DOCSLIB.ORG

Analysis of Boolean Functions and Its Applications to Topics Such As Property Testing, Voting, Pseudorandomness, Gaussian Geometry and the Hardness of Approximation

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Analysis of Boolean Functions and Its Applications to Topics Such As Property Testing, Voting, Pseudorandomness, Gaussian Geometry and the Hardness of Approximation Analysis of Boolean Functions Notes from a series of lectures by Ryan O’Donnell Guest lecture by Per Austrin Barbados Workshop on Computational Complexity February 26th – March 4th, 2012 Organized by Denis Th´erien Scribe notes by Li-Yang Tan arXiv:1205.0314v1 [cs.CC] 2 May 2012 Contents 1 Linearity testing and Arrow’s theorem 3 1.1 TheFourierexpansion ............................. 3 1.2 Blum-Luby-Rubinfeld. .. .. .. .. .. .. .. .. 7 1.3 Votingandinfluence .............................. 9 1.4 Noise stability and Arrow’s theorem . ..... 12 2 Noise stability and small set expansion 15 2.1 Sheppard’s formula and Stabρ(MAJ)...................... 15 2.2 Thenoisyhypercubegraph. 16 2.3 Bonami’slemma................................. 18 3 KKL and quasirandomness 20 3.1 Smallsetexpansion ............................... 20 3.2 Kahn-Kalai-Linial ............................... 21 3.3 Dictator versus Quasirandom tests . ..... 22 4 CSPs and hardness of approximation 26 4.1 Constraint satisfaction problems . ...... 26 4.2 Berry-Ess´een................................... 27 5 Majority Is Stablest 30 5.1 Borell’s isoperimetric inequality . ....... 30 5.2 ProofoutlineofMIST ............................. 32 5.3 Theinvarianceprinciple . 33 6 Testing dictators and UGC-hardness 37 1 Linearity testing and Arrow’s theorem Monday, 27th February 2012 Rn Open Problem [Guy86, HK92]: Let a with a 2 = 1. Prove Prx 1,1 n [ a, x • ∈ k k ∈{− } |h i| ≤ 1] 1 . ≥ 2 Open Problem (S. Srinivasan): Suppose g : 1, 1 n 2 , 1 where g(x) 2 , 1 if • {− } →± 3 ∈ 3 n x n and g(x) 1, 2 if n x n . Prove deg( f)=Ω(n). i=1 i ≥ 2 ∈ − − 3 i=1 i ≤− 2 P P In this workshop we will study the analysis of boolean functions and its applications to topics such as property testing, voting, pseudorandomness, Gaussian geometry and the hardness of approximation. Two recurring themes that we will see throughout the week are: The noisy hypercube graph is a small set expander. • Every boolean function has a “junta part” and a “Gaussian part”. • 1.1 The Fourier expansion Broadly speaking, the analysis of boolean functions is concerned with properties of boolean functions f : 1, 1 n 1, 1 viewed as multilinear polynomials over R. Consider the {− } → {− } majority function over 3 variables MAJ3(x) = sgn(x1 + x2 + x3). It is easy to check that MAJ (x) = 1 x + 1 x + 1 x 1 x x x , and this can be derived by summing 23 = 8 3 2 1 2 2 2 3 − 2 1 2 3 polynomials p : 1, 1 3 1, 0, 1 , one for each y 1, 1 3, where p (x) takes value y {− } → {− } ∈ {− } y MAJ3(x) when y = x and 0 otherwise. For example, 1 x1 1+ x2 1 x3 p( 1,1, 1)(x)= − − MAJ3( 1, 1, 1). − − 2 2 2 · − − Note that the final polynomial that results from expanding and simplifying the sum of py’s is indeed always multilinear (i.e. no variable xi occurs squared, or cubed, etc.) since x2 = 1 for bits x 1, 1 . The same interpolation procedure can be carried out for any i i ∈ {− } f : 1, 1 n R: {− } → Theorem 1 (Fourier expansion) Every f : 1, 1 n R can be uniquely expressed as {− } → a multilinear polynomial R, f(x)= c x , where each c R. S i S ∈ SX[n] iYS ⊆ ∈ ˆ We will write f(S) to denote the coefficient cS and χS(x) for the function i S xi, and call ˆ ∈ f(x)= S [n] f(S)χS (x) the Fourier expansion of f. We adopt the conventionQ that χ 1, ∅ ≡ ⊆ ˆ the identicallyP 1 function. We will write deg(f) to denote maxS [n] S : f(S) = 0 , and ⊆ {| | 6 } call this quantity the Fourier degree of f. 3 We will sometimes refer to χ (x) : 1, 1 n 1, 1 as the “parity-on-S” function, since S {− } → {− } it takes value 1 if there are an even number of 1 coordinates in x and 1 otherwise. − − Using the notation of Theorem 1, we have that MAJ\( 1 ) = 1 , MAJ\( 1, 2, 3 ) = 1 , 3 { } 2 3 { } − 2 MAJ\( 1, 2 ) = 0, and deg(MAJ ) = 3. 3 { } 3 We have already seen that every function f : 1, 1 n R can be expressed as a multilinear {− } → polynomial over R (via the interpolation procedure described for MAJ3); to complete the proof of Theorem 1 it remains to show uniqueness. Let V be the vector space of all functions n f : 1, 1 n R. Here we are viewing f as a 2n-dimensional vector in R2 , with each {− } → coordinate being the value of f on some input x 1, 1 n; if f is a boolean function this ∈ {− } is simply the truth table of f. Note that the parity functions χS(x) are all elements of V , and furthermore every f V can be expressed as a linear combination of them (i.e. the ∈ n n χS(x)’s are a spanning set for V ). Since there are 2 parity functions and dim(V ) = 2 , it follows that χ : S [n] is a basis for V , and this establishes the uniqueness of the { S ⊆ } Fourier expansion. Definition 2 (inner product) Let f, g : 1, 1 n 1, 1 . We define the inner prod- {− } → {− } uct between f and g as f(x) g(x) f, g := · = E [f(x)g(x)]. h i 2n x 1,1 n x X1,1 n ∈{− } ∈{− } n Note that this is simply the dot product between f and g viewed as vectors in R2 , normal- n n ized by a factor of 2− . Given this definition, every boolean function f : 1, 1 1, 1 n {− } → {− } is a unit vector in R2 since f,f = 1. We will also write f 2 to denote f,f , and more h i k k2 h i generally, f := E[ f(x) p]1/p. k kp | | Theorem 3 (orthonormality) The set of parity functions χS(x) : S [n] is an or- n { ⊆ } thonormal basis for R2 . That is, for every S, T [n], ⊆ 1 if S = T χ ,χ = h S T i 0 otherwise. Proof. First note that χ χ = χ since x x = x x2 = S · T S∆T i S i j T j i S∆T i j S T j x , where the final equality uses the factQ ∈ thatQx2∈= 1 for Qx ∈ 1, 1Q.∈ Next,∩ we i S∆T i i i ∈ {− } Qclaim∈ that 1 if U = E[χ ]= ∅ U 0 otherwise. noting that this implies the theorem since S∆T = iff S = T . Recall that we have defined ∅ χ to be the identically 1 function, and if U = then exactly half the inputs x 1, 1 n ∅ 6 ∅ ∈ {− } have χ (x) = 1 and the other half χ (x)= 1. U U − Proposition 4 (Fourier coefficient) Let f : 1, 1 n R. Then fˆ(S) = f,χ = {− } → h Si E[f(x)χS(x)]. 4 Proof. To see that this holds, we check that f,χ = fˆ(T )χ ,χ = fˆ(T ) χ ,χ = fˆ(S). h Si T S · h T Si D T [n] E TX[n] P⊆ ⊆ Here we have used the Fourier expansion of f for the first equality, linearity of the inner product for the second, and orthonormality of parity functions (Theorem 3) for the last. Next we have Plancherel’s theorem, which states that the inner product of f and g is precisely the dot product of their vectors of Fourier coefficients. Theorem 5 (Plancherel) Let f, g : 1, 1 n R. Then, f, g = fˆ(S)ˆg(S). {− } → h i S [n] P ⊆ Proof. Again we use the Fourier expansions of f and g to check that f, g = fˆ(S)χ , gˆ(T )χ = fˆ(S)ˆg(T ) χ ,χ = fˆ(S)ˆg(S). h i S T · h S T i D S [n] T [n] E XS XT SX[n] P⊆ P⊆ ⊆ The second equality holds by linearity of inner product, and the last by orthonormality. An important corollary of Plancherel’s theorem is Parseval’s identity: if f : 1, 1 n R, {− } → then f 2 = f,f = fˆ(S)2 (i.e. the Fourier transform preserves L -norm). In k k2 h i S [n] 2 P ⊆ ˆ 2 2 particular, if f is a boolean function then S [n] f(S) = E[f(x) ] = 1, which we may view as a probability distribution over the 2n possibleP ⊆ subsets S of [n]. Note that if f and g are boolean functions then f(x)g(x) = 1 iff f(x)= g(x), and so E[f(x)g(x)] = 1 2 dist(f, g), − · where dist(f, g)= Pr[f(x) = g(x)] is the normalized Hamming distance between f and g. 6 One of the advantages of analyzing f via its Fourier expansion is that this polynomial encodes a lot of combinatorial information about f, and these combinatorial quantities can be “read off” its Fourier coefficients easily. We give two basic examples now. Recall that for functions f : 1, 1 n R, the mean of f is E[f(x)] and its variance is Var(f) := {− } → E[f(x)2] E[f(x)]2. Note that if f is a boolean function then E[f(x)] measures the bias of − f towards 1 or 1, and Var(f) = 4 Pr[f(x) = 1] Pr[f(x) = 1]. If f has mean 0 and − · · − variance 1 we say that f is balanced, or unbiased. Proposition 6 (expectation and variance) Let f : 1, 1 n R. Then E[f(x)] = {− } → fˆ( ), and Var(f)= fˆ(S)2. ∅ S= P 6 ∅ Proof. For the first equality, we check that fˆ( ) = E[f(x)χ (x)] = E[f(x)].
Load more

Recommended publications
  • Laced Boolean Functions and Subset Sum Problems in Finite Fields
    Laced Boolean functions and subset sum problems in finite fields David Canright1, Sugata Gangopadhyay2 Subhamoy Maitra3, Pantelimon Stanic˘ a˘1 1 Department of Applied Mathematics, Naval Postgraduate School Monterey, CA 93943{5216, USA; fdcanright,[email protected] 2 Department of Mathematics, Indian Institute of Technology Roorkee 247667 INDIA; [email protected] 3 Applied Statistics Unit, Indian Statistical Institute 203 B. T. Road, Calcutta 700 108, INDIA; [email protected] March 13, 2011 Abstract In this paper, we investigate some algebraic and combinatorial properties of a special Boolean function on n variables, defined us- ing weighted sums in the residue ring modulo the least prime p ≥ n. We also give further evidence to a question raised by Shparlinski re- garding this function, by computing accurately the Boolean sensitivity, thus settling the question for prime number values p = n. Finally, we propose a generalization of these functions, which we call laced func- tions, and compute the weight of one such, for every value of n. Mathematics Subject Classification: 06E30,11B65,11D45,11D72 Key Words: Boolean functions; Hamming weight; Subset sum problems; residues modulo primes. 1 1 Introduction Being interested in read-once branching programs, Savicky and Zak [7] were led to the definition and investigation, from a complexity point of view, of a special Boolean function based on weighted sums in the residue ring modulo a prime p. Later on, a modification of the same function was used by Sauerhoff [6] to show that quantum read-once branching programs are exponentially more powerful than classical read-once branching programs. Shparlinski [8] used exponential sums methods to find bounds on the Fourier coefficients, and he posed several open questions, which are the motivation of this work.
    [Show full text]
  • Probabilistic Boolean Logic, Arithmetic and Architectures
    PROBABILISTIC BOOLEAN LOGIC, ARITHMETIC AND ARCHITECTURES A Thesis Presented to The Academic Faculty by Lakshmi Narasimhan Barath Chakrapani In Partial Fulfillment of the Requirements for the Degree Doctor of Philosophy in the School of Computer Science, College of Computing Georgia Institute of Technology December 2008 PROBABILISTIC BOOLEAN LOGIC, ARITHMETIC AND ARCHITECTURES Approved by: Professor Krishna V. Palem, Advisor Professor Trevor Mudge School of Computer Science, College Department of Electrical Engineering of Computing and Computer Science Georgia Institute of Technology University of Michigan, Ann Arbor Professor Sung Kyu Lim Professor Sudhakar Yalamanchili School of Electrical and Computer School of Electrical and Computer Engineering Engineering Georgia Institute of Technology Georgia Institute of Technology Professor Gabriel H. Loh Date Approved: 24 March 2008 College of Computing Georgia Institute of Technology To my parents The source of my existence, inspiration and strength. iii ACKNOWLEDGEMENTS आचायातर् ्पादमादे पादं िशंयः ःवमेधया। पादं सॄचारयः पादं कालबमेणच॥ “One fourth (of knowledge) from the teacher, one fourth from self study, one fourth from fellow students and one fourth in due time” 1 Many people have played a profound role in the successful completion of this disser- tation and I first apologize to those whose help I might have failed to acknowledge. I express my sincere gratitude for everything you have done for me. I express my gratitude to Professor Krisha V. Palem, for his energy, support and guidance throughout the course of my graduate studies. Several key results per- taining to the semantic model and the properties of probabilistic Boolean logic were due to his brilliant insights.
    [Show full text]
  • Predicate Logic
    Predicate Logic Laura Kovács Recap: Boolean Algebra and Propositional Logic 0, 1 True, False (other notation: t, f) boolean variables a ∈{0,1} atomic formulas (atoms) p∈{True,False} boolean operators ¬, ∧, ∨, fl, ñ logical connectives ¬, ∧, ∨, fl, ñ boolean functions: propositional formulas: • 0 and 1 are boolean functions; • True and False are propositional formulas; • boolean variables are boolean functions; • atomic formulas are propositional formulas; • if a is a boolean function, • if a is a propositional formula, then ¬a is a boolean function; then ¬a is a propositional formula; • if a and b are boolean functions, • if a and b are propositional formulas, then a∧b, a∨b, aflb, añb are boolean functions. then a∧b, a∨b, aflb, añb are propositional formulas. truth value of a boolean function truth value of a propositional formula (truth tables) (truth tables) Recap: Boolean Algebra and Propositional Logic 0, 1 True, False (other notation: t, f) boolean variables a ∈{0,1} atomic formulas (atoms) p∈{True,False} boolean operators ¬, ∧, ∨, fl, ñ logical connectives ¬, ∧, ∨, fl, ñ boolean functions: propositional formulas (propositions, Aussagen ): • 0 and 1 are boolean functions; • True and False are propositional formulas; • boolean variables are boolean functions; • atomic formulas are propositional formulas; • if a is a boolean function, • if a is a propositional formula, then ¬a is a boolean function; then ¬a is a propositional formula; • if a and b are boolean functions, • if a and b are propositional formulas, then a∧b, a∨b, aflb, añb are
    [Show full text]
  • Satisfiability 6 the Decision Problem 7
    Satisfiability Difficult Problems Dealing with SAT Implementation Klaus Sutner Carnegie Mellon University 2020/02/04 Finding Hard Problems 2 Entscheidungsproblem to the Rescue 3 The Entscheidungsproblem is solved when one knows a pro- cedure by which one can decide in a finite number of operations So where would be look for hard problems, something that is eminently whether a given logical expression is generally valid or is satis- decidable but appears to be outside of P? And, we’d like the problem to fiable. The solution of the Entscheidungsproblem is of funda- be practical, not some monster from CRT. mental importance for the theory of all fields, the theorems of which are at all capable of logical development from finitely many The Circuit Value Problem is a good indicator for the right direction: axioms. evaluating Boolean expressions is polynomial time, but relatively difficult D. Hilbert within P. So can we push CVP a little bit to force it outside of P, just a little bit? In a sense, [the Entscheidungsproblem] is the most general Say, up into EXP1? problem of mathematics. J. Herbrand Exploiting Difficulty 4 Scaling Back 5 Herbrand is right, of course. As a research project this sounds like a Taking a clue from CVP, how about asking questions about Boolean fiasco. formulae, rather than first-order? But: we can turn adversity into an asset, and use (some version of) the Probably the most natural question that comes to mind here is Entscheidungsproblem as the epitome of a hard problem. Is ϕ(x1, . , xn) a tautology? The original Entscheidungsproblem would presumable have included arbitrary first-order questions about number theory.
    [Show full text]
  • Problems and Comments on Boolean Algebras Rosen, Fifth Edition: Chapter 10; Sixth Edition: Chapter 11 Boolean Functions
    Problems and Comments on Boolean Algebras Rosen, Fifth Edition: Chapter 10; Sixth Edition: Chapter 11 Boolean Functions Section 10. 1, Problems: 1, 2, 3, 4, 10, 11, 29, 36, 37 (fifth edition); Section 11.1, Problems: 1, 2, 5, 6, 12, 13, 31, 40, 41 (sixth edition) The notation ""forOR is bad and misleading. Just think that in the context of boolean functions, the author uses instead of ∨.The integers modulo 2, that is ℤ2 0,1, have an addition where 1 1 0 while 1 ∨ 1 1. AsetA is partially ordered by a binary relation ≤, if this relation is reflexive, that is a ≤ a holds for every element a ∈ S,it is transitive, that is if a ≤ b and b ≤ c hold for elements a,b,c ∈ S, then one also has that a ≤ c, and ≤ is anti-symmetric, that is a ≤ b and b ≤ a can hold for elements a,b ∈ S only if a b. The subsets of any set S are partially ordered by set inclusion. that is the power set PS,⊆ is a partially ordered set. A partial ordering on S is a total ordering if for any two elements a,b of S one has that a ≤ b or b ≤ a. The natural numbers ℕ,≤ with their ordinary ordering are totally ordered. A bounded lattice L is a partially ordered set where every finite subset has a least upper bound and a greatest lower bound.The least upper bound of the empty subset is defined as 0, it is the smallest element of L.
    [Show full text]
  • Logic and Bit Operations
    Logic and bit operations • Computers represent information by bits. • A bit has two possible values, namely zero and one. This meaning of the word comes from binary digit, since zeroes and ones are the digits used in binary representations of numbers. The well-known statistician John Tuley introduced this terminology in 1946. (There were several other suggested words for a binary digit, including binit and bigit, that never were widely accepted.) • A bit can be used to represent a truth value, since there are two truth values, true and false. • A variable is called a Boolean variable if its values are either true or false. • Computer bit operations correspond to the logical connectives. We will also use the notation OR, AND and XOR for _ , ^ and exclusive _. • A bit string is a sequence of zero or more bits. The length of the string is the number of bits in the string. 1 • We can extend bit operations to bit strings. We define bitwise OR, bitwise AND and bitwise XOR of two strings of the same length to be the strings that have as their bits the OR, AND and XOR of the corresponding bits in the two strings. • Example: Find the bitwise OR, bitwise AND and bitwise XOR of the bit strings 01101 10110 11000 11101 Solution: The bitwise OR is 11101 11111 The bitwise AND is 01000 10100 and the bitwise XOR is 10101 01011 2 Boolean algebra • The circuits in computers and other electronic devices have inputs, each of which is either a 0 or a 1, and produce outputs that are also 0s and 1s.
    [Show full text]
  • Propositional Logic
    Reasoning About Programs Panagiotis Manolios Northeastern University March 22, 2012 Version: 58 Copyright c 2012 by Panagiotis Manolios All rights reserved. We hereby grant permission for this publication to be used for personal or classroom use. No part of this publication may be stored in a retrieval system or transmitted in any form or by any means other personal or classroom use without the prior written permission of the author. Please contact the author for details. 2 Propositional Logic The study of logic was initiated by the ancient Greeks, who were concerned with analyzing the laws of reasoning. They wanted to fully understand what conclusions could be derived from a given set of premises. Logic was considered to be a part of philosophy for thousands of years. In fact, until the late 1800’s, no significant progress was made in the field since the time of the ancient Greeks. But then, the field of modern mathematical logic was born and a stream of powerful, important, and surprising results were obtained. For example, to answer foundational questions about mathematics, logicians had to essentially create what later became the foundations of computer science. In this class, we’ll explore some of the many connections between logic and computer science. We’ll start with propositional logic, a simple, but surprisingly powerful fragment of logic. Expressions in propositional logic can only have one of two values. We’ll use T and F to denote the two values, but other choices are possible, e.g., 1 and 0 are sometimes used. The expressions of propositional logic include: 1.
    [Show full text]
  • CS40-S13: Functional Completeness 1 Introduction
    CS40-S13: Functional Completeness Victor Amelkin [email protected] April 12, 2013 In class, we have briefly discussed what functional completeness means and how to prove that a certain system (a set) of logical operators is functionally complete. In this note, I will review what functional completeness is, how to prove that a system of logical operators is complete as well as how to prove that a system is incomplete. 1 Introduction When we talk about logical operators, we mean operators like ^, _, :, !, ⊕. Logical operators work on propositions, i.e. if we write a ^ b, then both a and b are propositions. The mentioned propositions may be either simple or compound, as in a ^ :(x ! y) { here, a is simple, while b is compound. | {z } b For a logical operator it does not matter whether it operates on simple or compound propositions; the only thing that matters is that all the arguments are propositions, i.e., either True of False. Since logical operators apply to propositions and \return" either True or False, they can be seen as boolean functions. For example, conjunction ^(x1; x2) = x1 ^ x2 is a binary boolean function, i.e., it accepts two arguments (hence, \binary") that can be either True of False and returns either True of False (hence, \boolean"). As another example, negation can be seen as a unary (accepts only one argument) boolean function :(x) = :x. We, however, rarely use the functional notation ^(x1; x2) (like in f(x)); instead, we usually use the operator notation, e.g., x1 _ x2. (Similarly, when you do algebra, you probably write a + b instead of +(a; b), unless you do algebra in Lisp.) Further, instead of talking about logical operators, we will talk about boolean functions, but, having read the material above, you should understand that these two are the same.
    [Show full text]
  • Binary Recursion
    Binary recursion 1 Unate functions ! Theorem If a cover C(f) is unate in xj, then f is unate in xj. ! Theorem If f is unate in xj, then every prime implicant of f is unate in xj. 2 1 Why are unate functions so special? ! Special Boolean algorithms for unate functions are not only fast, but also give the guaranteed minimum result. ! The algorithms presented perform the Boolean operations and simplify the result in the same time. ! In practice, many functions are partially unate. The divide and conquer strategy aimed at unateness produces very shallow recursions; algorithms are fast in practice. 3 Unate recursive paradigm works for: ! Complementation ! Irredundant cover ! Reduce ! Essential primes = + ! Example f x 1 x 2 x 2 x 3 is unate 4 2 Important facts about unate functions Proposition 1. A logic function f is monotone increasing (decreasing) in x j if and only if no prime implicant of f has a 0 (1) in the j-th position. Proof: (if) If no prime implicant has a 0 in the jth position, any prime cover C(f) is unate in x j . So f is unate in x j. (only if) Let c i is a prime with 0 at jth position -> the cube c~i obtained by replacing 0 in jth position with a 1 contains at least one vector in f . For the input v corresponding to that vertex, f(v)=0. If we change x j from 1 to 0 in v, the value of f changes from 0 to 1, so f is not monotone increasing.
    [Show full text]
  • Boolean Algebra
    Boolean Algebra Definition: A Boolean Algebra is a math construct (B,+, . , ‘, 0,1) where B is a non-empty set, + and . are binary operations in B, ‘ is a unary operation in B, 0 and 1 are special elements of B, such that: a) + and . are communative: for all x and y in B, x+y=y+x, and x.y=y.x b) + and . are associative: for all x, y and z in B, x+(y+z)=(x+y)+z, and x.(y.z)=(x.y).z c) + and . are distributive over one another: x.(y+z)=xy+xz, and x+(y.z)=(x+y).(x+z) d) Identity laws: 1.x=x.1=x and 0+x=x+0=x for all x in B e) Complementation laws: x+x’=1 and x.x’=0 for all x in B Examples: (B=set of all propositions, ∨, ∧, ¬, T, F) (B=2A, U, ∩, c, Φ,A) Theorem 1: Let (B,+, . , ‘, 0,1) be a Boolean Algebra. Then the following hold: a) x+x=x and x.x=x for all x in B b) x+1=1 and 0.x=0 for all x in B c) x+(xy)=x and x.(x+y)=x for all x and y in B Proof: a) x = x+0 Identity laws = x+xx’ Complementation laws = (x+x).(x+x’) because + is distributive over . = (x+x).1 Complementation laws = x+x Identity laws x = x.1 Identity laws = x.(x+x’) Complementation laws = x.x +x.x’ because + is distributive over . = x.x+0 Identity laws = x.x b) x+1 =x+(x+x’) Complementation laws = (x+x)+x’ + is associative = x+x’ using (a) = 1 Complementation laws 0.x =(x’.x).x Complementation laws = x’.(x.x) .
    [Show full text]
  • Boolean Logic
    Boolean Logic Lecture Notes Gert Smolka Saarland University November 22, 2016 Abstract. We study satisfiability and equivalence of boolean expressions. We show that prime trees (reduced and ordered decision trees) are a unique normal form for boolean expressions. We verify a SAT solver and a prime tree normaliser. We show that assignment equivalence agrees with the equivalence obtained from the axioms of boolean algebra. The material is presented using type-theoretic language. There is an accompa- nying Coq development complementing the mathematical presentation given in the notes. The notes also serve as introduction to formal syntax. Expressions are modelled with an inductive type and recursive functions and inductive predicates on expres- sions are studied. This includes notions like substitution and equational deduction. Advice to the student. To master the material, you want to understand both the notes and the accompanying Coq development. It is important that you can switch between the informal mathematical presentation of the notes and the formal de- velopment in Coq. You may think of the notes as the design of the theory and the Coq development as the implementation of the theory. The goal is to understand both the design and the implementation of the theory. The notes occasionally omit details that are spelled out in the Coq development (e.g., the precise definition of substitution). Make sure you can fill in the missing details. Contents 1 Boolean Operations . 3 2 SAT Solver . 4 3 Equivalence of Boolean Expressions . 7 4 Shallow and Deep Embedding . 9 5 Decision Trees . 12 6 Prime Tree Normal Form . 12 7 Significant Variables .
    [Show full text]
  • CSE 1400 Applied Discrete Mathematics Boolean Logic Department of Computer Sciences College of Engineering Florida Tech Fall 2011
    CSE 1400 Applied Discrete Mathematics Boolean Logic Department of Computer Sciences College of Engineering Florida Tech Fall 2011 Control Structures 1 Boolean Logic 3 Normal Forms 4 The Conditional Operator 5 Complex Propositions 6 Truth Tables 8 Logical Equivalence 10 Satisfiable and Valid Propositions 11 Constructing Boolean Functions 12 Satisfiability 13 Counting Boolean Expressions 14 Consistency and Completeness of Boolean Algebra 14 Problems on Boolean Logic 15 Abstract Logic controls the action of a computer. Control Structures The Böhm-Jacopini theorem shows that only three funda- mental control structures are necessary to implement any Turing algorithm. 1. Execute instructions sequentially one after another as in figure 1 2. Select one of two instruction to execute according to the value of a Boolean variable as in figure 2 3. Iterate through a sequence of instructions until a Boolean variable changes value as in figure 3 cse 1400 applied discrete mathematics boolean logic 2 Figure 1: Sequential execution of statements. Statement Next Statement Figure 2: Conditional execution of statements. Boolean Condition Then Else Statement(s) Statement(s) Figure 3: Loop (iterative) execution of statements. Boolean Condition Loop Statement(s) cse 1400 applied discrete mathematics boolean logic 3 Boolean Logic Boolean logic provides the basis to control the execution of algorithms.Propositional calculus studies the behav- ior of formulas constructed using Boolean variables. The domain Boolean variables are typically named of these variables is the set of truth values B = fFalse, Trueg.A p, q, r, s, . .. Boolean variable is also called a proposition and is often inter- preted as the name for a declarative natural language sentence: A proposition is a sentence that can only be True or False.
    [Show full text]