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This is a Boolean formula in conjunctive normal form (CNF) It is a collection of clauses (the parentheses), each consisting of the disjunction (logical or, denoted ) of several literals, where a literal is either a Boolean variable (such as x) or the negation of one (such as x) A satisfying truth assignment is an assignment of false or true to each variable so that every clause contains a literal whose value is true The SAT problem is the following: given a Boolean formula in conjunctive normal form, either nd a satisfying truth assignment or else report that none exists In the instance shown previously, setting all variables to true, for example, satis es every clause except the last Is there a truth assignment that satis es all clauses With a little thought, it is not hard to argue that in this particular case no such truth assignment exists (Hint: The three middle clauses constrain all three variables to have the same value) But how do we decide this in general Of course, we can always search through all truth assignments, one by one, but for formulas with n variables, the number of possible assignments is exponential, 2n S AT is a typical search problem We are given an instance I (that is, some input data specifying the problem at hand, in this case a Boolean formula in conjunctive normal form), and we are asked to nd a solution S (an object that meets a particular speci cation, in this case an assignment that satis es each clause) If no such solution exists, we must say so More speci cally, a search problem must have the property that any proposed solution S to an instance I can be quickly checked for correctness What does this entail For one thing, S must at least be concise (quick to read), with length polynomially bounded by that of I This is clearly true in the case of SAT, for which S is an assignment to the variables To formalize the notion of quick checking, we will say that there is a polynomial-time algorithm that takes as input I and S and decides whether or not S is a solution of I For SAT, this is easy as it just involves checking whether the assignment speci ed by S indeed satis es every clause in I Later in this chapter it will be useful to shift our vantage point and to think of this ef cient algorithm for checking proposed solutions as de ning the search problem Thus: A search problem is speci ed by an algorithm C that takes two inputs, an instance I and a proposed solution S, and runs in time polynomial in |I| We say S is a solution to I if and only if C(I, S) = true.

Problem 15-5 Suppose you have a capacitor of 100 pF at a frequency of 300 MHz What is BC First, find XC by the formula for capacitive reactance:

Deriving the unit diagonal form Our remaining task is simple and clean We can divide the top row by 3 and the middle row by 2, obtaining the unit diagonal matrix

1 0 0 0 1 0 0 0 1 2 1 3

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Given the importance of the SAT search problem, researchers over the past 50 years have tried hard to nd ef cient ways to solve it, but without success The fastest algorithms we have are still exponential on their worst-case inputs Yet, interestingly, there are two natural variants of SAT for which we do have good algorithms If all clauses contain at most one positive literal, then the Boolean formula is called a Horn formula, and a satisfying truth assignment, if one exists, can be found by the greedy algorithm of Section 53 Alternatively, if all clauses have only two literals, then graph theory comes into play, and SAT can be solved in linear time by nding the strongly connected 231

XC = 1/(628f C ) Note that 100 pF = 0000100 F Therefore: XC = 1/(628 300 0000100) = 1/0001884 = 531 The imaginary value of XC is equal to j531 The susceptance, BC, is equal to 1/X C Thus, BC = 1/( j531) = j000188, rounded to three significant figures The general formula for capacitive susceptance in siemens, in terms of frequency in hertz and capacitance in farads, is: BC = 628f C This formula also works for frequencies in megahertz and capacitances in microfarads

Stating and reducing the solution The solution to the system is now clear We have no fractions to reduce, because we kept the sizes of the numbers down as we went along (Otherwise we d have fractions with large numerators and denominators, although they would divide out cleanly and leave us with integers) Evidently,

5 3 1 3 4 6 2 3 3 4

If you think we ve finished solving this problem, you re indeed confused We must check our work! Only then can we be totally confident that our solution is correct, because it s easy to make mistakes in the matrix morphing game When we plug the values for x, y, and z into the first original equation, here s what happens: 3x + z = 2y + 11 3 2 + 3 = 2 ( 1) + 11 6 + 3 = 2 + 11 9=9 Check one Now for the second equation: 4y + 2z = x 4 ( 1) + 2 3 = 2 4 + 6 = 2 2=2 Check two Finally, the third equation: 5x + y = 3z 20 5 2 + ( 1) = 3 3 20 10 1 = 9 20 11 = 11 Check three Mission accomplished

Problem 15-6 Suppose an inductor has L = 163 H at a frequency of 887 kHz What is BL Note that 887 kHz = 0887 MHz You can calculate XL from the formula for inductive reactance:

Here s an extra-credit challenge!

components of a particular graph constructed from the instance (recall Exercise 328) In fact, in 9, we ll see a different polynomial algorithm for this same special case, which is called 2 SAT On the other hand, if we are just a little more permissive and allow clauses to contain three literals, then the resulting problem, known as 3 SAT (an example of which we saw earlier), once again becomes hard to solve!

Play the matrix morphing game with the linear system we just got done solving, but take a different route this time Don t reduce the sizes of any of the integers Let the absolute values grow as large as they want !

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