# Finding Boolean/Logical Expressions for truth tables

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.

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

• I'm sorry, but how is what you gave meant to be interpreted? I would have guessed that each xxx y is meant to be that the row corresponding the the assigments xxx has y as the output, but you have four variables (P, Q, R, and S), and only three inputs. Commented May 10, 2011 at 21:20
• Thanks @Quanta. @Arturo, yes, I have 3 inputs and 1 output, S Commented May 10, 2011 at 21:26
• While boolean algebra and discrete mathematics would fit, logic is really the proper tag. Mathematical physics? Not at all... Commented May 10, 2011 at 21:54
• Karnaugh map (en.wikipedia.org/wiki/Karnaugh_map) is the simplest way to go.
– bzc
Commented May 10, 2011 at 22:14
• Looks to me like your expression is just $\text{if}(P, R, Q)$. Commented Dec 4, 2015 at 13:22

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.
• This can by easily simplified, for example the $$(P \land Q \land R)$$ and $$(P \land \neg Q \land R)$$ clauses only differ by the $$Q$$ and $$\neg Q$$ so they can be condensed to a single $$(P \land R)$$. This is the type of simplification that the Karnaugh map does for you. Commented May 10, 2011 at 23:19
• If you do the Karnaugh map the resulting expression is $$(P \land R) \lor (\neg P \land Q)$$ Commented May 10, 2011 at 23:15