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   "source": [
    "[Home](Home.ipynb)\n",
    "\n",
    "# Logarithms\n",
    "\n",
    "I got [this question](https://qr.ae/pG4Lso) on Quora the other day, asking how to distill a number into powers of 3, never using the same power more than once.  \n",
    "\n",
    "For example we would like to start with the biggest power of 3 that goes into it, and then treat the remainder to similar analysis, using ever smaller powers of 3.\n",
    "\n",
    "The question turns out to be asking how to convert a number to base 3.  We need a set of terms, each the product of a coefficent and a base (3) to a highest degree (power), descending to zero."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "id": "619b2f11",
   "metadata": {},
   "outputs": [],
   "source": [
    "from math import log, floor"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "id": "69c52391",
   "metadata": {},
   "outputs": [],
   "source": [
    "def distill(n, b=10):\n",
    "    \"\"\"\n",
    "    Assume the n starts in base 10, \n",
    "    where b is the target base\n",
    "    \"\"\" \n",
    "    terms = [] \n",
    "    while n > 0: \n",
    "        # b to what power gets closest to n without going over?\n",
    "        degree = floor(log(n, b))  # rounding down to nearest int\n",
    "        q, r = divmod(n, b**degree) \n",
    "        terms.append((q, degree))  # q * b ** degree term\n",
    "        n = r                      # what's still left to handle?\n",
    "    return terms "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "id": "af1dfa74",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "[(1, 8), (1, 7), (1, 6), (2, 5), (1, 3), (1, 2), (1, 0)]"
      ]
     },
     "execution_count": 3,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "distill(10000, 3)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "id": "0deafd2e",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "[(1, 3), (1, 1)]"
      ]
     },
     "execution_count": 4,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "distill(10, 2)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "a4b0d93d",
   "metadata": {},
   "source": [
    "$$\n",
    "(3^{8}) + (3^{7}) + (3^{6}) + 2(3^{5}) + (3^{3}) + (3^{2}) + (3^{0})\n",
    "$$\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "id": "f6914e3a",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "[(3, 2), (14, 1), (8, 0)]"
      ]
     },
     "execution_count": 5,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "distill(1000, 16)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "a15201f0",
   "metadata": {},
   "source": [
    "$$\n",
    "3(3^{2}) + 14(3^{1}) + 8(3^{0})\n",
    "$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "id": "5d4fe2f4",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "[(1, 4), (1, 3), (2, 0)]"
      ]
     },
     "execution_count": 6,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "distill(110, 3)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "73d11aff",
   "metadata": {},
   "source": [
    "$$\n",
    "(3^{4}) + (3^{3}) + 2(3^{0})\n",
    "$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "id": "e5d056d4",
   "metadata": {},
   "outputs": [],
   "source": [
    "def base10(tuples, base=10):\n",
    "    \"\"\"\n",
    "    base is the base used for the tuples, \n",
    "    return number in base 10\n",
    "    \"\"\"\n",
    "    total = 0\n",
    "    for t in tuples:\n",
    "        coeff, degree = t\n",
    "        total = total + coeff * base ** degree\n",
    "    return total"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "id": "4883dbc4",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "[(1, 3), (1, 1)]"
      ]
     },
     "execution_count": 8,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "tups = distill(10, 2)\n",
    "tups"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "3f4c2944",
   "metadata": {},
   "source": [
    "$$\n",
    "(2^{3}) + (2^{1})\n",
    "$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "id": "df8a3672",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "10"
      ]
     },
     "execution_count": 9,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "base10(tups, base=2)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "id": "d6e38c6f",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "[(2, 3), (7, 2), (1, 1)]"
      ]
     },
     "execution_count": 10,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "tups = distill(10000, 16)\n",
    "tups"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "1d627861",
   "metadata": {},
   "source": [
    "$$\n",
    "2(16^{3}) + 7(16^{2}) + 16^{1}\n",
    "$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "id": "c7a5810c",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "10000"
      ]
     },
     "execution_count": 11,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "base10(tups, 16)"
   ]
  }
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