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Re: BitAnd[True,False]
Posted:
Sep 26, 2013 3:45 AM


On Sep 25, 2013, at 2:36 AM, Alan <alan.isaac@gmail.com> wrote:
> I'd hoped BitAnd would work on Boolean Lists so I gave it a try. If it had simply failed completely I'd be disappointed but would understand. But oddly, it half succeeds. Why? BitOr fails the same way, and BitNot always fails. > > I just want to understand these results. I know there are ways to get the output I want. > > Thanks, > Alan Isaac > (using Mathematica 9) > > In[77]:= BooleanTable[{p,q,BitAnd[p,q]},{p,q}]//TableForm > Out[77]//TableForm= > True True True > True False BitAnd[False,True] > False True BitAnd[False,True] > False False False
There's a lot going on here.
To start with BooleanTable[expr,p,q] evaluates the truth value of expr, but {p,q,BitAnd[p,q]} is not a Boolean function and has no truth value. But BooleanTable's implementation is strange, it's actually written in Mathematica itself, try
Trace[BooleanTable[{stuff}, stuff]]
for example. The upshot of all this is rather than returning BooleanTable[expr,p,q] as most other functions that can't evaluate their arguments BooleanTable[{p,q},p,q] returns the expression with the variables replaced by True or False. Similarly consider:
(Debug) In[21]:= BooleanTable[expr, p, q] (Debug) Out[21]= {{expr, expr}, {expr, expr}}
So anyway that's a long winded way of saying BooleanTable's result is already a truth table but the values of the arguments to its first argument are implied and if you want to make a table with columns for the arguments you have to use a different method.
As for the behavior of BitAnd. First of all it simplifies its arguments so BitAnd[x,x] evaluates to BitAnd[x]:
(Debug) In[52]:= tRACe[BitAnd[x,x]] (Debug) Out[52]= {BitAnd[x,x],BitAnd[x],x}
Then BitAnd has the property OneIdentity so BitAnd[x] evaluates to x. Thus BitAnd[True,True]>True. This has the amusing side effect of meaning BitAnd sometimes will appear to evaluate for arguments of the wrong type, but BitAnd[True,False] is undefined since True and False are not integers which is what BitAnd expects for arguments. The Orderless attribute results in BitAnd[True,False]>BitAnd[False,True].
Regards, Sseziwa



