1. 定期管理 ht ——子句重复性所在的哈希表
1 //lbool Solver::solve_() 2 ... 3 4 if (dupl_db_size >= dupl_db_size_limit){ 5 printf("c Duplicate learnts added (Minimization) %i./n",duplicates_added_minimization); 6 printf("c Duplicate learnts added (conflicts) %i./n",duplicates_added_conflicts); 7 printf("c Duplicate learnts added (tier2) %i./n",duplicates_added_tier2); 8 printf("c Duptime: %i/n",duptime.count()); 9 printf("c Number of conflicts: %i/n",conflicts); 10 printf("c Core size: %i/n",learnts_core.size()); 11 12 uint32_t removed_duplicates = 0; 13 std::vector<std::vector<uint64_t>> tmp; 14 //std::map<int32_t,std::map<uint32_t,std::unordered_map<uint64_t,uint32_t>>> ht; 15 for (auto & outer_mp: ht){//variables 16 for (auto &inner_mp:outer_mp.second){//sizes 17 for (auto &in_in_mp: inner_mp.second){ 18 if (in_in_mp.second >= min_number_of_learnts_copies){ 19 tmp.push_back({
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2. 化简函数中考察当前子句集每一个子句c重复出现次数id,并做相应的处理
1 //bool Solver::simplifyLearnt_tier2() 2 ... 3 4 //duplicate learnts 5 int id = 0; 6 7 std::vector<uint32_t> tmp; 8 for (int i = 0; i < c.size(); i++) 9 tmp.push_back(c[i].x); // Lit定义文字对应的数据成员 int x; 10 id = is_duplicate(tmp); 11 12 13 //duplicate learnts 14 15 if (id < min_number_of_learnts_copies+2){ 16 attachClause(cr); 17 learnts_tier2[cj++] = learnts_tier2[ci]; 18 if (id == min_number_of_learnts_copies+1){ 19 duplicates_added_minimization++; 20 } 21 if ((c.lbd() <= core_lbd_cut)||
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3. search函数中,考察当新生学习子句c是否重复出现,如果重复出现则在函数is_duplicate(std::vector<uint32_t>&c)中
it2->second++; //将该子句重复出现计数值增加1.
1 //lbool Solver::search(int& nof_conflicts) 2 ... 3 4 if (learnt_clause.size() == 1){ 5 uncheckedEnqueue(learnt_clause[0]); 6 }else{ 7 CRef cr = ca.alloc(learnt_clause, true); 8 ca[cr].set_lbd(lbd); 9 //duplicate learnts 10 int id = 0; 11 if (lbd <= max_lbd_dup){ 12 std::vector<uint32_t> tmp; 13 for (int i = 0; i < learnt_clause.size(); i++) 14 tmp.push_back(learnt_clause[i].x); 15 id = is_duplicate(tmp); 16 if (id == min_number_of_learnts_copies +1){ 17 duplicates_added_conflicts++; 18 } 19 if (id == min_number_of_learnts_copies){ 20 duplicates_added_tier2++; 21 } 22 } 23 //duplicate learnts 24 25 if ((lbd <= core_lbd_cut) || (id == min_number_of_learnts_copies+1)){ 26 learnts_core.push(cr); 27 ca[cr].mark(CORE); 28 }else if ((lbd <= 6)||(id == min_number_of_learnts_copies)){ 29 learnts_tier2.push(cr); 30 ca[cr].mark(TIER2); 31 ca[cr].touched() = conflicts; 32 }else{ 33 learnts_local.push(cr); 34 claBumpActivity(ca[cr]); } 35 attachClause(cr); 36 37 uncheckedEnqueue(learnt_clause[0], backtrack_level, cr); 38 #ifdef PRINT_OUT 39 std::cout << "new " << ca[cr] << "/n"; 40 std::cout << "ci " << learnt_clause[0] << " l " << backtrack_level << "/n"; 41 #endif 42 } 43 44 ...
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4. 函数 is_duplicate(std::vector<uint32_t>&c) 及其主要数据量ht
1 //Solver.h 2 3 // duplicate learnts version 4 std::map<int32_t,std::map<uint32_t,std::unordered_map<uint64_t,uint32_t>>> ht; // duplicate learnts version uint32_t min_number_of_learnts_copies;
uint64_t duplicates_added_conflicts;
5 // duplicate learnts version
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//size_t dupl_db_size_limit = dupl_db_init_size; //dupl_db_size_limit*=1.1; |
1 int Solver::is_duplicate(std::vector<uint32_t>&c){ 2 auto time_point_0 = std::chrono::high_resolution_clock::now(); 3 dupl_db_size++; 4 int res = 0; 5 6 int sz = c.size(); 7 std::vector<uint32_t> tmp(c); 8 sort(tmp.begin(),tmp.end()); 9 10 uint64_t hash = 0; 11 12 for (int i =0; i<sz; i++) { 13 hash ^= tmp[i] + 0x9e3779b9 + (hash << 6) + (hash>> 2); 14 } 15 16 int32_t head = tmp[0]; 17 auto it0 = ht.find(head); 18 if (it0 != ht.end()){ 19 auto it1=ht[head].find(sz); 20 if (it1 != ht[head].end()){ 21 auto it2 = ht[head][sz].find(hash); 22 if (it2 != ht[head][sz].end()){ 23 it2->second++; 24 res = it2->second; 25 } 26 else{ 27 ht[head][sz][hash]=1; 28 } 29 } 30 else{ 31 ht[head][sz][hash]=1; 32 } 33 }else{ 34 ht[head][sz][hash]=1; 35 } 36 auto time_point_1 = std::chrono::high_resolution_clock::now(); 37 duptime += std::chrono::duration_cast<std::chrono::microseconds>(
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原创文章,作者:ItWorker,如若转载,请注明出处:https://blog.ytso.com/289920.html