add week03 assignment

week03/.ipynb_checkpoints/Chapter 03 Numerics and Error Analysis-checkpoint.ipynb

+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "slideshow": {
+     "slide_type": "slide"
+    }
+   },
+   "source": [
+    "# บทที่ 1 ตัวเลขและการวิเคราะห์ความคลาดเคลื่อน\n",
+    "\n",
+    "> Mathematically correct programming bug is **hard** to spot.\n",
+    "\n",
+    "> สมการถูกแต่...โปรแกรมผิด\n",
+    "\n",
+    ">> Paulgramming - 2017\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true,
+    "slideshow": {
+     "slide_type": "slide"
+    }
+   },
+   "source": [
+    "# ตัวเลขและความคลาดเคลื่อนในโปรแกรม\n",
+    "## C++\n",
+    "\n",
+    "```c++\n",
+    "#include <iostream>\n",
+    "using namespace std;\n",
+    "main() {\n",
+    "  float x = 1.0;    \n",
+    "  float y = x / 3.0;\n",
+    "  cout << ((x == y * 3.0)?\"\": \"Not\") << \" Equal\" << endl;\n",
+    "}\n",
+    "```\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true,
+    "slideshow": {
+     "slide_type": "subslide"
+    }
+   },
+   "source": [
+    "# ตัวเลขและความคลาดเคลื่อนในโปรแกรม\n",
+    "## Java\n",
+    "\n",
+    "```java\n",
+    "public class Test {\n",
+    "  public static void main(String[] a) {\n",
+    "    float x = 1.0;\n",
+    "    float y = x / 3.0;\n",
+    "    System.out.println( ((x == y * 3.0)?\"\": \"Not\") + \" Equal\" );\n",
+    "  }\n",
+    "}\n",
+    "```\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true,
+    "slideshow": {
+     "slide_type": "subslide"
+    }
+   },
+   "source": [
+    "# ตัวเลขและความคลาดเคลื่อนในโปรแกรม\n",
+    "## Python\n",
+    "\n",
+    "```python\n",
+    "x = 1.0\n",
+    "y = x / 3.0\n",
+    "print(\"Equal\" if x == y*3.0 else \"Not Equal\")\n",
+    "```"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": true,
+    "slideshow": {
+     "slide_type": "skip"
+    }
+   },
+   "outputs": [],
+   "source": [
+    "# มีหลายตัวเลขที่ไม่\n",
+    "x = 1.0\n",
+    "y = x / 3.0\n",
+    "#print(\"Equal\" if x == y*3.0 else \"Not Equal\")"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "slideshow": {
+     "slide_type": "slide"
+    }
+   },
+   "source": [
+    "# มีหลายตัวเลขที่คอมเก็บค่าไ่ม่ได้\n",
+    "\n",
+    "* ตัวเลขที่คอมพิวเตอร์ต้องประมาณการ (Inexact numbers)\n",
+    "    * ค่าอตรรกยะ (irrational numbers) $\\pi, \\e, ...$\n",
+    "    * ค่าตรรกยะ (rational numbers) ที่ไม่สามารถแปลงเป็นเลขฐานสองได้\n",
+    "```python\n",
+    "0.1 + 0.2 == 0.3\n",
+    "```\n",
+    "\n",
+    "* ตัวเลขที่คอมพิวเตอร์สามารถเก็บค่าได้ตรง (Exact numbers)\n",
+    "> ตัวเลขที่สามารถแปลงให้เป็นเลขฐานสองได้"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "slideshow": {
+     "slide_type": "slide"
+    }
+   },
+   "source": [
+    "## Floating-Point Binary\n",
+    "\n",
+    "> ตัวเลข(number) เขียนแทนด้วยสายของเลขเดี่ยว(digit) โดยจำนวนของเลขเดี่ยว ขึ้นอยู่กับ เลขฐาน(base)\n",
+    "\n",
+    "ในระบบจำนวนนับของคนเราจะเป็นเลขฐาน 10 โดยมี digit ที่ใช้ได้ได้แก่ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9\n",
+    "ยกตัวอย่างตัวเลขในเลขฐาน 10 เช่น  128  77  12100  เป็นต้น\n",
+    "\n",
+    "ตัวเลขที่ใช้ในคอมพิวเตอร์จะเป็นเลขฐาน 2 โดยมี digit ที่ใช้ได้ ได้แก่ 0 และ 1 \n",
+    "ยกตัวอย่างตัวเลขในเลขฐาน 2 เช่น  1, 101, 1101  เป็นต้น"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "slideshow": {
+     "slide_type": "slide"
+    }
+   },
+   "source": [
+    "# Terminology\n",
+    "** Bit **\n",
+    ": 0 or 1\n",
+    "\n",
+    "** Byte ** \n",
+    ": 8 bits\n",
+    "\n",
+    "** Real **\n",
+    ": 4 bytes **single precision**\n",
+    ": 8 bytes **double precision**\n",
+    "\n",
+    "** Integer **\n",
+    ": 1, 2, 4, or 8 byte signed\n",
+    ": 1, 2, 4, or 8 byte unsigned\n",
+    "\n",
+    "> ขึ้นอยู่กับภาษาโปรแกรมว่าจะใช้สายที่มีความยาวเท่าไหร่ 8, 16, 32 หรือ 64 เป็นต้น\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "slideshow": {
+     "slide_type": "slide"
+    }
+   },
+   "source": [
+    "# ข้อตกลง\n",
+    "\n",
+    "* เพื่อง่ายต่อการคำนวณเราจะใช้สายที่มีความยาว 8 และเขียนแยกเป็นกลุ่มละ 4 เพื่อง่ายต่อการคำนวน \n",
+    "\n",
+    "> เช่น  1 = 0000 0001\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "slideshow": {
+     "slide_type": "slide"
+    }
+   },
+   "source": [
+    "# Binary $\\rightarrow$ Integer\n",
+    "\n",
+    "|      |      |      |      |      |      |      |      |\n",
+    "|------|------|------|------|------|------|------|------|\n",
+    "|   1  |   1  |   1  |   0  |   0  |   1  |   1  |   1  |\n",
+    "|$2^7$ |$2^6$ |$2^5$ |$2^4$ |$2^3$ |$2^2$ |$2^1$ |$2^0$ |\n",
+    "\n",
+    "$1\\times2^7 + 1\\times2^6 + 1\\times2^5 + 0\\times2^4 + 0\\times2^3 + 1\\times2^2 + 1\\times2^1 + 1\\times2^0$\n",
+    "\n",
+    "$1\\times 128+ 1\\times 64 + 1\\times 32 + 0\\times 16 + 0\\times 8 + 1\\times 4 + 1\\times 2 + 1 \\times 1$\n",
+    "\n",
+    "\n",
+    "$128 + 64 + 32 + 0 + 0 + 4 + 2 + 1$\n",
+    "\n",
+    "**Answer**\n",
+    "> $231$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "slideshow": {
+     "slide_type": "slide"
+    }
+   },
+   "source": [
+    "# เขียนโปรแกรมทดสอบ\n",
+    "\n",
+    "* คำนวณหาผลลัพธ์\n",
+    "```python\n",
+    "1*2**7 + 1*2**6 + 1*2**5 + 0*2**4 + 0*2**3 + 1*2**2 + 1*2**1 + 1*2**0\n",
+    "```\n",
+    "\n",
+    "* คำสั่งแปลงเลขฐานต่างๆ\n",
+    "```python\n",
+    "int('11100111', 2)  # แปลง str '11100111' ฐาน 2 เป็น int\n",
+    "oct(35)             # แปลง 35 ฐานสิบ ให้เป็นฐาน 8\n",
+    "hex(35)             # แปลง 35 ฐานสิบ ให้เป็นฐาน 16\n",
+    "bin(35)             # แปลง 35 ฐานสิบ ให้เป็นฐาน 2\n",
+    "```"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": true,
+    "slideshow": {
+     "slide_type": "skip"
+    }
+   },
+   "outputs": [],
+   "source": [
+    "#int('11100111', 2)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "slideshow": {
+     "slide_type": "slide"
+    }
+   },
+   "source": [
+    "#### Exercise\n",
+    "จงแสดงการแปลงตัวเลขฐานสองต่อไปนี้เป็นจำนวนเต็ม\n",
+    "* 10011001\n",
+    "* 00010011\n",
+    "* 00011111"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "slideshow": {
+     "slide_type": "slide"
+    }
+   },
+   "source": [
+    "### Binary $\\rightarrow$ Float\n",
+    "กำหนดตำแหน่งที่เลขสองกำลังเป็นศูนย์ (fix-point representation)\n",
+    "\n",
+    "|      |      |      |      |      |      |      |      |\n",
+    "|------|------|------|------|------|------|------|------|\n",
+    "|   1  |   1  |   1  |   0  |   0  |   1  |   1  |   1  |\n",
+    "|$2^5$ |$2^4$ |$2^3$ |$2^2$ |$2^1$ |$2^0$ |$2^-1$ |$2^-2$ |\n",
+    "\n",
+    "$1\\times 2^5+1\\times 2^4+1\\times 2^3+0\\times 2^2+9\\times 2^1+1\\times 2^0+1\\times 2^{-1}+1\\times 2^{-2}$\n",
+    "\n",
+    "$1\\times2^5 + 1\\times2^4 + 1\\times2^3 +0\\times 2^2+0\\times2^1+1\\times 1+\\frac{1}{2}+\\frac{1}{4}$\n",
+    "\n",
+    "$32 + 16 + 8 + 0 + 0 + 1 + 0.5 + 0.25$\n",
+    "\n",
+    "**Answer**\n",
+    "> 57.75"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 26,
+   "metadata": {
+    "collapsed": true,
+    "slideshow": {
+     "slide_type": "skip"
+    }
+   },
+   "outputs": [],
+   "source": [
+    "#1*2**5 + 1*2**4 + 1*2**3 + 0*2**2 + 0*2**1 + 1*2**0 + 1*2**(-1) + 1*2**(-2)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "slideshow": {
+     "slide_type": "slide"
+    }
+   },
+   "source": [
+    "#### Binary Fixed-Point Arithmetic\n",
+    "|      |      |      |      |      |      |      |      |      |\n",
+    "|------|------|------|------|------|------|------|------|------|\n",
+    "|   1  |   1  |   1  | ...  |   0  |  0  |   ...  |   1  |   1  |\n",
+    "| sign ||$2^{m-1}$ |$2^{m-2}$ | ... |$2^0$ |$2^{-1}$ | ... |$2^{-n+1}$ |$2^{-n}$ |\n",
+    "\n",
+    "* Parameters: $m, n \\in Z$\n",
+    "* $m$ ความยาวของสาย(จำนวน bit) ที่ใช้เป็นจำนวนเต็ม(integer portion)\n",
+    "* $n$ ความยาวของสาย(จำนวน bit) ที่ใช้เป็นตัวหาร(fractional portion)\n",
+    "* ความยาวทั้งหมดของสายเป็น $m+n+1$ รวมกับ bit ที่ใช้บอกเครื่องหมาย (+/-)\n",
+    "* ความรู้เพิ่มเติม: [Fixed-Point Numbers](https://en.wikibooks.org/wiki/Floating_Point/Fixed-Point_Numbers)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "slideshow": {
+     "slide_type": "slide"
+    }
+   },
+   "source": [
+    "# Error in Binary Fixed-Point Arithmetic\n",
+    "\n",
+    "** สมมติว่ามีแค่ 2 bit ให้ใช้ ** \n",
+    "\n",
+    "> $m = 1, n=1$ โดยที่ไม่คิด sign bit\n",
+    "\n",
+    "$0.1 \\times 0.1 = 0.01 \\approx 0.0$ \n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 28,
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]"
+      ]
+     },
+     "execution_count": 28,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "v = [0 ,0,0,0,1 ,1,0,0]\n",
+    "\n",
+    "v = [0 , 0,0,0,0,0,0, 0,0,0,0,0]\n",
+    "\n",
+    "sign = -1 if v[0] == 1 else 1\n",
+    "value = 0\n",
+    "v_len = len(v)\n",
+    "\n",
+    "[ i for i in range(1,8)]\n",
+    "\n",
+    "# [  2 ** (4 - i) for i in range(1,8)]\n",
+    "# sum([ v[i] * 2 ** (4-i) for i in range(1,8) ])"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 23,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "-2.375\n"
+     ]
+    }
+   ],
+   "source": [
+    "def value(v):\n",
+    "    sign = -1 if v[0] == 1 else 1\n",
+    "    return sign * sum([ v[i] * 2 ** (4-i) for i in range(1,8) ])\n",
+    "\n",
+    "v = [1 ,0,0,1,0, 0,1,1]\n",
+    "\n",
+    "print(value(v))"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": true
+   },
+   "outputs": [],
+   "source": [
+    "def value(v , a, b):\n",
+    "    sign = -1 if v[0] == 1 else 1\n",
+    "    \n",
+    "    \n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "slideshow": {
+     "slide_type": "slide"
+    }
+   },
+   "source": [
+    "# Binary Floating-Point Arithmetic\n",
+    "1. IEEE single precision format\n",
+    "\n",
+    "|  0   |     1-8       |       9-31        |\n",
+    "|:----:|:-------------:|:-----------------:|\n",
+    "| $s$  |      $e$      |        $f$        |\n",
+    "| $0$  |   $00001001$  |   $0011011...01$  |\n",
+    "\n",
+    "$ = (-1)^s \\times 2^{e-127} \\times 1.f $\n",
+    "\n",
+    "โดยที่\n",
+    " * **sign** $s \\in {0, 1}$\n",
+    " * **biased exponent** $0 \\le e \\le 255$\n",
+    " * **exponent** $p=e-127$ so $-126 \\le p \\le 127$\n",
+    " * **significand** $1.f$ "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "slideshow": {
+     "slide_type": "slide"
+    }
+   },
+   "source": [
+    "# Binary Floating-Point Arithmetic \n",
+    "2. IEEE double precision format\n",
+    "\n",
+    "|  0   |     1-11      |       12-63       |\n",
+    "|:----:|:-------------:|:-----------------:|\n",
+    "| $s$  |      $e$      |        $f$        |\n",
+    "| $0$  | $00001000111$ |   $0011011...01$  |\n",
+    "\n",
+    "$ = (-1)^s \\times 2^{e-1023} \\times 1.f $\n",
+    "\n",
+    "โดยที่\n",
+    " * **sign** $s \\in {0, 1}$\n",
+    " * **biased exponent** $0 \\le e \\le 2047$\n",
+    " * **exponent** $p=e-1023$ so $-1022 \\le p \\le 1024$\n",
+    " * **significand** $x = 1.f$ "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "slideshow": {
+     "slide_type": "slide"
+    }
+   },
+   "source": [
+    "# Example \n",
+    "\n",
+    "จงเขียนคำสั่งเพื่อหาค่าของตัวเลขตามข้อต่อกำหนดต่อไปนี้ โดยใช้ IEEE single precision format\n",
+    "> 0 1111 0000 0000 1010 0000 0000 0000 000\n",
+    " \n",
+    "**Answer** \n",
+    "หาค่า $s, e, f$ จากโจทย์\n",
+    "> $s = 0$\n",
+    "\n",
+    "> $e$ = 11110000\n",
+    "\n",
+    "> $f$ = 0000101 \n",
+    "\n",
+    "```python\n",
+    "s = 0\n",
+    "e = int('11110000', 2)\n",
+    "x = 1 + 0*2**(-1) + 0*2**(-2) + 0*2**(-3) + 0*2**(-4) + 1*2**(-5) + 0*2**(-6) + 1*2**(-7)\n",
+    "print( (-1)**s * 2**(e-127) * x )\n",
+    "```"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "slideshow": {
+     "slide_type": "slide"
+    }
+   },
+   "source": [
+    "# Exercise \n",
+    "1. จงเขียนคำสั่งเพื่อหาค่าของตัวเลขตามข้อต่อกำหนดต่อไปนี้ โดยใช้ IEEE single precision format\n",
+    " * $s = 0$\n",
+    " * $e$ = 1111 0000 0000\n",
+    " * $f$ = 0000 1010 \n",
+    " \n",
+    "2. จงเขียนคำสั่งเพื่อหาค่าของตัวเลขตามข้อต่อกำหนดต่อไปนี้ โดยใช้ IEEE double precision format\n",
+    " * $s = 1$\n",
+    " * $e$ = 0000 0011 0000\n",
+    " * $f$ = 0110 1010 \n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "slideshow": {
+     "slide_type": "slide"
+    }
+   },
+   "source": [
+    "# Error Calculation\n",
+    "\n",
+    "* **Absolute Error** \n",
+    "    $\\rightarrow E_{abs} = \\| x_0 - x \\|$\n",
+    "* **Relative Error**\n",
+    "    $\\rightarrow E_{rel} = \\frac { E_{abs} }{x} $\n",
+    "* **Percentage Error**\n",
+    "    $\\rightarrow E_{per} = E_{rel} \\times 100$"
+   ]
+  }
+ ],
+ "metadata": {
+  "anaconda-cloud": {},
+  "celltoolbar": "Slideshow",
+  "kernelspec": {
+   "display_name": "Python 3",
+   "language": "python",
+   "name": "python3"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.6.3"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 1
+}

week03/Chapter 03 Numerics and Error Analysis.ipynb

@@ -235,6 +235,7 @@
    "cell_type": "code",
    "execution_count": 15,
    "metadata": {
+    "collapsed": true,
     "slideshow": {
      "slide_type": "skip"
     }
@@ -289,6 +290,7 @@
    "cell_type": "code",
    "execution_count": 26,
    "metadata": {
+    "collapsed": true,
     "slideshow": {
      "slide_type": "skip"
     }
@@ -336,6 +338,75 @@
     "$0.1 \\times 0.1 = 0.01 \\approx 0.0$ \n"
    ]
   },
+  {
+   "cell_type": "code",
+   "execution_count": 28,
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]"
+      ]
+     },
+     "execution_count": 28,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "v = [0 ,0,0,0,1 ,1,0,0]\n",
+    "\n",
+    "v = [0 , 0,0,0,0,0,0, 0,0,0,0,0]\n",
+    "\n",
+    "sign = -1 if v[0] == 1 else 1\n",
+    "value = 0\n",
+    "v_len = len(v)\n",
+    "\n",
+    "[ i for i in range(1,8)]\n",
+    "\n",
+    "# [  2 ** (4 - i) for i in range(1,8)]\n",
+    "# sum([ v[i] * 2 ** (4-i) for i in range(1,8) ])"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 23,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "-2.375\n"
+     ]
+    }
+   ],
+   "source": [
+    "def value(v):\n",
+    "    sign = -1 if v[0] == 1 else 1\n",
+    "    return sign * sum([ v[i] * 2 ** (4-i) for i in range(1,8) ])\n",
+    "\n",
+    "v = [1 ,0,0,1,0, 0,1,1]\n",
+    "\n",
+    "print(value(v))"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": true
+   },
+   "outputs": [],
+   "source": [
+    "def value(v , a, b):\n",
+    "    sign = -1 if v[0] == 1 else 1\n",
+    "    \n",
+    "    \n",
+    "\n"
+   ]
+  },
   {
    "cell_type": "markdown",
    "metadata": {
@@ -472,7 +543,7 @@
    "name": "python",
    "nbconvert_exporter": "python",
    "pygments_lexer": "ipython3",
-   "version": "3.5.2"
+   "version": "3.6.3"
   }
  },
  "nbformat": 4,

week03/assignment.py

+def value(v, p=4, n=3):
+    '''
+    รับค่า v มาเพื่อหาค่าเป็นเลขฐานสิบ
+    p เป็นจำนวนหลักที่ใช้เป็นบวก 4 หลัก
+    n เป็นจำนวนหลักที่ใช้เป็นลบ 3 หลัก
+    sign bit เป็นหลักแรกเสมอ
+    '''
+    tmp = [int(i) for i in v]
+    sign = -1 if tmp[0] == 1 else 1
+    return sign * sum([tmp[i] * 2 ** (p-i) for i in range(1, p+n+1)])
+
+
+v = [0, 0, 0, 0, 1, 1, 0, 0]
+print(v)
+print(value(v))
+
+v = [1, 0, 1, 0, 1, 0, 1, 1]
+print(v)
+print(value(v))
+
+v = [1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0]
+print(v)
+print(value(v, 6, 5))
+
+v = '100000110000'
+print(v)
+print(value(v, 6, 5))
+
+v = '00001100'
+print(v)
+print(value(v))
+
+v = '100000101010000'
+print(v)
+print(value(v, 9, 5))