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Conjunctive normal form first appears, in this context, in Davis and Putnam's A computing procedure for quantification theory, in which they describe a primitive form of the DPLL algorithm (which appears in a follow-up work of Davis, Logemann, and Loveland, and is the basis of all modern SAT solvers). They explain that one key property of CNF is that any ...

4

It's because this is close to how real-world problems tend to be expressed. SAT has a lot of practical applications, from controlling power grids to integrated circuit layout. In pretty much every case, the problem is that you have a bunch of complex constraints that all need to be satisfied. However you express each constraint, they ultimately get joined ...

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Clearly, $x\cdot y'\cdot z'≠ y'\cdot z'$ as it depends on the truth value of $x.$ However, in your specific case, the expression is $(\lnot x\land \lnot z)\lor (x\land \lnot y\land\lnot z).$ Note that if $x$ is false, then the $\lnot x$ literal in the left clause (i.e. $\lnot x\land \lnot z$) will become true, and if it's false then the one in the right ...

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The key here is graycode. I think the easiest way to see why Karnnaugh maps work is to go through an example: Consider the following truth table $\hskip2in$ We can easily find the logical formula describing this table in terms of the sums of products formula (I will the boolean algebraic notation): \begin{align*} E(A, B, C) &= \overline{A} \cdot ... 2 Here is one possible formula: \texttt{true}. Indeed, every graph admits a coloring that satisfies your requirements: simply color all vertices with the same color. 1 The relationship is that a short SOP expression implies a low Kolmogorov complexity, but not vice versa. If you have a short SOP expression, then that is a concise way to compress the string, hence Kolmogorov complexity is low. However, there might be strings that can be compressed in another way even though the SOP expression is long, and those strings ... 1 It's 4 because you're allowed to "wrap around" from the top edge to the bottom edge or from the right edge to the left edge (those moves still represent 1-bit changes because the Gray code is cyclic). So there is one PI of 3 minterms in the first column, one in the fourth column, one in the third row, and one of 2 minterms in the second row. 1 We like conjunctive normal form because all circuits can be transformed into conjunctive normal from in linear time via the Tseitin Transformation. It's a clean, normalized data structure for solver implementation. CNF conversion without Tseitin variables may lead to exponential expansion. That's the standard answer, but, as you see, there are nuances. ... 1 Its assumed you implemented a generalized Sudoku solver and not a 9x9 one. First you will need a parsimonious 1-to-1 reduction from 3SAT (SAT to 3SAT is easy) to Sudoku. This is because your sudoku solver may use assumptions of a unique solution. This leads into the another solution problem (ASP) but parsimoniety covers this. There are known reductions ... 1 Yes, absolutely. Every boolean function can be implemented by a circuit. In this case, it suffices to XOR bits 1 and n; bits 2 and n-1; and so on; and then take the NOR of all of the outputs of these XOR gates. 1 If f,g are two \pm1-valued functions and f \lor g is their maximum (this is the standard notation for maximum; \land stands for minimum), then f \lor g = \frac{1 + f + g - fg}{2}. $$Since \chi_S \chi_T = \chi_{S \Delta T}, this shows that$$ \chi_S \lor \chi_T = \frac{1}{2} + \frac{\chi_S}{2} + \frac{\chi_T}{2} - \frac{\chi_{S\Delta T}}{2}. $$... 1 I haven't edited my question, but in case others are here: this thesis addresses the question I was intending (https://core.ac.uk/download/pdf/52104064.pdf), finding e.g. 1189 average transitions required for a certain architecture for a 16-bit by 16-bit multiply that's completely random bits. \ 1 No, your answer is not correct, and for some reason you have four CNFs (I didn't understand why). I'll give you some hints, how to reduce the Four Queens Problem to a CNF. I'll use notation, which is common for boolean variables and expressions. A boolean variable x_{i,j}, where i,j \in [1,4], will be used to represent a queen in the cell (i,j). You ... 1 You can't. You have a QBF-SAT instance. That can't be efficiently resolved/reduced to SAT (given current conjectures in complexity theory). Instead, use a QBF-SAT solver on your formula. 1 When substituted for (x_1, ..., x_i) you have a formula in monadic predicate logic. The monadic class is the class of first-order formulas without function symbols, with unary (monadic) predicates only, but with arbitrary quantification. SAT for monadic predicate logic is NEXPTIME-complete in general Altogether this means in particular that Monadic-Sat ... 1 According to Wikipedia, A powerful and nontrivial metatheorem states that any theorem of 𝟐 holds for all Boolean algebras. Conversely, an identity that holds for an arbitrary nontrivial Boolean algebra also holds in 𝟐. Hence all the mathematical content of Boolean algebra is captured by 𝟐. What this means is that if \phi = \psi is an equation in ... 1 I am not used to this definition. Instead, the definition I am used to states that a system of boolean equations has the form$$\begin{align*} f_1(x_1,\dots,x_n)&=0\\ \vdots\\ f_m(x_1,\dots,x_n)&=0. \end{align*}$$where x_1,\dots,x_n are boolean variables and f_1,\dots,f_m are boolean functions. The values v_1,\dots,v_n are a solution to the ... 1$$ xy + \bar{x} z + yz = xy + \bar{x} z + (x+\bar{x})yz = xy + \bar{x} z + xyz + \bar{x}yz = xy(1 + z) + \bar{x} z(1 + y) = xy + \bar{x}z. 

1

Let $A,B$ be the two recursive repeating sequences, where $|A| = |B|$. If $|A| = |B| = 1$ then there is nothing to do. Otherwise, either $A$ is of the form $A_1A_2$ or of the form $(A')_k$, where $k > 1$. In the second case, we can write $A = A' (A')_{k-1}$, which is also of the form $A_1A_2$. Thus it suffices to explain how to compute $A_1 A_2 \land B$ (...

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