src/utils/next_combination.hpp

/*
 * Copyright (C) 2000-2004  Object Oriented Parallel Simulation Engine (OOPSE) project
 * 
 * Contact: oopse@oopse.org
 * 
 * This program is free software; you can redistribute it and/or
 * modify it under the terms of the GNU Lesser General Public License
 * as published by the Free Software Foundation; either version 2.1
 * of the License, or (at your option) any later version.
 * All we ask is that proper credit is given for our work, which includes
 * - but is not limited to - adding the above copyright notice to the beginning
 * of your source code files, and to any copyright notice that you may distribute
 * with programs based on this work.
 * 
 * This program is distributed in the hope that it will be useful,
 * but WITHOUT ANY WARRANTY; without even the implied warranty of
 * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
 * GNU Lesser General Public License for more details.
 * 
 * You should have received a copy of the GNU Lesser General Public License
 * along with this program; if not, write to the Free Software
 * Foundation, Inc., 59 Temple Place - Suite 330, Boston, MA  02111-1307, USA.
 *
 */

/**
 * @file GenerateCombination.hpp
 * @author    tlin
 * @date  10/27/2004
 * @version 1.0
 */ 

#ifndef UTILS_NEXT_COMBINATION_HPP
#define UTILS_NEXT_COMBINATION_HPP

#include <vector>
#include <iterator>
#include <iostream>
namespace oopse {

/**
 * @fn bool next_combination(IteratorContainer<RandomAccessIterator>& iterContainer,
 *                                           RandomAccessIterator first, RandomAccessIterator last)
 * @brief STL next_permuationtation like combination sequence generator.
 * Given the first and last iterator of a sequence, next_combination iteratively generates all 
 * possible combination.
 * @return if more combination is availiable, otherwise return false
 * @param iterContainer iterator container
 * @param first the first iterator
 * @param last the last iterator
 * @note first and last must be random access iterators and iterContainer must be the container of  
 * random access iterators . And all of the iteratos in iterContainer must be within the range [first, last)
 *
 * @code
 * std::vector<int> iv;
 * iv.push_back(1);
 * iv.push_back(8);
 * std::vector<std::vector<int>::iterator> ic;
 * while(next_combination(ic, iv.begin(), iv.end())) {
 *     for (i =  ic.begin(); i < ic.end(); ++i) {
 *         std::cout << **i << "\t";
 *     }
 *     std::cout << std::endl;
 * }
 * //output
 * //1
 * //8
 * //1  8
 */
template<class RandomAccessIterator, template<typename ELEM, typename = std::allocator<ELEM> > class IteratorContainer>
bool next_combination(IteratorContainer<RandomAccessIterator>& iterContainer, RandomAccessIterator first, RandomAccessIterator last) {
    if (first == last) {
        return false;
    }
    
    RandomAccessIterator endIter = --last;
    typename IteratorContainer<RandomAccessIterator>::iterator i = iterContainer.end();
    
    if (iterContainer.empty()) {
        //if sequence is empty, we insert the first iterator
        iterContainer.insert(iterContainer.end(), first);
        return true;
    } else if (*(--i) != endIter){
        //if the last iterator in iterContainer does not reaches the end, just increment it 
        ++(*i);
        return true;
    } else {// the last iterator in iterContainer does not reaches the end

        //starts at the end of the sequence and works its way towards the front, looking for two 
        //consecutive members of the sequence where the difference between them is greater 
        //than one. For example , if the sequence contains 1, 5, 8, 9 (total number is 10, begin
        //index is 0, therefore 9 is the end index, and the current size is 4). At the end of while
        //loop, j will point to 5, and i will point to 8, next combination should be 1, 6, 7, 8.
        //If j is less than zero, it means it already reaches the last combination of current size.
        //For instance, sequence may contain 6, 7, 8, 9 at this time, we need to increase the size
        // of combination to 5
        typename IteratorContainer<RandomAccessIterator>::iterator j = i;
        j--;
        while( j >= iterContainer.begin() && *i == *j + 1){
            i--;
            j--;
        };

        RandomAccessIterator raIter;
        if (j - iterContainer.begin() < 0) { //reaches the last combination of current size
            //half open range
            if (last  - first + 1  == iterContainer.size()) {
                //if the current size equals to total number, done
                return false;
            } else {

                //push the first currentSize+1 element into sequence 
                
                for(i = iterContainer.begin(), raIter= first; i  != iterContainer.end(); ++i, ++raIter) {
                    *i = raIter;
                }    
                iterContainer.insert(iterContainer.end(), raIter);
                
                return true; 
            }
            
        } else {
            ++(*j);
            raIter = *j;
            for(; i != iterContainer.end(); ++i) {
                ++raIter;
                *i = raIter;
            }
            return true;
        }
    }
} //end next_combination

/**
 * @fn bool replaceWildCard(std::vector<std::vector<std::string>::iterator>& cont,
 *                                        std::vector<std::string>& sequence, std::vector<std::string>& result,
 *                                        const std::string& wildCard)
 * @brief iteratively replace the sequence with wild cards
 * @return true if more combination sequence is avaliable, otherwise return true
 * @param cont iterator container, if expect whole series of combination, just pass an empty iterator 
 * container. The user should not modify the iterator container
 * @param sequence the whole sequence used to generate combination
 * @param result a possible combination sequence which is set on return
 * @wildCard the wild card string. Its value is "X" by default
 * @note since next_combination never returns an empty sequence, replaceWildCard will not generate 
 * one special combination, which is n identical wild cards (n is equal to the size of the passing sequence)
 * 
 * @code
 * std::vector<std::string> sv;
 * std::vector<std::vector<std::string>::iterator> sic;
 * std::vector<std::string> resultString;
 * sv.push_back("H");
 * sv.push_back("C");
 * sv.push_back("N");

 * while (replaceWildCard(sic, sv, resultString)) {   
 *     for(std::vector<std::string>::iterator i = resultString.begin(); i != resultString.end(); ++i) {
 *         std::cout << *i << "\t";
 *     }
 *     std::cout << std::endl;
 * }
 * //output
 * //H X X
 * //X C X
 * //X X N
 * //H C X
 * //H X N
 * //X C N
 * //H C N
 @endcode
 */
bool replaceWildCard(std::vector<std::vector<std::string>::iterator>& cont,
                                             std::vector<std::string>& sequence, std::vector<std::string>& result,
                                             const std::string& wildCard = "X") {
    if (cont.size() > sequence.size()) {
        std::cerr << "the size of iterator container is greater than the size of sequence";
    }

    bool hasMoreCombination = next_combination(cont, sequence.begin(), sequence.end());
    if (hasMoreCombination) {
        result.clear();
        result.insert(result.begin(), sequence.size(), wildCard);
        std::vector<std::vector<std::string>::iterator>::iterator i;
        for ( i = cont.begin(); i != cont.end(); i++){
            result[*i - sequence.begin()] = **i;
        }
    }

     return hasMoreCombination;
    
}

} //end namespace oopse
#endif //UTILS_NEXT_COMBINATION_HPP

Revision:	185
Committed:	Thu Oct 28 21:11:12 2004 UTC (21 years ago) by tim
File size:	7718 byte(s)
Log Message:	change documentation of next_combination.hpp
#	Content
1	/*
2	* Copyright (C) 2000-2004 Object Oriented Parallel Simulation Engine (OOPSE) project
3	*
4	* Contact: oopse@oopse.org
5	*
6	* This program is free software; you can redistribute it and/or
7	* modify it under the terms of the GNU Lesser General Public License
8	* as published by the Free Software Foundation; either version 2.1
9	* of the License, or (at your option) any later version.
10	* All we ask is that proper credit is given for our work, which includes
11	* - but is not limited to - adding the above copyright notice to the beginning
12	* of your source code files, and to any copyright notice that you may distribute
13	* with programs based on this work.
14	*
15	* This program is distributed in the hope that it will be useful,
16	* but WITHOUT ANY WARRANTY; without even the implied warranty of
17	* MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
18	* GNU Lesser General Public License for more details.
19	*
20	* You should have received a copy of the GNU Lesser General Public License
21	* along with this program; if not, write to the Free Software
22	* Foundation, Inc., 59 Temple Place - Suite 330, Boston, MA 02111-1307, USA.
23	*
24	*/
25
26	/**
27	* @file GenerateCombination.hpp
28	* @author tlin
29	* @date 10/27/2004
30	* @version 1.0
31	*/
32
33	#ifndef UTILS_NEXT_COMBINATION_HPP
34	#define UTILS_NEXT_COMBINATION_HPP
35
36	#include <vector>
37	#include <iterator>
38	#include <iostream>
39	namespace oopse {
40
41	/**
42	* @fn bool next_combination(IteratorContainer<RandomAccessIterator>& iterContainer,
43	* RandomAccessIterator first, RandomAccessIterator last)
44	* @brief STL next_permuationtation like combination sequence generator.
45	* Given the first and last iterator of a sequence, next_combination iteratively generates all
46	* possible combination.
47	* @return if more combination is availiable, otherwise return false
48	* @param iterContainer iterator container
49	* @param first the first iterator
50	* @param last the last iterator
51	* @note first and last must be random access iterators and iterContainer must be the container of
52	* random access iterators . And all of the iteratos in iterContainer must be within the range [first, last)
53	*
54	* @code
55	* std::vector<int> iv;
56	* iv.push_back(1);
57	* iv.push_back(8);
58	* std::vector<std::vector<int>::iterator> ic;
59	* while(next_combination(ic, iv.begin(), iv.end())) {
60	* for (i = ic.begin(); i < ic.end(); ++i) {
61	* std::cout << **i << "\t";
62	* }
63	* std::cout << std::endl;
64	* }
65	* //output
66	* //1
67	* //8
68	* //1 8
69	*/
70	template<class RandomAccessIterator, template<typename ELEM, typename = std::allocator<ELEM> > class IteratorContainer>
71	bool next_combination(IteratorContainer<RandomAccessIterator>& iterContainer, RandomAccessIterator first, RandomAccessIterator last) {
72	if (first == last) {
73	return false;
74	}
75
76	RandomAccessIterator endIter = --last;
77	typename IteratorContainer<RandomAccessIterator>::iterator i = iterContainer.end();
78
79	if (iterContainer.empty()) {
80	//if sequence is empty, we insert the first iterator
81	iterContainer.insert(iterContainer.end(), first);
82	return true;
83	} else if (*(--i) != endIter){
84	//if the last iterator in iterContainer does not reaches the end, just increment it
85	++(*i);
86	return true;
87	} else {// the last iterator in iterContainer does not reaches the end
88
89	//starts at the end of the sequence and works its way towards the front, looking for two
90	//consecutive members of the sequence where the difference between them is greater
91	//than one. For example , if the sequence contains 1, 5, 8, 9 (total number is 10, begin
92	//index is 0, therefore 9 is the end index, and the current size is 4). At the end of while
93	//loop, j will point to 5, and i will point to 8, next combination should be 1, 6, 7, 8.
94	//If j is less than zero, it means it already reaches the last combination of current size.
95	//For instance, sequence may contain 6, 7, 8, 9 at this time, we need to increase the size
96	// of combination to 5
97	typename IteratorContainer<RandomAccessIterator>::iterator j = i;
98	j--;
99	while( j >= iterContainer.begin() && i == j + 1){
100	i--;
101	j--;
102	};
103
104	RandomAccessIterator raIter;
105	if (j - iterContainer.begin() < 0) { //reaches the last combination of current size
106	//half open range
107	if (last - first + 1 == iterContainer.size()) {
108	//if the current size equals to total number, done
109	return false;
110	} else {
111
112	//push the first currentSize+1 element into sequence
113
114	for(i = iterContainer.begin(), raIter= first; i != iterContainer.end(); ++i, ++raIter) {
115	*i = raIter;
116	}
117	iterContainer.insert(iterContainer.end(), raIter);
118
119	return true;
120	}
121
122	} else {
123	++(*j);
124	raIter = *j;
125	for(; i != iterContainer.end(); ++i) {
126	++raIter;
127	*i = raIter;
128	}
129	return true;
130	}
131	}
132	} //end next_combination
133
134	/**
135	* @fn bool replaceWildCard(std::vector<std::vector<std::string>::iterator>& cont,
136	* std::vector<std::string>& sequence, std::vector<std::string>& result,
137	* const std::string& wildCard)
138	* @brief iteratively replace the sequence with wild cards
139	* @return true if more combination sequence is avaliable, otherwise return true
140	* @param cont iterator container, if expect whole series of combination, just pass an empty iterator
141	* container. The user should not modify the iterator container
142	* @param sequence the whole sequence used to generate combination
143	* @param result a possible combination sequence which is set on return
144	* @wildCard the wild card string. Its value is "X" by default
145	* @note since next_combination never returns an empty sequence, replaceWildCard will not generate
146	* one special combination, which is n identical wild cards (n is equal to the size of the passing sequence)
147	*
148	* @code
149	* std::vector<std::string> sv;
150	* std::vector<std::vector<std::string>::iterator> sic;
151	* std::vector<std::string> resultString;
152	* sv.push_back("H");
153	* sv.push_back("C");
154	* sv.push_back("N");
155
156	* while (replaceWildCard(sic, sv, resultString)) {
157	* for(std::vector<std::string>::iterator i = resultString.begin(); i != resultString.end(); ++i) {
158	* std::cout << *i << "\t";
159	* }
160	* std::cout << std::endl;
161	* }
162	* //output
163	* //H X X
164	* //X C X
165	* //X X N
166	* //H C X
167	* //H X N
168	* //X C N
169	* //H C N
170	@endcode
171	*/
172	bool replaceWildCard(std::vector<std::vector<std::string>::iterator>& cont,
173	std::vector<std::string>& sequence, std::vector<std::string>& result,
174	const std::string& wildCard = "X") {
175	if (cont.size() > sequence.size()) {
176	std::cerr << "the size of iterator container is greater than the size of sequence";
177	}
178
179	bool hasMoreCombination = next_combination(cont, sequence.begin(), sequence.end());
180	if (hasMoreCombination) {
181	result.clear();
182	result.insert(result.begin(), sequence.size(), wildCard);
183	std::vector<std::vector<std::string>::iterator>::iterator i;
184	for ( i = cont.begin(); i != cont.end(); i++){
185	result[i - sequence.begin()] = *i;
186	}
187	}
188
189	return hasMoreCombination;
190
191	}
192
193	} //end namespace oopse
194	#endif //UTILS_NEXT_COMBINATION_HPP
195