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Added clarification to the proof of the extmatch length heuristic

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Snaipe 2015-09-09 02:56:54 +02:00
parent f7d4d7616a
commit 9896377a47
1 changed files with 8 additions and 8 deletions
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16
src/extmatch.c
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@ -248,21 +248,21 @@ static int transform(const char *pattern, char *result, const char **errmsg) {
* We must now find the maximal length L such as ∀s, L >= length(T(s))
*
* It is immediately apparent that the largest string will depend on the number
* of occurrences of '!()'. Hence, let u be a string that is a repeating
* sequence of '!()' padded by '.' to a multiple of 3,
* of occurrences of '!()'. Hence, ∀s, let u(s) be a string that is a repeating
* sequence of '!()' padded by at most two '.', such as length(u(s)) == length(s),
*
* let N = floor(length(u) / 3),
* let Q = length(u) mod 3,
* let N = floor(length(u(s)) / 3),
* let Q = length(u(s)) mod 3,
* hence num('!()') = N.
*
* ∀s | lenght(s) = length(u),
* length(T(s)) <= length(T(u))
* <= length(u) | the original length
* ∀s | lenght(s) = length(u(s)),
* length(T(s)) <= length(T(u(s)))
* <= length(u(s)) | the original length
* + 4 * N | the expansion of all '!()'
* + Q * diff('.') | the expansion of Q '.'
* <= 3 * N + Q + 4 * N + Q
* <= 7 * N + 4
* <= 7 * floor(length(u) / 3) + 4
* <= 7 * floor(length(u(s)) / 3) + 4
* <= 7 * floor(length(s) / 3) + 4
*
*/