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**Hardness Results for Problems**

Example Problems P: Class of “easy to solve” problems Absolute hardness results Relative hardness results Reduction technique

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**Example: Clique Problem**

Input: Undirected graph G = (V,E), integer k Y/N Question: Does G contain a clique of size ≥ k? k=4 k=5

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**Example: Vertex Cover k=3 k=2**

Input: Undirected graph G = (V,E), integer k Y/N Question: Does G contain a vertex cover of size ≤ k? Vertex cover: A set of vertices C such that for every edge (u,v) in E, either u is in C or v is in C (or both are in C) k=3 k=2

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**Example: Satisfiability**

Input: Set of variables X and set of clauses C over X Y/N Question: Is there a satisfying truth assignment T for the variables in X such that all clauses in C are true?

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**Hardness Results for Problems**

Example Problems P: Class of “easy to solve” problems Absolute hardness results Relative hardness results Reduction technique

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Fundamental Setting When faced with a new problem P, we alternate between the following two goals Find a “good” algorithm for solving P Use algorithm design techniques Prove a “hardness result” for problem P No “good” algorithm exists for problem P

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Complexity Class P P is the set of problems that can be solved using a polynomial-time algorithm Sometimes we focus only on decision problems The task of a decision problem is to answer a yes/no question If a problem belongs to P, it is considered to be “efficiently solvable” If a problem is not in P, it is generally considered to be NOT “efficiently solvable” Looking back at previous slide, our goals are to: Prove that P belongs to P Prove that P does not belong to P

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**Hardness Results for Problems**

Example Problems P: Class of “easy to solve” problems Absolute hardness results Relative hardness results Reduction technique

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**Absolute Hardness Results**

Fuzzy Definition A hardness result for a problem P without reference to another problem Examples Solving the clique problem requires W(n) time in the worst-case Solving the clique problem requires W(2n) time in the worst-case. The clique problem is not in P.

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**Proof Techniques Diagonalization Information Theory argument**

Can be used to prove some problems are not in P Information Theory argument W(nlog n) lower bound for sorting Typically not a superpolynomial lower bounds “Size of input” argument Prove that solving the graph connectivity problem requires W(V2) time Prove that solving the maximum clique problem requires W(V2) time Typically not a superpolynomial lower bound

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**Status Many natural problems can be shown to be in P**

Graph connectivity Shortest Paths Minimum Spanning Tree Very few natural problems have been proven to NOT be in P Variants of halting problem are one example Many natural problems cannot be placed in or out of P Satisfiability Longest Path Problem Clique Problem Vertex Cover Problem

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**Hardness Results for Problems**

Example Problems P: Class of “easy to solve” problems Absolute hardness results Relative hardness results Reduction technique

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**Relative Hardness Results**

Fuzzy Definition A hardness result for a problem P with reference to another problem Examples Satisfiability is at least as hard as Clique to solve If Satisfiability is unsolvable, then Clique is unsolvable. If Satisfiability is in P, then Clique is in P If Clique is not in P, then Satisfiability is not in P

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**Important Observation**

We are interested in relative hardness results BECAUSE of our inability to prove absolute hardness results That is, if we could prove strong absolute hardness results, we would not be as interested in relative hardness results Example If I could prove “Satisfiability is not in P”, then I would be less interested in proving “If Clique is not in P, then Satisfiability is not in P”.

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**Relative Hardness Proof Technique**

We show that P2 is at least as hard as P1 in the following way Informal: We show how to solve problem P1 using a procedure P2 that solves P2 as a subroutine

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**Examples Multiplication and Squaring square(x) = mult(x,x)**

Proves multiplication is at least as hard as squaring mult(x,y) = (square(x+y) – square(x-y))/4 Prove squaring is at least as hard as multiplication Assumes that addition, subtraction, and division by 4 can be done with no substantial increase in complexity Specific complexity of multiplication may be higher as there are two calls to square, but the difference is polynomially bounded

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**Hardness Results for Problems**

Example Problems P: Class of “easy to solve” problems Absolute hardness results Relative hardness results Reduction technique

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**Decision Problems We restrict our attention to decision problems**

Key characteristic: 2 types of inputs Yes input instances No input instances Almost all natural problems can be converted into an equivalent decision problem without changing the complexity of the problem One technique: add an extra input variable that represents the solution for the original problem

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**Optimization to Decision**

Example using clique problem Optimization Problems Input: Graph G=(V,E) Task: Output size of maximum clique in G Task 2: Output a maximum sized clique of G Decision Problem Input: Graph G=(V,E), integer k ≤ |V| Y/N Question: Does G contain a clique of size k? Your task Show that if we can solve decision clique in polynomial-time, then we can solve the optimization clique problems in polynomial-time.

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**Polynomial-time Reduction Technique**

Can describe as a polynomial-time “Answer-preserving input transformation” Consider two problems П1 and П2, and suppose I want to show If П2 is in P, then П1 is in P If П1 is not in P, then П2 is not in P The basic idea Develop a function (reduction) R that maps input instances of problem П1 to input instances of problem П2 The function R should be computable in polynomial time x is a yes input to П1 ↔ R(x) is a yes input to П2 Notation: П1 ≤p П2

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**What П1 ≤p П2 Means If R exists, then: If П2 is in P, then П1 is in P**

Yes Input to П2 P2 solves П2 in poly time Yes Input to П1 Yes R No Input to П2 No No Input to П1 P1 solves П1 in polynomial-time If R exists, then: If П2 is in P, then П1 is in P If П1 is not in P, then П2 is not in P

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**Showing П1 ≤p П2 For any x input for П1, specify what R(x) will be**

Show that R(x) has polynomial size relative to x You should show that R runs in polynomial time; I only require the size requirement above Show that if x is a yes instance for П1, then R(x) is a yes instance for П2 Show that if x is a no instance for П1, then R(x) is a no instance for П2 Often done by showing that if R(x) is a yes instance for П2, then x must have been a yes instance for П1

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