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I am very new to Mathematica, and need to know how to plug values into existing functions. For example, Math World supplies the following code for building a truth table of n levels of operator op:

TruthTable[op_, n_] := 
Module[{l = 
 Flatten[Outer[List, Sequence @@ Table[{True, False}, {n}]], 
  n - 1], a = Array[A, n]}, 
DisplayForm[
GridBox[Prepend[Append[#, op @@ #] & /@ l, Append[a, op @@ a]], 
 RowLines -> True, ColumnLines -> True]]]

How do I set the operator and the value of n?

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  • $\begingroup$ You've seen BooleanTable[] already? $\endgroup$
    – J. M.'s missing motivation ♦
    Commented Oct 27, 2012 at 7:39
  • $\begingroup$ You say "Mathematica supplies the following code for building a truth table of n levels of operator op". It it not known to my copy of Mathematica. I found it on Math World, which is not same as "supplied by Mathematica". $\endgroup$
    – m_goldberg
    Commented Oct 27, 2012 at 7:44
  • $\begingroup$ @J.M. I have seen BooleanTable[] and the related functions and did not have problems with those--I'm only unsure about the notation used above. $\endgroup$
    – QuietThud
    Commented Oct 27, 2012 at 7:53
  • $\begingroup$ @m_goldberg: I meant "supplied by Wolfram". $\endgroup$
    – QuietThud
    Commented Oct 27, 2012 at 7:55

1 Answer 1

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Here is some results of my trying out this function on my copy of Mathematica:

notebook selection

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  • $\begingroup$ I see you've used an '@' in your definition of the op--is this a standard way of avoiding nested parentheses? Also, can you tell me about the use of the '&' at the end of the definition? $\endgroup$
    – QuietThud
    Commented Oct 27, 2012 at 8:23
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    $\begingroup$ @Quiet, @ is a prefix method of applying a function; that is, f @ x and f[x] are the same thing. There's a postfix version, x // f. &, on the other hand, is shorthand for the Function[] construct, which allows you to use pure functions. Look at the docs for Function[] for more detail on this. $\endgroup$
    – J. M.'s missing motivation ♦
    Commented Oct 27, 2012 at 8:28
  • $\begingroup$ @J.M: Thanks a lot, that's very helpful.=) $\endgroup$
    – QuietThud
    Commented Oct 28, 2012 at 5:51

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