euler/ipython/EulerProblem003.ipynb

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   "source": [
    "# Euler Problem 003\n",
    "\n",
    "The prime factors of 13195 are 5, 7, 13 and 29.\n",
    "\n",
    "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."
   ]
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   "execution_count": 1,
   "metadata": {
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   "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."
   ]
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  {
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   "execution_count": 2,
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   "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": 3,
   "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))\n",
    "print(6857) # computed the previously but remove it so that we can reexecute the complete kernel"
   ]
  },
  {
   "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": 4,
   "metadata": {
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   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "6857\n"
     ]
    }
   ],
   "source": [
    "def is_prime(n, smaller_primes):\n",
    "    for s in smaller_primes:\n",
    "        if n % s == 0:\n",
    "            return False\n",
    "        if s * s > n:\n",
    "            return True\n",
    "    return True\n",
    "\n",
    "def prime_generator_function():\n",
    "    primes = [2, 3, 5, 7]\n",
    "    for p in primes:\n",
    "        yield p\n",
    "    while True:\n",
    "        p += 2\n",
    "        if is_prime(p, primes):\n",
    "            primes.append(p)\n",
    "            yield p\n",
    "            \n",
    "def get_prime_factors(number):\n",
    "    prime_generator = prime_generator_function()\n",
    "    remainder = number\n",
    "    factors = []\n",
    "    for p in prime_generator:\n",
    "        while remainder % p == 0:\n",
    "            remainder /= p\n",
    "            factors.append(p)\n",
    "        if remainder == 1 or p * p > number:\n",
    "            break\n",
    "    if remainder != 1:\n",
    "        factors.append(remainder)\n",
    "    return factors\n",
    "\n",
    "print(get_largest_prime(600851475143))\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Here we go. Obviously much better than precalculation primes that we never need."
   ]
  }
 ],
 "metadata": {
  "completion_date": "Tue, 19 Aug 2014, 23:35",
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  "tags": [
   "prime"
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