{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# 附录 E:Lagrange 乘子" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## 等式约束下的拉格朗日乘子" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "拉格朗日乘子(`Lagrange multipliers`),又叫不定乘子(`undetermined multipliers`),用于找到多变量函数在有约束条件下的驻点。\n", "\n", "考虑最大化函数 $f(x_1, x_2)$,约束条件为\n", "\n", "$$\n", "g(x_1, x_2) = 0\n", "$$\n", "\n", "一个可能的方法是通过这个约束求出 $x_2 = h(x_1)$ 再带入函数消去 $x_2$,但很多时候函数 $h$ 是很难找到的。\n", "\n", "一个更普遍和简单的方法是引入一个拉格朗日乘子 $\\lambda$。\n", "\n", "现在考虑一个 $D$ 维变量 $\\mathbf x$ 的优化问题:\n", "\n", "$$\n", "\\begin{align}\n", "\\max_{\\mathbf x}~& f(\\mathbf x) \\\\\n", "s.t.~& g(\\mathbf x) = 0\n", "\\end{align}\n", "$$\n", "\n", "$g(\\mathbf x) = 0$ 定义了空间中一个 $D-1$ 维的超平面。\n", "\n", "我们注意到,梯度 $\\triangledown g(\\mathbf x)$ 平行于这个超平面 $g(\\mathbf x) = 0$。\n", "\n", "考虑超平面上的两点 $\\mathbf x, \\mathbf{x+\\epsilon}$,做泰勒展开有:\n", "\n", "$$\n", "g(\\mathbf{x+\\epsilon}) \\approx g(\\mathbf{x}) + \\mathbf\\epsilon^{\\text{T}} \\triangledown g(\\mathbf x) \n", "$$\n", "\n", "因为它们都在超平面上,所以有 $\\mathbf\\epsilon^{\\text{T}} \\triangledown g(\\mathbf x) \\approx 0$,令 $\\|\\mathbf \\epsilon\\|\\to 0$,我们有 $\\mathbf\\epsilon^{\\text{T}} \\triangledown g(\\mathbf x) = 0$,而 $\\epsilon$ 平行于超平面,所以 $\\triangledown g(\\mathbf x)$ 平行于超平面。\n", "\n", "在这个超平面上,最大化 $f(\\mathbf x)$ 的点 $\\mathbf x^\\star$ 也需要满足 $\\triangledown f(\\mathbf x^\\star)$ 垂直于超平面 $g(\\mathbf x) = 0$,否则,我们总可以在这个超平面上移动一点距离使得 $f(\\mathbf x)$ 增大。\n", "\n", "因此 $\\triangledown f(\\mathbf x^\\star)$ 与 $\\triangledown g(\\mathbf x^\\star)$ 应当平行,这意味着存在 $\\lambda \\neq 0$,使得:\n", "\n", "$$\n", "\\triangledown f + \\lambda \\triangledown g = 0\n", "$$\n", "\n", "此时我们可以引入一个拉格朗日函数:\n", "\n", "$$\n", "L(\\mathbf x, \\lambda) \\equiv f(\\mathbf x) + \\lambda g(\\mathbf x)\n", "$$\n", "\n", "其驻点满足 $\\triangledown_x L = 0$ 即 $\n", "\\triangledown f + \\lambda \\triangledown g = 0\n", "$,以及偏导 $\\frac{\\partial L}{\\partial \\lambda}$ 使得 $g(\\mathbf x) = 0$。\n", "\n", "因此,问题转化为求这个拉格朗日函数的驻点 $(x^\\star, \\lambda)$。\n", "\n", "由于拉格朗日函数是无约束的,我们很容易通过求偏导得到驻点的值。" ] }, { "cell_type": "code", "execution_count": 1, "metadata": { "collapsed": false }, "outputs": [ { "data": { "image/png": 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"text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "import matplotlib.pyplot as plt\n", "import numpy as np\n", "import scipy as sp\n", "\n", "%matplotlib inline\n", "\n", "fig, ax = plt.subplots()\n", "\n", "an = np.linspace(0, 2*np.pi, 100)\n", "\n", "xx, yy = -20 / (3 - np.cos(an)), -7 / (2 + np.sin(an)) + np.cos(an)\n", "\n", "ax.plot(xx, yy, 'r')\n", "\n", "ax.set_xlim(-11, -3)\n", "ax.set_ylim(-8, -1)\n", "\n", "x, y = xx[30], yy[30]\n", "\n", "ax.annotate(\"\",\n", " xy=(x, y), \n", " xytext=(x+0.4, y+1),\n", " arrowprops=dict(arrowstyle=\"<-\",\n", " connectionstyle=\"arc3\")\n", " )\n", "\n", "ax.annotate(\"\",\n", " xy=(x, y), \n", " xytext=(x-0.4, y-1),\n", " arrowprops=dict(arrowstyle=\"<-\",\n", " connectionstyle=\"arc3\")\n", " )\n", "\n", "ax.text(-5, -7, \"$g(\\mathbf{x}) = 0$\", fontsize=\"xx-large\")\n", "\n", "ax.text(x+0.5, y+1, r\"$\\triangledown f(\\mathbf{x})$\", fontsize=\"xx-large\")\n", "ax.text(x-0.8, y-1.5, r\"$\\triangledown g(\\mathbf{x})$\", fontsize=\"xx-large\")\n", "ax.text(x+0.2, y, '$x_A$', fontsize=\"xx-large\")\n", "\n", "ax.set_axis_off()\n", "\n", "ax.plot([x], [y], 'k.', linewidth=30)\n", "\n", "plt.show()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## 例子" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "考虑这样的问题:\n", "\n", "$$\n", "\\begin{align}\n", "\\max_{x_1, x_2}~& f(x_1, x_2)=1-x_1^2-x_2^2 \\\\\n", "s.t.~& g(x_1,x_2) = x_1 + x_2 - 1 = 0\n", "\\end{align}\n", "$$\n", "\n", "对应的拉格朗日函数为\n", "\n", "$$\n", "L(\\mathbf x, \\lambda) = 1-x_1^2-x_2^2 + \\lambda(x_1 + x_2 - 1)\n", "$$\n", "\n", "驻点满足:\n", "\n", "$$\n", "\\begin{align}\n", "-2x_1 + \\lambda & = 0\\\\\n", "-2x_2 + \\lambda & = 0\\\\\n", "x_1 + x_2 - 1 & = 0 \\\\\n", "\\end{align}\n", "$$\n", "\n", "因此得到 $(x_1^\\star, x_2^\\star)=(\\frac 1 2, \\frac 1 2)$,$\\lambda = 1$。" ] }, { "cell_type": "code", "execution_count": 2, "metadata": { "collapsed": false }, "outputs": [ { "data": { "image/png": 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"text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "an = np.linspace(0, 2*np.pi, 100)\n", "\n", "xx = np.linspace(-1, 2)\n", "yy = 1 - xx\n", "\n", "fig, ax = plt.subplots()\n", "\n", "for i in xrange(1, 4):\n", " ax.plot(i / np.sqrt(8.0) * np.cos(an), i / np.sqrt(8.0) * np.sin(an), 'b')\n", "\n", "ax.plot(xx, yy, 'r') \n", "\n", "ax.axis('equal')\n", "\n", "ax.spines['right'].set_color('none')\n", "ax.spines['top'].set_color('none')\n", "\n", "ax.xaxis.set_ticks_position('bottom')\n", "ax.spines['bottom'].set_position(('data',0))\n", "\n", "ax.yaxis.set_ticks_position('left')\n", "ax.spines['left'].set_position(('data',0))\n", "\n", "ax.set_xticks([])\n", "ax.set_yticks([])\n", "\n", "ax.text(2.6, -0.2, \"$x_1$\", fontsize=\"xx-large\")\n", "ax.text(-0.3, 1.8, \"$x_2$\", fontsize=\"xx-large\")\n", "ax.text(1, -1.3, \"$g(x_1, x_2) = 0$\", fontsize=\"xx-large\")\n", "\n", "ax.annotate(r\"$(x_1^\\star, x_2^\\star)$\", \n", " fontsize=\"xx-large\",\n", " xytext=(1, 1), \n", " xy=(0.5, 0.5),\n", " arrowprops=dict(arrowstyle=\"->\",\n", " connectionstyle=\"arc3\")\n", " )\n", "\n", "plt.show()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## 带不等式约束的拉格朗日乘子" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "现在考虑不等式约束的问题:\n", "\n", "$$\n", "\\begin{align}\n", "\\max_{\\mathbf x}~& f(\\mathbf x) \\\\\n", "s.t.~& g(\\mathbf x) \\geq 0\n", "\\end{align}\n", "$$\n", "\n", "分成两种情况考虑:\n", "\n", "- $g(\\mathbf x) < 0$,此时不等式约束不起作用(`inactive`),驻点的条件为 $\\triangledown f(\\mathbf x)=0$,因此必须有 $\\lambda=0$。\n", "\n", "\n", "- $g(\\mathbf x) = 0$,此时不等式约束起作用(`active`),驻点的条件为 $\\triangledown f(\\mathbf x) + \\lambda \\triangledown g(\\mathbf x)=0$,且 $\\lambda \\neq 0$。除此之外,$\\triangledown f(\\mathbf x)$ 应当与 $\\triangledown g(\\mathbf x)$ 的方向(指向 $g(\\mathbf x)>0$ 的区域)相反才能得到最大值(即 $f$ 的梯度方向要远离 $g(\\mathbf x)>0$ 的区域),因此我们必然有 $\\lambda > 0$。\n", " \n", "在这两种情况下,我们都有 $\\lambda g(\\mathbf x) = 0$,因此问题转换为求解拉格朗日函数\n", "\n", "$$\n", "L(\\mathbf x, \\lambda) \\equiv f(\\mathbf x) + \\lambda g(\\mathbf x)\n", "$$\n", "\n", "在约束\n", "\n", "$$\n", "\\begin{align}\n", "g(\\mathbf x)&\\geq 0\\\\\n", "\\lambda &\\geq 0\\\\\n", "\\lambda g(\\mathbf x) &= 0\n", "\\end{align}\n", "$$\n", "\n", "下的驻点,这些条件就是著名的 KKT 条件。\n", "\n", "对于最小化问题,我们只需要将拉格朗日函数变为:\n", "\n", "$$\n", "L(\\mathbf x, \\lambda) \\equiv f(\\mathbf x) - \\lambda g(\\mathbf x)\n", "$$" ] }, { "cell_type": "code", "execution_count": 3, "metadata": { "collapsed": false }, "outputs": [ { "data": { "image/png": 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"text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "fig, ax = plt.subplots()\n", "\n", "an = np.linspace(0, 2*np.pi, 100)\n", "\n", "xx, yy = -20 / (3 - np.cos(an)), -7 / (2 + np.sin(an)) + np.cos(an)\n", "\n", "ax.plot(xx, yy, 'r')\n", "\n", "ax.fill(xx, yy, color='gold')\n", "\n", "ax.set_xlim(-11, -3)\n", "ax.set_ylim(-8, -1)\n", "\n", "x, y = xx[30], yy[30]\n", "\n", "ax.annotate(\"\",\n", " xy=(x, y), \n", " xytext=(x+0.4, y+1),\n", " arrowprops=dict(arrowstyle=\"<-\",\n", " connectionstyle=\"arc3\")\n", " )\n", "\n", "ax.annotate(\"\",\n", " xy=(x, y), \n", " xytext=(x-0.4, y-1),\n", " arrowprops=dict(arrowstyle=\"<-\",\n", " connectionstyle=\"arc3\")\n", " )\n", "\n", "ax.text(-5, -7, \"$g(\\mathbf{x}) = 0$\", fontsize=\"xx-large\")\n", "ax.text(-9, -3.5, \"$g(\\mathbf{x}) > 0$\", fontsize=\"xx-large\")\n", "\n", "ax.text(x+0.5, y+1, r\"$\\triangledown f(\\mathbf{x})$\", fontsize=\"xx-large\")\n", "ax.text(x-0.8, y-1.5, r\"$\\triangledown g(\\mathbf{x})$\", fontsize=\"xx-large\")\n", "ax.text(x+0.2, y, '$x_A$', fontsize=\"xx-large\")\n", "\n", "ax.plot([x], [y], 'k.', linewidth=30)\n", "\n", "ax.set_axis_off()\n", "\n", "plt.show()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## 多约束问题" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "考虑这个问题:\n", "\n", "$$\n", "\\begin{align}\n", "\\max_{\\mathbf x}~& f(\\mathbf x) &\\\\\n", "s.t.~& g_j(\\mathbf x) \\geq 0, &&j = 1,\\dots,J \\\\\n", "& h_k(\\mathbf x) \\geq 0, &&k = 1,\\dots,K \\\\\n", "\\end{align}\n", "$$\n", "\n", "很容易推广得到,拉格朗日函数为:\n", "\n", "$$\n", "\\begin{align}\n", "& L(\\mathbf {x, \\lambda, \\mu}) \\equiv f(\\mathbf x) + \\sum_{j=1}^J \\lambda_j g(\\mathbf x) + \\sum_{k=1}^K \\mu_k h(\\mathbf x) \\\\\n", "s.t.~& \\mu_k \\geq 0, \\mu_k h(\\mathbf x) = 0, k = 1,\\dots,K \n", "\\end{align}\n", "$$" ] } ], "metadata": { "kernelspec": { "display_name": "Python 2", "language": "python", "name": "python2" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 2 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython2", "version": "2.7.6" } }, "nbformat": 4, "nbformat_minor": 0 }