analysis

How many recursions are made before every single root is found? Also, which ones are the roots? Here's my code: e=0.000001; f1=@(x) 14.*x.*exp(x-2)-12.*exp(x-2)-7.*x.^3+20.*x.^2-26.*x+12; a=0; c=3; while abs(c-a)>e b=(c+a)/2; if f1(a)*f1(b)<0 c=b; else a=b; end disp(b); end Bisection works by taking endpoints of some initial interval [a,b] and finding which half of the interval must contain the root (it evaluates the midpoint, and identifies which half has the sign change). Then bisection repeats the process on the identified half. Bisection converges upon only one possible root, and if your

I am having an issue with Regular expression, I need the most efficient regex that match IP address and in range of 255 only. I tried this one " ip_pattern = '\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3}' " , but it does match even numbers over 255, such as 321.222.11.4 This should do it: ^(?:(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.){3}(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)$ Use this Regex. It will match and check the IP range within 255. \b(?:25[0-5]|2[0-4][0-9]|[0-1]?[0-9]?[0-9]).(?:25[0-5]|2[0-4][0-9]|[0-1]?[0-9]?[0-9]).(?:25[0-5]|2[0-4][0-9]|[0-1]?[0-9]?[0-9]).(?:25[0-5]|2[0-4][0-9]|[0-1]?[0-9]?

I am trying to index strings that contain hyphens but do not contain spaces, periods or any other punctuation. I do not want to split up the words based on hyphens, instead I would like to have the hyphens be part of the indexed text. For example, my 6 text strings would be: magazineplayon magazineofhorses online-magazine best-magazine friend-of-magazines magazineplaygames I would like to be able to search these string for the text containing "play" or for the text starting with "magazine" . I have been able to use ngram to make the text containing "play" work properly. However, the hyphen is

I have a problem and I can not get the purpose of lines (14,15,16,17) of this site for Select algorithm. The site which had this question was located here . EDITED: Also, is this correct to write these lines for the part "partition and recur by using the pivot" ? ("m" is my pivot and "i" is the input of this algorithm) arrOne<--{a of arr : a<m} arrTwo<--{a of arr : a>m} if (i < m ) then return Select(arrOne,i) else if (i > m) then return Select(arrTwo,i-m) else return m This is where it selects the partition in which to recurse. For the sake of illustration, let's assume you're looking for the

In my Computer Science II class, the professor considers ++,--,*=, etc. to be 2 operations. However, at the Assembly level this is not really two operations. Can someone explain or is this just for the sake of simplicity? I'd actually consider it to be 3 operations: read, increment (or whatever), write. That's assuming it's reading from some sort of shared memory into some sort of local storage (e.g. register or stack), operating on the local storage, then writing back. How many operations it is at assembly level will depend on what you're incrementing, the platform, the hardware etc. Because

I am a newbie in r. I have a set of data about online videos and their tags. The data looks like film tag1 tag2 tag3 tag4.... 1 A B C D 2 A C F G 3 B D C X I want to create a matrix which tells me the co-occurrence of the tags, such as: A B C D ..... A 10 13 B 15 2 C 3 16 D 9 20 How should I do it? If I understand what you want here is one way: dat <- read.table(text='film tag1 tag2 tag3 tag4 1 A B C D 2 A C F G 3 B D C X', header=T) library(qdapTools) crossprod(as.matrix(mtabulate(as.data.frame(t(dat[, -1]))))) Giving: A B C D F G X A 2 1 2 1 1 1 0 B 1 2 2 2 0 0 1 C 2 2 3 2 1 1 1 D 1 2 2 2 0