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I need to find the Boolean expression for the truth table below where $P$, $Q$, $R$ are inputs, and $S$ is the output. Does anyone have a cool easy way of solving such problems please? Your help will be appreciated.

$\begin{array}{c|c|c||c} P & Q & R & S\\ \hline 1 & 1 & 1 & 1\\ \hline 1 & 1 & 0 & 0\\ \hline 1 & 0 & 1 & 1\\ \hline 1 & 0 & 0 & 0\\ \hline 0 & 1 & 1 & 1\\ \hline 0 & 1 & 0 & 1\\ \hline 0 & 0 & 1 & 0\\ \hline 0 & 0 & 0 & 0 \end{array}$

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    Looks to me like your expression is just $\text{if}(P, R, Q)$.2015-12-04

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A mechanical way of getting an expression that has a desired truth table is to take the disjunction of the formulas that determine the rows you want with 1s.

For example, for a truth table on P, Q, and R that has $\begin{array}{c|c|c||c} P & Q & R & \text{Table}\\ \hline 1 & 1 & 1 & 1\\ \hline 1 & 1 & 0 & 0\\ \hline 1 & 0 & 1 & 1\\ \hline 1 & 0 & 0 & 0\\ \hline 0 & 1 & 1 & 1\\ \hline 0 & 1 & 0 & 1\\ \hline 0 & 0 & 1 & 0\\ \hline 0 & 0 & 0 & 0 \end{array}$ we note that the rows corresponding to $1$s are: $P\land Q\land R$, $P\land \neg Q\land R$, $\neg P\land Q\land R$, and $\neg P\land Q\land\neg R$. So a formula that has the desired truth table is $(P\land Q\land R)\lor (P\land\neg Q\land R)\lor (\neg P\land Q\land R) \lor (\neg P\land Q\land \neg R).$ This is, of course, unlikely to be the simplest formula that works, and may be simplified later.

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    Note: this is also a very convenient way to write a formula in disjunctive normal form.2011-05-11
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You can just render the thing as a bunch of OR-ed together clauses, one checking for each case where the output is 1. If you want a simpler expression, you can use a Karnaugh map.

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    If you do the Karnaugh map the resulting expression is $(P \land R) \lor (\neg P \land Q)$2011-05-10
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Try a free program called Logic Friday...works standalone...you push new truth table, the number of inputs, then it creates a generic tt with all zeros out. Then you double click on each output you want to change to 1. Then you push Truth Table-->Submit and it will give you the Boolean. Then you push Equation-->Factor to get the factored Boolean.