pt-1.ipynb

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   "cell_type": "markdown",
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   "source": [
    "### Prep for Unit Test-I\n",
    "\n",
    "23-September-2024.\n",
    "\n",
    "Using this notebook to prep/check/test for concepts.\n",
    "\n",
    "---"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Range Check"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "for i in range(0, 4):\n",
    "    print(i)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "The `range` function outputs stuff from `x` to `y-1` where `range (x, y)`.\n",
    "\n",
    "You can bypass that by incrementing the placeholder variable `i` by `1`.\n",
    "\n",
    "---"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Printing outside loop"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "for i in range(len([0,2,3,4,5])):\n",
    "    i = i**2\n",
    "print(i) # <- check indentation"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "If you a print the variable _after_ the for loop runs, it will pick the **last output** the for loop has given.\n",
    "\n",
    ">Note: it stores the variable even after running the loop. which is *bad* for memory management?  \n",
    "\n",
    "---"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Problem Statement\n",
    "\n",
    "Write a python program which reads string from user and count the number of vowels present in it."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "vowels = ['a', 'e', 'i', 'o', 'u']\n",
    "\n",
    "# inp = \"bazooka dick fucker cunt\"\n",
    "\n",
    "v = 0\n",
    "for i in input(\"bazooka? \"):\n",
    "    if i in vowels:\n",
    "        v += 1\n",
    "print(\"Vowels:\", v)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "##### How this works:\n",
    "\n",
    "1. specify the vowels in a list\n",
    "2. add a string (or ask for input)\n",
    "3. specify var which stores the no. of vowels\n",
    "4. run the string through a for loop\n",
    "5. if `i` in `vowels` (the list), then increment the vowel-storing-variable.\n",
    "6. print the vowels *after* the loop as finished (the last output.)\n",
    "\n",
    "---"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Assignment Operators"
   ]
  },
  {
   "cell_type": "code",
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   "metadata": {},
   "outputs": [],
   "source": [
    "i = 1\n",
    "w = 2\n",
    "t = 3\n",
    "c = 4\n",
    "\n",
    "for n in range(0, 69):\n",
    "    i += 2 #add\n",
    "    w *= 2 #multiply\n",
    "    t -= 3 #subtract\n",
    "    c /= 2 #divide\n",
    "\n",
    "print(i, w, t, c)\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "---"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### PEMDAS Check\n",
    "\n",
    "To check if order of operations hold place or not."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "print(3**6/3)\n",
    "#vs\n",
    "print(3**(6/3))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Order of operations: (), **, [ *, / ] , [ + , - ]\n",
    "\n",
    ">The [] indicate equal \"preference\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Identity Operators"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "cable = \"dump the brain\"\n",
    "non_cable = \"dump the brain\"\n",
    "\n",
    "print(\"yes\") if cable is non_cable else print(\"no\")"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "This returns \"no\" because even though the variables have the same \"value\", they are *different*. They are differently initialised and all."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "cable = \"dump the brain\"\n",
    "non_cable = cable\n",
    "\n",
    "print(\"yes\") if cable is non_cable else print(\"no\")"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "This returns \"yes\" because `non_cable` **is** assigned to `cable`. \n",
    "\n",
    "---"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Several Problems"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Write the pseudocode to print all multiples of 5 between 10 and 25 (including both 10 and 25)."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "1. for loop for range 10 to 26\n",
    "2. if the varible % 5 equals to 0\n",
    "3. then print the variable"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "#not pseudocode\n",
    "for i in range(10, 26):\n",
    "    if i%5 == 0:\n",
    "        print(i)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Write an algorithm to find the greatest among two different numbers"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "# long way\n",
    "\n",
    "a = 6\n",
    "b = 3\n",
    "\n",
    "if a>b:\n",
    "    print(a)\n",
    "else:\n",
    "    print(b)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "# short way\n",
    "\n",
    "print(a) if (a:=6) > (b:=3) else print(b)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Write a pseudocode to calculate the factorial of a number (incomplete)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "1. given number `a`\n",
    "2. "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "a = 5\n",
    "\n",
    "i = 1\n",
    "while True:\n",
    "    b *= i\n",
    "    i += 1\n",
    "\n",
    "    print(b, i)\n",
    "    if i == a:\n",
    "        break \n",
    "\n",
    "print(b)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Write an algorithm to find greatest among three numbers"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 58,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "34\n"
     ]
    }
   ],
   "source": [
    "#long way\n",
    "\n",
    "a = 2\n",
    "b = 7\n",
    "c = 34\n",
    "\n",
    "if a > b and a > c:\n",
    "    print(a)\n",
    "elif a > b and a < c:\n",
    "    print(c)\n",
    "elif a < b and b < c:\n",
    "    print(c)\n",
    "elif a < b and b > c:\n",
    "    print(b)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 61,
   "metadata": {},
   "outputs": [],
   "source": [
    "a = 69\n",
    "b = 7\n",
    "c = 34\n",
    "\n"
   ]
  }
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