Path: news.mathworks.com!not-for-mail From: <HIDDEN> Newsgroups: comp.soft-sys.matlab Subject: Re: bitxor for n elements Date: Mon, 29 Dec 2008 16:40:18 +0000 (UTC) Organization: The MathWorks, Inc. Lines: 18 Message-ID: <gjauhi$4j2$1@fred.mathworks.com> References: <gjafh2$6kq$1@fred.mathworks.com> <gjag5n$fl8$1@fred.mathworks.com> <f791474c-8fa9-4300-a86e-7163970ab1e4@z6g2000pre.googlegroups.com> Reply-To: <HIDDEN> NNTP-Posting-Host: webapp-05-blr.mathworks.com Content-Type: text/plain; charset="ISO-8859-1" Content-Transfer-Encoding: 8bit X-Trace: fred.mathworks.com 1230568818 4706 172.30.248.35 (29 Dec 2008 16:40:18 GMT) X-Complaints-To: news@mathworks.com NNTP-Posting-Date: Mon, 29 Dec 2008 16:40:18 +0000 (UTC) X-Newsreader: MATLAB Central Newsreader 1187260 Xref: news.mathworks.com comp.soft-sys.matlab:509099 ImageAnalyst <imageanalyst@mailinator.com> wrote in message <f791474c-8fa9-4300-a86e-7163970ab1e4@z6g2000pre.googlegroups.com>... > On the odd chance that you're looking to xor three numbers.... > Since xor is normally "this or that but not both" then you might > define one for three numbers as > "number1 or number 2 or number 2 but not any two nor all three" > In that case, you can just sum the numbers (if they're binary 0/1) and > check to see if the sum is exactly 1. If the total of all three > numbers is 1 then it's true, else it's false. > Regards, > ImageAnalyst That's not quite right, ImageAnalyst. For three booleans it should be either any one of them or else all three. If summing any number of 0/1 booleans, the sum should be an odd number for a true answer to 'xor'. However, Suzi mentioned 'bitxor' and such summation wouldn't work in a bitwise manner. As David states looping is undoubtedly the best way. Roger Stafford