Solved 3 and fixed bug in publish script.

ipython/EulerProblem003.ipynb

@@ -11,6 +11,110 @@
     "What is the largest prime factor of the number 600851475143?"
    ]
   },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "We start by writing a function which calculates all primes till a certain value using the Sieve of Eratosthenes."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [],
+   "source": [
+    "def get_primes_smaller(number):\n",
+    "    primes = []\n",
+    "    prospects = [n for n in range(2, number)]\n",
+    "    while prospects:\n",
+    "        p = prospects[0]\n",
+    "        prospects = [x for x in prospects if x % p != 0]\n",
+    "        primes.append(p)\n",
+    "    return primes\n",
+    "\n",
+    "assert(get_primes_smaller(0) == [])\n",
+    "assert(get_primes_smaller(10) == [2, 3, 5, 7,])"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Now we create a function which does prime factorization. It is very important that we only test primes smaller than the squre root. Otherwise the complexity becomes too big."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [],
+   "source": [
+    "import math\n",
+    "\n",
+    "def get_prime_factors(number):\n",
+    "    possible_factors = get_primes_smaller(math.ceil(math.sqrt(number)))\n",
+    "    remainder = number\n",
+    "    factors = []\n",
+    "    for p in possible_factors:\n",
+    "        while remainder % p == 0:\n",
+    "            remainder /= p\n",
+    "            factors.append(p)\n",
+    "        if remainder == 1:\n",
+    "            break\n",
+    "    if remainder != 1:\n",
+    "        factors.append(remainder)\n",
+    "    return factors\n",
+    "\n",
+    "assert(get_prime_factors(7) == [7])\n",
+    "assert(get_prime_factors(12) == [2, 2, 3])    \n",
+    "assert(get_prime_factors(88) == [2, 2, 2, 11]) \n",
+    "assert(get_prime_factors(13195) == [5, 7, 13, 29])\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Now we can go ahead an brute force the solution."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "6857\n"
+     ]
+    }
+   ],
+   "source": [
+    "def get_largest_prime(number):\n",
+    "    return get_prime_factors(number)[-1]\n",
+    "\n",
+    "assert(get_largest_prime(13195) == 29)\n",
+    "print(get_largest_prime(600851475143))"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "Okay, actually we can brute force, but it is really slow. A better solution is to write a prime number generator which calculates the next number on demand."
+   ]
+  },
   {
    "cell_type": "code",
    "execution_count": null,

ipython/publish.py

@@ -40,7 +40,7 @@ def render_solutions(solutions):
 
 def convert_solutions_to_html(solutions):
     for s in solutions:
-        if getmtime(s.html) < getmtime(s.ipynb):
+        if not os.path.isfile(s.html) or getmtime(s.html) < getmtime(s.ipynb):
             args = ["ipython", "nbconvert", s.ipynb]
             subprocess.call(args)