6' The Most Difficult NP Problems: The Class NPC

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6. The Most Difficult NP Problems The Class NPC

• Polynomial transformation
• Given Xi (Di , Fi), i 1, 2
• ? function g D1 ? D2 such that g(d) ? F2 iff
  d ?F1.
• If the function g is computable in time that is
  polynomial in the length of the encoding of d,
  then X1 is said to be polynomially transformable
  to X2.
• (notation X1 ? X2 , L1 ? L2 ( in language
  term))
• Note that X1 ? X2 means that X2 is no easier
  than X1 in terms of polynomial time solvability.
  If we can solve X2 in polynomial time (find
  yes, no answer using deterministic
  algorithm), then we can answer whether any given
  instance of X1 is yes or no in polynomial
  time. But, the converse may not hold.

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• Polynomial reduction
• X1 is polynomially reducible to X2 if ?
  algorithm for X1 that use algorithm for X2 as a
  subroutine and runs in polynomial time assuming
  each call to the subroutine takes unit time.
• ( recall Opt and Feas for 0-1 IP problem)
• ( transformation is the special case of
  reduction)
• Polynomial transformation also called Karp
  reduction, ? (GJ).
• Polynomial reduction called Turing reduction, ?T
  (GJ).
• Prop If X1 is polynomially transformable
  (reducible) to X2 , and X2 ? P, then X1 ? P.
• Def X1 is a special case of X2 if D1 ? D2 and
  F1 D1 ? F2.
• ( Use identity transformation g(d) d.)

3

• Def X ? NP is said to be NP-complete if all
  problems in NP can be polynomially reduced
  (transformed) to X.
• The set of NP-complete problems is denoted by
  NPC.
• NP-complete problem is the hardest problem in NP
  since the existence of a polynomial time
  algorithm for any NP-complete problem implies
  that all the problems in NP can be solved in
  polynomial time.
• Prop 6.3 If X is NP-complete, then P NP iff X
  ? P.
• ( P ? NPC ? ? ? P NP )
• Pf) X?NP and P NP ? X?P.
• X ? P ? NPC ? ? poly time alg. for any
  problem in NP.
• Hence NP ? P. Also P?NP. So PNP. ?
• Prop 6.4 If X1 is NP-complete, X1 ? X2 and X2 ?
  NP, then X2 is NP-complete.
• ( Be careful about the direction of the
  transformation.)

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• Is NPC ? ? ?
• First NP-complete problem found
• Satisfiability problem (Cook, 1971)
• Given a set U of Boolean variables and a
  collection C of clauses over U, is there a
  satisfying truth assignment for C?
• ( Is there a truth assignment to ?i ? C ?j
  1ni ( uj or uj ) ? )
• ( ? and, ? or )
• ex) Given ( u1 ? u3 ) ? (u1 ? u2 ? u3 ) ? ( u2
  ? u3 ),
• a truth assignment is u1 true, u2 false, u3
  false
• Thm 6.5 (Cook) The satisfiability problem (SAT)
  is NP-complete.
• Pf) polynomial transformation of any NDTM into
  the satisfiability problem.

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• Prop 6.6 0-1 IP feasibility problem is
  NP-complete.
• Pf) clearly in NP.
• Let each clause be Ci ( Ci, Ci- ). Then
  (SAT) is satisfiable iff
• is feasible.
• We have a function that transforms any instance
  of (SAT) to an in stance of 0-1 IP feasibility.
  ?
• Note that we transform an arbitrary instance of
  (SAT) to a (specific) instance of 0-1 IP
  feasibilty, which depends on the given instance
  of (SAT). The rationale is that if there exists
  an efficient algorithm to solve any instance of
  0-1 IP feasibility, we can apply the algorithm to
  the transformed instance to solve any arbitrary
  instance of (SAT). So it is unlikely that such
  efficient algorithm for 0-1 IP feasibility exists
  since the existence of such algorithm implies
  that we can solve (SAT) easily and solve all
  problems in NP easily.

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• Boundary between P and NPC
• x ? Bn Ax ? b, cx ? z ? ? ? A m ? n
  0-1 matrix.
• two 1s in each column of A (node-edge incidence
  matrix)
• matching problem ? P
• Three 1s in each column ? NPC
• One 1 in each row ? P
• Two 1s in each row (edge-node incidence matrix)
  
• node packing problem ? NPC
• Be careful about the distinction between the
  feasibility problem itself and the 0-1 IP
  formulation.

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• Unweighted node packing problem
• Given a graph G (V, E), ? U ? V such that
  U ? k and U is a node packing?
• Prop 6.7 The lower bound feasibility problem
  for unweighted node packing is NP-complete.
• Pf) in NP. transform from (SAT)
• ex) Ci Ci-
• 1 1, 2 3
• 2 2, 3 4
• 3 4 1, 2
• 4 3 ?
• ? clauses m, ? vars n
• set k m
• 1, 3 true, 2, 4 false

(1, 1)
(5, 3)
(2, 1)
(7, 1)
(6, 3)
(3, 4)
(3, 2)
(4, 3)
(8, 2)
(2, 2)
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• Then the given instance of (SAT) has truth
  assignment iff ? node packing of size k.
• Note that k is set depending on the given
  instance of (SAT). It is not an arbitrary
  number.
• If we want to claim that IP feasibility
  formulation of node packing lower bound
  feasibility is NP-complete, we need to transform
  the node packing to the IP formulation.

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• Clique Given G (V, E), U ? V is called a
  clique if v, w ? U ? (v, w) ? E. (complete
  subgraph of G)
• Clique problem Given G (V, E) and positive
  integer k, ? a clique of size ? k ?
• Prop Clique problem is NP-complete
• Pf) in NP. Independent set (node packing) ?
  clique
• Given an instance of node packing (G, k),
  construct Gc (V, E), E (i, j) (i, j) ?E,
  i, j ? V . Set k k
• Then U ? V is an independent set for G iff U
  is a clique for Gc.
• Hence G has an independent set of size k iff Gc
  has a clique of size k. ?

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• Vertex cover Given G (V, E), U ? V is called a
  vertex cover if all edges in G are incident to at
  least one vertex in U.
• Vertex cover problem Given G (V, E), ? a
  vertex cover of size ? k ? (NP-complete)
• Pf) in NP. clique ? vertex cover
• Given an instance of clique (G, k), construct an
  instance of vertex cover. Construct Gc and set k
  V - k.
• Then U ? V is a clique of size k in G ? V\U is
  a vertex cover of size n-k in Gc.
• ?) nodes in U not connected by edges in Gc.
• ? edges in Gc incident to at least on node in
  V\U.
• ? V\U vertex cover in Gc.
• ?) every edge in Gc is incident to at least one
  node in V\U
• ? no edge in Gc connects nodes in U.
• ? U clique in G. ?

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Six Basic NP-complete Problems

• 3-Satisfiability (3SAT)
• Instance Collection C c1, c2, , cm of
  clauses on a finite set U of variables such that
  ci 3 for 1 ? i ? m.
• Question Is there a truth assignment for U
  that satisfies all the clauses in C ?
• 3-Dimensional Matching (3DM)
• Instance A set M ? W ? X ? Y, where W, X, and
  Y are disjoint sets having the same number q of
  elements.
• Question Does M contain a matching, that is,
  a subset M ? M such that M q and no two
  elements of M agree in any coordinate ?
• (generalization of marriage problem.)

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• Vertex Cover
• Clique
• Hamiltonian Circuit
• Instance A graph G (V, E)
• Question Does G contain a Hamiltonian circuit,
  that is, an ordering lt v1, v2, , vn gt of the
  vertices of G, where n V, such that vn, v1
  ? E and vi, vi1 ? E for all i, 1 ? i lt n ?
• Partition
• Instance A finite set A and a size s(a) ?
  Z for each a ? A.
• Question Is there a subset A ? A such that
• ?a ? A s(a) ?a ? A - A s(a) ?

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Proving NP-completeness

• Set partitioning feasibility problem
• Given an m ? n 0-1 matrix A, is x ? Bn Ax 1
  ? ? ?
• Prop 6. 8 Set partitioning feasibility problem
  is NP-complete.
• Pf) In NP. Lower bound feasibility problem
  for unweighted node packing ? Set partitioning
  feasibility problem
• Given an instance of node packing (G, k), let A
  be ( Ek) ? ( EkV) matrix defined as
  follows.

B0, B1, B2, , Bk ( AG is edge-node
incidence matrix of G.)
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• Ax 1 feasible ? pick k distinct columns of
  AGs ? node packing of size k. Similarly, for
  converse.
• Hence k node packing of G ? set partitioning
  feasible with k columns (variables) 1 ?
• Prop 6. 9 The set partitioning feasibility
  feasibility problem in which matrix A has, at
  most, three 1s per column is NP-complete.

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2

• Ex)

e1
e3
e4
1
3
e5
e2
4
A
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• Subset sum problem
• Given an integer n, integral n-vector (a1, ,
  an ), and integer b, is x ? Bn ?j ? N aj xj
  b ? ? ?
• Prop 6.10 The subset sum problem is NP-complete.
• Pf) in NP. Set partitioning feasibility
  problem ? subset sum.
• Given an m ? n 0-1 matrix A, construct an
  instance of subset sum as follows.

( j-th column of A is representation of aj using
(n1) symbols.)
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• 0-1 knapsack lower bound feasibility problem
• Given an integer n, integral n-vectors (a1, ,
  an) and (c1, , cn), and integers b and z, is
• x ? Bn ?j ? N aj xj ? b, ?j ? N cj xj ? z
  ? ? ? (NP-complete)
• Pf) in NP.
• Subset sum problem can be reformulated as
• x ? Bn ?j ? N aj xj ? b, ?j ? N aj xj ? b
  .
• Hence it is a special case of 0-1 knapsack lower
  bound feasibility. Since special case is
  NP-complete, more general case is NP-complete. (
  recall the definition of special case. It is
  identity transformation ( or any obvious
  one-to-one correspondence between the instances.)
• (special case also called restriction and one
  of the easiest and widely used proving
  techniques.)
• Note that 0-1 knapsack is not a special case of
  integer knapsack.

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More examples of Restriction

• Variants of Hamiltonian Circuit
• Hamiltonian Path Same as HC except that we drop
  the requirement that the first and the last
  vertices in the sequence be joined by an edge.
• Hamiltonian Path between Two Points same as HP
  except that two points u, v are given as input
  and the question is whether G contains a HP
  beginning with u and ending with v.
• All three problems NP-complete
• For directed graph ?
• transform each undirected problem to a directed
  problem by replacing each edge of G by two
  parallel arcs of opposite direction. Undirected
  version is a special case of directed version.

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• Bounded Degree Spanning Tree
• Instance A graph G (V, E) and a positive
  integer k ? V - 1.
• Question Is there a spanning tree for G in
  which no vertex has degree exceeding k, that is,
  a subset E ? E such that E V - 1, the
  graph G (V, E) is connected, and no vertex in
  V is included in more than k edges from E ?
• Restrict to Hamiltonian Path by allowing only
  instances in which k 2. HP is a special case
  of BDST with k 2.

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• Multiprocessor Scheduling
• Instance A finite set A of tasks, a length
  l(a) ? Z for each a ? A, a number m ? Z of
  processors, and a deadline D? Z.
• Question Is there a partition A A1 ? A2 ?
  ? Am of A into m disjoint sets such that
• max ?a ? Ai l(a) 1 ? i ? m ? D ?
• Restrict to partition by allowing only instances
  in which m 2 and
• D ½ ?a ? A l(a)

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• Longest Path
• Instance Graph G (V, E), positive integer k
  ? V
• Question Does G contain a simple path ( that
  is, a path encountering no vertex more that once)
  with k or more edges ?
• Restrict to Hamiltonian Path, i. e. set k
  V-1
• Set Packing
• Instance Collection C of finite sets, positive
  integer k ? C
• Question Does C contain k disjoint sets ?
• Restrict to Exact Cover by 3-Sets, i. e.
  restrict c 3, for all c ?C and k q.

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• Exact Cover by 3-Sets (X3C) (NP-complete)
• Instance A finite set X with X 3q and a
  collection C of 3-element subsets of X.
• Question Does C contain an exact cover for X,
  that is, a subcollection of C ? C such that
  every element of X occurs in exactly one member
  of C ?
• Note that 3DM is a restricted version of X3C,
  hence X3C is NP-complete.
• Compare to NW p115, 117 in representing the
  problems.

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• Proving Traveling Salesman Problem is
  NP-complete.
• Instance Set C of m cities, distance d(ci , cj
  ) ? Z for each pair of cities ci , cj ? C, a
  positive integer B.
• Question Is there a tour of C having length B
  or less, i.e., a permutation lt c?(1), c?(2), ,
  c?(m) gt of C such that
• ?i 1m-1 d( c?(i), c?(i1) ) d( c?(m),
  c?(1) ) ? B ?
• Pf) In NP. HC ? TSP
• Given G (V, E), construct an instance of TSP
  as follows.
• Let ci vi , vi ?V. Let d(ci , cj ) 1 if
  (vi , vj ) ? E and 2 if (vi , vj ) ? E. Set B
  V.
• Then G has a HC iff TSP has a tour ? B. ?

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• If P ? NP, then NPC ? P ? NP. ( i.e. there are
  problems of intermediate difficulty.)
• NP \ ( P ? NPC) is called NPI (Intermediate)
• Note
• Members in NPI not equivalent in terms of
  polynomial transformation. ( infinite number of
  equivalent classes.)
• Only artificial examples known. Have been looking
  for natural problems in NPI.
• Composite number, LP regarded as candidates in
  NPI.
• Composite Number Given positive integer k, are
  there integers m, n gt 1 such that k mn

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• Note that composite number is in NP and its
  complementary problem (Primes) is to check the
  primality of a given prime number. There exists
  a short proof to check primality. Hence
  Composite Number is in NP ? CoNP. So it is
  unlikely that composite number is NP-complete by
  Prop 6.12.
• Composite number was proven to be in P in 2002.

NP
NPC
NPI
P
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CoNP
NP
NPC
P

• Prop 6. 12 If NPC ? CoNP ? ?, then NP CoNP (
  It is unlikely that any problem in NP ? CoNP is
  NP-complete)
• Pf) If X ? NP ? there exists polynomial
  transformation g such that X ? Y for Y ? NPC ?
  CoNP
• Since Y ? CoNP, Y ? NP. For an instance in X,
  g maintains yes no answer after the
  transformation.
• If a no instance in X is given, we can apply g
  to it, and obtain an instance in Y which has no
  answer. Since Y ? NP, we can verify the no by
  polynomial time NDTM.
• So no instance in X can be verified ? X?CoNP
  ? NP ?CoNP
• Similarly, for the other direction. ?

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Status of Open Problems in GJ

• Source Pulleyblank, MPS 2000 tutorial
• 1. Graph Isomorphism open
• 2. Subgraph homeomorphism for a fixed graph P
• 3. Graph genus NP-complete
• 4. Chordal graph completion NP-complete
• 5. Chromatic index NP-complete
• 6. Spanning tree parity P
• 7. Partial Order Dimension NP-complete
• 8. Precedence constrained 3-processor
  schedu open
• 9. Linear Programming P
• 10. Total Unimodularity P
• 11. Composite number open (P)
• 12. Minimum length triangulation open

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NP-complete in the strong sense (strongly
NP-complete)

• Refer GJ. p90 -
• Recall that ? O(nb) algorithm for 0-1 knapsack
  problem (also subset sum). Though 0-1 knapsack
  feasibility is NP-complete, O(nb) is less
  formidable than O(2n). If b is small (polynomial
  of n), the algorithm runs in polynomial time for
  the restricted case.
• called pseudo-polynomial time algorithm.
• Q) ? pseudo-polynomial time algorithms for other
  NP-complete problems involving numbers like b?
• e.g.) TSP feasibility, Multiprocessor
  scheduling.
• Problems like clique, vertex cover do not have
  pseudo-polynomial time algorithms.
• Some problems involving numbers do not have even
  pseudo-polynomial time algorithms.

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• For each instance I of decision problem D?,
  define the functions
• Length I length of encoding of I.
• Max I magnitude of largest number in I.
  (not the size of encoding)
• Def An algorithm for ? is called a
  pseudo-polynomial time algorithm for ? if its
  time complexity function is bounded above by a
  polynomial function of Length I and Max I .
• (In NW, polynomial running time in unary
  encoding )
• Def A problem ? is called a number problem if
  there exists no polynomial p such that Max I ?
  p( Length I ) for all I ? D?
• e.g) knapsack, partition

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• Let ?p be subproblem of ? restricted to instances
  such that Max I ? p( Length I ) .
• Then if ? pseudo-polynomial time algorithm for
  ? ? ?p solvable in polynomial time.
• Def A decision problem ? is NP-complete in the
  strong sense (strongly NP-complete) if ? is in NP
  and there exists a polynomial function p for
  which ?p is NP-complete.
• It is the problem class for which there might
  not exist even pseudo-polynomial time algorithms.
• If a problem is not a number problem and
  NP-complete, then it is automatically strongly
  NP-complete.
• NW Feasibility problem X is called strongly
  NP-complete if the existence of a
  pseudo-polynomial time algorithm for it implies P
  NP.

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• Ex)
• 0-1 IP feasibility with coefficients 0, 1, -1
  (transformation from satisfiability, size of
  numbers in r.h.s. also small)
• TSP HC ? TSP. Edge weights 1 or 2 B V.
  
• Hence, all instances with this transformation
  satisfy Max I ? p(Length I ) and
  NP-complete. So TSP is strongly NP-complete (
  Length I V logV ? log d(I, j) )
• 3-Partition
• Instance A finite set A of 3m elements, a bound
  B ? Z, and a size s(a)?Z for each a?A, such
  that s(a) satisfies B/4 lt s(a) lt B/2 and such
  that ?a?A s(a) mB.
• Question Can A be partitioned into m disjoint
  sets S1, S2, , Sm such that, for 1 ? i ? m,
  ?a?Si s(a) B ?

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• Multiprocessor Scheduling
• Sequencing within intervals
• Instance A finite set T of tasks and, for
  each t ?T, an integer release time r(t) ? 0, a
  deadline d(t) ? Z, and a length l(t) ? Z.
• Question Does there exist a feasible schedule
  for T, that is, a function ? T ? Z such that,
  for each t ?T, ?(t) ? r(t), ?(t) l(t) ? d(t),
  and, if t ? T-t, then either ?(t) l(t) ?
  ?(t) or ?(t) ? ?(t) l(t) ?

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• Proving strong NP-completeness by considering
  restricted problem may be tedious. Use
  pseudo-polynomial transformation from a strongly
  NP-complete problem.
• Def A pseudo-polynomial transformation from ?
  to ? is a function f D? ? D? such that
• For all I ? D? , I ? Y? if and only if f(I) ?
  Y? ,
• f can be computed in time polynomial in the two
  variables Max I and Length I ,
• ? a polynomial q1 such that, for all I ? D? ,
• q1 (Length f(I) ) ? Length I
• ( size of transformed length does not shrink too
  much.)
• ? a two variable polynomial q2 such that, for
  all I ? D? ,
• Max f(I) ? q2 ( Max I , Length I )
• ( magnitude of the largest number does not blow
  up exponentially)

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• Lemma If ? is NP-complete in the strong sense,
  ? ? NP, and there exists a pseudo-polynomial
  transformation from ? to ?, then ? is
  NP-complete in the strong sense.
• Note that the transformation from set
  partitioning feasibility (which is strongly
  NP-complete) to subset sum is not a
  pseudo-polynomial transformation.
• Max f(I) ( b (n1)m 1/n ) is not
  bounded by any poly function of Length I ( mn
  ).
• Hence the existence of pseudo-polynomial time
  algorithm for subset sum cannot be ruled out.
• Also, if X is NP-complete and X ? Y, and if
  there exists a pseudo-polynomial time algorithm
  for Y, it does not necessarily imply the
  existence of a pseudo-polynomial time algorithm
  for X.

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NP-hard Problems

• GJ Chapter 5
• Search problem ?
• Given an instance I ? D? , return string s, such
  that s ? S? I or decide no such string s
  exists. (S? I set of solutions for I)
• ( Decision problem may be regarded as a special
  case of search problem by defining S? I
  yes if I ? Y? and S? I ? if I ? Y? )
• Def A polynomial time Turing reduction (or
  simply Turing reduction) from a search problem ?
  to a search problem ? is an algorithm A that
  solves ? by using a hypothetical subroutine S for
  solving ? such that, if S were a polynomial time
  algorithm for ?, then A would be a polynomial
  time algorithm for ? (the subroutine may be
  used many times and ? and ? need not be
  feasibility problems)

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• Def A search problem ? is NP-hard if there
  exists an NP-complete problem ? such that ?
  ?T ? (Turing reducible)
• ( ? is at least as hard as NP-complete problem
  ? and ? is search problem (optimization problem)
• Recall (0-1 IP FEAS) ?T (0-1 IP OPT)
• Def A search problem ? is called NP-easy if
  there exists a ? ? NP such that ? ?T ?
• ( search problem ? is no harder than NP-complete
  problems.
• Recall (0-1 IP OPT) ?T (0-1 IP FEAS) )

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• A search problem ? is called NP-equivalent if
  both NP-hard and NP-easy. (has the same degree
  of difficulty as an NP-complete problem in terms
  of polynomial time solvability.)
• To show that an optimization problem is
  difficult to solve (NP-hard), it is enough to
  show that the corresponding feasibility problem
  is NP-complete and the optimization problem is
  Turing reducible from the feasibility problem
  (usually trivial).
• Some people use the term NP-complete for
  optimization problems too.

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Optimization and Separation

• Relates the complexity of a class of optimization
  problem over certain polyhedra with the
  separation problem for the polyhedra.
• Optimization, separation trivial if A m?n is
  given explicitly. But if polyhedron is given as
  conv(X) where X is set of feasible solutions, the
  story is different.
• ex) x ? conv(X)? for node packing is
  nontrivial if x is fractional.
• Separation problem
• Given x ? Rn, is x ? conv(X)? If not, find an
  inequality ?x ? ?0 satisfied by all points in X,
  but violated by the point x.

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• Thm Optimization problem over a class of
  polyhedra is polynomially solvable if and only if
  the separation problem for the polyhedra is
  polynomiallly solvable.