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numer-60-2
Commit 4205dd1c authored by Littichai Buddaken's avatar Littichai Buddaken
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Edited week13/Initial-Value.ipynb

parent f7f9d76a
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week13/Initial-Value.ipynb
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...@@ -14,14 +14,25 @@ ...@@ -14,14 +14,25 @@
"\n", "\n",
"$g(x), h(x)$ - constraint functions\n", "$g(x), h(x)$ - constraint functions\n",
"\n", "\n",
"minimize $F(x)$ = maximize $-F(x)$\n" "minimize $F(x)$ = maximize $-F(x)$\n",
"\n",
"> จงหาค่า $x$ ในช่วง $[a,b]$ ที่ทำ $f(x)$ มีค่าน้อยที่สุด"
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"![image](./images/initialvalue.png)"
] ]
}, },
{ {
"cell_type": "markdown", "cell_type": "markdown",
"metadata": {}, "metadata": {},
"source": [ "source": [
"> จงหาค่า $x$ ในช่วง $[a,b]$ ที่ทำ $f(x)$ มีค่าน้อยที่สุด\n", "\n",
"\n", "\n",
"## Bracketing\n", "## Bracketing\n",
"- $x = a$\n", "- $x = a$\n",
...@@ -29,6 +40,84 @@ ...@@ -29,6 +40,84 @@
"- กำหนดให้ $x = x + dx$ ไปเรื่อยๆ จนกว่า $f(x+dx) > f(x)$" "- กำหนดให้ $x = x + dx$ ไปเรื่อยๆ จนกว่า $f(x+dx) > f(x)$"
] ]
}, },
{
"cell_type": "code",
"execution_count": 99,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"[-2.0, -1.9, -1.8, -1.7, -1.6, -1.5, -1.4, -1.3, -1.2, -1.1, -1.0, -0.9, -0.8, -0.7, -0.6, -0.5, -0.4, -0.3, -0.2, -0.1, 0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9]\n",
"[9.0, 8.41, 7.84, 7.289999999999999, 6.760000000000001, 6.25, 5.76, 5.29, 4.84, 4.41, 4.0, 3.6100000000000003, 3.24, 2.8899999999999997, 2.56, 2.25, 1.96, 1.69, 1.44, 1.21, 1.0, 0.81, 0.64, 0.49, 0.36, 0.25, 0.16000000000000003, 0.09000000000000008, 0.040000000000000036, 0.010000000000000009, 0.0, 0.010000000000000009, 0.040000000000000036, 0.09000000000000008, 0.15999999999999992, 0.25, 0.3600000000000003, 0.48999999999999977, 0.6400000000000001, 0.81]\n"
]
},
{
"data": {
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"text/plain": [
"<matplotlib.figure.Figure at 0x7f0d1b7d10b8>"
]
},
"metadata": {},
"output_type": "display_data"
}
],
"source": [
"def f(x) :return x**2-2*x+1\n",
"from matplotlib import pyplot as plt\n",
"import numpy as np\n",
"x = [x/10 for x in range(-20,20,1)]\n",
"y = [f(xi) for xi in x]\n",
"print(x)\n",
"print(y)\n",
"a,b=-2,2\n",
"fa,fb =f(a),f(b)\n",
"plt.plot(x, y, 'r-')\n",
"plt.grid(True)\n",
"plt.show()\n",
"#x = a\n",
"#h = 0.01\n",
"#f(x+h),f(x+2h),f(x+3h)... \n",
"#f(x+i*h)<f(x+(i+1)*h)\n",
"\n"
]
},
{
"cell_type": "code",
"execution_count": 48,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"solution xi: 1.0000000000000022 0.0\n"
]
}
],
"source": [
"\n",
"#def bracket(f,a=-2,b=2,h=0.01):\n",
"def bracket(f,**kwargs): #**kwargs input is key word agrument,dicinary\n",
" a = float (kwargs['a']) \n",
" b = float (kwargs['b'])\n",
" h = float (kwargs['h'])\n",
" xi = a \n",
" oldf = f(xi)\n",
" xi += h\n",
" newf =f(xi)\n",
" while newf < oldf :\n",
" oldf=newf\n",
" xi+=h\n",
" newf =f(xi)\n",
" xi -= h\n",
" print('solution xi: ',xi,f(xi))\n",
"#bracket(f)\n",
"bracket(f,a=-2,b=2,h=0.01)"
]
},
{ {
"cell_type": "code", "cell_type": "code",
"execution_count": null, "execution_count": null,
...@@ -36,14 +125,88 @@ ...@@ -36,14 +125,88 @@
"collapsed": true "collapsed": true
}, },
"outputs": [], "outputs": [],
"source": [] "source": [
"def Identity():\n",
" "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Golden Section Search"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"![image](./images/goldenratio.png)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"$$\n",
"(1) \\to x_2 - x1 = 2Rh - h \\\\\n",
"(2) \\to x_1 - a = h' - Rh' \n",
"$$\n",
"แก้สมการหา $R$ จะได้\n",
"$$\n",
"2R - 1 = R(1-R) \\\\\n",
"R = \\frac{-1 + \\sqrt{5}}{2}\n",
"$$\n",
"จำนวนครั้งที่จะต้องใช้เพื่อหาคำตอบคือ\n",
"$$\n",
"n = -ln(R) ln(\\frac{\\epsilon}{|b-a|}\n",
"$$"
]
},
{
"cell_type": "code",
"execution_count": 56,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": [
"import math "
]
},
{
"cell_type": "code",
"execution_count": 90,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"46"
]
},
"execution_count": 90,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"from math import log as ln\n",
"R = (-1+5**0.5)/2\n",
"eps = 10**-9\n",
"a,b=-2,2\n",
"def n(a=-2,b=2,eps = 10**-9):\n",
" from math import log as ln\n",
" from math import ceil\n",
" R = (-1+5**0.5)/2\n",
" return int(ceil(-2.078087*ln(eps/abs(b-a))))\n",
"n()"
]
}, },
{ {
"cell_type": "markdown", "cell_type": "markdown",
"metadata": {}, "metadata": {},
"source": [ "source": [
"## Golden Section Search\n",
"\n",
"```python\n", "```python\n",
"def search(f,a,b,tol=1.0e-9):\n", "def search(f,a,b,tol=1.0e-9):\n",
" nIter = int(math.ceil(-2.078087*math.log(tol/abs(b-a))))\n", " nIter = int(math.ceil(-2.078087*math.log(tol/abs(b-a))))\n",
...@@ -68,6 +231,91 @@ ...@@ -68,6 +231,91 @@
"```" "```"
] ]
}, },
{
"cell_type": "code",
"execution_count": 110,
"metadata": {},
"outputs": [],
"source": [
"def f(x):return 4.8*x**2+6*x-2\n",
"def search(f,a,b,tol=10**-9):\n",
" N =n(a,b)\n",
" R = 0.618033989\n",
" C = 1.0 - R\n",
" # First telescoping\n",
" x1 = R*a + C*b; x2 = C*a + R*b\n",
" f1 = f(x1); f2 = f(x2)\n",
" # Main loop\n",
" for i in range(N):\n",
" if f1 > f2:\n",
" a = x1\n",
" x1 = x2; f1 = f2\n",
" x2 = C*a + R*b; f2 = f(x2)\n",
" else:\n",
" b = x2\n",
" x2 = x1; f2 = f1\n",
" x1 = R*a + C*b; f1 = f(x1)\n",
"\n",
" if f1 < f2: return x1,f1\n",
" return x2,f2\n"
]
},
{
"cell_type": "code",
"execution_count": 111,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"(3.203906095884046e-10, -1.9999999980776564)"
]
},
"execution_count": 111,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"search(f,-2,2)\n",
"def f(x):return 4.8*x**2+6*x-2\n",
"search(f,0,100)"
]
},
{
"cell_type": "code",
"execution_count": 112,
"metadata": {},
"outputs": [
{
"data": {
"image/png": 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Nx4uCXiQxnnsO/vu/oWHDcLu/44+PXZHkCI3Ri+S7TZvg+uvhnHOgZcswHq+QlwrUoxfJ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\nlPmWAAAAAElFTkSuQmCC\n",
"text/plain": [
"<matplotlib.figure.Figure at 0x7f0d1b5943c8>"
]
},
"metadata": {},
"output_type": "display_data"
}
],
"source": [
"\n",
"x = [x/10 for x in range(10,100,1)]\n",
"y = [f(xi) for xi in x]\n",
"plt.plot(x, y, 'r-')\n",
"plt.grid(True)\n",
"plt.show()"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Exercise\n",
"1. จงเขียนโปรแกรมหา $x$ ที่จะทห้ให้ฟังก์ชัน $f(x) = 1.6x^3 + 3x^2 -2x$ มีค่าน้อยที่สุด ในช่วงค่า $x \\ge 0$\n",
"2. จงเขียนโปรแกรมหา $x$ ที่จะทห้ให้ฟังก์ชัน $f(x) = 4.8x^2 + 6x - 2$ มีค่าน้อยที่สุด ในช่วงค่า $x \\ge 0$\n"
]
},
{ {
"cell_type": "markdown", "cell_type": "markdown",
"metadata": {}, "metadata": {},
...@@ -90,9 +338,20 @@ ...@@ -90,9 +338,20 @@
" * $v_i = v_{i+1}$ ($v_1$ is discarded; the other vectors are reused).\n", " * $v_i = v_{i+1}$ ($v_1$ is discarded; the other vectors are reused).\n",
" * End loop\n", " * End loop\n",
" * $x_0 = x_{n+1}$\n", " * $x_0 = x_{n+1}$\n",
"* End cycle\n", "* End cycle\n"
"\n", ]
"\n", },
{
"cell_type": "markdown",
"metadata": {},
"source": [
"![Powell](./images/powell.png)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"```python\n", "```python\n",
"import numpy as np\n", "import numpy as np\n",
"import math\n", "import math\n",
......
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