1 NP –HARD AND NP – COMPLETE PROBLEMS 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 (Contd..) :

2 NP –HARD AND NP – COMPLETE PROBLEMS (Contd..) 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.

NP –HARD AND NP – COMPLETE PROBLEMS (Contd..) :

3 NP –HARD AND NP – COMPLETE PROBLEMS (Contd..) 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 contain many problems from a wide variety of disciplines.

DETERMINISTIC and NONDETERMINISTIC ALGORITHMS :

4 DETERMINISTIC and NONDETERMINISTIC ALGORITHMS Algorithms with the property that the result of every operation is uniquely defined are termed deterministic.
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 is not uniquely defined but is limited to a specified set of possibilities.

Deterministic and Nondeterministic algorithms (contd..) :

5 Deterministic and Nondeterministic algorithms (contd..) The machine executing such operations are allowed to choose any one of these outcomes subject to a termination condition.
This leads to the concept of non deterministic algorithms.
To specify such algorithms in SPARKS, we introduce three statements
i) choice (s) ……… arbitrarily chooses one of the
elements of the set S.
ii) failure …. Signals an unsuccessful completion.
iii) Success : Signals a successful completion.

Deterministic and Nondeterministic algorithms (contd..) :

6 Deterministic and Nondeterministic algorithms (contd..) 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 nondeterministic algorithm terminates unsuccessfully if and only if there exists no set of choices leading to a successful signal.
A machine capable of executing a non deterministic algorithm is called a non 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.

Example of a non deterministic algorithm :

7 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)

Deterministic and Nondeterministic algorithms (contd..) :

8 Deterministic and Nondeterministic algorithms (contd..) 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 .
EXAMPLE : [0/1 knapsack] :
The decision is to determine if there is a 0/1 assignment of values to xi 1? i ? n such that ? pixi ? R, and ? wixi ? M, R, M are given numbers pi, wi ?0, 1? i ? n.
It is easy to obtain polynomial time non deterministic algorithms for many problems that can be deterministically solved by a systematic search of a solution space of exponential size.

SATISFIABILITY :

9 SATISFIABILITY Let x1,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 satiability problem is to determine if a formula is true for some assignment of truth values to the variables.

SATISFIABILITY (Contd..) :

10 SATISFIABILITY (Contd..) It is easy to obtain a polynomial time non determination algorithm that terminates 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 assignments of truth values to (x1,x2…xn) and verify that E(x1,x2…xn) is true for that assignment.

Definition of the classes NP-hard and NP-complete :

11 Definition of the classes NP-hard and NP-complete P is the set of all decision problems solvable by a deterministic algorithm in polynomial time.
NP is the set of all decision problems solvable by a nondeterministic algorithm in polynomial time.
Since deterministic algorithm are a special case of non deterministic ones, we can conclude that P?NP.
The most famous unsolved problem in computer science is whether P = NP or P ? NP.
Is it possible that for all the problems in NP their exist polynomial time deterministic algorithms which have remained undiscovered ?

Definition of the classes NP-hard and NP-complete (contd..) :

12 Definition of the classes NP-hard and NP-complete (contd..) In considering this problem S.COOK formulated the following question :
Is there any single problem in NP such that if we showed it to be in P then that would imply that P = NP.
COOK’s Theorem : Satisfiability is in P if and only if P = NP.

Definition of the classes NP-hard and NP-complete (contd..) :

13 Definition of the classes NP-hard and NP-complete (contd..) Definition : Reducibility
Let L1 and L2 be problems. L1 reduces to L2 (L1 ? L2) if and only if there is a deterministic polynomial time algorithm to solve L1 that solves L2 in polynomial time.
If L1 ? L2 and L2 ? L3 then L1 ? L3.
Definition : NP-Hard Problem : A problem L is NP-hard if any only if satisfiability reduces to L.

Definition of the classes NP-hard and NP-complete (contd..) :

14 Definition of the classes NP-hard and NP-complete (contd..) Definition : NP-complete Problem :
A problem L is NP-complete if and only if L is NP-hard and L ? NP.
There are NP-hard problems that are not NP-complete.
Example :
Halting problem is NP-hard decision problem, but it is not NP-complete.
Halting Problem :
To determine for an arbitrary deterministic algorithm A and an input I whether algorithm A with input I ever terminates (or enters an infinite loop).

Halting problem is not NP-complete; but NP-hard :

15 Halting problem is not NP-complete; but NP-hard Halting problem is un-decidable.
- Hence there exists no algorithm to solve this problem.
- So, it is not in NP.
- So, it is not NP-complete.
Halting problem is NP-hard
To show that Halting problem is NP-hard, we show that satisfiability is ? halting problem.
For this let us construct an algorithm A whose input is a prepositional formula X.
- Suppose X has n variables.
- Algorithm A tries out all 2n possible truth assignments and verifies if X is satisfiable.

Halting problem is NP-hard (Contd..) :

16 Halting problem is NP-hard (Contd..) - If it is then A stops.
- If X is not satisfiable, then A enters an infinite loop.
- Hence A halts on input iff X is satisfiable.
- If we had a polynomial time algorithm for the halting problem, then we could solve the satisfiability problem in polynomial time using A and X as input to the algorithm for the halting problem .
- Hence the halting problem is an NP-hard problem which is not in NP.
- So it is not NP-complete.

NP-HARD GRAPH AND SCHEDULING PROBLEMS :

17 NP-HARD GRAPH AND SCHEDULING PROBLEMS Some NP-hard Graph Problems :
The strategy to show that a problem L2 is NP-hard is
Pick a problem L1 already known to be NP-hard.
Show how to obtain an instance I1 of L2 from any instance I of L1 such that from the solution of I1
- We can determine (in polynomial deterministic
time) the solution to instance I of L1.

NP-HARD GRAPH AND SCHEDULING PROBLEMS (contd..) :

18 NP-HARD GRAPH AND SCHEDULING PROBLEMS (contd..) Conclude from (ii) that L1 ? L2.
Conclude from (i),(ii), and the transitivity of ? that
Satisfiability ? L1
L1 ? L2
? Satisfiability L2
? L2 is NP-hard

NP-HARD GRAPH AND SCHEDULING PROBLEMS (CONTD..) :

19 NP-HARD GRAPH AND SCHEDULING PROBLEMS (CONTD..) Chromatic Number Decision Problem (CNP)
A colouring of a graph G = (V,E) is a function
f : V ? { 1,2, …, k} ? i ? V .
If (U,V) ?E then f(u) ? f(v).
The CNP is to determine if G has a colouring for a given K.
Satisfiability with at most three literals per clause ? chromatic number problem.
? CNP is NP-hard.

NP-HARD GRAPH AND SCHEDULING PROBLEMS (contd..) :

20 NP-HARD GRAPH AND SCHEDULING PROBLEMS (contd..) Directed Hamiltonian Cycle (DHC)
Let G = (V,E) be a directed graph and length n = 1V1
The DHC is a cycle that goes through every vertex exactly once and then returns to the starting vertex.
The DHC problem is to determine if G has a directed Hamiltonian Cycle.
Theorem : CNF (Conjunctive Normal Form) satisfiability ? DHC
? DHC is NP-hard.

NP-HARD GRAPH AND SCHEDULING PROBLEMS (CONTD..) :

21 NP-HARD GRAPH AND SCHEDULING PROBLEMS (CONTD..) Travelling Salesperson Decision Problem (TSP) :
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
But from problem (2) satisfiability ? DHC
? Satisfiability ? TSP
? TSP is NP-hard.

NP-HARD SCHEDULING PROBLEMS :

22 NP-HARD SCHEDULING PROBLEMS 1. Sum of subsets
The problem is to determine if A={a1,a2,…….,an} (a1,a2,………,an are positive integers) has a subset S that sums to a given integer M.
2. Scheduling identical processors
Let Pi 1=i=m be identical processors or machines Pi.
Let Ji 1=i=n be n jobs.
Jobs Ji requires ti processing time.

NP-HARD SCHEDULING PROBLEMS (Contd..) :

23 NP-HARD SCHEDULING PROBLEMS (Contd..) A schedule S is an assignment of jobs to processors.
For each job Ji ,S specifies the time intervals and the processors on which this job is to be processed.
A job can not be processed by more than one processor 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.

SOME SIMPLIIFIED NP-HARD PROBLEMS :

24 SOME SIMPLIIFIED NP-HARD PROBLEMS 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 polynomially solvable problem.
Examples.
(i) CNF- Satisfiability with at most three literals per clause is NP-hard.
If each clause is restricted to have at most two literals then CNF-satisfiability is polynomially solvable.

SOME SIMPLIIFIED NP-HARD PROBLEMS :

25 SOME SIMPLIIFIED NP-HARD PROBLEMS (ii)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.
(iii)Generating optimal code for level one directed a- cyclic graphs is NP-hard but optimal code for trees can be generated in polynomial time.
(iv)Determining if a planner graph is three colourable is NP-Hard
- To determine if it is two colorable is a polynomial complexity problem. (We only have to see if it is bipartite)

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