Summary

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Summary

• NP-hard and NP-complete
• NP-completeness proof
• Polynomial time reduction
• List of NP-complete problems
• Knapsack problem
• Isomprphisim is an open problem, network flow
  problem where capacities are nodes, node capacity
  reduced to edge capacity problem.

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Reduction-1

• Isomorphism is an open problem.
• Network flow problem where capacities are nodes,
  node capacity reduced to edge capacity problem.
  Polynomial Prbl.
• Reduction from independent set to a recipe. IS to
  (node cover) VC.
• Example of nodes and edges..
• Traffic police
• VC(G, k)? G(V, E), k, is there a node cover of
  size ? k, that touches every edge, every edge
  touches one of the nodes,
• IS, no two nodes in that subset are adjacent.
• IS?VC. Idea is IS-gtconvert-gtsolver for VC-gt?
• Question if there is an independent set IS of G,
  size k??
• This means if there is a VC of size ?k,
• (G, k) ?IS-gt(G, k)?VC-gt
• (G, k) ?IS?VC?
• how to do ? (G, k)? GG, k ? k n-k ?VC?
• (G,k)?IS? reduction? VC?
• I have to issue a theorem
• Theorem A subset of nodes S is a nodecover iff
  V-S is an independent set.
• IS G, k given in to the box of VC

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?p Reduction takes polytime

• IS G, k given in to the box of VC
• Reduction runs in polytime at the worst case, any
  instance of IS, reduced to VC, by, gg, kn-k,
  (one clock time)..
• IS ?p VC
• The otherway around is not always true in
  general.
• Question1? If X ?p Y and if Y has a polytime
  solution, does X has a polytime solution? ..
• Proof Poly of Poly is poly.. O(p'(s))p(s)
  ?O(p) bounded by a poly solution for X.
• Question2? If X ?p Y and if X does not have a
  polytime solution (cannot be solved with a
  polytime question), does Y has a polytime
  solution? ..
• Cannot Poly of Poly is poly.. O(p'(bs)p(s)
  ?O(bs) solution.
• Clauses related together by ANDS.. (x1 v ?x2)
  (?x1 v ?x3) (x2 v x3) how to satisfy this.
• Not all can be made true, example (x2 ) (?x2)
• The maximum number of literals in a clause
  determines k-sat.
• Example 3-sat at most 3 literals. 2-sat at most 2
  literals..
• 3sat ?p IS

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?p Reduction takes polytime

• 3sat ?p IS
• A boolean formula?3sat?reducer?input for IS?(G,
  k)?IS?
• If formula is satisfiable then there is an IS.
• Tree boxes,, input, X, Reducer and Y
  solver-output yes or no.
• Decision YES OR NO

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3-SAT ?p IS - Reduction in polytime

• A boolean formula?3-SAT?reducer?input for IS?(G,
  k)?IS?
• If formula is satisfiable then there is an IS.
• Tree boxes, input, X, Reducer and Y solver-output
  yes or no.
• Decision YES OR NO
• If y runs in polytime, and result will be ploy of
  poly(x).
• If there is no poly solver for X, X reduced to Y,
  then there cannot be a polysolver for Y.
• 3-SAT ?p IS no two nodes adjacent.
• (x1, x2, x3)(?x1,?x4, x5)(x1, ?x2,
  x4)(x3,?x4,?x5)
• At least one of literals must be true,
• Satisfying this formula definitely corresponds to
  an IS, with size equal to number of clauses..
  Question make clauses true..
• Conversely if there is an independent set, then
  formula must be true, that means there must be at
  least one literal to be true.
• 3-SAT although seems straight forward but
  presently we dont know an algorithm to solve it
  in polytime algorithm..

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Class of NP problems

• Def A Decision Prbl X is associated with a set
  of strings or input, to a problem X language,
  such that x?X.
• Def An algorithm A, solves prbl x, iff A(x) says
  yes iff x?X.
• Def If A runs in O(px), for some polyn
  function p, then all X is poly solvable.
• P poly time solvable. Network flow, that is not
  a decision question. Maximize flow. Decision
  question would if there is a flow less than some
  value ie 20, yes or no.
• For 3-SAT we dont know yet if it is in NP, 3
  sat, since we dont know yet if there is a
  polynomial time to solve 3sat.
• CHECKING Verifying. If an input is in the
  language or not. Instance, certificate, witness
  proof, hint f -gtB X checker algorithm A,
  polytime-gt..

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• Def B is an efficient certifier for language X,
  if
• 1) B is a polytime algorithm with respect to its
  inputs x, t. outputs yes or no,
• 2) ? exists a polyfunct p, such that for any
  input x?X (target, k of IS), iff ? some x in
  language X, certificate C (bounded by a
  polynomial program size!), witness, t. p(t).
  and B(x, t) yes.
• NP is the class of problems that have efficient
  certifiers B as given above.
• X is in NP iff there is an efficient certifier
  for X.
• ie for VC reduces to IS and IS reduces to VC, but
  certifier is SAT, the instances of problem G and
  k, there exists a program t, and a B(x, t), x(G,
  k), t is a subset of nodes which is VC of G, size
  k, if there is one, and B(xm t), check whether
  this subset is actually a nodecover of G.

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• ie for VC reduces to IS and IS reduces to VC, but
  certifier is SAT, the instances of problem G and
  k, there exists a program t, and a B(x, t), x(G,
  k), t is a subset of nodes which is VC of G, size
  k, if there is one, and B(xm t), check whether
  this subset is actually a nodecover of G.
• Algorithm checker is simple, checks the size of
  proposed nodes smaller than k, and checks if
  these nodes are actually are nodecover, VC. If so
  it says yes. All these takes polynomial time. If
  input size is n, total time is n2. Yes. Otherwise
  it outputs not verified.. No!.
• To be in NP, Checker checks if proposed solution
  is a solution. May be a list or other
  constraints.
• P?NP.. Because all X?P. X can be certified
  (instance ans A(t, x)) in the order of a
  poly-operations but no need since easy to
  compute. There can never be a NO since solver
  solves the problem In polytime, easily. P?NP..
  is a basic elementary fact.

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NP-hard and NP-complete

• Def Problem X is NP-C if
• X ? NP
• For any Y ? NP, Y ?p X with a diagram,
• y?Y?Y ?p X ?x?solver for X poly solver?yes
  or no
• xy
• Every known reduces to unknown.
• Consequence of this is
• Corollary if at least one NPC problem X is in P
  (runs in polynomial time) then NPP.
• Even one problem X is X ? NP (1), P is already
  contained on NP, P?NP, this would mean NP ? P
• From second statement and diagram (2), that is
  running in polytime. The words swapping..
• Then the roof of NP is collapsing into P.
• Equivalently if P ? NP, then no any NPC is in P.
• Therefore if X is NPC, there can be no polynomial
  time solution for problem X.

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NP-hard and NP-complete

• Then the roof of NP is collapsing into P.
• Equivalently if P ? NP, then no any NPC is in P.
• Therefore if X is NPC, there can be no polynomial
  time solution for problem X.
• There were not NPC classified problems, Cookss
  theorem Satisfiabilility..
• tedious to prove but tractable.. Levin.
• Then how to establish other problems are npc,
• Poly reduction of Y to X, and show X is in NP.
• IF X is known to be NPC, we can proove Y to be
  NPC by show by a polytime reduction of instances
  of X to instances of Y.
• Poly of poly,.
• For each NP problem there is a poly reduction to
  X and from X to Y, (by transitivity every
  problem in NP reduces to Y)
• Problem X is called NP-hard, not the first part
  but the second part is satisfied, in other words,
  if every problem in NP reduces to X in ploytime.

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• From
• www.ece.mtu.edu/faculty/shiyan/EE5900Spring10/NP-c
  omplete.ppt
• www.cs.dartmouth.edu/ac/Teach/CS39Winter09/Slides
  AndNotes/lec28npcomplete.pdf

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Nondeterministic algorithms

• A nondeterminstic algorithm consists of
• phase 1 guessing
• phase 2 checking
• If the checking stage of a nondeterministic
  algorithm is of polynomial time-complexity, then
  this algorithm is called an NP (nondeterministic
  polynomial) algorithm.
• NP problems
• e.g. searching, MST
• sorting
• satisfiability problem (SAT)
• traveling salesperson problem (TSP)

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Nondeterministic searching algorithm

• Search for x in an array A
• Choice(S) arbitrarily chooses one of the
  elements in set S
• Failure an unsuccessful completion
• Success a successful completion
• Nonderministic searching algorithm
• j ? choice(1 n) / guessing /
• if A(j) x then success / checking /
• else failure

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• A nondeterministic algorithm terminates
  unsuccessfully iff there exist not a set of
  choices leading to a success signal.
• A deterministic interpretation of a
  non-deterministic algorithm can be made by
  allowing unbounded parallelism in computation.
• The runtime required for choice(1 n) is O(1).
• The runtime for nondeterministic searching
  algorithm is also O(1)

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Nondeterministic sorting

• B ? 0
• / guessing /
• for i 1 to n do
• j ? choice(1 n)
• if Bj ? 0 then failure
• Bj Ai
• / checking /
• for i 1 to n-1 do
• if Bi gt Bi1 then failure
• success

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• NP the class of decision problem which can be
  solved by a non-deterministic polynomial
  algorithm.
• P the class of problems which can be solved by a
  deterministic polynomial algorithm.
• NP-hard the class of problems to which every NP
  problem reduces.
• NP-complete (NPC) the class of problems which
  are NP-hard and belong to NP.

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Some concepts of NPC

• Definition of reduction Problem A reduces to
  problem B (A ? B) iff A can be solved by a
  deterministic polynomial time algorithm using a
  deterministic algorithm that solves B in
  polynomial time. B is harder.
• Up to now, none of the NPC problems can be solved
  by a deterministic polynomial time algorithm in
  the worst case.
• It does not seem to have any polynomial time
  algorithm to solve the NPC problems.

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• The theory of NP-completeness always considers
  the worst case.
• The lower bound of any NPC problem seems to be in
  the order of an exponential function.
• Not all NP problems are difficult. (e.g. the MST
  problem is an NP problem.)
• If A, B ? NPC, then A ? B and B ? A.
• Theory of NP-completeness
• If any NPC problem can be solved in polynomial
  time, then all NP problems can be solved in
  polynomial time. (NP P)

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Decision problems

• The solution is simply Yes or No.
• Optimization problems are more difficult.
• e.g. the traveling salesperson problem
• Optimization version
• Find the shortest tour
• Decision version
• Is there a tour whose total length is less than
  or equal to a constant c ?

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Solving an optimization problem by a decision
algorithm

• Solving TSP optimization problem by decision
  algorithm
• Give c1 and test (decision algorithm)
• Give c2 and test (decision algorithm)
• ?
• Give cn and test (decision algorithm)
• We can find the smallest ci

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The satisfiability problem

• The satisfiability problem Given a Boolean
  statement, determine whether it is satisfiable or
  not.
• The logical formula
• x1 v x2 v x3
• x1 x2 x3
• DNF, CNF, literals and clause. Conjunctive nrml
  form
• If there is at least one assignment which
  satisfies a formula in DNF, then we say that this
  formula is satisfiable otherwise, it is
  unsatisfiable.
• (F, F, T) will make statement true (satisfied) or
  (-x1, -x2 , x3)
• Unsatisfiable
• (x1 v x2) (x1 v -x2) (-x1 v x2) (-x1 v -x2)

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Cooks theorem

• NP P iff the satisfiability problem is a P
  problem.
• SAT is NP-complete.
• It is the first NP-complete problem.
• Every NP problem reduces to SAT.

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Toward NP-Completeness

• Once we have found an NP-complete problem,
  proving that other problems are also NP-complete
  becomes easier.
• Given a new problem Y, it is sufficient to prove
  that Cooks problem, or any other NP-complete
  problems, is polynomially reducible to Y.
• Known problem -gt unknown problem
• Reduction here is from unknown to every known
  prbl.

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NP-Completeness Proof

• Assume that CLIQUE problem is NP-complete, to
  prove that vertex cover (VC) problem is
  NP-complete.
• Definition
• A vertex cover of G(V, E) is V?V such that
  every edge in E is incident to some v?V.
• Vertex Cover(VC) Given undirected G(V, E) and
  integer k, does G have a vertex cover with ?k
  vertices?
• CLIQUE Does G contain a clique of size ?k?

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NP-Completeness Proof Vertex Cover(VC)

• Problem Given undirected G(V, E) and integer k,
  does G have a vertex cover with ?k vertices?
• Theorem the VC problem is NP-complete.
• Proof (Reduction from CLIQUE)
• VC is in NP. This is trivial since we can check
  it easily in polynomial time.
• Goal Transform arbitrary CLIQUE instance into VC
  instance such that CLIQUE answer is yes iff VC
  answer is yes.

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NP-Completeness Proof Vertex Cover(VC)

• Claim CLIQUE(G, k) has same answer as VC ( ,
  n-k), where n V.
• Observe There is a clique of size k in G iff
  there is a VC of size n-k in .

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NP-Completeness Proof Vertex Cover(VC)

• Observe If D is a VC in , then has no
  edge between vertices in V-D.
• So, we have k-clique in G n-k VC in
• Can transform in polynomial time.

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NP-Completeness Proof CLIQUE

• Assume that SAT problem is NP-complete, to prove
  that CLIQUE problem is NP-complete
• Problem Does G(V,E) contain a clique of size k?
• Theorem Clique is NP-Complete. (reduction from
  SAT)
• Idea Make column for each of k clauses.
• No edge within a column.
• All other edges present except between x and x
  

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NP-Completeness Proof CLIQUE

• Example
• G
• G has m-clique (m is the number of clauses in E),
  iff E is satisfiable.
• (Assign value 1 to all variables in clique)

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Graph coloring problem

• Def A coloring of a graph G(V, E) is a function
  f V ? 1, 2, 3,, k such that if (u, v) ? E,
  then f(u)?f(v). The graph coloring problem is to
  determine if G has a coloring for k.
• E.g.

3-colorable f(a)1, f(b)2, f(c)1 f(d)2,
f(e)3
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Set cover problem

• Def F Si S1, S2, , Sk
• Si u1, u2, , un
• T is a set cover of F if T ? F and Si
  Si
• The set cover decision problem is to determine if
  F has a cover T containing no more than c sets.
• Example
• F (a1, a3), (a2, a4), (a2, a3), (a4),
  (a1, a3 , a4)
• s1 s2 s3 s4
  s5
• T s1, s3, s4 set cover

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Subset sum problem

• Def A set of positive numbers A a1, a2, ,
  an
• a constant C
• Determine if ? A? ? A s.t. ai C
• e.g. A 7, 5, 19, 1, 12, 8, 14
• C 21, A? 7, 14
• C 11, no solution

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Partition problem

• Def Given a set of positive numbers A
  a1,a2,,an ,
• determine if ? a partition P, s.t. ?ai ?ai
• 
  
  i?p i?p
• e. g. A 3, 6, 1, 9, 4, 11
• partition 3, 1, 9, 4 and 6, 11

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Bin packing problem

• Def n items, each of size ci , ci gt 0
• bin capacity C
• Determine if we can assign the items into
• k bins, s.t. ?ci ? C , 1?j?k.
• i?binj

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Hamiltonian cycle problem

• Def A Hamiltonian cycle is a round trip path
  along n edges of G which visits every vertex once
  and returns to its starting vertex.
• e.g.
• Hamiltonian cycle 1, 2, 8, 7, 6, 5, 4, 3, 1.

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Traveling salesperson problem

• Def A tour of a directed graph G(V, E) is a
  directed cycle that includes every vertex in V.
  The problem is to find a tour of minimum cost.

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0/1 knapsack problem

• Def n objects, each with a weight wi gt 0
• a profit pi gt 0
• capacity of knapsack M
• Maximize ?pixi
• 
  1?i?n
• Subject to ?wixi ? M
• 
  1?i?n
• xi 0 or 1, 1? i ?n
• Decision version
• Given K, ? ?pixi ? K ?
• 
  1?i?n
• Knapsack problem 0 ? xi ? 1, 1? i ?n.

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Subset sum problem

• Reduce from 3SAT problem to show that subset sum
  problem is NP-complete.

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Three dimensional matching problem

• Reduce from 3SAT problem to show that three
  dimensional matching (3DM) problem is NP-complete.