interactive-coding-challenges/recursion-dynamic/fibonacci.ipynb

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   "source": [
    "<small><i>This notebook was prepared by [Donne Martin](http://donnemartin.com). Source and license info is on [GitHub](https://bit.ly/code-notes).</i></small>"
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    "## Problem: Implement fibonacci recursively, dynamically, and iteratively.\n",
    "\n",
    "* [Constraints and Assumptions](#Constraints-and-Assumptions)\n",
    "* [Test Cases](#Test-Cases)\n",
    "* [Algorithm](#Algorithm)\n",
    "* [Code: Recursive](#Code:-Recursive)\n",
    "* [Code: Dynamic](#Code:-Dynamic)\n",
    "* [Code: Iterative](#Code:-Iterative)\n",
    "* [Unit Test](#Unit-Test)"
   ]
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   "source": [
    "## Constraints and Assumptions\n",
    "\n",
    "*Problem statements are often intentionally ambiguous.  Identifying constraints and stating assumptions can help to ensure you code the intended solution.*\n",
    "\n",
    "* None"
   ]
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   "source": [
    "## Test Cases\n",
    "\n",
    "* n = 0\n",
    "* n = 1\n",
    "* n > 1"
   ]
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   "source": [
    "## Algorithm\n",
    "\n",
    "* Fibonacci is as follows: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34...\n",
    "* If n = 0 or 1, return n\n",
    "* Else return fib(n-1) + fib(n+2)\n",
    "\n",
    "Complexity:\n",
    "* Time: O(2^n) if recursive or iterative, O(n) if dynamic\n",
    "* Space: O(n) if recursive, O(1) if iterative, O(1) if dynamic"
   ]
  },
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   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Code: Recursive"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
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   "outputs": [],
   "source": [
    "def fib_recursive(n):\n",
    "    if n == 0 or n == 1:\n",
    "        return n\n",
    "    else:\n",
    "        return fib_recursive(n-1) + fib_recursive(n-2)"
   ]
  },
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   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Code: Dynamic"
   ]
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   "outputs": [],
   "source": [
    "num_items = 10\n",
    "cache = [None] * (num_items + 1)\n",
    "\n",
    "def fib_dynamic(n):\n",
    "    if n == 0 or n == 1:\n",
    "        return n\n",
    "    if cache[n] != None:\n",
    "        return cache[n]\n",
    "    cache[n] = fib_dynamic(n-1) + fib_dynamic(n-2)\n",
    "    return cache[n]"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Code: Iterative"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
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   "outputs": [],
   "source": [
    "def fib_iterative(n):\n",
    "    a = 0        \n",
    "    b = 1\n",
    "    for _ in xrange(n):\n",
    "        a, b = b, a + b\n",
    "    return a"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Unit Test\n",
    "\n",
    "*It is important to identify and run through general and edge cases from the [Test Cases](#Test-Cases) section by hand.  You generally will not be asked to write a unit test like what is shown below.*"
   ]
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  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": false
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   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Success: test_fib\n",
      "Success: test_fib\n",
      "Success: test_fib\n"
     ]
    }
   ],
   "source": [
    "from nose.tools import assert_equal\n",
    "\n",
    "class Test(object):\n",
    "    def test_fib(self, func):\n",
    "        result = []\n",
    "        for i in xrange(num_items):\n",
    "            result.append(func(i))\n",
    "        fib_seq = [0, 1, 1, 2, 3, 5, 8, 13, 21, 34]\n",
    "        assert_equal(result, fib_seq)\n",
    "        print('Success: test_fib')\n",
    "\n",
    "if __name__ == '__main__':\n",
    "    test = Test()\n",
    "    test.test_fib(fib_recursive)\n",
    "    test.test_fib(fib_dynamic)\n",
    "    test.test_fib(fib_iterative)"
   ]
  }
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