9. chapter 8 np hard and np complete problems

Chapter-8
NP hard and NP Complete problems
8.1 Basic Concepts
The computing times of algorithms fall into two groups.
Group1– consists of problems whose solutions are bounded by the polynomial of small
degree.
Example – Binary search o (log n) , sorting o(n log n), matrix multiplication 0(n 2.81).
NP –HARD AND NP – COMPLETE PROBLEMS
Group2 – contains problems whose best known algorithms are non polynomial.
Example –Traveling salesperson problem 0(n22n), knapsack problem 0(2n/2) etc.
There are two classes of non polynomial time problems
1. NP- hard
2. NP-complete
A problem which is NP complete will have the property that it can be solved in polynomial
time iff all other NP – complete problems can also be solved in polynomial time.
The class NP (meaning non-deterministic polynomial time) is the set of problems that might
appear in a puzzle magazine: ``Nice puzzle.''
What makes these problems special is that they might be hard to solve, but a short answer can
always be printed in the back, and it is easy to see that the answer is correct once you see it.
Example... Does matrix A have LU decomposition?
No guarantee if answer is ``no''.

Another way of thinking of NP is it is the set of problems that can solved efficiently by a
really good guesser.
The guesser essentially picks the accepting certificate out of the air (Non-deterministic
Polynomial time). It can then convince itself that it is correct using a polynomial time
algorithm. (Like a right-brain, left-brain sort of thing.)
Clearly this isn't a practically useful characterization: how could we build such a machine?
Exponential Upper bound
Another useful property of the class NP is that all NP problems can be solved in exponential
time (EXP).
This is because we can always list out all short certificates in exponential time and check all
O (2nk) of them.
Thus, P is in NP, and NP is in EXP. Although we know that P is not equal to EXP, it is
possible that NP = P, or EXP, or neither. Frustrating!
NP-hardness
As we will see, some problems are at least as hard to solve as any problem in NP. We call
such problems NP-hard.
How might we argue that problem X is at least as hard (to within a polynomial factor) as
problem Y?
If X is at least as hard as Y, how would we expect an algorithm that is able to solve X to
behave?
NP –HARD and NP – Complete Problems Basic Concepts
If an NP-hard problem can be solved in polynomial time, then all NP-complete problems can
be solved in polynomial time.
All NP-complete problems are NP-hard, but all NP- hard problems are not NP-complete.
The class of NP-hard problems is very rich in the sense that it contains many problems from a
wide variety of disciplines.
P: The class of problems which can be solved by a deterministic polynomial algorithm.

NP: The class of decision problem which can be solved by a non-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.
NP-Competence
• How we would you define NP-Complete
• They are the “hardest” problems in NP
8.2 Deterministic and Nondeterministic Algorithms
• Algorithmswiththepropertythattheresultofeveryoperationisuniquelydefinedaretermedd
eterministic
• Such algorithms agree with the way programs are executed on a computer.
• In a theoretical framework, we can allow algorithms to contain operations whose
outcome are not uniquely defined but are limited to a specified set of possibilities.
• Themachineexecutingsuchoperationsareallowedtochooseanyoneoftheseoutcomessubje
cttoaterminationcondition.
• This leads to the concept of non deterministic algorithms.
• To specify such algorithms in SPARKS, we introduce three statements
Choice (s) ……… arbitrarily chooses one of the elements of the set S.
Failure …. Signals an unsuccessful completion.
Success: Signals a successful completion.
• Whenever there is a set of choices that leads to a successful completion then one such
set of choices is always made and the algorithm terminates.
• A non deterministic algorithm terminates unsuccessfully if and only if there exists no
set of choices leading to a successful signal.
• A machine capable of executing an on deterministic algorithm is called an un
deterministic machine.
• While non deterministic machines do not exist in practice they will provide strong
intuitive reason to conclude that certain problems cannot be solved by fast
deterministic algorithms.
Nondeterministic algorithms
A non deterministic algorithm consists of
Phase 1: guessing
Phase 2: checking

• If the checking stage of a non deterministic algorithm is of polynomial time-
complexity, then this algorithm is called an NP (nondeterministic polynomial)
algorithm.
• NP problems : (must be decision problems)
–e.g. searching, MST
Sorting
Satisfy ability problem (SAT)
travelling salesperson problem (TSP)
Example of a non deterministic algorithm
// The problem is to search for an element x //
// Output j such that A(j) =x; or j=0 if x is not in A //
j choice (1 :n )
if A(j) =x then print(j) ; success endif
print (‘0’) ; failure
complexity 0(1);
Non-deterministic decision algorithms generate a zero or one
as their output.
Deterministic search algorithm complexity. (n)
• Many optimization problems can be recast into decision problems with the property
that the decision problem can be solved in polynomial time iff the corresponding
optimization problem can.
• The decision is to determine if there is a 0/1 assignment of values to xi 1≤ i ≤ n such
that ∑pi xi ≥ R, and ∑ wi xi ≤ M, R, M are given numbers pi, wi ≥ 0, 1 ≤ i ≤ n.
• It is easy to obtain polynomial time nondeterministic algorithms for many problems
that can be deterministically solved by a systematic search of a solutions pace of
exponential size.
8.3 Satisfiability
• Letx1, x2, x3…. xn denotes Boolean variables.
• Let xi denotes the relation of xi.
• A literal is either a variable or its negation.
• A formula in the prepositional calculus is an expression that can be constructed using
literals and the operators and ^ or v.
• A clause is a formula with at least one positive literal.
• The satisfy ability problem is to determine if a formula is true for some assignment of
truth values to the variables.
• It is easy to obtain a polynomial time non determination algorithm that terminates s
successfully if and only if a given prepositional formula E(x1, x2……xn) is satiable.
• Such an algorithm could proceed by simply choosing (non deterministically) one of
the 2n possible assignment so f truth values to (x1, x2…xn) and verify that E(x1,x2…xn)
is true for that assignment.
The satisfy ability problem
The logical formula:

x1v x2 v x3
& -x1
& -x2
the assignment : x1 ← F , x2 ← F , x3 ← T will make the above formula true .
(-x1, -x2, x3) represents x1 ← F, x2 ← F, x3 ← T
If there is at least one assignment which satisfies a formula, then we say that this
formula is satisfiable; otherwise, it is un satisfiable.
An un satisfiable formula:
x1vx2
&x1v-x2
&-x1vx2
&-x1v-x2
Definition of the satisfiability problem:
Given a Boolean formula, determine whether this formula is satisfiable or not.
Aliteral: xi or-xi
Aclause:x1vx2v-x3Ci
A formula: conjunctive normal form (CNF)
C1&C2&…&Cm
8.4 Some NP-hard Graph Problems
ThestrategytoshowthataproblemL2isNP-hardis
1. Pick a problem L1 already known to be NP-hard.
2. Show how to obtain an instance I1 of L2m from any instance I of L1 such that from
the solution of I1 We can determine (in polynomial deterministic time)
thesolutiontoinstanceIofL1
3. Conclude from (ii) that L1L2.
4. Conclude from (1),(2), and the transitivity of that
Satisfiability L1 L1L2
Satisfiability L2
L2is NP-hard
1. Chromatic Number Decision Problem (CNP)
a. A coloring of a graph G = (V,E) is a function f : V { 1,2, …, k} i V
b. If (U, V) E then f(u) f(v).
c. The CNP is to determine if G has a coloring for a given K.
d. Satisfiability with at most three literals per clause chromatic number problem.
CNP is NP-hard.
2. Directed Hamiltonian Cycle(DHC)
Let G=(V,E) be a directed graph and length n=1V1
TheDHCisacyclethatgoesthrougheveryvertexexactlyonceandthenreturnstothestartingv
ertex.
The DHC problem is to determine if G has a directed Hamiltonian Cycle.
Theorem: CNF (Conjunctive Normal Form) satisfiability DHC
DHC is NP-hard.
3. Travelling Salesperson Decision Problem (TSP) :
1. The problem is to determine if a complete directed graph G = (V,E) with edge
costs C(u,v) has a tour of cost at most M.

Theorem: Directed Hamiltonian Cycle (DHC) TSP
2. But from problem (2) satisfiability DHC Satisfiability TSP
TSP is NP-hard.
8.5 Sum of subsets
TheproblemistodetermineifA={a1,a2,…….,an}(a1,a2,………,anarepositiveintegers) has a
subset S that sum s to a given integer M.
Scheduling identical processors
Let Pi1 ≤ i ≤m be identical processors or machines Pi.
Let Ji 1≤ i ≤ n be n jobs.
Jobs Ji requires ti processing time
A schedule S is an assignment of jobs to processors.
For each job Ji, S specifies the time interval s and the processors on which this job i is to be
processed.
A job cannot be processed by more than one process or at any given time.
The problem is to find a minimum finish time non-preemptive schedule. The finish time of S
is FT(S) = max {Ti}1≤i≤m. Where Ti is the time at which processor Pi finishes processing all
jobs (or job segments) assigned to it
An NP-hard problem L cannot be solved in deterministic polynomial time.
By placing enough restrictions on any NP hard problem, we can arrive at a polynomials
solvable problem.
Examples
CNF-Satisfy ability with at most three literals per clause is NP-hard. If each clause is
restricted to have at most two literals then CNF-satisfy ability is polynomial solvable.
Generating optimal code for a parallel assignment statement is NP-hard, However if the
expressions are restricted to be simple variables, then optimal code can be generated in
polynomial time.
Generating optimal code for level one directed a-cyclic graphs is NP-hard but optimal code
for trees can be generated in polynomial time.
Determining if a planner graph is three color able is NP-Hard
Todetermineifitistwocolorableisapolynomialcomplexityproblem.
(Weonlyhavetoseeifitisbipartite)
General definitions - P, NP, NP-hard, NP-easy, and NP-complete... - Polynomial-time
reduction • Examples of NP-complete problems

P - Decision problems (decision problems) that can be solved in polynomial time - can be
solved “efficiently”
NP - Decision problems whose “YES” answer can be verified in polynomial time, if we
already have the proof (or witness)
Co-NP - Decision problems whose “NO” answer can be verified in polynomial time, if we
already have the proof (or witness)
E.g. the satisfy ability problem (SAT) - Given a Boolean formula is it possible to assign the
input x1...x9, so that the formula evaluates to TRUE?
If the answer is YES with a proof (i.e. an assignment of input value), then we can check
the proof in polynomial time (SAT is in NP). We may not be able to check the NO answer in
polynomial time. (Nobody really knows.)
•NP-hard
A problem is NP-hard iff an polynomial-time algorithm for it implies a
polynomial-time algorithm for every problem in NPNP-hard problems are at least as hard as
NP problems
•NP-complete
A problem is NP-complete if it is NP-hard, and is an element of NP (NP-easy)

Relationship between decision problems and optimization problems every optimization
problem has a corresponding decision problem
Optimization: minimize x, subject to constraints yes/no: is there a solution, such that x is less
than c? an optimization problem is NP-hard (NP-complete) if its corresponding decision
problem is NP-hard (NP-complete)
Polynomial-time reductions
How to know another problem, A, is NP-complete?
To prove that A is NP-complete, reduce a known NP-complete problem to A
Requirement for Reduction
Polynomial time
YES to A also implies YES to SAT, while
NO to A also implies No to SAT (Note that A must also have short proof for YES answer)
An example of reduction 3CNF
3SAT: is a boolean formula in 3CNF has a feasible assignment of inputs so that it evaluates
to TRUE?
reduction from 3SAT to SAT (3SAT is NP-complete)
Examples of NP-complete problems
Vertex cover
Independent set
Set cover
Steiner tree
Vertex cover
Given a graph G = (V, E), find the smallest number of vertexes that cover each edge

Decision problem: is the graph has a vertex cover of size K?
Reduction
Vertex cover
An example of the constructive graph
Vertex cover
We must prove: the graph has a n+2c vertex cover, if and only if the 3SAT is satisfiable (to
make the two problem has the same YES/NO answer!)
Vertex cover
- If the graph has a n+2c vertex cover
1) There must be 1 vertex per variable gadget, and 2 per clause gadget
2) In each clause gadget, set the remaining one literal to be true

• Vertex cover
If the 3SAT is satisfiable
1) Choose the TURE literal in each variable gadget
2) Choose the remaining two literal in each clause gadget
Independent set
Independent set: a set of vertices in the graph with no edges between each pairof nodes.
given a graph G=(V,E), find the largest independent set
reduction from vertex cover:
Independent set
If G has a vertex cover S, then V/S is an independent set
Proof: consider two nodes in V/S: if there is an edge connecting them, then one of them must
be in S, which means one of them is not in V/S
If G has an independent set I, then V/I is a vertex cover
Proof: consider one edge in G:
If it is not covered by any node in V/I, then its two end vertices must be
both in I, which means I is not an independent set
Given a universal set U, and several subsets S1...Sn
find the least number of subsets that contains each elements in the universal set
vertex cover is a special case of set cover:

1) the universal set contains all the edges
2) each vertex corresponds to a subset, containing the edges it covers
Steiner tree
Given a graph G = (V, E), and a subset C of V
find the minimum tree to connect each vertex in C reduction
Steiner tree
- G’ is a complete graph
- for every node u in G, create a node u in G’
- for every edge (u, v) in G, create a node (u, v) in G’
- in G’, every node (u, v) is connected to u and v with distance 1
- in G’, every node u and v is connected with distance 1
- other edges in G’ are of distance 2
In the Steiner tree problem for G’, choose C to be the set of all nodes (u, v)
G’ has a minimum Steiner tree of close m+k-1 iff G has a minimum vertex cover of size k
Examples of NP-complete problems

9. chapter 8 np hard and np complete problems

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