Programmatic solution to the "Impossible puzzle"

ImpossiblePuzzle.ipynb

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   "source": [
    "from collections import defaultdict"
   ]
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   "cell_type": "code",
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     "text": [
      "96 1179\n"
     ]
    }
   ],
   "source": [
    "# For every product and sum, list all the pairs corresponding to it\n",
    "productset=defaultdict(list)\n",
    "sumset=defaultdict(list)\n",
    "for y in range(2,100):\n",
    "    for x in range(2,min(y,101-y)):\n",
    "        productset[x*y].append((x,y))\n",
    "        sumset[x+y].append((x,y))\n",
    "print (len(sumset), len(productset))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "574\n"
     ]
    }
   ],
   "source": [
    "# P should be unable to find x and y\n",
    "# This means we can immediately discard all products with a single way of forming\n",
    "productset = dict((p,productset[p]) for p in productset if len(productset[p]) > 1)\n",
    "print (len(productset))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
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     "text": [
      "10\n"
     ]
    }
   ],
   "source": [
    "# S' knowledge of inability of P to find x and y means that only those sums need to be retained\n",
    "# of which every possible (x,y) pair results in a still valid product\n",
    "sumset = dict((s,sumset[s]) for s in sumset if sum(x[0]*x[1] not in productset for x in sumset[s]) == 0)\n",
    "print (len(sumset))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
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     "text": [
      "86\n"
     ]
    }
   ],
   "source": [
    "# This information allows P to find x and y\n",
    "# Means that we need to only consider those products with exactly one (x,y) pair giving a still valid sum\n",
    "productset=dict((p,[xy for xy in productset[p] if xy[0]+xy[1] in sumset]) for p in productset if sum(xy[0]+xy[1] in sumset for xy in productset[p]) == 1)\n",
    "print (len(productset))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "1\n"
     ]
    }
   ],
   "source": [
    "# And further, this information allows S to find x and y\n",
    "# Means that only those sums with exactly one (x,y) pair with a valid product need to be retained\n",
    "sumset=dict((s,[xy for xy in sumset[s] if xy[0]*xy[1] in productset]) for s in sumset if sum(xy[0]*xy[1] in productset for xy in sumset[s]) == 1)\n",
    "print (len(sumset))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "{17: [(4, 13)]}\n"
     ]
    }
   ],
   "source": [
    "# We have only one element remaining in the set of sums\n",
    "print (sumset)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  }
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