ in the middle, and ${I}_{4}$ on the right, as shown in the middle part of Figure 2 (otherwise, we could renumber the vertices).
- Next, consider ${I}_{1}$. Since vertex 1 is adjacent to vertices 2, 3, and 4, ${I}_{1}$ should overlap with all three intervals ${I}_{2}$, ${I}_{3}$, and ${I}_{4}$. More precisely, the left endpoint of ${I}_{1}$ cannot be larger than the right endpoint of ${I}_{2}$ since, otherwise, the two intervals would not overlap. Similarly, the right endpoint of ${I}_{1}$ cannot be smaller than the left endpoint of ${I}_{3}$. So, the situation looks as shown in the upper part of Figure 2. Consequently, ${I}_{3}$ must be a subset of ${I}_{1}$.
- Finally, we consider ${I}_{6}$. Since vertices 3 and 6 are adjacent, ${I}_{6}$ should overlap with ${I}_{3}$, as shown in the lower part of Figure 2. But since ${I}_{3}$ is a subset of ${I}_{1}$, it means that ${I}_{6}$ must also overlap with ${I}_{1}$. This is a contradiction with the fact that vertices 1 and 6 are not adjacent.

## 3. Complexity Issues

Instance:a list of positive integers ${a}_{1},{a}_{2},\dots ,{a}_{n}$.

Question:is there a subset of indices $I\subset \{1,2,\dots ,n\}$ such that ${\sum}_{i\in I}{a}_{i}={\sum}_{i\notin I}{a}_{i}$ ?

**Theorem**

**1.**

**Proof.**

- The root has weight 0.
- In the i-th vertical segment, the upper vertex has weight ${a}_{i}$, and the lower vertex has weight 0.

- The root has weight 0.
- In the i-th vertical segment, the lower vertex has weight ${a}_{i}$, and the upper vertex has weight 0.

- If the selected independent set contains many upper vertices, it will have a large weight under the first scenario but a small weight under the second scenario. Thus, its minimum weight over scenarios will be small, so that it will not be an absolute robust solution.
- If the selected independent set contains many lower vertices, it will have a large weight under the second scenario but a small weight under the first scenario. Again, its minimum weight over scenarios will be small, so that it again will not produce an absolute robust solution.
- An absolute robust solution (max-min) is achieved when the sum of weights of the chosen upper vertices is approximately equal to the sum of weights of the chosen lower vertices, since then the minimum over scenarios is as large as possible.

**Theorem**

**2.**

**Proof.**

- The optimal conventional solution under the first scenario is obtained by choosing the upper vertex within each vertical segment from Figure 3. The weight of that solution is T.
- The optimal conventional solution under the second scenario is obtained by choosing the lower vertex within each vertical segment from Figure 3, and its weight is again T.
- Let us consider any independent set X. Let I be the set of indices of vertical segments from Figure 3 where X has chosen the upper vertex. Then, the “regret” for X under the first scenario is equal to $T-{\sum}_{i\in I}{a}_{i}$, and the regret for X under the second scenario is $T-{\sum}_{i\notin I}{a}_{i}$.
- If X contains many upper vertices, it will have a small regret under the first scenario but a large regret under the second scenario, so that its maximal regret over both scenarios will be large.
- If X contains many lower vertices, it will have a large regret under the first scenario and a small regret under the second scenario. Again, its maximal regret over both scenarios will be large.
- A robust deviation solution (min-max regret) is achieved when ${\sum}_{i\in S}{a}_{i}$ is approximately equal to ${\sum}_{i\notin S}{a}_{i}$, since then the maximal regret over both scenarios is as small as possible and $\approx T/2$.

**Theorem**

**3.**

**Proof.**

- Let us consider any independent set X, and let I be the set of indices of vertical segments from Figure 3 where X has chosen the upper vertex. Then, the relative regret for X under the first scenario is $(T-{\sum}_{i\in I}{a}_{i})/T$, while the relative regret under the second scenario is $(T-{\sum}_{i\notin I}{a}_{i})/T$. Division with T is legal since T is a sum of positive integers; thus, $>0$.
- A relative robust deviation solution (min-max relative regret) is achieved again when ${\sum}_{i\in S}{a}_{i}$ is approximately equal to ${\sum}_{i\notin S}{a}_{i}$, since then the maximal relative regret over both scenarios is as small as possible and $\approx (T-T/2)/T=1/2$.

**Corollary**

**1.**

**Corollary**

**2.**

**Corollary**

**3.**

## 4. Algorithms

- An independent set for the subtree rooted at ${v}_{i}$ that has the greatest weight and that contains ${v}_{i}$. Such set will be called the inclusive independent set.
- An independent set for the subtree rooted at ${v}_{i}$ that has the greatest weight, and that does not contain ${v}_{i}$. Such set will be called the exclusive independent set.

- The inclusive independent set for ${v}_{i}$ is obtained as the union of exclusive independent sets for children of ${v}_{i}$, plus ${v}_{i}$ itself.
- The exclusive independent set for ${v}_{i}$ is obtained as a union of either inclusive or exclusive independent sets for children of ${v}_{i}$. For each child, the one with greater weight is chosen.

- A collection of independent sets for the subtree rooted at ${v}_{i}$ that are considered “good” according to the chosen robustness criterion. Thereby, each of those independent sets contains ${v}_{i}$. Such collection is called the inclusive population.
- A collection of independent sets for the subtree rooted at ${v}_{i}$ that are considered “good” according to the chosen robustness criterion. Thereby, none of those independent sets contains ${v}_{i}$. Such collection is called the exclusive population.

- A member of the inclusive population for ${v}_{i}$ is assembled as a union, where for each child of ${v}_{i}$ one member of its exclusive population is added. Vertex ${v}_{i}$ is also put into the union. The selection of particular members from children populations is done randomly but not with equal probability. Independent sets that are better according to the used robustness criterion have more chance to be chosen. In more detail, all independent sets are represented by sub-segments in which the lengths correspond to the respective objective-function values. Next, a uniformly distributed random number is generated in which the range covers the whole segment. The random number will more likely fall into a sub-segment with greater length.
- A member of the exclusive population for ${v}_{i}$ is created as a union, where for each child of ${v}_{i}$ one member of its inclusive or exclusive population is added. Again, independent sets from children’s populations are chosen randomly, but those sets that are better according to the used robustness criterion are more likely to be selected. Random choice is implemented in the same way as for the members of inclusive population.

- It requires less memory. Moreover, it would be almost impossible to store simultaneously optimal solutions for all subtrees in case of a large tree.
- Our aim is to find the global robustly optimal solution, hence comparing “partial” solutions to global conventional solutions would hopefully lead more directly to the desired goal.

- An inclusive population must have a member that is obtained as a union of best members from the respective children’s exclusive populations. The comparison of members within any child population is done according to the involved robustness criterion.
- An inclusive population must have the so-called average-best member. It is the optimal solution of a conventional (non-robust) problem instance where the weight of any vertex is set to the average of its weights over all scenarios. According to the exact algorithm from Reference [2], the average-best member is easily obtained as a union of average-best members from the respective children’s exclusive populations.

- (1)
- Dynamic programming. We calculate a table of partial solutions. We use results of previous calculations within subsequent calculations.
- (2)
- Greedy approach. We implicitly assume that an independent set will be a “good” solution for the whole tree if it is assembled from parts that are “good” solutions for subtrees of that tree.
- (3)
- Evolutionary computing. We use populations of solutions instead of single solutions. Our populations evolve during algorithm execution. Better solutions have more chance to survive and evolve.

## 5. Testing and Results

## 6. Conclusions

## Author Contributions

## Funding

## Conflicts of Interest

## References

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**Figure 2.**Demonstrating by contradiction that the tree from Figure 1 cannot be an interval graph.

**Figure 3.**A tree that reduces a 2-partition problem instance to a maximum weighted independent set (MWIS) problem instance.

**Table 1.**The inclusive and exclusive populations for vertices of the tree from Figure 4 obtained by our population algorithm according to the absolute robustness criterion.

Vertex | Inclusive Population Members | Exclusive Population Members |
---|---|---|

13 | $\left\{13\right\}$ | ∅ |

12 | $\left\{12\right\}$ | ∅ |

11 | $\left\{11\right\}$ | ∅ |

10 | $\left\{10\right\}$ | ∅ |

9 | $\left\{9\right\}$ | ∅ |

8 | $\left\{8\right\}$ | ∅ |

7 | $\left\{7\right\},\left\{7\right\},\left\{7\right\},\left\{7\right\},\left\{7\right\}$ | $\{12,13\},\{12,13\},\{12,13\},\{12,13\},\left\{13\right\}$ |

6 | $\left\{6\right\},\left\{6\right\},\left\{6\right\},\left\{6\right\},\left\{6\right\}$ | $\{10,11\},\{10,11\},\{10,11\},\left\{10\right\},\{10,11\}$ |

5 | $\left\{5\right\},\left\{5\right\},\left\{5\right\},\left\{5\right\},\left\{5\right\}$ | $\left\{9\right\},\left\{9\right\},\left\{9\right\},\left\{9\right\},\left\{9\right\}$ |

4 | $\left\{4\right\},\left\{4\right\},\left\{4\right\},\left\{4\right\},\left\{4\right\}$ | $\left\{8\right\},\left\{8\right\},\left\{8\right\},\left\{8\right\},\left\{8\right\}$ |

3 | $\{3,12,13\},\{3,12,13\},\{3,12,13\},$ | $\left\{7\right\},\{12,13\},\left\{7\right\},\left\{7\right\},\{12,13\}$ |

$\{3,12,13\},\{3,12,13\}$ | ||

2 | $\{2,9,10,11\},\{2,9,10,11\},\{2,9,10,11\},$ | $\{5,10,11\},\{9,10,11\},\{9,10,11\},$ |

$\{2,9,10,11\},\{2,9,10,11\}$ | $\{9,10,11\},\{6,9\}$ | |

1 | $\{1,8\},\{1,8\},\{1,8\},\{1,8\},\{1,8\}$ | $\left\{4\right\},\left\{4\right\},\left\{8\right\},\left\{8\right\},\left\{4\right\}$ |

0 | $\{0,4,5,7,10,11\},\{0,4,9,10,11,12,13\},$ | $\{1,2,3,8,9,10,11,12,13\},$ |

$\{0,4,7,9,10,11\},\{0,4,7,9,10,11\},$ | $\{1,2,3,8,9,10,11,12,13\},$ | |

$\{0,4,5,7,10,11\}$ | $\{3,4,9,10,11,12,13\},$ | |

$\{2,3,4,9,10,11,12,13\},$ | ||

$\{1,2,3,8,9,10,11,12,13\}$ |

Average Relative Errors: | |||
---|---|---|---|

5 Children | 10 Children | 15 Children | |

30,000 vertices | 0.25% | 0.16% | 0.12% |

60,000 vertices | 0.21% | 0.13% | 0.08% |

90,000 vertices | 0.17% | 0.11% | 0.08% |

Best Relative Errors: | |||

5 Children | 10 Children | 15 Children | |

30,000 vertices | 0.24% | 0.16% | 0.12% |

60,000 vertices | 0.21% | 0.13% | 0.08% |

90,000 vertices | 0.17% | 0.11% | 0.08% |

Average Relative Errors: | |||
---|---|---|---|

5 Children | 10 Children | 15 Children | |

30,000 vertices | 5.71% | 6.37% | 8.05% |

60,000 vertices | 4.18% | 5.78% | 6.72% |

90,000 vertices | 4.05% | 5.55% | 5.37% |

Best Relative Errors: | |||

5 Children | 10 Children | 15 Children | |

30,000 vertices | 5.70% | 6.37% | 8.05% |

60,000 vertices | 4.16% | 5.78% | 6.70% |

90,000 vertices | 4.04% | 5.53% | 5.37% |

Average Relative Errors: | |||
---|---|---|---|

5 Children | 10 Children | 15 Children | |

30,000 vertices | 5.73% | 6.35% | 7.85% |

60,000 vertices | 4.19% | 5.78% | 6.53% |

90,000 vertices | 4.07% | 5.54% | 5.32% |

Best Relative Errors: | |||

5 Children | 10 Children | 15 Children | |

30,000 vertices | 5.59% | 6.20% | 7.59% |

60,000 vertices | 4.17% | 5.67% | 6.34% |

90,000 vertices | 4.00% | 5.46% | 5.20% |

Population Algorithm: | |||
---|---|---|---|

5 Children | 10 Children | 15 Children | |

30,000 vertices | 0.176 | 0.154 | 0.134 |

60,000 vertices | 0.357 | 0.292 | 0.268 |

90,000 vertices | 0.552 | 0.452 | 0.408 |

CPLEX: | |||

5 Children | 10 Children | 15 Children | |

30,000 vertices | 17.417 | 32.336 | 33.582 |

60,000 vertices | 25.671 | 28.138 | 37.576 |

90,000 vertices | 37.576 | 34.802 | 41.975 |

Population Algorithm: | |||
---|---|---|---|

5 Children | 10 Children | 15 Children | |

30,000 vertices | 1.154 | 0.909 | 0.760 |

60,000 vertices | 2.557 | 1.918 | 1.648 |

90,000 vertices | 3.864 | 2.895 | 2.469 |

CPLEX: | |||

5 Children | 10 Children | 15 Children | |

30,000 vertices | 163.743 | 119.776 | 109.281 |

60,000 vertices | 851.177 | 137.453 | 127.226 |

90,000 vertices | 3294.592 | 188.199 | 142.982 |

Population Algorithm: | |||
---|---|---|---|

5 Children | 10 Children | 15 Children | |

30,000 vertices | 1.681 | 1.167 | 0.986 |

60,000 vertices | 3.467 | 2.496 | 2.101 |

90,000 vertices | 5.205 | 3.809 | 3.187 |

CPLEX: | |||

5 children | 10 children | 15 children | |

30,000 vertices | 250.002 | 104.695 | 80.759 |

60,000 vertices | 1426.761 | 150.37 | 106.375 |

90,000 vertices | 2217.743 | 199.58 | 142.479 |

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**MDPI and ACS Style**

Klobučar, A.; Manger, R. Solving Robust Variants of the Maximum Weighted Independent Set Problem on Trees. *Mathematics* **2020**, *8*, 285.
https://doi.org/10.3390/math8020285

**AMA Style**

Klobučar A, Manger R. Solving Robust Variants of the Maximum Weighted Independent Set Problem on Trees. *Mathematics*. 2020; 8(2):285.
https://doi.org/10.3390/math8020285

**Chicago/Turabian Style**

Klobučar, Ana, and Robert Manger. 2020. "Solving Robust Variants of the Maximum Weighted Independent Set Problem on Trees" *Mathematics* 8, no. 2: 285.
https://doi.org/10.3390/math8020285