Discrete Optimization 2010 Lecture 8 Lagrangian Relaxation ...

Lagrangian TSP & 1-Trees P, NP, co-NP

Discrete Optimization 2010Lecture 8

Lagrangian Relaxation / P , NP and co-NP

Marc UetzUniversity of Twente

[email protected]

Lecture 8: sheet 1 / 32 Marc Uetz Discrete Optimization

mailto:[email protected]


Outline

1 Lagrangian Relaxation

2 TSP and the 1-Tree Relaxation

3 Decision Problems, P, NP, and co-NP



LP and Lagrangian Relaxation

max ctx

s.t. Ax ≤ b (← complicating constraints)

Bx ≤ d (IP)

x integer

Lagrangian relaxation (λ ≥ 0)

L(λ) = maxx ctx − λt(Ax − b)

or L(λ) = λtb + maxx (ct − λtA)x

s.t. Bx ≤ d (LR(λ))

x integer

Idea: penalize violation of constraints instead of enforcing themExercise: L(λ) is an upper bound on optimum IP solution.



LP and Lagrangian Relaxation

Best possible Lagrangian upper bound is Langragian Dual:

LD = minλ≥0L(λ) a

anote: λ is unrestricted in sign if we relax equality constraints Ax = b

LP upper bound:

LP = maxxctx | Ax ≤ b,Bx ≤ d

denote by ILP the optimum solution value of the integer program

Theorem (for maximization problems)

ILP ≤ LD ≤ LP

(i.e., best Lagrangian bound is never worse than the LP relaxation)



Proof (of 2nd inequality)

Use (strong) linear programming duality twice:

LD = minλ≥0btλ+ maxx(ct − λtA)x | Bx ≤ d , x integer

≤ minλ≥0btλ+ maxx(ct − λtA)x | Bx ≤ d

=︸︷︷︸LP duality

minλ≥0btλ+ minyd ty | Bty = (c − Atλ), y ≥ 0

= minλ,ybtλ+ d ty | Atλ+ Bty = c , y ≥ 0, λ ≥ 0

=︸︷︷︸LP duality

maxxctx | Ax ≤ b,Bx ≤ d = LP

Note: ‘=’ holds in 2nd step for example if B is totally unimodular



An Example: TSP

TSP

Given (w.l.o.g. complete)undirected graph G = (V ,E ),|V | = n, edge lengths ce ≥ 0∀ e ∈ E , find shortestHamiltonian cycle T (= tour).

Example

a short but non-optimal tourfor all 15,112 cities in D



IP Formulation for TSP I

variables xe = 1 if e ∈ T , 0 otherwise

min∑e∈E

cexe∑e∈δ(v)

xe = 2 v ∈ V

∑e∈S

xe ≤ |S | − 1 ∅ ⊂ S ⊂ V

x ∈ 0, 1n

Subtour elimination constraints (SEC), need to forbid solution:



IP Formulation for TSP II


min∑e∈E

cexe

s.t.∑e∈E

xe = n∑e∈δ(S)

xe ≥ 2 ∅ ⊂ S ⊂ V

x ∈ 0, 1n



IP Formulation for TSP III (why valid?)


min∑e∈E

cexe

s.t.∑e∈E

xe = n∑e∈δ(v1)

xe = 2,∑

e∈δ(v)

xe = 2 v ∈ V \ v1∑e∈S

xe ≤ |S | − 1 ∅ ⊂ S ⊂ V \ v1

x ∈ 0, 1n

[Held & Karp 1970] Lagrange-relax all but one of the constraints∑e∈δ(v) xe = 2 (all but for node v1)



TSP & One-Tree Relaxation

Claim

Lagrange-relaxing the degree constraints yields a spanning tree onG \ v1 + two edges incident with v1

Proof.∑e∈E xe = n, so we have n − 2 edges on G \ v1 (with n − 1

nodes), and∑

e∈δ(S) xe ≥ 1 for all ∅ ⊂ S ⊂ V \ v1, so thesolution on G \ v1 is a connected subgraph

But we know how to compute a minimum spanning tree inpolynomial time



The One-Tree Relaxation (given λ)

min∑e∈E

cexe −∑v∈V

λv (∑

e∈δ(v)

xe − 2) (let λv1 = 0)

= min 2∑v∈V

λv +∑

e=u,v∈E

(ce − λv − λu)xe (let λv1 = 0)

s.t.∑e∈E

xe = n,∑

e∈δ(v1)

xe = 2,

∑e∈S

xe ≤ |S | − 1, ∅ ⊂ S ⊂ V \ v1

x ∈ 0, 1n

Solution

Pick the two cheapest edges leaving v1, and compute a minimumspanning tree (edge weights ce − λu − λv ) on the graph G \ v1.



Held & Karp Bound for TSP

To get a good lower bound for the TSP need values for λ. . .

Procedure for Computing λ

Start with λv = 0 ∀ v ∈ V in iteration 1

in iteration k , compute optimal one-tree x∗ and update λ:

if k :=∑

e∈δ(v) x∗e = 1, let λv = λv + ∆k(2− k)

[making edges incident with v cheaper, i.e., more ‘attractive’]if k :=

∑e∈δ(v) x∗e ≥ 3, let λv = λv + ∆k(2− k)

[making edges incident with v less ‘attractive’]

for some “step size” ∆k ≥ 0

iterate until improvement of bound becomes marginal

By “right”choice of step size ∆k (∆k ↓ 0,∑

k ∆k =∞) thisconverges to the optimal solution of the Lagrangian dual, LD.



Subgradient Optimization for LD

For the TSP (minimization problem) we have LD = maxλ L(λ)

L(λ) is linear in λ for given one-tree x∗

L(λ, x∗) =∑e∈E

cex∗e −

∑v∈V

λv (∑

e∈δ(v)

x∗e − 2)

hence L(λ) = minx L(λ, x) is a piecewise linear, concavefunction

regions where for varying λ, same x is optimal

λ

L(λ)



Subgradient Optimization for LD (cont’d)

Now for any given λ and the corresponding optimal one-tree x∗,consider

dv :=dL(λ, x∗)

dλv=

(2−

∑e∈δ(v)

x∗e

)Then dv coincides with our update step for λ, as we defined

λv = λv + ∆k

(2−

∑e∈δ(v)

x∗e

)= λv + ∆kdv

Exercise: vector d = dL(λ,x∗)dλ ∈ Rn is a subgradient of L( · ) at λ,

i.e., L(λ′) ≤ L(λ) + d t(λ′ − λ) ∀λ′ ∈ Rn

Note: Previous “intuitive” update of λs is just a special case ofwhat is known as subgradient optimization



More Generally

Given λ, let x∗ be the optimal solution to a Lagrangian L(λ) with

L(λ) = minx

ctx − λt (Ax − b)

then b − Ax∗ is a subgradient of L( · ) at λ, that is,

L(λ′) ≤ L(λ) + (b − Ax∗)t(λ′ − λ) ∀λ′

Proof: Exercise

Remark: For any L : Rn → R concave, the set

dL(λ) := d ∈ Rn | L(λ′) ≤ L(λ) + d t(λ′ − λ) ∀λ′ is called

subdifferential of L at λ. If L is differentiable at λ, then

dL(λ) = ∇L(λ). If 0 ∈ dL(λ), L attains its minimum in λ



Remarks on the Held & Karp bound

Very strong lower bound in practical implementations

Reasonably fast to compute, so feasible within B & B

Gives same lower bound as the LP relaxation, LP = LD[fractional 1-tree relaxation always has integer optimalsolution, follows from a theorem by Edmonds 1971]

There are examples where LD = maxλ L(λ)< IP[to really solve TSP to optimality, can use Lagrangianrelaxation only as a lower bound, e.g. within B & B]



Example for Suboptimality of 1-Tree Bound

1

1

1 1

1

1

0

0

0Exercise: Optimal tour costs 4, but there is always a 1-tree withcost ≤ 3, no matter what the values of the λv ’s are



One-Trees: Trees With a Cycle!



Outline

1 Lagrangian Relaxation

2 TSP and the 1-Tree Relaxation

3 Decision Problems, P, NP, and co-NP



Decision Problems

Definition

A decision (recognition) problem P is a set of instances I that canbe partitioned into ‘Yes’ and ‘No’ instances IY , IN , such that

I = IY ∪ IN

IY ∩ IN = ∅

Examples

All graphs G = (V ,E ) with edge lengths ce , and k ∈ N.Graphs with TSP tour of length ≤ k, and those without.

All graphs G = (V ,E ), and k ∈ N. Graphs with a matchingM ⊆ E of cardinality ≥ k , and those without.



Optimization and Decision Problems

Optimization (OPT)

Given an instance I = (S , f ) of discrete optimization problem P,find a solution s∗ ∈ S minimizing / maximizing f (s).

Decision (DEC)

Given an instance I = (S , f ) of discrete optimization problem P,and integer k , decide if ∃ solution s with f (s) ≤ / ≥ k.

Clearly, if we can solve OPT, we can also solve DEC.

What about the other direction?



Decision Problems are often General Enough

Proposition

(For quite many discrete optimization problems) if we can solveDEC in poly-time, we can also solve OPT in poly-time.

There may be cases where OPT is provably more difficult (an openresearch question), but in many cases it isn’t.

How to solve OPT using an algorithm for DEC?

1 First compute optimal solution value ξ for OPT, by binarysearch on possible optimal values

2 Given ξ, construct a corresponding optimal solution



Example: Maximum Matching

Given G = (V ,E ), find a matching of maximal cardinality.

Assume algorithm A(k) that correctly tells us:Is there a matching of size ≥ k in a graph G?

1 Binary Search for optimal value ξ (note, ξ ∈ [0,m], integer)

Let [a, b] = [0,m]While (b > a)

k = a + d b−a2e

If A(k) = ‘Yes’ on G , then a = kOtherwise b = k − 1

return ξ := a(= b) O( log m ) × A(k)

2 Finding the solutionFor all edges (e ∈ E )

if A(ξ) = ‘Yes’ on graph (G − e) then delete e from G

return remaining edges E (G ) O( m ) × A(k)



Problem class P : Deterministic Polynomial Time

Given: Decision problem P, |I | encoding length of instance I ∈ P

Definition

Algorithm A is a deterministic polynomial time algorithm for P ifthere exists a polynomial p such that,

A(I ) =‘Yes’ ⇔ I is ‘Yes’ instance

tA(I ) ≤ p(|I |) ∀ instances I of P (A is a poly-time algorithm)

Definition

Problem P ∈ P :⇔ P has deterministic poly-time algorithm

Examples: decision problems for MST, ShortestPath, MaxFlow,MinCostFlow, Matching ∈ P



Certificates

Given: TSP instance, a complete undirected graph with integeredge lengths ce ≥ 0, e ∈ E , integer k : ∃ tour ≤ k?

Can we give a short certificate, that is, a short proof that I is a‘Yes’ instance?

This is simple: The tour T ⊆ E itself (n edges, |T | ∈ O( n ))

!"#$%&'(")*(+,$-."

!"#$%&'&()*+ ,$-./%"0".%&"%10'%-$2&!&3&45676896&"%0$/$,&:;&

<'%&=$&/"#$&'&1>.,0&?,..@&&1>.,0&?,..@&&0>'0&"0&"1&'&AB7)C "%10'%-$D

“Easy”: !?,..@2 0.E,&(&"01$F@

!?,..@&#$,"@"-'0".%2

(&"1&"%8$$8&'&0.E,&! 84(9&! :&!

G.0$2&H"%8"%/H"%8"%/ 0>$&?,..@&40>$&0.E,9&I'J&K$&LE"0$&>',86&

KE0&#$,"@"-'0".%#$,"@"-'0".% .@&0>$&?,..@&"1&$'1J&4?.FJ%.I"'F&0"I$9&

Verifying the certificate

check that T is indeed a tour

check its length∑

e∈T ce ≤ k?

Possible in polynomial time, O( n2 )

Note: Finding the certificate maybe hard, but verifying itscorrectness is easy (in polynomial time)



Problem class NP : Nondeterministic Polynomial Time

Given: Decision problem P, |I | encoding length of instance I ∈ P

Definition

Algorithm A is nondeterministic polynomial time algorithm for P if

For all ‘Yes’ instances I , ∃ poly-length certificate z(I ) (i.e.,|z(I )| ≤ p(|I |) for some polyn. p), and A(I , z(I )) =‘Yes’

For all ‘No’ instances I , and any poly-length input z (‘fakecertificate’), A(I , z) =‘No’

A(I , z) is a poly-time algorithm for any poly-length z

Definition

P ∈ NP :⇔ P has nondeterministic poly-time algorithm⇔ ‘Yes’ instances of P have a polynomial certificate

that can be verified in polynomial time

Example: We have just argued that TSP∈ NPLecture 8: sheet 26 / 32 Marc Uetz Discrete Optimization


Why ‘nondeterministic’?

Definition uses existence of a polynomial length certificate z(I ),but not how to obtain it!

We can interpret this equivalently as:

Consider all possible certificates z of polynomial length,

If the instance I =‘Yes’, at least one yields the algorithm toterminate with ‘Yes’

Could we ‘guess’ the right certificate z(I ) for any given ‘Yes’instance I , we could solve the problem

↑guessing = nondeterminism



What about ‘No’ Instances

Given: TSP instance I , so a complete undirected graph withinteger edge lengths ce ≥ 0, e ∈ E , integer k .

Can we give a short certificate that I is a ‘No’ instance?

???All proofs seem to require exponential length. . .

!"#$%&'()*+,-.&/ 01&2$+#"

!"#$%&'&()*+ ,$-./%"0".%&"%10'%-$2&!&3&45676896&"%0$/$,&:;&

<'%&=$&/"#$&'&1>.,0&?,..@&&1>.,0&?,..@&&0>'0&"0&"1&'&ABCD "%10'%-$E

?).F$>.=6&'GG&?,..@1&1$$F&0.&,$HI",$&$J?.%$%0"'G&G$%/0>$J?.%$%0"'G&G$%/0>KK

…

LGG&>'#$&G$%/0>&! :MN&check all (n−1)!2 tours, see they all have length ≥ k + 1



‘No’ Instances with Certificates

Given: An integer matrix A ∈ Zm×n: is A totally unimodular?

Short certificate for ‘No’ instances: square submatrix B of A

encoding length |B| ≤ |A|, so certainly polynomial

verifying the certificate: verify that det(B) 6∈ 0,±1

Is certificate verification possible in polynomial time?

Laplace formula det(B) =∑k

j=1 bij(−1)i+jdet(B−i ,−j)requires Ω(n!) time

but O( n3 ) is possible, computing LU decomposition of B(B = LU, with L = lower triangular, U = upper triangular)



Problem class co-NP : Complement-NPGiven: Decision problem P, |I | encoding length of instance I ∈ P

Definition

Algorithm A is nondeterministic polynomial time algorithm for thecomplement of P if

For all ‘No’ instances I , ∃ poly-length certificate z(I ) (i.e.,|z(I )| ≤ p(|I |) for some polyn. p), and A(I , z(I )) =‘Yes’

For all ‘Yes’ instances I , and any poly-length input z (‘fakecertificate’), A(I , z) =‘No’

A(I , z) is a poly-time algorithm for any poly-length z

Definition

P ∈ co-NP:⇔ complement of P has nondet. poly-time algorithm⇔ ‘No’ instances of P have a polynomial certificate

that can be verified in polynomial time



P , NP , and co-NP

Theorem

P ⊆ NP and P ⊆ co-NP

PNP co-NP

TSP TUM

MST

Open conjectures

P 6= NP, worth 1.000.000$ claymath.org/millennium/P vs NP/

P = NP∩co-NP, worth world-fame and recognition as scientist


http://www.claymath.org/millennium/P_vs_NP/


Proof P ⊆ NP ,co-NP

Given Problem Q ∈ P.

To show: Q ∈ NP i.e., for all ‘Yes’ instances I ∈ Q existspolynomial length certificate z(I ), and an algorithm A that verifiesthe certificate in poly-time, so A(I , z(I )) = ‘Yes’⇔ I = ‘Yes’.

Proof: We know Q ∈ P, so Q has a poly-time algorithm AQ withAQ(I ) =‘Yes’ ⇔ I =‘Yes’

Now use z(I ) = ∅, and define A(I , ∅) := AQ(I )

To see P ⊆co-NP, replace all underlined ‘Yes’ by ‘No’, and letA(I , ∅) := ¬AQ(I )


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