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week03/Chapter 03 Numerics and Error Analysis.ipynb.ipynb

@@ -139,6 +139,15 @@
     "ยกตัวอย่างตัวเลขในเลขฐาน 2 เช่น  1, 101, 1101  เป็นต้น"
    ]
   },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    ""
+   ]
+  },
   {
    "cell_type": "markdown",
    "metadata": {
@@ -233,7 +242,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 15,
+   "execution_count": null,
    "metadata": {
     "slideshow": {
      "slide_type": "skip"
@@ -336,6 +345,49 @@
     "$0.1 \\times 0.1 = 0.01 \\approx 0.0$ \n"
    ]
   },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "v = [0, 0, 0, 0, 1, 1, 0, 0]\n",
+    "sign = -1 if v[0] == 1 else 1\n",
+    "value = v[1]*2**3\n",
+    "value += v[2]*2**2\n",
+    "value += v[3]*2**1\n",
+    "value += v[4]*2**0\n",
+    "\n",
+    "value += v[5]*2**(-1)\n",
+    "value += v[6]*2**(-2)\n",
+    "value += v[7]*2**(-3)\n",
+    "\n",
+    "print(sign*value)\n",
+    "[(i, 4-i) for i in range(1, 8)]\n",
+    "[v[i]*2**(4-i) for i in range(1, 8)]\n",
+    "sum([v[i]*2**(4-i) for i in range(1, 8)])\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "def value(v):\n",
+    "    sign = -1 if v[0] == a else 1\n",
+    "    return  sign*sum([ v[i]*2**(4-i) for i in range(1,8)])"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "v = [1, 0, 0, 1, 0, 0, 1, 1]"
+   ]
+  },
   {
    "cell_type": "markdown",
    "metadata": {

week03/assignment.py

+def value(v, pos=4, neg=3):
+    """รับ v มาเพื่อหาค่าเป็นเลขฐานสิบ binary --> difital.
+    โดยค่ากำหนกเริ่มต้น
+    pos เป็น 4 หลัก
+    neg เป็น 3 หลัก
+    sign bit เป็นหลักแรกเสมอ
+    """
+    sign = -1 if v[0] == 1 else 1
+    return sign*sum([v[i] * 2 ** (pos-i) for i in range(1, 1+pos+neg)])
+
+def values(v, pos, neg):
+    sign = -1 if int(v[0]) == 1 else 1
+    return sign * sum([int(v[i]) * 2 ** (pos - i) for i in range(1, 1 + pos + neg)])
+
+
+
+