DOCSLIB.ORG

Lecture 4: Space and Time

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Lecture 4: Space and Time Computational Complexity Theory, Fall 2010 September 3 Lecture 4: Space and Time Lecturer: Kristoffer Arnsfelt Hansen Scribe: Camilla Gravgaard How to halt a space bounded TM Definition 1 (Configuration). A configuration of a Turing machine consists of • Internal state. • Positions of all tape heads. • Content of all writable tapes. Theorem 2. Suppose S(n) ≥ log(n). Let M be an S(n) space-bounded TM. Then there is a O(S(n)) space bounded TM M 0 that always halts with L(M) = L(M 0). Proof. First we give an upper bound on the number of possible configurations of a TM M. • Number of configurations of M ≤ jQj · (n + 2)S(n) · jΓjS(n) • In general, number of configurations of M with using less than k cells is at most k jQj · (n + 2) · k · jΓj ≤ alog2(n)+k+b for some constants a and b. Operation of M 0 : • Keeping a counter in a-ary, we simulate of M, and increase the counter with each step of M. • Whenever M uses a new cell on the work tape for the first time, we reset the counter to 0. • After each simulated step, we see if the counter uses more than log2(n) + k + b digits, where k is the number of worktape cells currently in use, then halt and reject. By our calculation, if the counter in fact uses more than log2(n) + k + b digits, then we know that the machine has repeated a configuration, and has thus entered an infinite loop. We can thus safely reject. Observation 3. The above proof does not work for sublogarithmic space-bounds. The result, how- ever, still remains true, by a result of Sipser (1978). Definition 4 (Space constructability). We say S(n) is space constructable iff there is a O(S(n)) space-bounded Turing machine that on input 1n halts, leaving the string 1S(n) on the work tape. Definition 5 (Complexity classes). For any function S(n) we define the following complexity classes. • DSPACE(S(n)) = class of language accepted by a O(S(n)) space bounded Turing machine. • NSPACE(S(n)) = class of language accepted by a O(S(n)) space bounded nondeternimistic Tuting machine. 1 The Space Hierarchy theorem Theorem 6. Let S2(n) ≥ log(n) be space-constructable. If S1(n) = o(S2(n)), then DSPACE(S1(n)) ( DSPACE(S2(n)) Proof. We construct TM M, such that L(M) 2 DSPACE(S2(n)) n DSPACE(S1(n)). On input x, jxj = n, the machine operates as follows. 1. Use space constructibility of S2 to measure out S2(n) cells of worktape. We use this in the following ways: • enforce that the remaining computation stays within the S2(n) cells. Halt and reject if simulation exceeds the bound • Use it for a binay counter to run the remaining computation for at most 2log2(n)+S2(n) steps. If time runs out halt and accept. 2. Check if x = 0i1w, where w is a encoding of a TM with read-only input and a single worktape. If not, halt and reject. 3. Simulate M 0 on input x. If M 0 accepts, then reject. If M 0 reject, then accept. Analysis: 0 Clearly M is O(S2(n)) space bounded, by construction. Assume now that M is a O(S1(n)) space bounded TM. Let w be an encoding of M 0 and look at the computation of M with inputs of the form x = 0i1w. 0 Then M will simulate M on input x. Since S1(n)=o(S2(n)) then simulation will not ex- 0 log (n)+O(S (n)) ceed spece bound for all sufficiently large i. M has 2 2 1 configurations. Since S1(n) log (n)+S (n) =o(S2(n)), for sufficiently large i this is less than 2 2 2 . Thus for all sufficiently large i, if the simulation runs for 2log2(n)+S2(n) steps, we know for sure that M 0 has entered an infinite loop. In this case we halt and accept. If M 0 halts, M 0 will give the opposite answer of M. Thus in conlusion, for all sufficiently large i, M and M 0 differ on input 0i1w. The reachability method Theorem 7. For T (n) ≥ n, S(n) ≥ log(n), (a) NTIME(T (n)) ⊆ DSPACE(T (n)), (b) NSPACE(S(n)) ⊆ DTIME(2O(S(n))). Proof. a) Observe first that DTIME(T (n)) ⊆ DSPACE(T (n)), since the Turing machine cannot use more tape cells than the number of computation steps. Now, assume first T (n) is time constructible, and let M be a T (n) time bounded TM. Then we first measure out T (n) tape cells. Let c be a constant such that M has at most c nondeterministic choices at every step. We then iterate through all strings in f1; : : : ; cgT (n) on the work tape, reusing space each time. For each string we simulate M using the string as the nondeterministic choices. If M accepts, then accept. Otherwise continue with the next string. If M did not accept during any of these runs, we reject. The space usage is clearly O(T (n)). 2 To get rid of the time constructibility assumption, we can simply try T = 1;T = 2;T = 3;::: , as a substitute for T (n) until all simulations of length T suceeds. Alternative view: We can see the computation of a non-deterministic Turing machine as a tree of depth T (n), and we are basically doing a depth first search in the tree searching for an accepting configuration at a leaf. b) Recall that a configuration of a Turing machine consists of the following information: The internal state, position of all tape heads, and content of all writable tapes. If M is S(n) ≥ log(n) space bounded then there are 2O(S(n)) configurations, and we can encode these by strings of length O(S(n)). We now define the configuration graph of M as the directed graph G = (V; E) where V is all the configurations and there is directed edges E between configurations if the machine can move between them in one step. Now, assume that S(n) is space-constructible, then G can be generated in time 2O(S(n)). We need to check if there is a path from the initial configuration to an accepting configuration doing graph traversing (for ex. using depth first search). This takes polynomial time in the size of G, and therefore the total time usage is 2O(S(n)). To get rid of the space-constructibility assumption, we run the algorithm for S = 0; 1; 2;::: , restricting to configurations using S cells of work tape, until it works. That is, until for all configurations generated, M does not move to a configuration that uses more than S cells of work tape. The total time for doing this remains 2O(S(n)). Now we have a hierarchy of classes: L = DSPACE(log(n)) ⊆ b NL = NSPACE(log(n)) ⊆ PTIME = DTIME(nO(1)) ⊆ a NP = NTIME(nO(1)) ⊆ b PSPACE = DSPACE(nO(1)) ⊆ O(1) EXP = DTIME(2n ) ⊆ O(1) a NEXP = NTIME(2n ) ⊆ O(1) EXPSPACE = DSPACE(2n ): Savitch's Theorem Theorem 8. Let S(n) ≥ log(n): Then NSPACE(S(n)) ⊆ DSPACE(S(n)2): 3 Proof. We want to find a path in the configuration graph of M from the initial configuration s to an accepting configuration t. Now assume that S(n) is space constructible and encode the configurations by strings of length O(S(n)). We define the predicate reach(u; v; i) = Yes iff there is a path from u to v in at most 2i steps. We now want to compute reach(s; t; c · S(n)) for some constant c. Now note that we can write: reach(u; v; i) = Yes iff 9w : reach(u; w; i − 1) = Yes ^ reach(w; v; i − 1) = Yes: The implementation of reach by a recursive algorithm has the idea of reusing space. To do the recursion, we run over all string w encoding configurations and call recusively twice. Doing this we get O(S(n)) levels of recursion. Doing this, in each level we only need to store 3 configurations u, v and w and the depth i, taking space O(S(n)). The space used is then: S(n) levels of recursion times S(n) space which is a total of O(S(n)2) space usage. Again, we can get rid of the assumption of space-constructibility, by trying all S = 0; 1; ::: until the computation is successful. 4.
Load more

Recommended publications
  • CS601 DTIME and DSPACE Lecture 5 Time and Space Functions: T, S
    CS601 DTIME and DSPACE Lecture 5 Time and Space functions: t, s : N → N+ Definition 5.1 A set A ⊆ U is in DTIME[t(n)] iff there exists a deterministic, multi-tape TM, M, and a constant c, such that, 1. A = L(M) ≡ w ∈ U M(w)=1 , and 2. ∀w ∈ U, M(w) halts within c · t(|w|) steps. Definition 5.2 A set A ⊆ U is in DSPACE[s(n)] iff there exists a deterministic, multi-tape TM, M, and a constant c, such that, 1. A = L(M), and 2. ∀w ∈ U, M(w) uses at most c · s(|w|) work-tape cells. (Input tape is “read-only” and not counted as space used.) Example: PALINDROMES ∈ DTIME[n], DSPACE[n]. In fact, PALINDROMES ∈ DSPACE[log n]. [Exercise] 1 CS601 F(DTIME) and F(DSPACE) Lecture 5 Definition 5.3 f : U → U is in F (DTIME[t(n)]) iff there exists a deterministic, multi-tape TM, M, and a constant c, such that, 1. f = M(·); 2. ∀w ∈ U, M(w) halts within c · t(|w|) steps; 3. |f(w)|≤|w|O(1), i.e., f is polynomially bounded. Definition 5.4 f : U → U is in F (DSPACE[s(n)]) iff there exists a deterministic, multi-tape TM, M, and a constant c, such that, 1. f = M(·); 2. ∀w ∈ U, M(w) uses at most c · s(|w|) work-tape cells; 3. |f(w)|≤|w|O(1), i.e., f is polynomially bounded. (Input tape is “read-only”; Output tape is “write-only”.
    [Show full text]
  • Interactive Proof Systems and Alternating Time-Space Complexity
    Theoretical Computer Science 113 (1993) 55-73 55 Elsevier Interactive proof systems and alternating time-space complexity Lance Fortnow” and Carsten Lund** Department of Computer Science, Unicersity of Chicago. 1100 E. 58th Street, Chicago, IL 40637, USA Abstract Fortnow, L. and C. Lund, Interactive proof systems and alternating time-space complexity, Theoretical Computer Science 113 (1993) 55-73. We show a rough equivalence between alternating time-space complexity and a public-coin interactive proof system with the verifier having a polynomial-related time-space complexity. Special cases include the following: . All of NC has interactive proofs, with a log-space polynomial-time public-coin verifier vastly improving the best previous lower bound of LOGCFL for this model (Fortnow and Sipser, 1988). All languages in P have interactive proofs with a polynomial-time public-coin verifier using o(log’ n) space. l All exponential-time languages have interactive proof systems with public-coin polynomial-space exponential-time verifiers. To achieve better bounds, we show how to reduce a k-tape alternating Turing machine to a l-tape alternating Turing machine with only a constant factor increase in time and space. 1. Introduction In 1981, Chandra et al. [4] introduced alternating Turing machines, an extension of nondeterministic computation where the Turing machine can make both existential and universal moves. In 1985, Goldwasser et al. [lo] and Babai [l] introduced interactive proof systems, an extension of nondeterministic computation consisting of two players, an infinitely powerful prover and a probabilistic polynomial-time verifier. The prover will try to convince the verifier of the validity of some statement.
    [Show full text]
  • Arxiv:Quant-Ph/0202111V1 20 Feb 2002
    Quantum statistical zero-knowledge John Watrous Department of Computer Science University of Calgary Calgary, Alberta, Canada [email protected] February 19, 2002 Abstract In this paper we propose a definition for (honest verifier) quantum statistical zero-knowledge interactive proof systems and study the resulting complexity class, which we denote QSZK. We prove several facts regarding this class: The following natural problem is a complete promise problem for QSZK: given instructions • for preparing two mixed quantum states, are the states close together or far apart in the trace norm metric? By instructions for preparing a mixed quantum state we mean the description of a quantum circuit that produces the mixed state on some specified subset of its qubits, assuming all qubits are initially in the 0 state. This problem is a quantum generalization of the complete promise problem of Sahai| i and Vadhan [33] for (classical) statistical zero-knowledge. QSZK is closed under complement. • QSZK PSPACE. (At present it is not known if arbitrary quantum interactive proof • systems⊆ can be simulated in PSPACE, even for one-round proof systems.) Any honest verifier quantum statistical zero-knowledge proof system can be parallelized to • a two-message (i.e., one-round) honest verifier quantum statistical zero-knowledge proof arXiv:quant-ph/0202111v1 20 Feb 2002 system. (For arbitrary quantum interactive proof systems it is known how to parallelize to three messages, but not two.) Moreover, the one-round proof system can be taken to be such that the prover sends only one qubit to the verifier in order to achieve completeness and soundness error exponentially close to 0 and 1/2, respectively.
    [Show full text]
  • Spring 2020 (Provide Proofs for All Answers) (120 Points Total)
    Complexity Qual Spring 2020 (provide proofs for all answers) (120 Points Total) 1 Problem 1: Formal Languages (30 Points Total): For each of the following languages, state and prove whether: (i) the lan- guage is regular, (ii) the language is context-free but not regular, or (iii) the language is non-context-free. n n Problem 1a (10 Points): La = f0 1 jn ≥ 0g (that is, the set of all binary strings consisting of 0's and then 1's where the number of 0's and 1's are equal). ∗ Problem 1b (10 Points): Lb = fw 2 f0; 1g jw has an even number of 0's and at least two 1'sg (that is, the set of all binary strings with an even number, including zero, of 0's and at least two 1's). n 2n 3n Problem 1c (10 Points): Lc = f0 #0 #0 jn ≥ 0g (that is, the set of all strings of the form 0 repeated n times, followed by the # symbol, followed by 0 repeated 2n times, followed by the # symbol, followed by 0 repeated 3n times). Problem 2: NP-Hardness (30 Points Total): There are a set of n people who are the vertices of an undirected graph G. There's an undirected edge between two people if they are enemies. Problem 2a (15 Points): The people must be assigned to the seats around a single large circular table. There are exactly as many seats as people (both n). We would like to make sure that nobody ever sits next to one of their enemies.
    [Show full text]
  • Some Hardness Escalation Results in Computational Complexity Theory Pritish Kamath
    Some Hardness Escalation Results in Computational Complexity Theory by Pritish Kamath B.Tech. Indian Institute of Technology Bombay (2012) S.M. Massachusetts Institute of Technology (2015) Submitted to Department of Electrical Engineering and Computer Science in partial fulfillment of the requirements for the degree of Doctor of Philosophy in Electrical Engineering & Computer Science at Massachusetts Institute of Technology February 2020 ⃝c Massachusetts Institute of Technology 2019. All rights reserved. Author: ............................................................. Department of Electrical Engineering and Computer Science September 16, 2019 Certified by: ............................................................. Ronitt Rubinfeld Professor of Electrical Engineering and Computer Science, MIT Thesis Supervisor Certified by: ............................................................. Madhu Sudan Gordon McKay Professor of Computer Science, Harvard University Thesis Supervisor Accepted by: ............................................................. Leslie A. Kolodziejski Professor of Electrical Engineering and Computer Science, MIT Chair, Department Committee on Graduate Students Some Hardness Escalation Results in Computational Complexity Theory by Pritish Kamath Submitted to Department of Electrical Engineering and Computer Science on September 16, 2019, in partial fulfillment of the requirements for the degree of Doctor of Philosophy in Computer Science & Engineering Abstract In this thesis, we prove new hardness escalation results in computational complexity theory; a phenomenon where hardness results against seemingly weak models of computation for any problem can be lifted, in a black box manner, to much stronger models of computation by considering a simple gadget composed version of the original problem. For any unsatisfiable CNF formula F that is hard to refute in the Resolution proof system, we show that a gadget-composed version of F is hard to refute in any proof system whose lines are computed by efficient communication protocols.
    [Show full text]
  • Computational Complexity Theory
    Computational Complexity Theory Markus Bl¨aser Universit¨atdes Saarlandes Draft|February 5, 2012 and forever 2 1 Simple lower bounds and gaps Lower bounds The hierarchy theorems of the previous chapter assure that there is, e.g., a language L 2 DTime(n6) that is not in DTime(n3). But this language is not natural.a But, for instance, we do not know how to show that 3SAT 2= DTime(n3). (Even worse, we do not know whether this is true.) The best we can show is that 3SAT cannot be decided by a O(n1:81) time bounded and simultaneously no(1) space bounded deterministic Turing machine. aThis, of course, depends on your interpretation of \natural" . In this chapter, we prove some simple lower bounds. The bounds in this section will be shown for natural problems. Furthermore, these bounds are unconditional. While showing the NP-hardness of some problem can be viewed as a lower bound, this bound relies on the assumption that P 6= NP. However, the bounds in this chapter will be rather weak. 1.1 A logarithmic space bound n n Let LEN = fa b j n 2 Ng. LEN is the language of all words that consists of a sequence of as followed by a sequence of b of equal length. This language is one of the examples for a context-free language that is not regular. We will show that LEN can be decided with logarithmic space and that this amount of space is also necessary. The first part is easy. Exercise 1.1 Prove: LEN 2 DSpace(log).
    [Show full text]
  • Computational Complexity
    Computational complexity Plan Complexity of a computation. Complexity classes DTIME(T (n)). Relations between complexity classes. Complete problems. Domino problems. NP-completeness. Complete problems for other classes Alternating machines. – p.1/17 Complexity of a computation A machine M is T (n) time bounded if for every n and word w of size n, every computation of M has length at most T (n). A machine M is S(n) space bounded if for every n and word w of size n, every computation of M uses at most S(n) cells of the working tape. Fact: If M is time or space bounded then L(M) is recursive. If L is recursive then there is a time and space bounded machine recognizing L. DTIME(T (n)) = fL(M) : M is det. and T (n) time boundedg NTIME(T (n)) = fL(M) : M is T (n) time boundedg DSPACE(S(n)) = fL(M) : M is det. and S(n) space boundedg NSPACE(S(n)) = fL(M) : M is S(n) space boundedg . – p.2/17 Hierarchy theorems A function S(n) is space constructible iff there is S(n)-bounded machine that given w writes aS(jwj) on the tape and stops. A function T (n) is time constructible iff there is a machine that on a word of size n makes exactly T (n) steps. Thm: Let S2(n) be a space-constructible function and let S1(n) ≥ log(n). If S1(n) lim infn!1 = 0 S2(n) then DSPACE(S2(n)) − DSPACE(S1(n)) is nonempty. Thm: Let T2(n) be a time-constructible function and let T1(n) log(T1(n)) lim infn!1 = 0 T2(n) then DTIME(T2(n)) − DTIME(T1(n)) is nonempty.
    [Show full text]
  • DTIME, NTIME, DSPACE, Measured on Multi-Tape Turing Machines. Th
    CMPSCI 601: Recall From Last Time Lecture 9 Def: DTIME, NTIME, DSPACE, measured on Multi-tape Turing Machines. ¡¤£¥§ © £¡¤£¥§§© Th: DTIME ¢¡¤£¦¥¨§ © RAM-TIME DTIME ¥© L DSPACE + ¥! #"%$'&(© ¥ © ) * P DTIME +-, DTIME $ + #"%$'& ¥ © ¥ © * ) NP NTIME +-, NTIME $ + ¥! ."/$ &0© ¢¥ © ) * PSPACE DSPACE +-, DSPACE $ ¥43657£¦¥§81 ¥ Th: For ¡¤£¥§21 , ¢¡¤£¦¥§:© ¡9£¦¥§ © DTIME ¡9£¦¥§ © NTIME DSPACE #">"@?9&%& =< © DSPACE ;57£¦¥§:© DTIME Cor: L P NP PSPACE 1 CMPSCI 601: NTIME and NP Lecture 9 ¢¡¤£¦¥¨§ © £¦¡¤£¦¥§§ NTIME probs. accepted by NTMs in time + ."/$ & ¢¥ © ¥ © ) * NP NTIME +-, NTIME $ Theorem 9.1 For any function ¡¤£¦¥§ , ¢¡¤£¦¥§:© ¡9£¦¥§ © DTIME ¡9£¦¥§ © NTIME DSPACE Proof: The first inclusion is obvious. For the second, £¡¤£¥§§ note that in space we can simulate all computa- £¡¤£¦¥¨§§ tions of length , so we will find the shortest ac- cepting one if it exists. ¡ =< #"£¢ " ?9&%&(© Recall: DSPACE ¡¤£¥§ © DTIME Corollary 9.2 L P NP PSPACE 2 So we can simulate NTM’s by DTM’s, at the cost of an exponential increase in the running time. It may be pos- sible to improve this simulation, though no essentially better one is known. If the cost could be reduced to poly- nomial, we would have that P NP. There is probably such a quantitative difference between the power of NTM’s and DTM’s. But note that qualita- tively there is no difference. If ¡ is the language of some NTM ¢ , it must also be r.e. because there is a DTM that £ searches through all computations of ¢ on , first of one ¡ ¤ ¥ step, then of two steps, and so on. If £ , will eventually find an accepting computation. If not, it will search forever. What about an NTM-based definition of “recursive” or “Turing-decidable” sets? This is less clear because NTM’s don’t decide – they just have a range of possible actions.
    [Show full text]
  • Advanced Complexity
    Advanced Complexity TD n◦3 Charlie Jacomme September 27, 2017 Exercise 1 : Space hierarchy theorem Using a diagonal argument, prove that for two space-constructible functions f and g such that f(n) = o(g(n)) (and as always f; g ≥ log) we have SPACE(f(n)) ( SPACE(g(n)). Solution: We dene a language which can be recognized using space O(g(n)) but not in f(n). L = f(M; w)jM reject w using space ≤ g(j(M; w)j andjΓj = 4g Where Γis the alphabet of the Turing Machine We show that L 2 SPACE(g(n)) by constructing the corresponding TM. This is where we need the fact that the alphabet is bounded. Indeed, we want to construct one xed machine that recognizes L for any TM M, and if we allow M to have an arbitrary size of alphabet, the xed machine might need a lot of space in order to represent the alphabet of M, and it might go over O(g(n)). On an input x, we compute g(x) and mark down an end of tape marker at position f(x), so that we reject if we use to much space. If x is not of the form (M; w), we reject, else we simulate M on w for at most jQj × 4g(x) × n steps. If we go over the timeout, we reject. Else, if w is accepted, we reject, and if w is rejected, we accept. This can be done in SPACE(O(g(n))), and we conclude with the speed up theorem.
    [Show full text]
  • A Study of Statistical Zero-Knowledge Proofs Salil Pravin
    A Study of Statistical Zero-Knowledge Proofs by Salil Pravin Vadhan A.B., Mathematics and Computer Science, Harvard University (1995) Certificate of Advanced Study in Mathematics, Cambridge University (1996) Submitted to the Department of Mathematics in partial fulfillment of the requirements for the degree of Doctor of Philosophy at the MASSACHUSETTS INSTITUTE OF TECHNOLOGY September 1999 @ Salil Pravin Vadhan, 1999. All rights reserved. The author hereby grants to MIT permission to reproduce and distribute publicly paper and electronic copies of this thesis document in whole or in part. A uthor ........... .'v --~-- g -. - --- ---. - .---------------------------- Dtopartment of Mathematics August 6, 1999 ,/1 I; Certified by ....... .. .. - .. -. .... - ...- .. .-- fi Gold asser RSA Professor of Computer Science Thesis Supervisor .. V... ...... V J .. ................ Accepted by ......... c alS p. e Michael Sipser Chairman, Applied Mathematics Committee A A ccepted by ................. ...--.. - -- -- - Tomasz Mrowka Chairman, Departmental Committee on Graduate Students MASSACHUSETTS INSTITUTE LIBRARIES A Study of Statistical Zero-Knowledge Proofs by Sail Pravin Vadhan Submitted to the Department of Mathematics on August 6, 1999, in partial fulfillment of the requirements for the degree of Doctor of Philosophy Abstract Zero-knowledge interactive proofs, introduced by Goldwasser, Micali, and Rackoff, are fas- cinating constructs which enable one party (the "prover") to convince another party (the "verifier") of an assertion, with the property that the verifier learns nothing other than the fact that the assertion being proven is true. In addition to being powerful tools for construct- ing secure cryptographic protocols, zero-knowledge proofs yield rich classes of computational problems that are of both complexity-theoretic and cryptographic interest. This thesis is a detailed investigation of statisticalzero-knowledge proofs, which are zero- knowledge proofs in which the condition that the verifier "learns nothing" is interpreted in a strong statistical sense.
    [Show full text]
  • The NP-Completeness of Some Lesser Known Logic Puzzles
    The NP-completeness of some lesser known logic puzzles Author: Mieke Maarse - 5750032 Supervisor: Dr. Benjamin Rin Second evaluator: Prof. dr. Michael Moortgat 7.5 ECTS Bachelor Kunstmatige Intelligentie Utrecht University June 28, 2019 1 Abstract Logic puzzles have been gaining popularity and many puzzles have been proved to be NP-complete. When a puzzle is NP-complete, it is not feasible to try to compute a solution. In this case algorithms that approximate the solution, or programs that are limited in input size can be used. In this paper we use the Hamiltonian path and cycle problems in grid graphs, and the Latin square completion problem to show that the puzzles unequal and adjacent, towers, chains and linesweeper are NP- complete. 2 Contents 1 Introduction4 1.1 The Hamiltonian path and cycle problem in grid graphs.....5 1.2 The Latin square completion problem...............6 1.3 Structure of this paper........................6 2 Trivial proofs7 2.1 unequal and adjacent .........................7 2.1.1 NP...............................8 2.1.2 NP-hardness.........................8 2.2 towers .................................8 2.2.1 NP...............................9 2.2.2 NP-hardness.........................9 3 chains 9 3.1 NP-completeness........................... 10 4 linesweeper 13 4.1 NP-completeness........................... 13 5 Conclusion 18 3 1 Introduction Logic puzzles are puzzles that can be solved through deductive reasoning, like sudoku or kakuro. These puzzles are satisfying to solve because, when finished, it is immediately clear if a solution is correct, but solving one is a challenge. This type of puzzle is gaining popularity, and so more research on the topic of these puzzles is being done [1][2].
    [Show full text]
  • Dspace Cover Page
    Efficient Semantic Communication via Compatible Beliefs The MIT Faculty has made this article openly available. Please share how this access benefits you. Your story matters. Citation Juba, Brendan and Madhu Sudan. "Efficient Semantic Communication via Compatible Beliefs." Proceedings of the Second Symposium on Innovations in Computer Science - ICS 2011, Tsinghua University, Beijing, China, January 7-9, 2011. As Published http://conference.itcs.tsinghua.edu.cn/ICS2011/content/ paper/19.pdf Publisher Institute for Computer Science, Tsinghua University Version Author's final manuscript Citable link http://hdl.handle.net/1721.1/64417 Terms of Use Creative Commons Attribution-Noncommercial-Share Alike 3.0 Detailed Terms http://creativecommons.org/licenses/by-nc-sa/3.0/ Efficient Semantic Communication via Compatible Beliefs∗ Brendan Juba Madhu Sudan Harvard University and MIT CSAIL Microsoft Research [email protected] [email protected] October 14, 2010 Abstract In previous works, Juba and Sudan [6] and Goldreich, Juba and Sudan [4] considered the idea of \semantic communication", wherein two players, a user and a server, attempt to communicate with each other without any prior common language (or communication) protocol. They showed that if communication was goal-oriented and the user could sense progress towards the goal (or verify when it has been achieved), then meaningful communication is possible, in that the user's goal can be achieved whenever the server is helpful. A principal criticism of their result has been that it is inefficient: in order to determine the \right" protocol to communicate with the server, the user enumerates protocols and tries them out with the server until it finds one that allows it to achieve its goal.
    [Show full text]