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Coping withNP-completeness:
Special Cases
Alexander S. KulikovSteklov Institute of Mathematics at St. Petersburg
Russian Academy of Sciences
Advanced Algorithms and ComplexityData Structures and Algorithms
The fact that a problem is NP-completedoes not exclude an efficient algorithm forspecial cases of the problem.
Outline
1 2-Satisfiability
2 Independent Sets in Trees
This part
Striking connection between stronglyconnected components of a graph andformulas in 2-CNFA linear time algorithm for 2-SAT
2-Satisfiability (2-SAT)
Input: A set of clauses, each containing atmost two literals (that is, a 2-CNFformula).
Output: Find a satisfying assignment(if exists).
Example
(x ∨ y)(z)(z ∨ x) is satisfied byx = 0, y = 1, z = 0
(x ∨ y)(z)(z ∨ x)(y) is unsatisfiable(x ∨ y)(x ∨ y)(x ∨ y)(x ∨ y) isunsatisfiable
Example
(x ∨ y)(z)(z ∨ x) is satisfied byx = 0, y = 1, z = 0(x ∨ y)(z)(z ∨ x)(y) is unsatisfiable
(x ∨ y)(x ∨ y)(x ∨ y)(x ∨ y) isunsatisfiable
Example
(x ∨ y)(z)(z ∨ x) is satisfied byx = 0, y = 1, z = 0(x ∨ y)(z)(z ∨ x)(y) is unsatisfiable(x ∨ y)(x ∨ y)(x ∨ y)(x ∨ y) isunsatisfiable
Consider a clause (ℓ1 ∨ ℓ2)
Essentially, it says that ℓ1 and ℓ2 cannotbe both equal to 0In other words, if ℓ1 = 0, then ℓ2 = 1and if ℓ2 = 0, then ℓ1 = 1
Consider a clause (ℓ1 ∨ ℓ2)
Essentially, it says that ℓ1 and ℓ2 cannotbe both equal to 0
In other words, if ℓ1 = 0, then ℓ2 = 1and if ℓ2 = 0, then ℓ1 = 1
Consider a clause (ℓ1 ∨ ℓ2)
Essentially, it says that ℓ1 and ℓ2 cannotbe both equal to 0In other words, if ℓ1 = 0, then ℓ2 = 1and if ℓ2 = 0, then ℓ1 = 1
DefinitionImplication is a binary logical operationdenoted by ⇒ and defined by the followingtruth table:
x y x ⇒ y
0 0 10 1 11 0 01 1 1
DefinitionFor a 2-CNF formula, its implication graph isconstructed as follows:
for each variable x , introduce twovertices labeled by x and x ;for each 2-clause (ℓ1 ∨ ℓ2), introducetwo directed edges ℓ1 → ℓ2 and ℓ2 → ℓ1
for each 1-clause (ℓ), introduce an edgeℓ→ ℓ
Encodes all implications imposed by theformula.
DefinitionFor a 2-CNF formula, its implication graph isconstructed as follows:
for each variable x , introduce twovertices labeled by x and x ;for each 2-clause (ℓ1 ∨ ℓ2), introducetwo directed edges ℓ1 → ℓ2 and ℓ2 → ℓ1
for each 1-clause (ℓ), introduce an edgeℓ→ ℓ
Encodes all implications imposed by theformula.
(x ∨ y)(y ∨ z)(x ∨ z)(z ∨ y)
(x ∨ y)(y ∨ z)(x ∨ z)(z ∨ y)
z
z
x
x
y
y
(x ∨ y)(y ∨ z)(x ∨ z)(z ∨ y)
z
z
x
x
y
y
(x ∨ y)(y ∨ z)(x ∨ z)(z ∨ y)
z
z
x
x
y
y
(x ∨ y)(y ∨ z)(x ∨ z)(z ∨ y)
z
z
x
x
y
y
(x ∨ y)(y ∨ z)(x ∨ z)(z ∨ y)
z
z
x
x
y
y
(x ∨ y)(y ∨ z)(x ∨ z)(z ∨ y)
z
z
x
x
y
y
x = 1, y = 1, z = 1
(x ∨ y)(y ∨ z)(x ∨ z)(z ∨ y)
z
z
x
x
y
y
x = 0, y = 0, z = 0
Thus, our goal is to assign truth values tothe variables so that each edge in theimplication graph is “satisfied”, that is, thereis no edge from 1 to 0.
Skew-Symmetry
The graph is skew-symmetric: if there isan edge ℓ1 → ℓ2, then there is an edgeℓ2 → ℓ1
This generalizes to paths: if there is apath from ℓ1 to ℓ2, then there is a pathfrom ℓ2 to ℓ1
Skew-Symmetry
The graph is skew-symmetric: if there isan edge ℓ1 → ℓ2, then there is an edgeℓ2 → ℓ1
This generalizes to paths: if there is apath from ℓ1 to ℓ2, then there is a pathfrom ℓ2 to ℓ1
TransitivityLemmaIf all the edges are satisfied by an assignmentand there is a path from ℓ1 to ℓ2, then itcannot be the case that ℓ1 = 1 and ℓ2 = 0.
Proof
TransitivityLemmaIf all the edges are satisfied by an assignmentand there is a path from ℓ1 to ℓ2, then itcannot be the case that ℓ1 = 1 and ℓ2 = 0.
Proof
ℓ1
1ℓ2
0
TransitivityLemmaIf all the edges are satisfied by an assignmentand there is a path from ℓ1 to ℓ2, then itcannot be the case that ℓ1 = 1 and ℓ2 = 0.
Proof
ℓ1
1ℓ2
0
TransitivityLemmaIf all the edges are satisfied by an assignmentand there is a path from ℓ1 to ℓ2, then itcannot be the case that ℓ1 = 1 and ℓ2 = 0.
Proof
ℓ1
1ℓ2
01 0
TransitivityLemmaIf all the edges are satisfied by an assignmentand there is a path from ℓ1 to ℓ2, then itcannot be the case that ℓ1 = 1 and ℓ2 = 0.
Proof
ℓ1
1ℓ2
01 0
Strongly Connected Components
All variables lying in the same SCC ofthe implication graph should be assignedthe same value
In particular, if a SCC contains avariable together with its negation, thenthe formula is unsatisfiableIt turns out that otherwise the formulais satisfiable!
Strongly Connected Components
All variables lying in the same SCC ofthe implication graph should be assignedthe same valueIn particular, if a SCC contains avariable together with its negation, thenthe formula is unsatisfiable
It turns out that otherwise the formulais satisfiable!
Strongly Connected Components
All variables lying in the same SCC ofthe implication graph should be assignedthe same valueIn particular, if a SCC contains avariable together with its negation, thenthe formula is unsatisfiableIt turns out that otherwise the formulais satisfiable!
2SAT(2-CNF F)construct the implication graph Gfind SCC’s of Gfor all variables x:
if x and x lie in the same SCC of G:return “unsatisfiable”
find a topological ordering of SCC’sfor all SCC’s C in reverse order:
if literals of C are not assigned yet:set all of them to 1set their negations to 0
return the satisfying assignment
Running time: O(|F |)
2SAT(2-CNF F)construct the implication graph Gfind SCC’s of Gfor all variables x:
if x and x lie in the same SCC of G:return “unsatisfiable”
find a topological ordering of SCC’sfor all SCC’s C in reverse order:
if literals of C are not assigned yet:set all of them to 1set their negations to 0
return the satisfying assignment
Running time: O(|F |)
LemmaThe algorithm 2SAT is correct.
ProofWhen a literal is set to 1, all the literalsthat are reachable from it have alreadybeen set to 1 (since we process SCC’s inreverse topological order).
When a literal is set to 0, all the literalsit is reachable from have already beenset to 0 (by skew-symmetry).
LemmaThe algorithm 2SAT is correct.
ProofWhen a literal is set to 1, all the literalsthat are reachable from it have alreadybeen set to 1 (since we process SCC’s inreverse topological order).When a literal is set to 0, all the literalsit is reachable from have already beenset to 0 (by skew-symmetry).
Outline
1 2-Satisfiability
2 Independent Sets in Trees
Planning a company party
You are organizing a company party. Youwould like to invite as many people aspossible with a single constraint: no personshould attend a party with his or her directboss.
Maximum independent set in a tree
Input: A tree.Output: An independent set (i.e., a subset
of vertices no two of which areadjacent) of maximum size.
Example
Example
Example
Example
Safe moveFor any leaf, there exists an optimal solutionincluding this leaf.
It is safe to take all the leaves.
Safe moveFor any leaf, there exists an optimal solutionincluding this leaf.
It is safe to take all the leaves.
PartyGreedy(T )
while T is not empty:take all the leaves to the solutionremove them and their parents from T
return the constructed solution
Running time: O(|T |) (for each vertex, maintainthe number of its children).
PartyGreedy(T )
while T is not empty:take all the leaves to the solutionremove them and their parents from T
return the constructed solution
Running time: O(|T |) (for each vertex, maintainthe number of its children).
Planning a company party
You are organizing a company party again.However this time, instead of maximizing thenumber of attendees, you would like tomaximize the total fun factor.
Maximum weighted independent setin trees
Input: A tree T with weights on vertices.Output: An independent set (i.e., a subset
of vertices no two of which areadjacent) of maximum total weight.
Example
3
5 1 6
2 3 7 2
1 2 1
3
5 1 6
2 3 7 2
1 2 1total weight: 17
3
5 1 6
2 3 7 2
1 2 1total weight: 18
Example
3
5 1 6
2 3 7 2
1 2 1
3
5 1 6
2 3 7 2
1 2 1total weight: 17
3
5 1 6
2 3 7 2
1 2 1total weight: 18
Example
3
5 1 6
2 3 7 2
1 2 1
3
5 1 6
2 3 7 2
1 2 1total weight: 17
3
5 1 6
2 3 7 2
1 2 1total weight: 18
Subproblems
D(v) is the maximum weight of anindependent set in a subtree rooted at v
Recurrence relation: D(v) is
max
⎧⎪⎪⎨⎪⎪⎩w(v) +∑︁
grandchildrenw of v
D(w),∑︁
childrenw of v
D(w)
⎫⎪⎪⎬⎪⎪⎭ .
Subproblems
D(v) is the maximum weight of anindependent set in a subtree rooted at vRecurrence relation: D(v) is
max
⎧⎪⎪⎨⎪⎪⎩w(v) +∑︁
grandchildrenw of v
D(w),∑︁
childrenw of v
D(w)
⎫⎪⎪⎬⎪⎪⎭ .
Function FunParty(v)if D(v) =∞:
if v has no children:D(v)← w(v)
else:m1 ← w(v)for all children u of v:
for all children w of u:m1 ← m1 + FunParty(w)
m0 ← 0for all children u of v:
m0 ← m0 + FunParty(u)D(v)← max(m1,m0)
return D(v)
Function FunParty(v)if D(v) =∞:
if v has no children:D(v)← w(v)
else:m1 ← w(v)for all children u of v:
for all children w of u:m1 ← m1 + FunParty(w)
m0 ← 0for all children u of v:
m0 ← m0 + FunParty(u)D(v)← max(m1,m0)
return D(v)
Function FunParty(v)if D(v) =∞:
if v has no children:D(v)← w(v)
else:m1 ← w(v)for all children u of v:
for all children w of u:m1 ← m1 + FunParty(w)
m0 ← 0for all children u of v:
m0 ← m0 + FunParty(u)
D(v)← max(m1,m0)return D(v)
Function FunParty(v)if D(v) =∞:
if v has no children:D(v)← w(v)
else:m1 ← w(v)for all children u of v:
for all children w of u:m1 ← m1 + FunParty(w)
m0 ← 0for all children u of v:
m0 ← m0 + FunParty(u)D(v)← max(m1,m0)
return D(v)
Example
318
55 13 610
22 33 77 22
11 22 11
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