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Chapter 2: The Theory of NP-completeness. Problems vs. Languages Encoding schemes P, NP. Problem vs. Language. As computer scientists, we are typically concerned with solving problems, not recognizing languages (unless your last name is Stansifer).
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Chapter 2: The Theoryof NP-completeness • Problems vs. Languages • Encoding schemes • P, NP
Problem vs. Language • As computer scientists, we are typically concerned with solving problems, not recognizing languages (unless your last name is Stansifer). • Ultimately we want to make precise claims about problems: • P, NP or NP-complete • 2-SAT can be solved by a deterministic polynomial time algorithm • TSP can be solved by a non-deterministic polynomial time algorithm • Such claims raise several questions: • What is an algorithm? • What do deterministic and non-deterministic mean? • What does it mean to “solve” a problem?” • What are P, NP, NP-complete? • How is time measured, and what is input length? • In order to answer such questions precisely, we will us a formal model of computation, i.e., the Turing Machine. • Note that some other model could be used, but the TM is easy to use in this context.
The Language Level • Symbol – An atomic unit, such as a digit, character, lower-case letter, etc. Sometimes a word. • Alphabet – A finite set of symbols, usually denoted by Σ. Σ = {0, 1} Σ = {0, a, 4} Σ = {a, b, c, d} • String – A finite length sequence of symbols, presumably from some alphabet. w = 0110 y = 0aa x = aabcaa z = 111 special string: ε concatenation: wz = 0110111 length: |w| = 4 |ε| = 0 |x| = 6 reversal: yR = aa0
The Language Level • Some special sets of strings: Σ* All strings of symbols from Σ (Kleene closure) Σ+ Σ* - {ε} (positive closure) • Example: Σ = {0, 1} Σ* = {ε, 0, 1, 00, 01, 10, 11, 000, 001,…} Σ+ = {0, 1, 00, 01, 10, 11, 000, 001,…} • A (formal) language is: 1) A (finite or infinite) set of strings from some alphabet. 2) Any subset L of Σ* • Some special languages: {} The empty set/language, containing no strings {ε} A language containing one string, the empty string.
The Language Level • Language Examples: Σ = {0, 1} L = {x | x Σ* and x contains an even number of 0’s} Σ = {0, 1, 2,…, 9, .} L = {x | x Σ* and x forms a finite length real number} = {0, 1.5, 9.326,…} Σ = {a, b, c,…, z, A, B,…, Z} L = {x | x Σ* and x is a Java reserved word} = {while, for, if,…} Σ = {Java reserved words} U { (, ), ., :, ;,…} U {Legal Java identifiers} L = {x | x is in Σ* and x is a syntactically correct Java program} Σ = {English words} L = {x | x is in Σ* and x is a syntactically correct English sentence}
Problem vs. Language • An analogy between languages and problems: • L partitions Σ* into two sets: L and Σ* - L • Ypartitions D into two sets: Yand D - Y • Additionally: • DTM vs. NDTM • deterministic vs. nondeterministic algorithms • How we use these analogies: • The formal language level will be used to define our terms, i.e., algorithm, deterministic, non-deterministic mean, solve, problem, P, NP, NP-complete, time, input length. • These terms will then be applied in the context of problems. • Encoding schemes will form the “bridge” between the formal and informal levels that will allow this.
Encoding Schemes • Purpose: • Bridges between the problem level and the more formal language level, thereby enabling the application of formal terms, definitions and concepts to be applied to problems. • What is an encoding scheme? • Suppose you wrote a program that inputs a graph G = (V, E) • You need to come up with an input (or file) format • That format is an encoding scheme => see page 10 • Notes: • A particular problem has many different encoding schemes • Length is now precise - the number of input symbols • A problem and an associated encoding scheme define a language
Encoding Schemes, cont. • Let be a problem, e an encoding scheme that uses symbols from Σ. Then partitions Σ*into 3 subsets: • { x | x Σ* and x D} • { x | x Σ*, x D and x Y} • { x | x Σ*, x D and x Y (x D - Y)} • The second of the above is the language associated with and e: L[, e] = {x Σ* | Σ is the alphabet used by e, and x is the encoding under e of an instance I Y} (page 20) • How does this allow application of formal terms, concepts and definitions at the problem level? • See page 20, second paragraph • Terms such as “NP-complete” are defined formally in terms of languages • Saying that a problem is NP-complete is a somewhat informal way of saying that the language L[, e] is NP-complete, for some encoding scheme e.
Encoding Schemes, cont. • A problem can have many encoding schemes • The previous page suggests our results for a problem are encoding dependent • But is this true? • Could a property hold for a problem under one encoding scheme, but not another? • Yes! • Example #1: • Input a graph G = (V, E) and for each pair of vertices output an indication of whether or not the two vertices are adjacent. • Adjacency matrix: O(n) (n = input length) • Vertex-edge list: O(n2) (no edges) • Note that the encoding scheme does make a difference
Encoding Schemes, cont. • Example #2: • Input a positive integer n and output all binary numbers between 0 and n-1 • Input n in binary: O(m2m) (m = input length) • Input n in unary: O(log(m)2log(m)) = O(mlogm) • Note that the encoding scheme makes a dramatic difference in running time, i.e., polynomial vs. exponential • The second encoding scheme is unnecessarily padding the input, and is therefore said to be unreasonable • Reasonable encoding schemes are: • Concise (relative to how a computer stores things) • Decodable (in polynomial time, pages 10, 21) • We do not want the fact that a problem is NP-complete or solvable in polynomial time to depend on a specific encoding scheme • Hence, we restrict ourselves to reasonable encoding schemes • Comes naturally, unreasonable encoding schemes will look suspicious
Encoding Schemes and Input Length. • The running time of an algorithm, or the computational complexity of a problem is typically expressed as a function of the length of a given instance. • O(n2) • Ω (n3) • In order to talk about the computational complexity of a problem or the running time of an algorithm, we need to associate a length with each instance. • Every problem will have an associated length function: LENGTH: D Z+ which must be encoding-scheme independent, i.e., “polynomial related” to the input lengths we would obtain from a reasonable encoding scheme (see page 19 for a definition). • This ensures that the length function is also reasonable and that complexity results for a problem are consistent with those for an associated encoded language.
Input Length Function • Recall the CLIQUE problem: CLIQUE INSTANCE: A Graph G = (V, E) and a positive integer J <= |V|. QUESTION: Does G contain a clique of size J or more? • What would be a reasonable length function? • # of vertices • # of edges • # of vertices + # of edges • Which would be polynomial related to a reasonable encoding scheme? • Note that the first and last differ quite substantially, although not exponentially.
Deterministic, One-Tape TM …….. …….. • Two-way, infinite tape, broken into cells, each containing one symbol. • Two-way, read/write tape head. • Finite control, i.e., a program, containing the position of the read head, current symbol being scanned, and the current state. • An input string is placed on the tape, padded to the left and right infinitely with blanks, read/write head is positioned at the left end of input string. • In one move, depending on the current state and the current symbol being scanned, the TM 1) changes state, 2) prints a symbol over the cell being scanned, and 3) moves its’ tape head one cell left or right. Finite Control
Deterministic, One-Tape TM • A DTM is a seven-tuple: M = (Q, Σ, Γ, δ, q0, B, F) Q A finite set of states Γ A finite tape alphabet B A distinguished blank symbol, which is in Γ Σ A finite input alphabet, which is a subset of Γ– {B} q0 The initial/starting state, q0 is in Q qn, qy Halt states, which are in Q δ A next-move function, which is a mapping (i.e., may be undefined) from (Q - {qn, qy}) x Γ –> Q x Γ x {-1,+1} Intuitively, δ(q,s) specifies the next state, symbol to be written, and the direction of tape head movement by M after reading symbol s while in state q.
DTM Definitions • Let M be a DTM. • M acceptsx Σ* IFF M halts in qy when give x as input. • The language recognized by M is: LM = { x | x Σ* and M accepts x} • M is an algorithm IFF it always halts • M solves a decision problem under encoding scheme e if M always halts and LM = L[,e]
DTM Definitions, cont. • The time used by M on x Σ*is the number of steps that occur up until a halt state is entered. • In what sense does this represent “time?” • Suppose M always halts. Then the time complexity function TM : Z+ Z+ is denoted by TM(n) • See page 26 • M is a polynomial time DTM program if there exists a polynomial p such that: TM(n) <= p(n) for all n Z+
Definition of P • Definition: P = { L : There exists a polynomial-time DTM M such that L = LM } • If L[, e] P then we say that belongs to P under e. • What are the implications for if L[, e] P? • The problem can be solved in polynomial time
Non-DeterministicOne-Tape TM (Version #1) • A NDTM is a seven-tuple: M = (Q, Σ, Γ, δ, q0, B, F) Q A finite set of states Γ A finite tape alphabet B A distinguished blank symbol, which is in Γ Σ A finite input alphabet, which is a subset of Γ– {B} q0 The initial/starting state, q0 is in Q qn, qy Halt states, which are in Q δ A next-move function, which is a mapping (i.e., may be undefined) from (Q - {qn, qy}) x Γ –> finite subsets of Q x Γ x {-1,+1} Intuitively, δ(q,s) specifies zero or more options, each of which includes the next state, symbol to be written, and the direction of tape head movement by M after reading symbol s while in state q.
NDTM Definitions • Let M be a NDTM. • M acceptsx Σ* IFF there exists some sequence of transitions that leads M to (halt in) qy. • The language recognized by M is: LM = { x | x Σ* and M accepts x} • M solves a decision problem under encoding scheme e if LM = L[, e] • Why is there no requirement, as there was with a DTM, for the NDTM to always halt?
NDTM Definitions • Why is there no requirement, as there is with a DTM, for a NDTM to always halt? • See slide 1.15 • Why is there no definition for the notion of a non-deterministic algorithm? • Although not stated explicitly, every NDTM is a non-deterministic algorithm • But under that definition, a non-deterministic algorithm for some inputs might only go into infinite loops…why do we tolerate this? • Basically, for convenience, generality and simplicity • Some authors do require all computations to terminate • Others require all non-accepting computations to enter an infinite loop
DTMs vs. NDTMs • A DTM is both a language defining device and an intuitive model of computation. • A computer program can solve a problem in polynomial time IFF and DTM can. • Anything computable by a DTM is computable by a computer program. • A NDTM is a language defining device, but not so much a good intuitive model of computation, i.e., it is not clear on the surface that anything “doable” by a NDTM is “doable” by a computer program. • In fact, the first is NOT clear at all! • How a NDTM “computes” is not even clear, at least in a real-world sense. => For out purposes, it is not as important to think of a NDTM as a “practical” computing device, so much as a language defining device.
NDTM Definitions, cont. • The time used by M to accept x LMis: min{ i | M accepts x in i transitions} • Then the time complexity function TM : Z+ Z+ is denoted by TM(n) • See page 31 • M is a polynomial time NDTM program if there exists a polynomial p such that: TM(n) <= p(n) for all n >= 1.
NDTM Definitions, cont. • Note that the time complexity function only depends on accepting computations. • In fact, if no inputs of a particular size n are accepted, the complexity function is 1, even if some computations go into an infinite loop. • Why? Aren’t we concerned about the time required for non-accepted strings? • The time complexity function could be modified to reflect all inputs • We could even require every path in the NDTM to terminate (some authors do this) • Neither of these changes would ultimately matter (I.e., we could convert a NDTM with the first type of polynomial complexity bound to a NDTM with the second type of polynomial complexity bound and visa versa) • The books model of a NDTM allows infinite computations for convenience - its easy to describe their version of a NDTM, and showing problems are in NP. • To accommodate this, the time complexity function only reflects accepting computations • Note that we could have defined P based only on accepting computations. • Altough equivalent, it would be less intuitive (some authors do this)
Definition of NP • Definition: NP = { L : There exists a polynomial-time NDTM program M such that L = LM } • If L[, e] NP then we say that belongs to NP under e. • How would one show that a language is in NP? • Give a polynomial time NDTM for it (wow!) • What are the implications for if L[, e] NP? • How useful or important are problems in NP? • Not as clear in this case…
DTMs vs. NDTMs Lemma: Every polynomial-time DTM is a polynomial-time NDTM. Proof: • Observe that every DTM is a NDTM • If DTM runs in TM(n) deterministic time, then it runs in TM(n) non-deterministic time, by definition (the reader can verify that the latter is less than the former, i.e., TNDM(n) <= TdM(n). • Therefore, if a DTM runs in deterministic polynomial time, then it runs in non-deterministic polynomial time. Theorem: P NP Proof: • Suppose L P • By definition, there exists a polynomial-time DTM M such that L = LM • By the lemma, M is also a polynomial-time NDTM such that L = LM • Hence, L NP
DTMs vs. NDTMs Theorem: If L NP then there exists a polynomial p and a DTM M such that L = LM and TM(n) <= O(2p(n)). Proof: • Let M be a NDTM where L = LM • Let q(n) be a polynomial-time bound on M • Let r be the maximum number of options provided by δ, i.e: r = max{ |S| : q Q, s Γ, and δ(q,s) = S } Proof: If x LM and |x| = n, then • There is a “road map” of length <= q(n) for getting to qy • There are rq(n) road maps of length <= q(n) • A (3-tape) DTM M’ generates and tries each road map • M’ operates in (approximately) O(q(n)rq(n)) time, which is O(2p(n)) for some p(n).
DTMs vs. NDTMs • Question: • Could this DTM simulation of a NDTM be improved to run in polynomial time? • Only if P = NP • Observations: • The preceding proof differs slightly from that given in the book, due to the different NDTM model • Non-deterministic algorithms are starting to look like they might be useful for realistic problems, i.e., those using deterministic exponential time algorithms. • Note the “exhaustive search” nature of the simulation • For many problem in NP, exhaustive search is the only non-deterministic algorithm • Additional question: • Why didn’t we just define P as problems solvable by polynomial time DTMS, and NP as problems solvable by exponential time DTMS, and then just observe that P is a subset of NP….hmmm….requires more thought…
Non-DeterministicOne-Tape TM (Version #2) • A NDTM is a seven-tuple: M = (Q, Σ, Γ, δ, q0, B, F) Q A finite set of states Γ A finite tape alphabet B A distinguished blank symbol, which is in Γ Σ A finite input alphabet, which is a subset of Γ– {B} q1 The initial/starting state, q0 is in Q qn, qy Halt states, which are in Q δ A next-move function, which is a mapping (i.e., may be undefined) from (Q - {qn, qy}) x Γ –> finite subsets of Q x Γ x {-1,+1} Exactly as before, however,…(this will make clear how NP problems are important)
write head read/write head Guessing Module Finite Control NDTM (Version #2) -3 -2 -1 0 1 2 3 …….. …….. • Just like a DTM, but with an additional non-deterministic guessing module • Initial configuration: • input string x is placed in squares 1 through |x| • write-only head is scanning square -1, read-write head is scanning square 1 • Computation on x takes place in two separate, distinct stages • Guessing stage: • at each step, the guessing module directs the write-only head to either write a symbol from Γ and move one square to the left, or to stop • guessing module decides when (or if!) to stop • Checking stage: • if the guessing module stops, then control is transferred to q0 the start state of the finite control • the finite control, which is deterministic, then computes using the same rules as a DTM • The finite control either halts and accepts, halts and rejects, or loops infinitely
NDTM (Version #2) Notes: • The guessing module can guess any string from Γ* • Hmmm…so what exactly does a guessing module look like? • The guessing module allows for an infinite # of computations • This explains the need for the time complexity function to only consider accepting computations. • We could put a bound on the guess to eliminate this, but its easier to describe the NDTM without it, and it doesn’t matter anyway (class NP stays the same). • The guessed string is usually used in the second stage • For a given input string x, there are an infinite number of computations of a NDTM
NDTM (Version #2) • M accepts x Σ*if there exists a guess x Γ*that brings M to qy • Note that this is basically the same as our our definition of string acceptance with Version #1 • The time to accept x Σ* … same as before (see slide 1.21) • All definitions and theorems proved about the first verson of NDTMs still hold: • LM, TM(n), NP • P NP (verify this one) • If L NP then there exists a polynomial p and a DTM such that L = LM and TM(n) <= O(2p(n))
NDTM (Version #2) • Are the two different versions of NDTMs equivalent? • There exists a NDTM version #1 that accepts L IFF there exist a NDTM version #2 that accepts L • Prove this as an exercise • But are the models equivalent with respect to time complexity? • Do the two models define the same class NP? • Yes! • Why do Garey and Johnson use Version #2? • Convenience • At the problem level, it is easer to show a problem (or language) is in NP • How does one show a problem (or language) is in P? • How does one show a problem (or language) is in NP?
Proving a Problem NP-Complete(overview) • To show a problem is NP-complete we must show: • NP • ’ , for all ’ NP • To show NP we need to give a NDTM, or rather, a non-deterministic algorithm for , that operates in polynomial time. • For most decision problems, Garey and Johnson’s NDTM model makes this easy.
Proving a Problem is in NP Another way to look at model #2: • NP IFF there exits a deterministic polynomial time algorithm A such that: • If I Y then there exists a “structure” S such that (I,S) is accepted by A • If I Y then for any “structure” S, it will always be the case that (I,S) is rejected by A
Example #1: TSP • Recall TSP: TRAVELING SALESMAN INSTANCE: Set C of m cities, distance d(ci, cj) Z+ for each pair of cities ci, cj C positive integer B. QUESTION: Is there a tour of C having length B or less, I.e., a permutation <c(1) , c(2),…, c(m)> of C such that: • A Non-Deterministic Polynomial Time Algorithm for TSP: • Guess (non-deterministically, in polynomial-time) a permutation of the cities. • Verify (deterministically, in polynomial-time) that the permutation is a valid tour of the cities having total length less than B (output “yes,” output “no” otherwise). • From this it follows that TSP NP
Example #2: Clique • Recall Clique: CLIQUE INSTANCE: A Graph G = (V, E) and a positive integer J <= |V|. QUESTION: Does G contain a clique of size J or more? • A Non-Deterministic Polynomial Time Algorithm for Clique: • Guess (non-deterministically, in polynomial-time) a set of vertices from G. • Verify (deterministically, in polynomial-time) that the set is a clique of size J or more. • From this it follows that Clique NP
Example #3: Graph K-Colorability • Recall Graph K-Colorability: GRAPH K-COLORABILITY INSTANCE: A Graph G = (V, E) and a positive integer K <= |V|. QUESTION: Is the graph GK-colorable? • A Non-Deterministic Algorithm Polynomial Time for Graph K-colorability: • Guess (non-deterministically, in polynomial-time) a coloring of G. • Verify (deterministically, in polynomial-time) that the coloring is a valid K-coloring of G. • From this it follows that Graph K-colorability NP
Example #4: SAT • Recall Satisfiability: SATISFIABILITY INSTANCE: Set U of variables and a collection C of clauses over U. QUESTION: Is there a satisfying truth assignment for C? • A Non-Deterministic Polynomial Time Algorithm for Satisfiability: • Guess (non-deterministically, in polynomial-time) a truth assignment for the variables in U. • Verify (deterministically, in polynomial-time) that the truth assignment satisfies C. • From this it follows that Satisfiability NP
Example #5: Non-Tautology NON-TAUTOLOGY INSTANCE: Boolean expression E (, , , ). QUESTION: Is E not a tautology? • A Non-Deterministic Polynomial Time Algorithm for Non-Tautology: • Guess (non-deterministically, in polynomial-time) a truth assignment for E. • Verify (deterministically, in polynomial-time) that the truth assignment does not satisfy E. • From this it follows that Non-Tautology NP
Example #6: Tautology TAUTOLOGY INSTANCE: Boolean expression E (, , , ). QUESTION: Is E a tautology? • A Non-Deterministic Polynomial Time Algorithm for Tautology: • Guess ? • Verify ? • Is Tautology NP? • We do not know!
Tentative View of the World of NP • If P = NP then everything in NP can be solved in deterministic polynomial time. • If P≠ NP then: • P NP • Everything in P can be solved in deterministic polynomial time • Everything in NP - P is intractable • Hence, given a problem we would like to know if P or NP - P • Unfortunately, we do not know if P = NP or P≠ NP • It does appear that P≠ NP since there are many problem in NP for which nobody has been able to find deterministic polynomial time algorithms
Tentative View of the World of NP • For some problems we can show that P • Unfortunately, we cannot not show that NP - P, at least not without resolving the P =? NP question, which appears to be very difficult • Hence, we will restrict ourselves to showing that for some problems, if P≠ NP then NP - P • Although the theory of NP-completeness assumes that P≠ NP, if it turned out that P = NP, then parts of the theory (i.e., transformations and reductions) can still be used to prove results concerning the relative complexity of problems.
Polynomial Transformation • A polynomial transformation from a language L1 Σ1*and to a language L2 Σ2* is a function f: Σ1* Σ2* that statisfies the following conditions: • There is a polynomial time DTM program that computes f • For all x Σ1*, x L1*iff and only if f(x) L2* • If there is a polynomial transformation from L1 to L2 then we write L1 L2, read “L1 transforms to L2“
Polynomial Transformation • The significance of polynomial transformations follows from Lemma 2.1 Lemma 2.1: If L1 L2 and L2 P then L1 P Proof: Let M be a polynomial time DTM that decides L2 , and let M’ be a polynomial time DTM that computes f. Then M’ and M can be composed to give a polynomial time DTM that decides . • Equivalently, if L1 L2 and L1 P then L2 P
Polynomial Transformation • Informally, if L1 L2 then it is frequently said that: • L1 Is no harder than L2 • L2Is no easier than L1 • L2Is at least as hard as L1 • L1 Is as easy as L2 • Note that “hard” and “easy” here are only with respect to polynomial v.s. exponential time • We could have L1 decidable in 2ntime, L2 decidable in time, but where L1 is not.
Polynomial Transformation • At the problem level we can regard a polynomial transformation from the decision problem 1 Σ1*and to a decision problem 2 Σ2* is a function f : D1 D2that statisfies the following conditions: • f is computable by a polynomial time algorithm • For all I D1, I Y1 if and only if f ( I) Y2 • We will deal with transformations only at the problem level
Polynomial Transformation • Another helpful factoid concerning transformations: Lemma 2.2: If L1 L2 and L2 L3 then L1 L3 • In other words, polynomial transformations are transitive.
Definition of NP-Complete • A decision problem is NP-complete if: • NP, and • For all other decision problems ’ NP, ’ • For many problems, proving the first requirement is easy. • On the surface, the second requirement would appear very difficult to prove, especially given that there are an infinite number of problems in NP. • As will be shown, the second requirement will be established using the following lemma: Lemma 2.3: If L1and L2 belong to NP, L1 is NP-complete, andL1 L2 then L2 is NP-complete.
Definition of NP-Complete, cont. • Let be an NP-complete problem • Observation: • P If and only if P = NP (only if) • If an NP-complete problem can be solved in polynomial time, then every problem in NP can be solved in polynomial time (should be clear, by definition). • If P then P = NP • If any problem in NP is intractable, then so are all NP-complete problems. • If P≠ NP then P (i.e., NP-P), for all NP-complete problems
Definition of NP-Complete, cont. • Let be an NP-complete problem • Observation: • P If and only if P = NP (if) • If every problem in NP can be solved in polynomial time, then clearly every NP-complete problem can be also, since every NP-complete problem, by definition, is in NP. • If P = NP then P, for all NP-complete problems • If any NP-complete problem in NP is intractable, then clearly P ≠ NP. • If P (i.e., NP-P) then P ≠ NP, for any NP-complete problem
