minisat数据成员ok的使用分析及两大类化简操作

 

定义	// Solver state: // bool ok; // If FALSE, the constraints are already unsatisfiable. No part of the solver state may be used!
初始化	, ok                 (true)
使用	在以下函数中数据成员ok被使用
	bool Solver::simplifyLearnt_x(vec<CRef>& learnts_x)； bool Solver::simplifyLearnt_core()； bool Solver::simplifyLearnt_tier2()； bool Solver::simplifyAll()； bool Solver::addClause_(vec<Lit>& ps); bool Solver::simplify(); lbool Solver::search(int& nof_conflicts){assert(ok);......} lbool Solver::solve_() {     // signal(SIGALRM, SIGALRM_switch);     // alarm(2500);     model.clear();     conflict.clear();     if (!ok) return l_False;     solves++;      ...... } void Solver::toDimacs(FILE* f, const vec<Lit>& assumps);
simplifyAll 化简的 含义: 移除 已满足 的子句	simplifyAll集中使用在每次重启（实际进行了条件限定）之后，见search函数的前部。 主要进行的工作是：和simplify函数只是移除sat子句不同，是真正利用自动推理相关原理进行化简。 simplifyAll直接关联的化简函数如下： ——void Solver::simpleAnalyze(CRef confl, vec<Lit>& out_learnt, vec<CRef>& reason_clause, bool True_confl) ——void Solver::simplifyLearnt(Clause& c) ——bool Solver::simplifyLearnt_x(vec<CRef>& learnts_x) ——bool Solver::simplifyLearnt_core() ——bool Solver::simplifyLearnt_tier2()   bool Solver::simplifyAll() { //// simplified_length_record = original_length_record = 0; if (!ok || propagate() != CRef_Undef) return ok = false; //// cleanLearnts(also can delete these code), here just for analyzing //if (local_learnts_dirty) cleanLearnts(learnts_local, LOCAL); //if (tier2_learnts_dirty) cleanLearnts(learnts_tier2, TIER2); //local_learnts_dirty = tier2_learnts_dirty = false; if (!simplifyLearnt_core()) return ok = false; if (!simplifyLearnt_tier2()) return ok = false; //if (!simplifyLearnt_x(learnts_local)) return ok = false; checkGarbage(); //// // printf("c size_reduce_ratio : %4.2f%%\n", // original_length_record == 0 ? 0 : //(original_length_record - simplified_length_record) * 100 / (double)original_length_record); return true; }
	赋值改变的操作体现在几个化简函数体中。主要代码段如下：       return ok = (propagate() == CRef_Undef);//传播没有遇到冲突返回值为CRef_Undef   if (c.size() == 1){ // when unit clause occur, enqueue and propagate uncheckedEnqueue(c[0]); if (propagate() != CRef_Undef){ ok = false; return false; } // delete the clause memory in logic c.mark(1); ca.free(cr); //#ifdef BIN_DRUP // binDRUP('d', c, drup_file); //#else // fprintf(drup_file, "d "); // for (int i = 0; i < c.size(); i++) // fprintf(drup_file, "%i ", (var(c[i]) + 1) * (-2 * sign(c[i]) + 1)); // fprintf(drup_file, "0\n"); //#endif } 。。。。。。。 }
simplify 化简的 含义: 移除 已满足 的子句	【前 提】：传播的0层中已经有若干文字（bool Solver::addClause_(vec<Lit>& ps)增加子句时,                 单文字子句直接进入传播的0层）   【simplify化简操作调用】：       1.在真正的CDCL开始之间已经被调用过，见派生类函数：bool SimpSolver::eliminate_()之中。       2.化简函数simplify其余的调用都在每次search函数被调用，见lbool Solver::search(int& nof_conflicts)之中。          在传播没有发现冲突阶段集中了若干操作：            —— 化简操作 　　　　　　// Simplify the set of problem clauses: 　　　　　　if (decisionLevel() == 0 && !simplify())   //首先决策层为0层时，才会进行simplif()的调用 　　　　　　    return l_False;             ——学习子句管理（删除）             ——是否“重启”的检查   /*________________________________________________________________________________________ | | simplify : [void] -> [bool] | | Description: | Simplify the clause database according to the current top-level assigment. Currently, the only | thing done here is the removal of satisfied clauses, but more things can be put here. |__________________________________________________________________________________________ @*/ bool Solver::simplify() { assert(decisionLevel() == 0); //首先决策层为0层时，才会进行simplif()的调用 if (!ok || propagate() != CRef_Undef) return ok = false; if (nAssigns() == simpDB_assigns || (simpDB_props > 0)) return true; // Remove satisfied clauses: removeSatisfied(learnts_core); // Should clean core first. safeRemoveSatisfied(learnts_tier2, TIER2); safeRemoveSatisfied(learnts_local, LOCAL); if (remove_satisfied) // Can be turned off. removeSatisfied(clauses); checkGarbage(); rebuildOrderHeap(); simpDB_assigns = nAssigns(); simpDB_props = clauses_literals + learnts_literals; // (shouldn't depend on stats really, but it will do for now) return true; }