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| From \url{https://learnxinyminutes.com/docs/bash/}. | ||
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| \inputminted[mathescape=false]{bash}{src/src/LearnBash.sh} |
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| #ifndef TIFALIBS_OPT_SIMPLEX | ||
| #define TIFALIBS_OPT_SIMPLEX | ||
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| #include "../util/fp_const.hpp" | ||
| #include "../util/util.hpp" | ||
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| namespace tifa_libs::opt { | ||
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| template <class T = double> | ||
| struct LPSolver { | ||
| #define ltj(X) \ | ||
| if (s == -1 || std::make_pair(X[j], N[j]) < std::make_pair(X[s], N[s])) s = j | ||
| static constexpr T inf = 1 / .0; | ||
| int m, n; // # m = contraints, # n = variables | ||
| vec<int> N, B; // N[j] = non-basic variable (j-th column), = 0 | ||
| vvec<T> D; // B[j] = basic variable (j-th row) | ||
| LPSolver(vvec<T> const& A, vec<T> const& b, vec<T> const& c) : m(sz(b)), n(sz(c)), N(n + 1), B(m), D(m + 2, vec<T>(n + 2)) { | ||
| for (int i = 0; i < m; ++i) | ||
| for (int j = 0; j < n; ++j) D[i][j] = A[i][j]; | ||
| for (int i = 0; i < m; ++i) B[i] = n + i, D[i][n] = -1, D[i][n + 1] = b[i]; | ||
| // B[i]: basic variable for each constraint | ||
| // D[i][n]: artificial variable for testing feasibility | ||
| for (int j = 0; j < n; ++j) N[j] = j, D[m][j] = -c[j]; | ||
| // D[m] stores negation of objective, | ||
| // which we want to minimize | ||
| N[n] = -1; | ||
| D[m + 1][n] = 1; // to find initial feasible | ||
| } // solution, minimize artificial variable | ||
| void pivot(int r, int s) { // swap B[r] (row) | ||
| T inv = 1 / D[r][s]; // with N[r] (column) | ||
| for (int i = 0; i <= m + 1; ++i) | ||
| if (i != r && abs(D[i][s]) > EPS<T>) { | ||
| T binv = D[i][s] * inv; | ||
| for (int j = 0; j < (n + 2); ++j) | ||
| if (j != s) D[i][j] -= D[r][j] * binv; | ||
| D[i][s] = -binv; | ||
| } | ||
| D[r][s] = 1; | ||
| for (int j = 0; j < (n + 2); ++j) D[r][j] *= inv; // scale r-th row | ||
| std::swap(B[r], N[s]); | ||
| } | ||
| bool simplex(int phase) { | ||
| int x = m + phase - 1; | ||
| while (1) { // if phase=1, ignore artificial variable | ||
| int s = -1; | ||
| for (int j = 0; j <= n; ++j) | ||
| if (N[j] != -phase) ltj(D[x]); | ||
| // find most negative col for nonbasic (NB) variable | ||
| if (D[x][s] >= -EPS<T>) return 1; | ||
| // can't get better sol by increasing NB variable | ||
| int r = -1; | ||
| for (int i = 0; i < m; ++i) { | ||
| if (D[i][s] <= EPS<T>) continue; | ||
| if (r == -1 || std::make_pair(D[i][n + 1] / D[i][s], B[i]) < std::make_pair(D[r][n + 1] / D[r][s], B[r])) r = i; | ||
| // find smallest positive ratio | ||
| } // -> max increase in NB variable | ||
| if (r == -1) return 0; // objective is unbounded | ||
| pivot(r, s); | ||
| } | ||
| } | ||
| T solve(vec<T>& x) { // 1. check if x=0 feasible | ||
| int r = 0; | ||
| for (int i = (1); i < m; ++i) | ||
| if (D[i][n + 1] < D[r][n + 1]) r = i; | ||
| if (D[r][n + 1] < -EPS<T>) { // if not, find feasible start | ||
| pivot(r, n); // make artificial variable basic | ||
| assert(simplex(2)); // I think this will always be true?? | ||
| if (D[m + 1][n + 1] < -EPS<T>) return -inf; | ||
| // D[m+1][n+1] is max possible value of the negation of | ||
| // artificial variable, optimal value should be zero | ||
| // if exists feasible solution | ||
| for (int i = 0; i < m; ++i) | ||
| if (B[i] == -1) { // artificial var basic | ||
| int s = 0; | ||
| for (int j = (1); j <= n; ++j) ltj(D[i]); // -> nonbasic | ||
| pivot(i, s); | ||
| } | ||
| } | ||
| bool ok = simplex(1); | ||
| x = vec<T>(n); | ||
| for (int i = 0; i < m; ++i) | ||
| if (B[i] < n) x[B[i]] = D[i][n + 1]; | ||
| return ok ? D[m][n + 1] : inf; | ||
| } | ||
| }; | ||
| #undef ltj | ||
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| } // namespace tifa_libs::opt | ||
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| #endif |
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| 设有 \(n\) 个球, \(m\) 个盒子 | ||
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| 类似 \fullref{sec:球与盒(球全部相同, 盒全部相同)}, 答案为 \(f(n-m,m)\) | ||
| 类似 \fullref{sec:球与盒(球相同,盒相同)}, 答案为 \(f(n-m,m)\) |
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| 设有 \(n\) 个球, \(m\) 个盒子 | ||
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| \fullref{sec:球与盒(球互不相同, 盒全部相同, 至多装一个球)} | ||
| \fullref{sec:球与盒(球相同,盒相同,至多装一个球)} |
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| 求解如下线性规划问题: | ||
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| \[ | ||
| \begin{array}{ll} | ||
| \max & c^T x. \\ | ||
| \textrm{s.t.} & Ax \le b, x \ge 0. | ||
| \end{array} | ||
| \] | ||
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| 若无解则返回 \verb|-inf|, 若无界则返回 \verb|inf|, 否则返回 \(c^T x\) 的最大值. 输入的 \verb|x| 会被修改为解. 不保证数值稳定性. \(x = 0\) 最好在可行域中. | ||
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| \paragraph{使用} | ||
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| \inputminted{cpp}{src/src/simplex_usage.txt} | ||
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| \paragraph{原问题与对偶问题} | ||
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| \begin{tabular}{|c|c|} | ||
| \hline | ||
| 原问题 & 对偶问题 \\ | ||
| \hline | ||
| \(\max_x c^T x\) & \(\min_y b^T y\) \\ | ||
| \hline | ||
| \(n\) 个变量 \(x_1,\dots,x_n\) & \(n\) 个限制条件 \(A^T y\gtreqqless c\) \\ | ||
| \(x_i\geq 0\) & 第 \(i\) 个限制条件为 \(\geq c_i\) \\ | ||
| \(x_i\leq 0\) & 第 \(i\) 个限制条件为 \(\leq c_i\) \\ | ||
| \(x_i\in \mathrm{R}\) & 第 \(i\) 个限制条件为 \(=c_i\) \\ | ||
| \hline | ||
| \(m\) 个限制条件 \(Ax\gtreqqless b\) & \(m\) 个变量 \(y_1,\dots,y_m\) \\ | ||
| 第 \(i\) 个限制条件为 \(\geq b_i\) & \(y_i\leq 0\) \\ | ||
| 第 \(i\) 个限制条件为 \(\leq b_i\) & \(y_i\geq 0\) \\ | ||
| 第 \(i\) 个限制条件为 \(=b_i\) & \(y_i\in \mathrm{R}\) \\ | ||
| \hline | ||
| \end{tabular} | ||
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| \begin{tabular}{|cc|cc|} | ||
| \hline | ||
| \multicolumn{2}{|c|}{原问题} & \multicolumn{2}{|c|}{对偶问题} \\ | ||
| \hline | ||
| \(\max\) & \(3x_1+4x_2\) & \(\min\) & \(7y_1\) \\ | ||
| \(\textrm{s.t.}\) & \(5x_1+6x_2=7\) & \(\textrm{s.t.}\) & \(5y_1\geq 3\) \\ | ||
| & & & \(6y_2\geq 4\) \\ | ||
| & \(x_1\geq 0,x_2\geq 0\) & & \(y_1\in\mathrm{R}\) \\ | ||
| \hline | ||
| \end{tabular} | ||
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| \paragraph{复杂度} | ||
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| \(O(NM \cdot \texttt{pivots})\), where a pivot may be e.g. an edge relaxation. \(O(2^N)\) in the general case. | ||
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| \paragraph{参考} \cite{bqi343cp} |
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