How do I convert a Boolean table to a Boolean expression using BooleanMinterms.
For example:
table = {{0, 0, 0} -> 0, {0, 0, 1} -> 0, {0, 1, 0} -> 0, {0, 1, 1} ->
1, {1, 0, 0} -> 0, {1, 0, 1} -> 1, {1, 1, 0} -> 1, {1, 1, 1} -> 0}
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2 Answers
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table = {{0, 0, 0} -> 0, {0, 0, 1} -> 0, {0, 1, 0} -> 0, {0, 1, 1} ->
1, {1, 0, 0} -> 0, {1, 0, 1} -> 1, {1, 1, 0} -> 1, {1, 1, 1} -> 0}
tab = Pick[table, table[[All, 2]], 1]
(* {{0, 1, 1} -> 1, {1, 0, 1} -> 1, {1, 1, 0} -> 1} *)
bMin = BooleanMinterms[#, {a, b, c}] &@tab[[All, 1]]
(* (a && b && ! c) || (a && ! b && c) || (! a && b && c) *)
Check the result
TableForm[Reverse@Boole@BooleanTable[{a, b, c, bMin}], TableHeadings -> {None, {a, b, c, "bMin"}}]
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$\begingroup$ but if we check result it's wrong: !a &&!b must give 1 but in the table it's 0 $\endgroup$– ْزَيْنَبMar 5, 2016 at 14:58
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$\begingroup$ @Zineb.l I have made an edit. You are very quick! $\endgroup$– user36273Mar 5, 2016 at 15:00
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$\begingroup$ Zineb.l You're welcome. $\endgroup$– user36273Mar 5, 2016 at 15:10
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$\begingroup$ we didn't pay attention to an easiest way: BooleanMinterms[{3, 5, 6}, {a, b, c}] where 3,5 and 6 represent the indice of the minterms with an output =1 $\endgroup$– ْزَيْنَبMar 29, 2016 at 0:37
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1$\begingroup$ @ Zineb.l You are right, however, you have only 3 pins (inputs). An 8 bit input has 256 rows, then your method becomes more difficult to act, while my method automatically determines "1". $\endgroup$– user36273Mar 29, 2016 at 12:16
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In addition to the answer above of @rewi there is a direct and simple method
BooleanMinterms[{3, 5, 6}, {a, b, c}]
where 3,5 and 6 represent the order of the minterms with an output =1, you can see the link below for more information on how to extract boolean expressions from truth table using Minterms. http://www.cs.ucr.edu/~ehwang/courses/cs120a/minterms.pdf
BooleanFunction[]instead… $\endgroup$