Path: news.mathworks.com!not-for-mail From: "Sadik " <sadik.hava@gmail.com> Newsgroups: comp.soft-sys.matlab Subject: Re: Programming Question Date: Sun, 21 Jun 2009 20:47:01 +0000 (UTC) Organization: The MathWorks, Inc. Lines: 28 Message-ID: <h1m685$150$1@fred.mathworks.com> References: <h1m34q$fod$1@fred.mathworks.com> Reply-To: "Sadik " <sadik.hava@gmail.com> NNTP-Posting-Host: webapp-02-blr.mathworks.com Content-Type: text/plain; charset="ISO-8859-1" Content-Transfer-Encoding: 8bit X-Trace: fred.mathworks.com 1245617221 1184 172.30.248.37 (21 Jun 2009 20:47:01 GMT) X-Complaints-To: news@mathworks.com NNTP-Posting-Date: Sun, 21 Jun 2009 20:47:01 +0000 (UTC) X-Newsreader: MATLAB Central Newsreader 1666517 Xref: news.mathworks.com comp.soft-sys.matlab:549342 You seem to have figured out. I am writing the equations in case you haven't: q = 0.4; n = 2; d = [0 1 1 2;1 0 2 1;1 2 0 1;2 1 1 0]; M=q^n * ((1-q)/(q))^d; Best. "Rafael " <rrodriguez1989@gmail.com> wrote in message <h1m34q$fod$1@fred.mathworks.com>... > Hello, I am having issues with inputing this equation into matlab. > > basic equation: > M=q^n * ((1-q)/(q))^d > > q and n are both inputs from the user > > d is a hamming distance that is calculated based on a length > > the length is the input n (the length is the number of points on the line) > > depending on the length, each point can have a 1 or 0 entry > > i.e. if n is 2, the length is 2, and each of the two points can have a 1 or 0 entry, therefore there are 4 different arrangements in a length of 2 units > > d is the hamming distance between each of these possible arrangements > > this may seem confusing, but any help would be greatly appreciated, i can clarify any questions that may be asked.