From mboxrd@z Thu Jan 1 00:00:00 1970 X-Spam-Checker-Version: SpamAssassin 3.4.4 (2020-01-24) on polar.synack.me X-Spam-Level: X-Spam-Status: No, score=0.1 required=5.0 tests=BAYES_05,INVALID_MSGID autolearn=no autolearn_force=no version=3.4.4 X-Google-Language: ENGLISH,ASCII-7-bit X-Google-Thread: 1108a1,9a0ff0bffdf63657 X-Google-Attributes: gid1108a1,public X-Google-Thread: 103376,4b06f8f15f01a568 X-Google-Attributes: gid103376,public X-Google-Thread: fac41,9a0ff0bffdf63657 X-Google-Attributes: gidfac41,public X-Google-Thread: f43e6,9a0ff0bffdf63657 X-Google-Attributes: gidf43e6,public From: Tim McDermott Subject: Re: Software landmines (loops) Date: 1998/09/03 Message-ID: <35EEF597.A1119EF4@draper.com>#1/1 X-Deja-AN: 387593027 Content-Transfer-Encoding: 7bit References: <35f51e53.48044143@ Content-Type: text/plain; charset=us-ascii Organization: CSDL-DC Mime-Version: 1.0 Newsgroups: comp.lang.eiffel,comp.object,comp.software-eng,comp.lang.ada Date: 1998-09-03T00:00:00+00:00 List-Id: Patrick Doyle wrote: snip > >you are talking about a 3-term boolean expression. There are only 6 ways to > put one of those together, and I have no trouble > >evaluating any of the forms. I know because I just listed them all, and ran > >through their evaluation with no problem. > > Could you explain how you got the 6? I was wearing my bithead, thinking about the textual form of a predicate. I did neglect negation. > If you want to talk about how many possible boolean expressions there > are with n terms, that's 2^(2^n). There are 2^n assignments for the > variables, and any combination of those assignments could make > the expression true. There are 4 possible 1-term boolean expressions? Could you list them?Perhaps this is a terminology problem. By 3-term, I don't mean 3 variables used as often as you like, I mean 3 occurances of any variable and 2 operators (again neglecting negation). I am counting "A & (!A)" as a 2-term expression. > If you're just talking about how many disjunctions there are, even > that is 2^n because any combination of terms can be negated. > > Maybe what you're talking about is these?... Yep, the very beasts. > A + B + C > A & B + C > A &(B + C) > A + B & C > (A + B)& C > A & B & C > > However, these aren't really 6 distinct expressions. The third > and fifth are the same, as are the second and fourth. Hmmm. With A and B false and C true, the second expresion is true, while the fourth is false. The third and fifth expressions differ when A is false and B and C are true. (assuming '&' has higher precedence than '+') I guess the moral of the story is that this stuff really is hard, even at 3 terms. Tim