EMIS 8373: Integer Programming

NP-Complete Problemsupdated 21 April 2009

The Class of NP-Complete Problems (NPC)

• A problem q is NP-complete if 1. It is in NP and,2. Any other problem p in NP is polynomial reducible

to q.

• Implications of q being NP-complete– If q is NP-complete then q is at least as hard as

every other problem in NP.– If there is an efficient (polynomial) algorithm for q,

then there is an efficient algorithm for every problem in NP (i.e., P = NP).

slide 2

Boolean Algebra

• A Boolean variable can either be true or false.

• There are three operators in Boolean algebra:

1. x y = true if x = true and y = true, and false otherwise.

2. x y = true if at least one of x and y is true.

3. x = true if x is false, and false if x is true.

slide 3

Boolean Formulae

• A Boolean formula is satisfiable if there exists at least one truth assignment of its variables such that it evaluates to true.– Example: f(x1,x2,x3) = x3 (x1 x2 x3)

• f(true,true,false) = true• f(false,false,false) = true• f is satisfiable.

– (x1 x2 x3) is an example of a clause containing the literals x1 , x2 , and x3 .

slide 4

An Unsatisfiable Formula

xxxxxxxxxxxxxxxf

321133221321

321),,(

Suppose x1 is true

=> x3 must also be true to satisfy clause 4 => x2 must also be true to satisfy clause 3

=> all three variables are true => clause 5 is false => the formula evaluates to false

Suppose x1 is false => x2 must also be false to satisfy clause 2

=> x3 must also be false to satisfy clause 3 => all three variables are false => clause 1 is false

=> the formula evaluates to false slide 5

The Satisfiability (SAT) Problem

• Input: A set of m clauses {C1, C2, …, Cm} involving n Boolean variables x1, x2, …, xn.– Note that a clause contains only and operators.

• Question: Is C1 C2 … Cm satisfiable?

• Cook’s Theorem (1971): SAT is NP-Complete.– Every problem in NP is polynomial reducible to SAT.

– If there is a polynomial-time algorithm for SAT, then there is a polynomial-time algorithm for every problem in NP.

slide 6

Binary Integer Programming (BIP)

• Instance:A set of n binary variables x1, x2, …, xn, and

m constraints: ai1x1 + ai2x2 + … + ainxn (=, or ) bi

• Question: Is the following problem feasible?

njx

mjbxa

j

j

n

jjij

,,2,11,0

,,2,1 ,1

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BIP is NP-Complete

1. Show that BIP is in NP• The size of a BIP instance is O(mn).• The complexity of checking a proposed

solution to an instance of BIP is O(mn). • Each constraint requires n multiplications, n-1

additions, and one comparison to verify.

2. Show that some other NP-complete problem can be “formulated” as BIP with a polynomial reduction.

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BIP is NP-Complete: Reduction from SAT

• Let x1, x2, …, xn, be the binary variables in the BIP instance.

• For each clause Ci in the SAT instance let ci be a constraint in the BIP instance:

– Let Bi be the set of literals in Ci with a operator and let Ai be the other literals in Ci

– Add the following constraint to the BIP instance:

• The reduction is clearly polynomial since the BIP has the same size as the SAT instance.

A B

xxi i

ii 11

slide 9

Example SAT to BIP Reduction

xxxxxxxxxxxx 321133221321

1,0,,

1111

11

11

11

1

321

321

31

32

21

321

xxx

xxx

xx

xx

xx

xxx

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Proof: “yes” SAT “yes” BIP

fromula. SAT thein true toevalues that assignment a truth be Let x

.in variablefalse oneleast at or

in variable trueoneleast at bemust there clauseIn

i

ii

B

AC

:constraint hesatisify t will variablefalseeach for 0 and variableeach truefor 1 Setting

i

ix

x

instance. BIP feasible

a yields instance SAT feasible a Thus,

A B

xxi i

ii 11

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Proof: “yes” BIP “yes” SAT

instance. BIP the tosolution feasiblea be Let x

Setting literal xi = true in the SAT formula if variable xi = 1 in the BIP solution and xi = false otherwise ensures that each clause evaluates to true since at least one Ai literal will be true or at least one Bi literal will be false.

satisfy to0 in variableoneleast at or

1variableoneleast at bemust there constraintIn

i

ii

B

AC

A B

xxi i

ii 11

slide 12

Complexity of BIP

• SAT is NP-complete (Cook’s theorem)

• BIP is in NP

• BIP is at least as hard as SAT (reduction)

• BIP is NP-complete

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The Clique Problem

• Input: An undirected graph G = (V, E), and an integer k.

• Question: Is there a k-clique in G?

• A k-clique is a subset U V of k nodes such that for every pair of nodes in i U and j U, the edge (i, j) E.

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Show that CLIQUE is NP-Complete

1. Show that CLIQUE is in NP• We can verify whether or not a given set of k nodes

is a clique in O(k2) time.

2. Show that some other NP-complete problem is polynomial reducible to CLIQUE.

• Give an efficient formulation of some other NP-complete problem as CLIQUE

• We will use a reduction from SAT

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Reduction from SAT to CLIQUE

• Let f = C1 C2 … Cm be the SAT instance where Ci = (i1 or i2 or … or i,s(i)) and s(i) is the number of literals in clause i.

• Say that a pair of literals x and x are complementary.

• Let G = (V, E) where– V = {ij: 1 i m and 1 j s(i)}– (ij, hk) is an edge in E if i h, and ij and hk are not

complementary – Let k = m

slide 16

Example Reduction of SAT to CLIQUE

zyzyzyxyxzyxf ),,(

xl 11 yl

12

xl 21

zl 23

zl 32

yl 31

yl 41

zl 42

yl 22

slide 17

Example Reduction SAT to CLIQUE

zyzyzyxyxzyxf ),,(

xl 11 yl

12

xl 21

zl 23

zl 32

yl 31

yl 41

zl 42

yl 22

slide 18

Example Reduction SAT to CLIQUE

zyzyzyxyxzyxf ),,(

xl 11 yl

12

xl 21

zl 23

zl 32

yl 31

yl 41

zl 42

yl 22

slide 19

Correctness of the Reduction

• The graph will have at most mn nodes and fewer than (mn)(mn-1)/2 edges. Therefore the time to construct the graph is polynomial in the number of variables and clauses of the SAT instance.

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Correctness of the Reduction: Feasible SAT Feasible Clique

1. Consider a truth assignment that makes the Boolean formula true.

2. At least one literal in each clause must be true.3. Select one such literal from each clause.4. This gives us k literals.5. Since no pair of these k literals can be

complementary, the corresponding set of k nodes in G forms a clique.

6. Thus, a “yes” instance of SAT yields a “yes” instance of Clique.

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Correctness of the Reduction: Feasible Clique Feasible SAT

1. Suppose that G contains a k-clique.2. Consider the k literals that correspond to the nodes

in the k-clique.3. Each literal must come from a different clause.4. No pair of the k literals can be complementary

(otherwise there would be no edge between the corresponding nodes).

5. Setting each of the k literals to true makes the formula evaluate to true.

6. Thus, the given SAT instance must be satisfiable.

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STABLE SET

• Input: A graph G = (V, E), and an integer k.

• Question: Is there a set of k vertices U V such that there are no edges between any pair of vertices in U?

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STABLE SET is NP-Complete

• Given a set of k nodes in V, we can verify that there are no edges between them in O(k2) time. Thus, STABLE SET is in NP.

• Reduction: Let {G =(V, E), k} be an instance of CLIQUE and let H = (V, F) be the complement of G where (i, j) F if (i, j) E, and (i, j) F if (i, j) E.– If {v1, v2, …, vk} is a clique in G, then it is a stable set in

H.– If {v1, v2, …, vk} is a stable set in H, then it is a clique in

G.

slide 24

Reducing CLIQUE to STABLE SET

1

32

5 4

1

32

5 4

G H

3-Clique: {2, 3, 4} Stable Set: {2, 3, 4}

slide 25

“Tree” of Reductions (Karp ’72)

SAT

3-SATBIP CLIQUE

STABLE SETTSP3D-Matching

3-Exact Cover

0-1 Knapsack Integer Knapsackslide 26

Cook’s theorem

SAT

BIP CLIQUE

STABLE SETTSP0-1 Knapsack

slide 27

Recognition vs. Optimization: TSP Example3

1

1

32

5 4

4

2

3

5

75

17

k = 35 yes

k = 5 no

k = (35 + 5)/2 =20 yes

k = (20+5)/2 =13

k = (13+5)/2 = 9

yes

no

k = (13+9)/2 =11 no

k = (13+11)/2 =12 yes

Is there a TSP tour of lengthk or less?

k =5

k =12 (optimal)k =11

k =35yes

no

slide 28

Optimization vs. Recognition

• An optimization problem is said to be NP-hard if its recognition problem is NP-complete.

• Many important optimization problems are NP-hard.

• NP-hard problems are considered theoretically intractable (difficult to solve).

• A problem can be NP-hard in general, but have special cases that are polynomial.– Example: MCNFP (easy) is a special case of ILP

(hard).

slide 29