3. Control flow.ipynb

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   "source": [
    "# Control flow statements"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# 1. Conditional statements"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "![](images/conditions.jpeg)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "![](images/conditional.png)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Hello\n",
      "None\n"
     ]
    }
   ],
   "source": [
    "print(print(\"Hello\"))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "end\n"
     ]
    }
   ],
   "source": [
    "a = 15\n",
    "if a == 10:\n",
    "    print(\"something\")\n",
    "elif a == 12:\n",
    "    print(\"jatin\")\n",
    "else:\n",
    "    print(\"end\")"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Challenge:\n",
    "[Python If-Else](https://www.hackerrank.com/challenges/py-if-else)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Nested if-else\n",
    "\n",
    "Write a python program to input an integer from user and print:\n",
    "- -1 if integer is negative.\n",
    "- 0 if integer is zero.\n",
    "- 1 if integer is positive.\n",
    "\n",
    "What's not cool with this approach???\n",
    "\n",
    "```python\n",
    "if x < 0:\n",
    "   sign = -1\n",
    "if x == 0\n",
    "   sign = 0\n",
    "if x > 0\n",
    "   sign = 1\n",
    "```\n",
    "\n",
    "![](images/nested.jpeg)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
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   "outputs": [],
   "source": [
    "a = int(input())\n",
    "\n",
    "if a < 0:\n",
    "    "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# 2. Looping statements"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## (i) while\n",
    "\n",
    "The while statement allows you to repeatedly execute a block of statements as long as a condition is true.\n",
    "\n",
    "![](images/while-loop.png)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "10\n",
      "9\n",
      "8\n",
      "7\n",
      "6\n",
      "5\n",
      "4\n",
      "3\n",
      "2\n",
      "1\n"
     ]
    }
   ],
   "source": [
    "a = 10\n",
    "while a > 0:\n",
    "    print(a)\n",
    "    a -= 1\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## (ii) for\n",
    "\n",
    "The for..in statement is another looping statement which iterates over a sequence of objects i.e. go through each item in a sequence.\n",
    "\n",
    "![](images/for_loop.jpg)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {},
   "outputs": [],
   "source": [
    "import dis\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "metadata": {},
   "outputs": [],
   "source": [
    "def some():\n",
    "    for i in range(10):\n",
    "        print(i)\n",
    "        i = 100"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": []
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Challenge:\n",
    "Solve using **for** as well as **while** loop:\n",
    "[Loops](https://www.hackerrank.com/challenges/python-loops)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Infinite loops\n",
    "\n",
    "Beware!\n",
    "\n",
    "![](images/infinite.jpg)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Nested loops\n",
    "\n",
    "Yes! Loops can also be nested.\n",
    "\n",
    "Lets do matrix addition!"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 28,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "5 5 5 5 5 \n",
      "5 4 4 4 5 \n",
      "5 4 3 4 5 \n",
      "5 4 4 4 5 \n",
      "5 5 5 5 5 \n"
     ]
    }
   ],
   "source": [
    "n = 5\n",
    "for i in range(n):\n",
    "    for j in range(n):\n",
    "        print(n - min(i, j, n-i-1, n-j-1), end = ' ')\n",
    "    print()"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 24,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "4 3 2 1 0 \n",
      "4 3 2 1 0 \n",
      "4 3 2 1 0 \n",
      "4 3 2 1 0 \n",
      "4 3 2 1 0 \n"
     ]
    }
   ],
   "source": [
    "for i in range(5):\n",
    "    for j in range(5):\n",
    "        print(5-j-1, end = ' ')\n",
    "    print()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Challenge for you!\n",
    "\n",
    "Given an integer n, print a nXn matrix with squares (outermost to innermost) having values n, then n-1 and so on.\n",
    "For example, given 5, output is:\n",
    "```\n",
    "5 5 5 5 5\n",
    "5 4 4 4 5\n",
    "5 4 3 4 5\n",
    "5 4 4 4 5 \n",
    "5 5 5 5 5\n",
    "```"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": []
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## break, continue and pass"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 41,
   "metadata": {},
   "outputs": [],
   "source": [
    "for i in range(10):\n",
    "    pass\n",
    "    # TODO"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "![](images/break.jpg)"
   ]
  }
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