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Commit 238259c3 authored by Kittipong Maneewong's avatar Kittipong Maneewong Committed by sumet
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ทำจาก Pycharm IDE เป็นครั้งแรก

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Week01/hello.py 0 → 100644
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print("Hello World")
Week02/hellomodule.py 0 → 100644
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#cookiecuter .....
print("we've done it.")
\ No newline at end of file
Week03/Chapter 03 Numerics and Error Analysis.ipynb 0 → 100644
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{
"cells": [
{
"cell_type": "markdown",
"metadata": {
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"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"
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"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": "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",
" $\\rightarrow E_{per} = E_{rel} \\times 100$"
]
},
{
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"execution_count": null,
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"source": []
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