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Independent domination in finitely defined classes of graphs:
Polynomial algorithms
Vadim Lozin, Raffaele Mosca, Christopher Purcell
Discrete Applied Mathematics 182 (2015) 2–14
報告者 : 陳政謙
Related worksJournal Topic Author Result
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Peter Damaschke, Haiko Müller, Dieter Kratsch
It is shown by a reduction from 3SAT that independent dominating set remains NP-complete when restricted to chordal bipartite graphs.
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The weighted independent domination problem is NP-completefor chordal graphs
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Algorithm: Generation-1
u({v3}) = v2
v1 v5v2 v3
u({v2}) = v1
S = {{v1, v3}, {v2}, {v3}}u({v1, v3}) = v2
v4
Algorithm: Generation-1
v1 v5v2 v3 v4
u({v3}) = v2u({v2}) = v1
S = {{v1, v3}, {v2}, {v3}}u({v1, v3}) = v2
Algorithm: Generation-1
v1 v5v2 v3 v4
u({v3}) = v2u({v2, v4}) = v1
S = {{v1, v3}, {v2, v4}, {v3}}u({v1, v3}) = v2
Algorithm: Generation-1
v1 v5v2 v3 v4
u({v3}) = v2u({v2, v4}) = v1
S = {{v1, v3}, {v2, v4}, {v3}}u({v1, v3}) = v2
Algorithm: Generation-1
v1 v5v2 v3 v4
u({v3}) = v2u({v2, v4}) = v1
S = {{v1, v3}, {v2, v4}, {v3}, {v1, v4}}u({v1, v3}) = v2 u({v1, v4}) = v3
Algorithm: Generation-1
v1 v2 v3 v4
u({v3}) = v2u({v2, v4}) = v3
S = {{v1, v3}, {v2, v4}, {v3}, {v1, v4}}u({v1, v3}) = v2 u({v1, v4}) = v3
v5
Algorithm: Generation-1
v1 v2 v3 v4
u({v3}) = v2u({v2, v4}) = v3
S = {{v1, v3}, {v2, v4}, {v3}, {v1, v4}}u({v1, v3}) = v2 u({v1, v4}) = v3
v5
Algorithm: Generation-1
v1 v2 v3 v4
u({v3}) = v2u({v2, v4}) = v3
S = {{v1, v3, v5}, {v2, v4}, {v3}, {v1, v4}}u({v1, v3, v5}) = v2 u({v1, v4}) = v3
v5
Algorithm: Generation-1
v1 v2 v3 v4
u({v3, v5}) = v2u({v2, v4}) = v3
S = {{v1, v3, v5}, {v2, v4}, {v3, v5}, {v1, v4}}u({v1, v3, v5}) = v2
u({v1, v4}) = v3
v5
Algorithm: Generation-1
v1 v2 v3 v4
u({v3, v5}) = v2u({v2, v4}) = v3
S = {{v1, v3, v5}, {v2, v4}, {v3, v5}, {v1, v4}}u({v1, v3, v5}) = v2
u({v1, v4}) = v3
v5
Algorithm: Generation-1
v1 v2 v3 v4
u({v3, v5}) = v2u({v2, v4}) = v3
S = {{v1, v3, v5}, {v2, v4}, {v3, v5}, {v1, v4}, {v1, v2, v5}}u({v1, v3, v5}) = v2
u({v1, v4}) = v3
v5
u({v1, v2 , v5}) = v4
Algorithm: Generation-1
v1 v2 v3 v4
u({v3, v5}) = v2u({v2, v4}) = v3
S = {{v1, v3, v5}, {v2, v4}, {v3, v5}, {v1, v4}, {v1, v2, v5}}u({v1, v3, v5}) = v4
u({v1, v4}) = v3
v5
u({v1, v2 , v5}) = v4
Lemma 3• For a graph G with n vertices, Algorithm Generation-1 runs in
time O(n5) and the family S produced by this algorithm contains O(n3) subsets of V(G).
If a set H S was created in Step 2.1∈
G
v
AG({v, u})u
We denote W the set of neighbors of u each of which is not dominatedby every maximal independent set in G[H].
W
H := {v} A∪ G ({v, u})
Proposition 4• A set H created in Step 2.1 contains an independent set
dominating G if and only if H – H0 contains an independent set dominating W.
The partition of cliques in G[H – H0]
• Lemma 5 shows that if Q = {q1, ... , qp} is a component (clique) in G[H – H0] and Wi = W ∩ AG(qi), then {W1, ... , Wp} is a partition of W. We denote this partition by P(Q).
q1
q2 q3
v1
v2
v3
W
P(Q) = {{v1, v2}, {v3}}
Q
Lemma 6• The set H – H0 contains a maximal independent set
dominating W if and only if there is an element (Y1, . . . , Yt) ∈P(Q1) × … × P(Qt) such that Y1 ∩ … ∩ Yt = .∅P(Q1) = {{v1}, {v2, v3}}P(Q2) = {{v1, v2}, {v3}}
{v1} ∩ {v3} = ∅
The proof of Lemma 6
• Therefore, I dominates W if and only if … ∪ ∪= W.
• By De Morgan’s law, this holds if and only if Y1 ∩ … ∩ Yt = .∅
Lemma 7• Given a set W of n elements and a number of partitions P1, . . .
,Pt of W, one can check if there is an element (Y1, . . . , Yt) P∈ 1 × …× Pt such Y1 ∩…∩ Yt = in O(n∅ 2) time.
P1
P2
{v1} {v2, v3}
W = {v1, v2, v3}P1 = {{v1}, {v2, v3}}P2 = {{v1, v2}, {v3}}
{v1, v2, v3}
{v1} { }∅ {v2} {v3}
Theorem 8• Given a P2 + P3-free graph G with n vertices, one can find an
independent dominating set of minimum cardinality in G in O(n5) time.
By Lemma 3, the time complexity of Algorithm Generation-1 is O(n5).
By Proposition 4, Lemmas 6 and 7, the problem of determining if H S contains a maximal independent set dominating G can be ∈solved in O(n2) time.
Therefore, an independent dominating set of minimum cardinality in G can be found in O(n5) time.
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