Triangles.ipynb

{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# A Funny Triangle Question"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "by Kunal Marwaha"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### The Question"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "How many triangles are in a triangle composed of smaller triangles?"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "![triangles!](http://mathworld.wolfram.com/images/eps-gif/TriangleTilings_551.gif)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### First Thoughts"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Suppose there are $n$ smaller triangles per side. Obviously, when $n=1$, then the answer is $1$."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "To compose a large triangle from smaller triangles, you can interleave rows of face-up triangles with rows of face-down triangles. \n",
    "\n",
    "You can build rows of $1,2,...,n$ face-up triangles interleaved with rows of $1,2,...,n-1$ face-down triangles. \n",
    "\n",
    "These stacked rows of triangles inspire the series $1, 1+2, 1+2+3, ...$. These numbers are known as the \"triangle numbers\". Thus, the $j^{th}$ triangle number is $T_j = 1+2+...+j = (j+1)(j/2)$."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "There are a total of $(1+2+...+n)+(1+2+...+n-1) = ((n+1)n/2) + (n(n-1)/2) = n^2$ of the smaller triangles in the large triangle. This is perfectly sensible, since an $n$-fold increase in linear proportion magnifies the area by $n^2$."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Intermediate Triangles: Face-Up"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "We can also count all the intermediate-sized triangles. (This is only relevant for $n\\ge 3$.)\n",
    "\n",
    "Let's inspect triangles with $k$ smaller triangles per side. You can certainly build rows of $1,2,...,n-k+1$ face-up triangles. So, we can immediately count the total number of face-up triangles by taking a partial sum of the triangle number series."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Analytic Solution"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Number of face-up triangles = $\\sum_{k=1}^{n} \\sum_{i=1}^{n-k+1} i = \\sum_{k=1}^{n} \\sum_{i=1}^{k} i = \\sum_{k=1}^n T_k = \\sum_{k=1}^n (k+1)(k/2) =\\sum_{k=1}^n k^2/2 + k/2$ .\n",
    "\n",
    "This simplifies to $(2n+1)(n+1)(n/12) + (n+1)(n/4)$ since both $\\sum_{i=1}^n i = T_n$ and $\\sum_{i=1}^n i^2$ have a closed-form expression. Thus, there are $(n/12)(n+1)(2n+1+3) = n(n+1)(n+2)/6$ face-up triangles in the large triangle."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Code and Verification"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "from __future__ import division"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [],
   "source": [
    "def tr(n):\n",
    "    return int(n*(n+1)/2)\n",
    "def num_faceup(n):\n",
    "    return int(n*(n+1)*(n+2)/6)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "n_max = 100\n",
    "assert [sum(map(tr, range(1,n+1))) for n in range(1,n_max+1)] == \\\n",
    "    [num_faceup(n) for n in range(1,n_max+1)], \"num_faceup function isn't correct\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "The above ```assert``` checks that the ```num_faceup``` closed-form expression exactly matches the partial sum of the triangle number series."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Result"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": false,
    "scrolled": true
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Number of face-up triangles when n=1..20:\n",
      "[1, 4, 10, 20, 35, 56, 84, 120, 165, 220, 286, 364, 455, 560, 680, 816, 969, 1140, 1330, 1540]\n"
     ]
    }
   ],
   "source": [
    "n_max=20\n",
    "print(\"Number of face-up triangles when n=1..\"+str(n_max)+\":\")\n",
    "print([num_faceup(n) for n in range(1,n_max+1)])"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Intermediate Triangles: Face-Down"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Counting the face-down triangles is trickier. When $k=1$, as discussed earlier, there are $T_{n-1} = n(n-1)/2$ face-down triangles. The biggest face-down triangle that fits in the large triangle is the largest integer no more than $n/2$ (let's call this $m$). If $n$ is even, it fits right in the middle.\n",
    "\n",
    "Inspecting triangles of $k$ smaller triangles per side, you can build overlapping rows of $1,2,...,n+1-2k$ face-down triangles (supposing $n\\ge 2k$). Now, we can count the total number of face-down triangles by taking a partial sum of odd or even terms in the triangle number series."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Number of face-down triangles = $\\sum_{k=1}^{m} \\sum_{i=1}^{n+1-2k} i  = \\sum_{k=1}^{m} T_{n+1-2k} = T_{n-1} + T_{n-3} + ...$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Analytic Solution (even $n$)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "This summation can be done analytically. First, note that $T_n - T_{n-1} = \\sum_{i=1}^n i - \\sum_{i=1}^{n-1} i = n$. Also, recall that $\\sum_{k=1}^n T_k = n(n+1)(n+2)/6$ is the number of face-up triangles in a large triangle of size $n$.\n",
    "\n",
    "Suppose $n$ is even. We can split up the triangle series into even and odd parts; that is, $\\sum_{k=1}^n T_k = \\sum_{k=1}^{n/2} (T_{2k} + T_{2k-1})$. By the above note, $\\sum_{k=1}^{n/2} (T_{2k} - T_{2k-1}) = \\sum_{k=1}^{n/2} (2k) = (n/2)(n+2)/2$. \n",
    "\n",
    "We can then determine the partial sum of odd terms in the triangle series, as $0.5[\\sum_{k=1}^{n/2} (T_{2k} + T_{2k-1}) - \\sum_{k=1}^{n/2} (T_{2k} - T_{2k-1})] = \\sum_{k=1}^{n/2} T_{2k-1} = \\sum_{k=1}^{n/2} T_{n+1-2k}$.\n",
    "\n",
    "Plugging all expressions in, the number of face-down triangles for even $n$ is $0.5[n(n+1)(n+2)/6  - n(n+2)/4] = n(n+2)(2n+2)/24 -3n(n+2)/24 = n(2n-1)(n+2)/24 = (2n^3 + 3n^2 - 2n)/24$."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Analytic Solution (odd $n$)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "To be exact, the above expression is a summation for the first $n/2$ odd terms in the triangle series, assuming $n$ is even. We can easily find the sum of the first $n/2$ even terms, since $\\sum_{k=1}^{n} T_k - \\sum_{k=1}^{n/2} T_{2k-1} = \\sum_{k=1}^{n/2} T_{2k}$. Plugging in: $n(n+1)(n+2)/6 - n(2n-1)(n+2)/24 = n(2n+5)(n+2)/24$.\n",
    "\n",
    "When $n$ is odd, the number of face-down triangles is the sum of even terms in the triangle series, from $2$ to $n-1$. That is exactly the summation above, except with $n$ relabeled to $n-1$. \n",
    "\n",
    "Precisely, let $x=n-1$ be odd. (Then, $n$ is even.) Then, $\\sum_{k=1}^{n/2} T_{2k} = \\sum_{k=1}^{(x-1)/2} T_{2k} = \\sum_{k=1}^{(x-1)/2} T_{x+1-2k}$ describes the number of face-down triangles in a large triangle of size $x$ (since in this case, $m=(x-1)/2$). This quantity is $n(2n+5)(n+2)/24 = (x-1)(2x+3)(x+1)/24 = (2x^3 + 3x^2 - 2x - 3)/24$."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "There are other solutions for the odd $n$ case, but you will arrive at the same answer. I encourage you to try a method similar to the even $n$ case with a pencil and paper. It is so rewarding to 'figure it out' yourself."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Code and Verification"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "def num_facedown_even(n):\n",
    "    return (2*n-1)*n*(n+2)/24\n",
    "def num_facedown_odd(n):\n",
    "    return (2*n+3)*(n*n-1)/24\n",
    "def num_facedown(n):\n",
    "    return int(num_facedown_even(n)) if n%2==0 else int(num_facedown_odd(n))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
    "collapsed": false
   },
   "outputs": [],
   "source": [
    "n_max = 100\n",
    "assert [sum(map(tr,range(n-1,0,-2))) for n in range(1,n_max+1)] == \\\n",
    "    [num_facedown(n) for n in range(1,n_max+1)], \"num_facedown function isn't correct\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "The above ```assert``` checks that the ```num_facedown``` closed-form expressions exactly match the subset sums of the triangle number series (from $T_{n-1}$, to zero, counting backwards by twos)."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Result"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {
    "collapsed": false,
    "scrolled": true
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Number of face-down triangles when n=1..20:\n",
      "[0, 1, 3, 7, 13, 22, 34, 50, 70, 95, 125, 161, 203, 252, 308, 372, 444, 525, 615, 715]\n"
     ]
    }
   ],
   "source": [
    "n_max=20\n",
    "print(\"Number of face-down triangles when n=1..\"+str(n_max)+\":\")\n",
    "print([num_facedown(n) for n in range(1,n_max+1)])"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### A fun fact"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Note that the solutions for even-sized and odd-sized triangles differ by $3/24=1/8$. The math works out. Strange!"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "[0.0, 0.875, 3.0, 6.875, 13.0, 21.875, 34.0, 49.875, 70.0, 94.875]"
      ]
     },
     "execution_count": 8,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "[num_facedown_odd(i) for i in range(1,10+1)]"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "[0.125, 1.0, 3.125, 7.0, 13.125, 22.0, 34.125, 50.0, 70.125, 95.0]"
      ]
     },
     "execution_count": 9,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "[num_facedown_even(i) for i in range(1,10+1)]"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Total Number of Triangles"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "By the symmetry of the drawing, all triangles are either face-up or face-down. So, the total number of triangles in a large triangle (with $n$ smaller triangles per side) can be calculated analytically:\n",
    "\n",
    "There are $n(n+1)(n+2)/6$  face-up triangles, and $n(2n-1)(n+2)/24$ face-down triangles (when $n$ is even). Thus, there are $n(6n+3)(n+2)/24 = n(2n+1)(n+2)/8 = (2n^3+5n^2+2n)/8$ triangles in a large triangle of even size $n$. (If $n$ is odd, subtract $1/8$ from this expression.)\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Result"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Total number of triangles when n=1..20:\n",
      "[1, 5, 13, 27, 48, 78, 118, 170, 235, 315, 411, 525, 658, 812, 988, 1188, 1413, 1665, 1945, 2255]\n"
     ]
    }
   ],
   "source": [
    "n_max=20\n",
    "print(\"Total number of triangles when n=1..\"+str(n_max)+\":\")\n",
    "print([num_faceup(n) + num_facedown(n) for n in range(1,n_max+1)])"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Another fun thought"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "A cheap way to calculate the number of triangles given $n$ is to round the expression for even $n$ (since $1/8 < 1/2$). The following ```assert``` checks that claim."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "def num_total(n):\n",
    "    return round((2*n+1)*(n+2)*n/8)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "metadata": {
    "collapsed": false
   },
   "outputs": [],
   "source": [
    "n_max=100\n",
    "assert [num_faceup(n) + num_facedown(n) for n in range(1,n_max+1)] == \\\n",
    "    [num_total(n) for n in range(1,n_max+1)], 'num_total rounding doesn\\'t work'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Analysis"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "You can see that this number grows very fast with $n$."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "%matplotlib inline\n",
    "import matplotlib.pyplot as plt"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "metadata": {
    "collapsed": false,
    "scrolled": true
   },
   "outputs": [
    {
     "data": {
      "image/png": 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MB3JrfDjnnHNbeDde55xz2Y6FJekGMztJ0mOEXledmNmhFcTonHOuQXTXBjIt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\nAvkT4L+TXL+zeTPceGMYk2r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      "text/plain": [
       "<matplotlib.figure.Figure at 0x63516a0>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "n_max=20\n",
    "plt.plot(range(1,n_max+1), [num_faceup(n) + num_facedown(n) for n in range(1,n_max+1)], 'b.', markersize=20)\n",
    "plt.plot(range(1,n_max+1), [num_total(n) for n in range(1,n_max+1)], 'r.', markersize=10)\n",
    "plt.xlabel(\"size of large triangle ($n$)\")\n",
    "plt.ylabel(\"number of triangles in the large triangle\")\n",
    "plt.title(\"triangles in a triangle composed of smaller triangles\")\n",
    "plt.grid()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "It is not exponential, by the way. (Why? Think about it, then read on.)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Scaling"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Remember that the expression for the solution (for even $n$) is $(2n^3+5n^2+2n)/8$. A-ha! So the asymptotic scaling for this question is $n^3$. This means that when $n$ is very large, there are about $n^3$ triangles contained in the large triangle. "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "You can visualize this with the slope of the log-log plot (it should be about 3 as $n$ gets large)."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "metadata": {
    "collapsed": false,
    "scrolled": false
   },
   "outputs": [
    {
     "data": {
      "image/png": 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JmiPpcUmnJw5tD9xpZt8AGsqN5ji1IBn/eOKJUPf978NNN3n8w6kvaWIimwE/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\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0x6521be0>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "n_max=2000\n",
    "plt.loglog(range(1,n_max+1), [num_faceup(n) + num_facedown(n) for n in range(1,n_max+1)], 'b.', markersize=2)\n",
    "plt.xlabel(\"size of large triangle ($n$)\")\n",
    "plt.ylabel(\"number of triangles in the large triangle\")\n",
    "plt.title(\"triangles in a triangle composed of smaller triangles\")\n",
    "plt.grid()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### The end"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "But now you know the answer. Go ahead, draw a small example and ask your friends: \"How many triangles?\" (I've found that $n=4$ is a great example.)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "![triangles!](http://mathworld.wolfram.com/images/eps-gif/TriangleTilings_551.gif)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "I hope you enjoyed reading this bit on triangles."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Here's the WolframAlpha page on [Tiling Triangles](http://mathworld.wolfram.com/TriangleTiling.html). Triangle image ($n=1,2,3,4$) belongs to WolframAlpha."
   ]
  }
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