Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

Wildcard string matching

Tags:

algorithm

What is the most efficient wildcard string matching algorithm? I am asking only about an idea, it is not necessary to provide actual code.

I'm thinking that such algorithm can be built with sorted suffix arrays, which can yield performance of O(log(n)).

Am I correct?

Edited:

I mean patterns like "A*B", "*sip*" or "A?B" where star means any number of symbols and question mark means single symbol.

like image 952
Vitaliy Liptchinsky Avatar asked Jan 19 '10 12:01

Vitaliy Liptchinsky


People also ask

What is wildcard pattern matching?

A wildcard pattern is a series of characters that are matched against incoming character strings. You can use these patterns when you define pattern matching criteria. Matching is done strictly from left to right, one character or basic wildcard pattern at a time.

What is wildcard string?

In software, a wildcard character is a kind of placeholder represented by a single character, such as an asterisk ( * ), which can be interpreted as a number of literal characters or an empty string. It is often used in file searches so the full name need not be typed.

How do you wildcard a string in Python?

the Asterisk * Wildcard in Python The * character or the asterisk can specify any number of characters. The asterisk * is mostly utilized at the end of the given root word and when there is a need to search for endings with several possibilities for the given root word.


4 Answers

There is a paper covering the fastest options here http://swtch.com/~rsc/regexp/regexp1.html in particular it allows you avoid naive algorithms that become pathologically slow when long patterns are used.

It covers generic regular expressions but you can limit your implementation to the subset you require.

like image 148
martinwguy Avatar answered Nov 16 '22 01:11

martinwguy


Hm, I think that normal pattern matching rules would apply here. Usually, since you have a stream of data and short patterns, you would not need to implement something more efficient than linear. However, the longer the pattern gets, the more room there is for optimization.

What kind of wildcard do you have in mind? a one-character-wildcard (e.g. . in regex), or a multiple-character-wildcard (e.g. .*)? Are there limitations? What is the expected pattern length, and do you have random or serial access to the data to be checked?

like image 32
Lucero Avatar answered Nov 16 '22 01:11

Lucero


I was looking for a simple wildcard matching algorithm which runs in polynomial time. E.g. this one is simple, but doesn't run in polynomial time when the pattern contains many stars (*): http://www.codeproject.com/Articles/188256/A-Simple-Wildcard-Matching-Function Below is the code which uses dynamic programming to reduce the time complexity to O(n*m) where n is the length of the text and m is the length of the pattern.

#include <string>
#include <vector>
#include <algorithm>
using namespace std;

const int UNKNOWN = -1;
const int NOMATCH = 0;
const int MATCHES = 1;

class Wildcard {
    string _text;
    string _pattern;
    vector<vector<int>> _mf;
    int F(int n, int m) {
        if (_mf[n][m] >= 0) return _mf[n][m];
        if (n == 0 && m == 0) {
            _mf[n][m] = MATCHES;
            return _mf[n][m];
        }
        if (n > 0 && m == 0) {
            _mf[n][m] = NOMATCH;
            return _mf[n][m];
        }
        // m > 0
        int ans = NOMATCH;
        if (_pattern[m - 1] == '*') {
            ans = max(ans, F(n, m-1));
            if (n > 0) {
                ans = max(ans, F(n - 1, m));
            }
        }
        if (n > 0) {
            if (_pattern[m - 1] == '?' || _pattern[m - 1] == _text[n - 1]) {
                ans = max(ans, F(n - 1, m - 1));
            }
        }
        _mf[n][m] = ans;
        return _mf[n][m];
    }
public:
    bool match(string text, string pattern) {
        _text = text;
        _pattern = pattern;
        _mf.clear();
        for (int i = 0; i <= _text.size(); i++) {
            _mf.push_back(vector<int>());
            for (int j = 0; j <= _pattern.size(); j++) {
                _mf[i].push_back(UNKNOWN); // not calculated
            }
        }
        int ans = F(_text.size(), _pattern.size());
        return ans == MATCHES;
    }
};
like image 27
Serge Rogatch Avatar answered Nov 16 '22 01:11

Serge Rogatch


If your pattern can contain only * wild cards, then the trivial implementation is as fast as possible. The main thing to realize in this case, is that you only need to look for each card once (card = segment between stars).

Here is an implementation (supporting only * wild cards):

#include <cstddef>
#include <cstring>
#include <algorithm>
#include <string>
#include <vector>
#include <iostream>

using namespace std;

class wildcard_pattern {
public:
    explicit wildcard_pattern(const string& text);

    bool match(const char* begin, const char* end) const;

    bool match(const char* c_str) const;

private:
    string m_text;

    struct card {
        size_t m_offset, m_size;
        card(size_t begin, size_t end);
    };

    // Must contain at least one card. The first, and the last card
    // may be empty strings. All other cards must be non-empty. If
    // there is exactly one card, the pattern matches a string if, and
    // only if the string is equal to the card. Otherwise, the first
    // card must be a prefix of the string, and the last card must be
    // a suffix.
    vector<card> m_cards;
};


wildcard_pattern::wildcard_pattern(const string& text):
    m_text(text)
{
    size_t pos = m_text.find('*');
    if (pos == string::npos) {
        m_cards.push_back(card(0, m_text.size()));
        return;
    }
    m_cards.push_back(card(0, pos));
    ++pos;
    for (;;) {
        size_t pos_2 = m_text.find('*', pos);
        if (pos_2 == string::npos)
            break;
        if (pos_2 != pos)
            m_cards.push_back(card(pos, pos_2));
        pos = pos_2 + 1;
    }
    m_cards.push_back(card(pos, m_text.size()));
}

bool wildcard_pattern::match(const char* begin, const char* end) const
{
    const char* begin_2 = begin;
    const char* end_2   = end;

    size_t num_cards = m_cards.size();
    typedef string::const_iterator str_iter;

    // Check anchored prefix card
    {
        const card& card = m_cards.front();
        if (size_t(end_2 - begin_2) < card.m_size)
            return false;
        str_iter card_begin = m_text.begin() + card.m_offset;
        if (!equal(begin_2, begin_2 + card.m_size, card_begin))
            return false;
        begin_2 += card.m_size;
    }

    if (num_cards == 1)
        return begin_2 == end_2;

    // Check anchored suffix card
    {
        const card& card = m_cards.back();
        if (size_t(end_2 - begin_2) < card.m_size)
            return false;
        str_iter card_begin = m_text.begin() + card.m_offset;
        if (!equal(end_2 - card.m_size, end_2, card_begin))
            return false;
        end_2 -= card.m_size;
    }

    // Check unanchored infix cards
    for (size_t i = 1; i != num_cards-1; ++i) {
        const card& card = m_cards[i];
        str_iter card_begin = m_text.begin() + card.m_offset;
        str_iter card_end   = card_begin + card.m_size;
        begin_2 = search(begin_2, end_2, card_begin, card_end);
        if (begin_2 == end_2)
            return false;
        begin_2 += card.m_size;
    }

    return true;
}

inline bool wildcard_pattern::match(const char* c_str) const
{
    const char* begin = c_str;
    const char* end   = begin + strlen(c_str);
    return match(begin, end);
}

inline wildcard_pattern::card::card(size_t begin, size_t end)
{
    m_offset = begin;
    m_size   = end - begin;
}


int main(int, const char* argv[])
{
    wildcard_pattern pat(argv[1]);
    cout << pat.match(argv[2]) << endl;
}
like image 23
Kristian Spangsege Avatar answered Nov 16 '22 01:11

Kristian Spangsege