NP – HARD JAYASRI JETTI CHINMAYA KRISHNA SURYADEVARA.
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Transcript of NP – HARD JAYASRI JETTI CHINMAYA KRISHNA SURYADEVARA.
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NP – HARD
JAYASRI JETTICHINMAYA KRISHNA SURYADEVARA
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P and NPP – The set of all problems solvable in polynomial time by a deterministic Turing Machine (DTM).Example: Sorting and searching.
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P and NP
• NP- the set of all problems solvable in polynomial time by non deterministic Turing Machine (NDTM)
• Example Partition.
P is a subset of NP
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NP complete
• NP – Complete (NP – C): the set of all problems in NP “at least as hard” as every other problem in NP.
• To prove a problem x is NP complete – Show that x is in NP– Show that some other NP - C problem reduces to x
• If an NP hard problem can be solved in polynomial time, then all NP complete problems can be solved in polynomial time.
• NP – H includes all NP - C problems
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Deterministic and non deterministic problems
• Algorithm is deterministic if for a given input the output generated is same for function.
• A mathematical function is deterministic hence the state is known at every step of algorithm.
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• Algorithm is nondeterministic if there are more than one path the algorithm can take. Due to this one cannot determine the next state of the machine running the algorithm.
• To specify such algorithms we introduce three statements
– Choice(s) - arbitrary choose one of the elements of the set S
– Failure - signals an unsuccessful completion– Success - signals a successful completion
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• Whenever there is a set of choices that leads to a successful completion than one search set of choices is always made and the algorithm terminates.
• A nondeterministic algorithm terminates unsuccessfully iff there exists no set of choices leading to a successful signal
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• A machine capable of executing a nondeterministic algorithm is called a nondeterministic machine.
• While nondeterministic machine do not exist in practice they will provide strong reason to conclude that certain problems cannot be solved by fast deterministic algorithms
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• Example of nondeterministic 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 end ifPrint (‘0’); failureComplexity O(1);
Nondeterministic decision algorithms generate a zero or 1 as their output.
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Definition of classes NP - hard and NP - complete
• P is a set of all decision problems solvable by a deterministic algorithm in polynomial time.
• NP is a set of all decision problems solvable by a nondeterministic algorithm in polynomial time.
• Since deterministic algorithms are special case of nondeterministic ones, we can conclude that PNP
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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 L3Definition: NP-Hard problem: a problem L is NP
hard iff satisfiability reduces to L
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Definition : NP-complete Problem : • A problem L is NP-complete if and only if L is NP-hard
and L є 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).
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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.
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Halting problem is NP-hard (Contd..)
- If it is satisfied 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.
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NP-HARD GRAPH AND SCHEDULING PROBLEMS (CONTD..)
1. Chromatic Number Decision Problem (CNP)• A coloring 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 coloring
for a given K.• Satisfiability with at most three literals per
clause chromatic number problem. CNP is NP-hard.
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NP-HARD GRAPH AND SCHEDULING PROBLEMS (contd..)
2. Directed Hamiltonian Cycle (DHC)• Let G = (V,E) be a directed graph and length n = V• 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.
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NP-HARD GRAPH AND SCHEDULING PROBLEMS (CONTD..)
3. 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 TSP is NP-hard.
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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 polynomial solvable problem.
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Examples.(i) CNF(conjuctive normal form)- 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 polynomial solvable
(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.
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(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 colorable is NP-Hard
- To determine if it is two colorable is a polynomial complexity problem.
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• https://www.youtube.com/watch?v=CY_klh5uD-E
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THANK YOU