add problem 85
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filifa
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{
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"cells": [
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{
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"cell_type": "markdown",
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"id": "f6ad1b47",
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"metadata": {},
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"source": [
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"# [Counting Rectangles](https://projecteuler.net/problem=85)\n",
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"\n",
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"Let $R(m, n)$ give the number of rectangles contained in an $m \\times n$ grid. We will derive a simple formula for $R$. First, observe that an $m \\times 1$ grid contains $\\frac{m(m+1)}{2}$ rectangles, i.e.\n",
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"$$R(m, 1) = \\frac{m(m+1)}{2}$$\n",
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"(You could prove this [inductively](https://en.wikipedia.org/wiki/Mathematical_induction) if you really wanted to).\n",
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"\n",
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"Then, we can derive the following recursive formula for $R$.\n",
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"$$R(m, n) = R(m, n - 1) + n R(m, 1)$$\n",
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"To make sense of this formula, consider that the number of rectangles in an $m \\times n$ grid must be the sum of the number of rectangles in an $m \\times (n-1)$ grid (i.e. $R(m, n-1)$) and the number of rectangles in an $m \\times n$ grid *that are at least partially in the bottom row* - these are [disjoint sets](https://en.wikipedia.org/wiki/Disjoint_sets), so there's no double counting with this approach. We already know the number of rectangles that lie *entirely* in the bottom row - it's just $R(m, 1)$. We can then make any of those rectangles any height we want between 1 and $n$, giving the formula $n R(m, 1)$.\n",
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"\n",
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"At this point, we could easily translate this into a recursive function to compute $R$, but we can also simplify this into a closed formula. Substituting the formula for $R(m,1)$ from above, we get\n",
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"$$R(m, n) = R(m, n - 1) + \\frac{mn(m+1)}{2}$$\n",
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"Then, by repeatedly substituting $R(m, n-1)$ with our recursive formula, we get\n",
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"$$R(m, n) = \\frac{mn(m+1)}{2} + \\frac{m(n-1)(m+1)}{2} + \\frac{m(n-2)(m+1)}{2} + \\cdots + \\frac{m(m+1)}{2} = \\sum_{k=1}^n \\frac{mk(m+1)}{2}$$\n",
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"Applying some [summation identities](https://en.wikipedia.org/wiki/Summation) (also see [problem 1](https://projecteuler.net/problem=1)), this simplifies to\n",
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"$$R(m, n) = \\frac{mn(m+1)(n+1)}{4}$$\n",
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"Interestingly, this formula is really just the product of $m$th and $n$th [triangular numbers](https://en.wikipedia.org/wiki/Triangular_number)."
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]
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},
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{
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"cell_type": "code",
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"execution_count": 1,
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"id": "2c2038a4",
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"metadata": {},
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"outputs": [],
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"source": [
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"def R(m, n):\n",
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" return m * n * (m + 1) * (n + 1) // 4"
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]
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},
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{
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"cell_type": "markdown",
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"id": "f29eaf82",
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"metadata": {},
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"source": [
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"Figuring out the formula's the hard part - now we can just iterate over grid sizes to calculate how many rectangles they have."
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]
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},
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{
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"cell_type": "code",
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"execution_count": 2,
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"id": "3409e88c",
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"metadata": {},
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"outputs": [],
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"source": [
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"totals = dict()\n",
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"\n",
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"for x in range(1, 2001):\n",
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" for y in range(1, 2001):\n",
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" r = R(x, y)\n",
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" totals[(x,y)] = r\n",
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" if r > 2000000:\n",
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" break"
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]
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},
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{
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"cell_type": "markdown",
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"id": "d9ce5480",
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"metadata": {},
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"source": [
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"We'll figure out which grid is the closest to containing 2000000 rectangles."
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]
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},
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{
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"cell_type": "code",
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"execution_count": 3,
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"id": "dbc72aaa",
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"(36, 77)"
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]
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},
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"execution_count": 3,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"m, n = min(totals, key=lambda x: abs(totals[x] - 2000000))\n",
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"m, n"
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]
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},
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{
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"cell_type": "markdown",
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"id": "e69c5c48",
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"metadata": {},
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"source": [
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"The area of that grid is our answer."
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]
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},
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{
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"cell_type": "code",
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"execution_count": 4,
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"id": "f230a0d0",
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"2772"
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]
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},
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"execution_count": 4,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"m * n"
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]
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},
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{
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"cell_type": "markdown",
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"id": "d30d6d4f",
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"metadata": {},
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"source": [
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"## Relevant sequences\n",
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"* Number of rectangles in an $m \\times n$ grid: [A098358](https://oeis.org/A098358)"
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]
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "SageMath 9.5",
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"language": "sage",
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"name": "sagemath"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.11.2"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 5
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}
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