A 0.5-Approximation Algorithm for max cut

Back to maxsat

• In maxsat, the φ_i’s are clauses.

• Hence p(φ_i) ≥ 1/2, which happens when φ_i contains a single literal.

• And the heuristic becomes a polynomial-time

²-approximation algorithm with ² = 1/2.^a

• If the clauses have k distinct literals, p(φ_i) = 1 − 2^−k.

• And the heuristic becomes a polynomial-time

²-approximation algorithm with ² = 2^−k.

– This is the best possible for k ≥ 3 unless P = NP.

aJohnson (1974).

max cut Revisited

• The NP-complete max cut seeks to partition the nodes of graph G = (V, E) into (S, V − S) so that there are as many edges as possible between S and V − S (p. 318).

• Local search starts from a feasible solution and

performs “local” improvements until none are possible.

• Next we present a local search algorithm for max cut.

A 0.5-Approximation Algorithm for max cut

1: S := ∅;

2: while ∃v ∈ V whose switching sides results in a larger cut do

3: Switch the side of v;

4: end while

5: return S;

• A 0.12-approximation algorithm exists.^a

• 0.059-approximation algorithms do not exist unless NP = ZPP.

aGoemans and Williamson (1995).

Analysis

V₃ V₄

V₂ V₁

Optimal cut

Our cut

e₁₂

e₁₃

e₂₄

e₃₄ e₁₄ e₂₃

Analysis (continued)

• Partition V = V₁ ∪ V₂ ∪ V₃ ∪ V₄, where

– Our algorithm returns (V₁ ∪ V₂, V₃ ∪ V₄).

– The optimum cut is (V₁ ∪ V₃, V₂ ∪ V₄).

• Let e_ij be the number of edges between V_i and V_j.

• For each node v ∈ V₁, its edges to V₁ ∪ V₂ are outnumbered by those to V₃ ∪ V₄.

– Otherwise, v would have been moved to V₃ ∪ V₄ to improve the cut.

Analysis (continued)

• Considering all nodes in V₁ together, we have 2e₁₁ + e₁₂ ≤ e₁₃ + e₁₄

– It is 2e₁₁ is because each edge in V₁ is counted twice.

• The above inequality implies

e₁₂ ≤ e₁₃ + e₁₄.

Analysis (concluded)

• Similarly,

e₁₂ ≤ e₂₃ + e₂₄ e₃₄ ≤ e₂₃ + e₁₃ e₃₄ ≤ e₁₄ + e₂₄

• Add all four inequalities, divide both sides by 2, and add the inequality e₁₄ + e₂₃ ≤ e₁₄ + e₂₃ + e₁₃ + e₂₄ to obtain

e₁₂ + e₃₄ + e₁₄ + e₂₃ ≤ 2(e₁₃ + e₁₄ + e₂₃ + e₂₄).

• The above says our solution is at least half the optimum.

Approximability, Unapproximability, and Between

• knapsack, node cover, maxsat, and max cut have approximation thresholds less than 1.

– knapsack has a threshold of 0 (p. 664).

– But node cover and maxsat have a threshold larger than 0.

• The situation is maximally pessimistic for tsp: It cannot be approximated unless P = NP (p. 662).

– The approximation threshold of tsp is 1.

∗ The threshold is 1/3 if the tsp satisfies the triangular inequality.

– The same holds for independent set.

Unapproximability of tsp

Theorem 77 The approximation threshold of tsp is 1 unless P = NP.

• Suppose there is a polynomial-time ²-approximation algorithm for tsp for some ² < 1.

• We shall construct a polynomial-time algorithm for the NP-complete hamiltonian cycle.

• Given any graph G = (V, E), construct a tsp with | V | cities with distances

d_ij =





1, if { i, j } ∈ E

|V |

1−², otherwise

aSahni and Gonzales (1976).

The Proof (concluded)

• Run the alleged approximation algorithm on this tsp.

• Suppose a tour of cost |V | is returned.

– This tour must be a Hamiltonian cycle.

• Suppose a tour with at least one edge of length _1−²^{|V |} is returned.

– The total length of this tour is > _1−²^{|V |} .

– Because the algorithm is ²-approximate, the optimum is at least 1 − ² times the returned tour’s length.

– The optimum tour has a cost exceeding | V |.

– Hence G has no Hamiltonian cycles.

knapsack Has an Approximation Threshold of Zero

Theorem 78 For any ², there is a polynomial-time

²-approximation algorithm for knapsack.

• We have n weights w₁, w₂, . . . , w_n ∈ Z⁺, a weight limit W , and n values v₁, v₂, . . . , v_n ∈ Z⁺.^b

• We must find an S ⊆ {1, 2, . . . , n} such that P

i∈S w_i ≤ W and P

i∈S v_i is the largest possible.

aIbarra and Kim (1975).

bIf the values are fractional, the result is slightly messier but the main conclusion remains correct. Contributed by Mr. Jr-Ben Tian (R92922045) on December 29, 2004.

The Proof (continued)

• Let

V = max{v₁, v₂, . . . , v_n}.

• Clearly, P

i∈S v_i ≤ nV .

• Let 0 ≤ i ≤ n and 0 ≤ v ≤ nV .

• W (i, v) is the minimum weight attainable by selecting some of the first i items with a total value of v.

• Set W (0, v) = ∞ for v ∈ { 1, 2, . . . , nV } and W (i, 0) = 0 for i = 0, 1, . . . , n.^a

aContributed by Mr. Ren-Shuo Liu (D98922016) and Mr. Yen-Wei Wu (D98922013) on December 28, 2009.

The Proof (continued)

• Then, for 0 ≤ i < n,

W (i + 1, v) = min{W (i, v), W (i, v − v_i+1) + w_i+1}.

• Finally, pick the largest v such that W (n, v) ≤ W .

• The running time is O(n²V ), not polynomial time.

• Key idea: Limit the number of precision bits.

The Proof (continued)

• Define

v_i⁰ = 2^b j v_i 2^b

k .

– This is equivalent to zeroing each v_i’s last b bits.

• From the original instance

x = (w₁, . . . , w_n, W, v₁, . . . , v_n), define the approximate instance

x⁰ = (w₁, . . . , w_n, W, v₁⁰ , . . . , v_n⁰ ).

The Proof (continued)

• Solving x⁰ takes time O(n²V /2^b).

– The algorithm only performs subtractions on the v_i-related values.

– So the b last bits can be removed from the calculations.

– That is, use v_i⁰ = ¥_v

2^bi

¦ in the calculations.

– Then multiply the returned value by 2^b.

• The solution S⁰ is close to the optimum solution S:

X

i∈S⁰

v_i ≥ X

i∈S⁰

v_i⁰ ≥ X

i∈S

v_i⁰ ≥ X

i∈S

(v_i − 2^b) ≥ X

i∈S

v_i − n2^b.

The Proof (continued)

• Hence X

i∈S⁰

v_i ≥ X

i∈S

v_i − n2^b.

• Without loss of generality, assume w_i ≤ W for all i.

– Otherwise, item i is redundant.

• V is a lower bound on opt.

– Picking an item with value V is a legitimate choice.

• The relative error from the optimum is ≤ n2^b/V : P

i∈S v_i − P

i∈S⁰ v_i P

i∈S v_i ≤

P

i∈S v_i − P

i∈S⁰ v_i

V ≤ n2^b

V .

The Proof (concluded)

• Suppose we pick b = blog₂ ^²V_n c.

• The algorithm becomes ²-approximate (see Eq. (10) on p. 640).

• The running time is then O(n²V /2^b) = O(n³/²), a polynomial in n and 1/².^a

aIt hence depends on the value of 1/². Thanks to a lively class dis- cussion on December 20, 2006. If we fix ² and let the problem size increase, then the complexity is cubic. Contributed by Mr. Ren-Shan Luoh (D97922014) on December 23, 2008.

Pseudo-Polynomial-Time Algorithms

• Consider problems with inputs that consist of a

collection of integer parameters (tsp, knapsack, etc.).

• An algorithm for such a problem whose running time is a polynomial of the input length and the value (not length) of the largest integer parameter is a

pseudo-polynomial-time algorithm.^a

• On p. 666, we presented a pseudo-polynomial-time algorithm for knapsack that runs in time O(n²V ).

• How about tsp (d), another NP-complete problem?

aGarey and Johnson (1978).

No Pseudo-Polynomial-Time Algorithms for tsp (d)

• By definition, a pseudo-polynomial-time algorithm becomes polynomial-time if each integer parameter is limited to having a value polynomial in the input length.

• Corollary 42 (p. 335) showed that hamiltonian path is reducible to tsp (d) with weights 1 and 2.

• As hamiltonian path is NP-complete, tsp (d) cannot have pseudo-polynomial-time algorithms unless P = NP.

• tsp (d) is said to be strongly NP-hard.

• Many weighted versions of NP-complete problems are strongly NP-hard.

Polynomial-Time Approximation Scheme

• Algorithm M is a polynomial-time approximation scheme (PTAS) for a problem if:

– For each ² > 0 and instance x of the problem, M runs in time polynomial (depending on ²) in | x |.

∗ Think of ² as a constant.

– M is an ²-approximation algorithm for every ² > 0.

Fully Polynomial-Time Approximation Scheme

• A polynomial-time approximation scheme is fully polynomial (FPTAS) if the running time depends polynomially on | x | and 1/².

– Maybe the best result for a “hard” problem.

– For instance, knapsack is fully polynomial with a running time of O(n³/²) (p. 664).

Square of G

• Let G = (V, E) be an undirected graph.

• G² has nodes {(v₁, v₂) : v₁, v₂ ∈ V } and edges

{{ (u, u⁰), (v, v⁰) } : (u = v ∧ { u⁰, v⁰ } ∈ E) ∨ { u, v } ∈ E}.

1

2

3

(1,1)

G

(1,2) (1,3)

(2,1) (2,2) (2,3)

(3,1) (3,2) (3,3)

G²

Independent Sets of G and G

Lemma 79 G(V, E) has an independent set of size k if and only if G² has an independent set of size k².

• Suppose G has an independent set I ⊆ V of size k.

• {(u, v) : u, v ∈ I} is an independent set of size k² of G².

1

2

3

(1,1)

G

(1,2) (1,3)

(2,1) (2,2) (2,3)

(3,1) (3,2) (3,3)

G²

The Proof (continued)

• Suppose G² has an independent set I² of size k².

• U ≡ {u : ∃v ∈ V (u, v) ∈ I²} is an independent set of G.

1

2

3

(1,1)

G

(1,2) (1,3)

(2,1) (2,2) (2,3)

(3,1) (3,2) (3,3)

G²

• | U | is the number of “rows” that the nodes in I² occupy.

The Proof (concluded)

• If | U | ≥ k, then we are done.

• Now assume | U | < k.

• As the k² nodes in I² cover fewer than k “rows,” there must be a “row” in possession of > k nodes of I².

• Those > k nodes will be independent in G as each “row”

is a copy of G.

aThanks to a lively class discussion on December 29, 2004.

Approximability of independent set

• The approximation threshold of the maximum independent set is either zero or one (it is one!).

Theorem 80 If there is a polynomial-time ²-approximation algorithm for independent set for any 0 < ² < 1, then there is a polynomial-time approximation scheme.

• Let G be a graph with a maximum independent set of size k.

• Suppose there is an O(nⁱ)-time ²-approximation algorithm for independent set.

• We seek a polynomial-time ²⁰-approximation algorithm with ²⁰ < ².

The Proof (continued)

• By Lemma 79 (p. 676), the maximum independent set of G² has size k².

• Apply the algorithm to G².

• The running time is O(n²ⁱ).

• The resulting independent set has size ≥ (1 − ²) k².

• By the construction in Lemma 79 (p. 676), we can obtain an independent set of size ≥ p

(1 − ²) k² for G.

• Hence there is a (1 − √

1 − ²)-approximation algorithm for independent set by Eq. (11) on p. 641.

The Proof (concluded)

• In general, we can apply the algorithm to G^{2`</sup> to obtain an (1 − (1 − ²)<sup>2−`})-approximation algorithm for

independent set.

• The running time is n^2`i.^a

• Now pick ` = dlog _log(1−²^log(1−²)0)e.

• The running time becomes n^{ilog(1−²0)log(1−²)} .

• It is an ²⁰-approximation algorithm for independent set.

aIt is not fully polynomial.

Comments

• independent set and node cover are reducible to each other (Corollary 39, p. 312).

• node cover has an approximation threshold at most 0.5 (p. 646).

• But independent set is unapproximable (see the textbook).

• independent set limited to graphs with degree ≤ k is called k-degree independent set.

• k-degree independent set is approximable (see the textbook).

On P vs. NP

Density

The density of language L ⊆ Σ^∗ is defined as dens_L(n) = |{x ∈ L : | x | ≤ n}|.

• If L = {0, 1}^∗, then dens_L(n) = 2ⁿ⁺¹ − 1.

• So the density function grows at most exponentially.

• For a unary language L ⊆ {0}^∗,

dens_L(n) ≤ n + 1.

– Because L ⊆ {², 0, 00, . . . ,

z }| {n

00 · · · 0, . . .}.

aBerman and Hartmanis (1977).

Sparsity

• Sparse languages are languages with polynomially bounded density functions.

• Dense languages are languages with superpolynomial density functions.

Self-Reducibility for sat

• An algorithm exhibits self-reducibility if it finds a certificate by exploiting algorithms for the decision version of the same problem.

• Let φ be a boolean expression in n variables x₁, x₂, . . . , x_n.

• t ∈ {0, 1}^j is a partial truth assignment for x₁, x₂, . . . , x_j.

• φ[ t ] denotes the expression after substituting the truth values of t for x₁, x₂, . . . , x_{| t |} in φ.

An Algorithm for sat with Self-Reduction

We call the algorithm below with empty t.

1: if | t | = n then

2: return φ[ t ];

3: else

4: return φ[ t0 ] ∨ φ[ t1 ];

5: end if

The above algorithm runs in exponential time, by visiting all the partial assignments (or nodes on a depth-n binary tree).

NP-Completeness and Density

Theorem 81 If a unary language U ⊆ {0}^∗ is NP-complete, then P = NP.

• Suppose there is a reduction R from sat to U .

• We use R to find a truth assignment that satisfies

boolean expression φ with n variables if it is satisfiable.

• Specifically, we use R to prune the exponential-time exhaustive search on p. 687.

• The trick is to keep the already discovered results φ[ t ] in a table H.

aBerman (1978).

1: if | t | = n then

2: return φ[ t ];

3: else

4: if (R(φ[ t ]), v) is in table H then

5: return v;

6: else

7: if φ[ t0 ] = “satisfiable” or φ[ t1 ] = “satisfiable” then

8: Insert (R(φ[ t ]), “satisfiable”) into H;

9: return “satisfiable”;

10: else

11: Insert (R(φ[ t ]), “unsatisfiable”) into H;

12: return “unsatisfiable”;

13: end if

14: end if

15: end if

The Proof (continued)

• Since R is a reduction, R(φ[ t ]) = R(φ[ t⁰ ]) implies that φ[ t ] and φ[ t⁰ ] must be both satisfiable or unsatisfiable.

• R(φ[ t ]) has polynomial length ≤ p(n) because R runs in log space.

• As R maps to unary numbers, there are only polynomially many p(n) values of R(φ[ t ]).

• How many nodes of the complete binary tree (of invocations/truth assignments) need to be visited?

• If that number is a polynomial, the overall algorithm runs in polynomial time and we are done.

The Proof (continued)

• A search of the table takes time O(p(n)) in the random access memory model.

• The running time is O(M p(n)), where M is the total number of invocations of the algorithm.

• The invocations of the algorithm form a binary tree of depth at most n.

The Proof (continued)

• There is a set T = {t₁, t₂, . . .} of invocations (partial truth assignments, i.e.) such that:

1. |T | ≥ (M − 1)/(2n).

2. All invocations in T are recursive (nonleaves).

3. None of the elements of T is a prefix of another.

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An Example

r

a c

d e f

g h i j

l k

1

2 3

4 5

T = { h, j }.

The Proof (continued)

• All invocations t ∈ T have different R(φ[ t ]) values.

– None of h, j ∈ T is a prefix of the other.

– The invocation of one started after the invocation of the other had terminated.

– If they had the same value, the one that was invoked second would have looked it up, and therefore would not be recursive, a contradiction.

• The existence of T implies that there are at least (M − 1)/(2n) different R(φ[ t ]) values in the table.

The Proof (concluded)

• We already know that there are at most p(n) such values.

• Hence (M − 1)/(2n) ≤ p(n).

• Thus M ≤ 2np(n) + 1.

• The running time is therefore O(M p(n)) = O(np²(n)).

• We comment that this theorem holds for any sparse language, not just unary ones.^a

aMahaney (1980).

coNP-Completeness and Density

Theorem 82 (Fortung (1979)) If a unary language U ⊆ {0}^∗ is coNP-complete, then P = NP.

• Suppose there is a reduction R from sat complement to U .

• The rest of the proof is basically identical except that, now, we want to make sure a formula is unsatisfiable.

Finis