# Questions tagged [boolean-algebra]

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### Classical Computation without NOT

Is it possible to do universal classical computation using bits and 2-bit gates when you cannot perform a NOT operation on a single bit (hence cant do CNOT and so on). If yes, what are the possible ...
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### Implementing a Boolean function with NOR gates

Implement f(a, b, c, d) = Σ m(3, 4, 5, 6, 7, 11, 15) as a 2-level gate circuit (a) Using OR gates and NOR gates. (b) Using NOR gates only. I have found that F=ab+d using Karnaugh map. I have also ...
65 views

### How does one calculate the block-sensitivity of a function?

I am looking at this paper : http://arxiv.org/pdf/1411.3419v1.pdf But somehow I am not being able to fish out a method to calculate this quantity called the "block-sensitivity". Can someone kindly ...
92 views

### Hardness of problem related to number of subsets that satisfy a particular property

I have the following algorithmic problem. I am given a set of elements. Each element has a set of properties. For example: ...
35 views

### Learning a small disjunction using an input distribution of our choice

I have a boolean function $f: \{0,1\}^n \to \{0,1\}$ that I know takes the form $$f(x_1,\dots,x_n) = x_{i_1} \lor x_{i_2} \lor \dots \lor x_{i_k}.$$ I don't know the values of $i_1,\dots,i_k$, but I ...
132 views

### Learning a small disjunction

I have a boolean function $f: \{0,1\}^n \to \{0,1\}$ that I know takes the form $$f(x_1,\dots,x_n) = x_{i_1} \lor x_{i_2} \lor \dots \lor x_{i_k},$$ but I don't know the values of $i_1,\dots,i_k$. ...
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### Issue understanding the reduction of SAT to 3-SAT in poly time

Reading this http://classes.soe.ucsc.edu/cmps102/Spring10/lect/17/SAT-3SAT-and-other-red.pdf, I came to know that reducing a clause $C_i$ from a $SAT$ instance containing more than 3 literals to a \$3-...
264 views

### Recognizing Horn clauses

I am currently studying model theory and I am trying to decide if a clause is a Horn Clause. I know that a Horn Clause is a clause with at most one positive literal, but there are some clauses that it ...
$$(\neg A \wedge \neg C) \vee (\neg A \wedge D) \vee (\neg A \wedge B) \vee (\neg B \wedge \neg C)$$ can simplify down to  $$(\neg A \wedge D) \vee (\neg A \wedge B) \vee (\neg B \wedge \neg C)$$ ...