import Foundation
let pattern = #"(\$[^\n$]*[^\s$])(-|=|\+)([^\s$][^\n$]*\$)"#
let regex = try! NSRegularExpression(pattern: pattern)
let testString = #"""
\end{theorem}
A $k$-periodic sequence has the property that $s_i = s_{i + k}$ for all $i = 0,1,\dots$.
Thus a $k$-periodic sequence $(s_i)_{i = 0}^\infty$ may be represented by any finite sequence $(s_i)_{i=a}^{a+k - 1}$, where $a$ is usually chosen to be $0$.
Sadly our Fibonacci sequence examples are not defined over a finite field but over the naturals and thus are not necessarily periodic.
Examples such as these may be interpreted to have a period of $\infty$.
The period and related stability of linear recurrence sequences in regard to linear complexity has a very rich and broadly studied background~\cite{DingZiaoShan1991}.
\begin{theorem}
\label{th: max period is m-sequence}
\cite[Theorem~6.33]{LidlNiederreiter1994}
A linear recurrence sequence $s$ over a finite field $\gf_2$ with linear complexity $n$ has a maximum possible period of $2^n-1$.
\end{theorem}
\begin{definition}
\label{de: m-sequence}
A sequence which has maximum period for giv
"""#
let stringRange = NSRange(location: 0, length: testString.utf16.count)
let matches = regex.matches(in: testString, range: stringRange)
var result: [[String]] = []
for match in matches {
var groups: [String] = []
for rangeIndex in 1 ..< match.numberOfRanges {
let nsRange = match.range(at: rangeIndex)
guard !NSEqualRanges(nsRange, NSMakeRange(NSNotFound, 0)) else { continue }
let string = (testString as NSString).substring(with: nsRange)
groups.append(string)
}
if !groups.isEmpty {
result.append(groups)
}
}
print(result)
Please keep in mind that these code samples are automatically generated and are not guaranteed to work. If you find any syntax errors, feel free to submit a bug report. For a full regex reference for Swift 5.2, please visit: https://developer.apple.com/documentation/foundation/nsregularexpression