Summary
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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
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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->[convert]->[solver for VC]->?
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]->(G’, k’)[VC]->
(G, k) [IS][VC]
how to do  (G, k) [G=G’, 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.
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IS {G, k} given in to the box of VC
p Reduction takes polytime
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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,
g=g’, k’=n-k, (one clock time)..
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IS p VC
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The otherway around is not always true in general.
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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'(b|s|)+p(|s|) O(b|s|) 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
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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
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A boolean formula3-SATreducerinput for IS(G, k)IS
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If formula is satisfiable then there is an IS.
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Tree boxes, input, X, Reducer and Y solver-output yes or no.
Decision YES OR NO
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If y runs in polytime, and result will be ploy of poly(|x|).
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If there is no poly solver for X, X reduced to Y, then there cannot be a polysolver
for Y.
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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 don’t know an
algorithm to solve it in polytime algorithm..
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Class of NP problems
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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(p|x|), 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 don’t know yet if it is in NP, 3 sat, since we
don’t 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} ->[B: X
checker algorithm A, polytime]->..
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Def: B is an efficient certifier for language X, if
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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.
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NP is the class of problems that have efficient
certifiers: B as given above.
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X is in NP iff there is an efficient certifier for X.
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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.
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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
1.
2.
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
|x|=|y|
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 NP=P.
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, Cooks’s 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/EE590
0Spring10/NP-complete.ppt
www.cs.dartmouth.edu/~ac/Teach/CS3
9Winter09/SlidesAndNotes/lec28npcom
plete.pdf
Nondeterministic algorithms
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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 :
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e.g.
searching, MST
sorting
satisfiability problem (SAT)
traveling salesperson problem (TSP)
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Nondeterministic searching
algorithm
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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 nondeterministic 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 B[j] ≠ 0 then failure
B[j] = A[i]
/* checking */
for i = 1 to n-1 do
if B[i] > B[i+1] 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
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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.
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Theory of NP-completeness
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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
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The solution is simply “Yes” or “No”.
Optimization problems are more difficult.
e.g. the traveling salesperson problem
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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 :
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Solving TSP optimization
problem by decision algorithm :
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Give c1 and test (decision algorithm)
Give c2 and test (decision algorithm)
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Give cn and test (decision algorithm)
We can find the smallest ci
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The satisfiability problem
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The satisfiability problem: Given a Boolean
statement, determine whether it is satisfiable or
not.
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The logical formula :
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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.
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x1 v x2 v x3
x1 & x2 & x3
(F, F, T) will make statement true (satisfied) or (-x1, -x2 , x3)
Unsatisfiable:
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(x1 v x2) & (x1 v -x2) & (-x1 v x2) & (-x1 v -x2)
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Cook’s theorem
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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
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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 Cook’s
problem, or any other NP-complete problems, is
polynomially reducible to Y.
Known problem -> 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)
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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)
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Claim: CLIQUE(G, k) has same answer as VC
( G , 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 G .
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NP-Completeness Proof:
Vertex Cover(VC)
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Observe: If D is a VC in G , then G has no edge
between vertices in V-D.
So, we have k-clique in G n-k VC in G
Can transform in polynomial time.
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NP-Completeness Proof:
CLIQUE
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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
E  ( x  y  z)  ( x  y  z)  ( y  z)
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Example:
x
G=
y
z
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x
y
y
z
z
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
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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
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Def: F = {Si} = { S1, S2, …, Sk }
 Si = { u1, u2, …, un }
T is a set cover of F if T  F and  Si =  Si
Si F
Si T
The set cover decision problem is to
a cover T containing no more than
 Example
F = {(a1, a3), (a2, a4), (a2, a3),
s1
s2
s3
T = { s1, s3, s4 } set cover
Si F
determine if F has
c sets.
(a4), (a1, a3 , a4)}
s4
s5
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Subset sum problem
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Def: A set of positive numbers A = { a1,
a2, …, an }
a constant C
Determine if  A  A s.t.

ai = C
a i A 
e.g. A = { 7, 5, 19, 1, 12, 8, 14 }
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C = 21, A = { 7, 14 }
C = 11, no solution
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Partition problem
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Def: Given a set of positive numbers A =
{ a1,a2,…,an },
determine if  a partition P, s.t. ai = ai
ip
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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
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Def: n items, each of size ci , ci > 0
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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
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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
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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
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Def: n objects, each with a weight wi > 0
a profit pi > 0
capacity of knapsack : M
Maximize pixi
1in
Subject to wixi  M
1in
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xi = 0 or 1, 1 i n
Decision version :
Given K,  pixi  K ?
1in
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Knapsack problem : 0  xi  1, 1 i n.
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Subset sum problem
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Reduce from 3SAT problem to show that subset sum
problem is NP-complete.
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Three dimensional matching
problem
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Reduce from 3SAT problem to show that three
dimensional matching (3DM) problem is NP-complete.
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