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refer to relevant problems instead of re-implementing

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filifa 2025-05-15 23:30:57 -04:00
parent 275bfae648
commit fc7c74c323
2 changed files with 34 additions and 58 deletions
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41
notebooks/problem0065.ipynb
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@ -36,7 +36,7 @@
"id": "396468b7", "id": "396468b7",
"metadata": {}, "metadata": {},
"source": [ "source": [
"If we wanted to compute the convergents ourselves, we could first make a generator for the partial denominators of the continued fraction of $e$." "To compute the convergents ourselves, we'll first make a generator for the partial denominators of the continued fraction of $e$."
] ]
}, },
{ {
@ -48,14 +48,9 @@
"source": [ "source": [
"from itertools import count, chain\n", "from itertools import count, chain\n",
"\n", "\n",
"def partial_denominators_e(n):\n", "def partial_denominators_e():\n",
" yield 2\n", " yield 2\n",
" denominators = chain.from_iterable((1, 2 * k, 1) for k in count(1))\n", " yield from chain.from_iterable((1, 2 * k, 1) for k in count(1))"
" for (i, b) in enumerate(denominators):\n",
" if i >= n:\n",
" break\n",
" \n",
" yield b"
] ]
}, },
{ {
@ -63,7 +58,7 @@
"id": "33288cd0", "id": "33288cd0",
"metadata": {}, "metadata": {},
"source": [ "source": [
"Then write a function for computing a continued fraction from a sequence of partial denominators (outside of SageMath, you might want to use a [fraction type](https://docs.python.org/3/library/fractions.html))." "Then we'll apply a simple algorithm for computing [convergents using the partial denominators](https://en.wikipedia.org/wiki/Simple_continued_fraction) (outside of SageMath, you might want to use a [fraction type](https://docs.python.org/3/library/fractions.html))."
] ]
}, },
{ {
@ -73,13 +68,21 @@
"metadata": {}, "metadata": {},
"outputs": [], "outputs": [],
"source": [ "source": [
"def cf(denominators):\n", "def convergents(partial_denoms):\n",
" a = next(denominators)\n", " h, hprev = 1, 0\n",
" \n", " k, kprev = 0, 1\n",
" try:\n", " for b in partial_denoms:\n",
" return a + 1 / cf(denominators)\n", " h, hprev = b * h + hprev, h\n",
" except StopIteration:\n", " k, kprev = b * k + kprev, k\n",
" return a" " yield h/k"
]
},
{
"cell_type": "markdown",
"id": "99b790f3",
"metadata": {},
"source": [
"Now just iterate until we reach the 100th convergent."
] ]
}, },
{ {
@ -100,7 +103,11 @@
} }
], ],
"source": [ "source": [
"sum(cf(partial_denominators_e(99)).numerator().digits())" "for (i, c) in enumerate(convergents(partial_denominators_e())):\n",
" if i == 99:\n",
" break\n",
"\n",
"sum(c.numerator().digits())"
] ]
}, },
{ {

51
notebooks/problem0066.ipynb
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@ -19,10 +19,10 @@
{ {
"data": { "data": {
"text/plain": [ "text/plain": [
"[(-sqrt(2)*(2*sqrt(2) + 3)^t + sqrt(2)*(-2*sqrt(2) + 3)^t - 3/2*(2*sqrt(2) + 3)^t - 3/2*(-2*sqrt(2) + 3)^t,\n", "[(sqrt(2)*(2*sqrt(2) + 3)^t - sqrt(2)*(-2*sqrt(2) + 3)^t + 3/2*(2*sqrt(2) + 3)^t + 3/2*(-2*sqrt(2) + 3)^t,\n",
" 3/4*sqrt(2)*(2*sqrt(2) + 3)^t - 3/4*sqrt(2)*(-2*sqrt(2) + 3)^t + (2*sqrt(2) + 3)^t + (-2*sqrt(2) + 3)^t),\n", " -3/4*sqrt(2)*(2*sqrt(2) + 3)^t + 3/4*sqrt(2)*(-2*sqrt(2) + 3)^t - (2*sqrt(2) + 3)^t - (-2*sqrt(2) + 3)^t),\n",
" (sqrt(2)*(2*sqrt(2) + 3)^t - sqrt(2)*(-2*sqrt(2) + 3)^t + 3/2*(2*sqrt(2) + 3)^t + 3/2*(-2*sqrt(2) + 3)^t,\n", " (-sqrt(2)*(2*sqrt(2) + 3)^t + sqrt(2)*(-2*sqrt(2) + 3)^t - 3/2*(2*sqrt(2) + 3)^t - 3/2*(-2*sqrt(2) + 3)^t,\n",
" -3/4*sqrt(2)*(2*sqrt(2) + 3)^t + 3/4*sqrt(2)*(-2*sqrt(2) + 3)^t - (2*sqrt(2) + 3)^t - (-2*sqrt(2) + 3)^t)]" " 3/4*sqrt(2)*(2*sqrt(2) + 3)^t - 3/4*sqrt(2)*(-2*sqrt(2) + 3)^t + (2*sqrt(2) + 3)^t + (-2*sqrt(2) + 3)^t)]"
] ]
}, },
"execution_count": 1, "execution_count": 1,
@ -97,51 +97,20 @@
"## Solving Pell equations\n", "## Solving Pell equations\n",
"Lagrange proved that if $(x_0, y_0)$ is a solution to\n", "Lagrange proved that if $(x_0, y_0)$ is a solution to\n",
"$$x^2 - dy^2 = 1$$\n", "$$x^2 - dy^2 = 1$$\n",
"then $\\frac{x_0}{y_0}$ is a [convergent of the continued fraction](https://en.wikipedia.org/wiki/Simple_continued_fraction) of $\\sqrt{d}$. This is great for us, since we can write generators for computing convergents of square roots (FYI, SageMath can do this with built-in methods: `continued_fraction(sqrt(d)).convergents()`)." "then $\\frac{x_0}{y_0}$ is a [convergent of the continued fraction](https://en.wikipedia.org/wiki/Simple_continued_fraction) of $\\sqrt{d}$. This is great for us, since there are algorithms to compute these convergents. We'll use SageMath here; see [problem 64](https://projecteuler.net/problem=64) for how to compute the partial denominators of the continued fraction of a square root, and see [problem 65](https://projecteuler.net/problem=65) for an algorithm that uses partial denominators to compute convergents of continued fractions.\n",
"\n",
"Here, we iterate over each convergent to see if its numerator and denominator are a solution to the Pell equation."
] ]
}, },
{ {
"cell_type": "code", "cell_type": "code",
"execution_count": 4, "execution_count": 4,
"id": "c51f9b7e",
"metadata": {},
"outputs": [],
"source": [
"def continued_fraction_sqrt(d):\n",
" x = sqrt(d)\n",
" while True:\n",
" b = floor(x)\n",
" yield b\n",
" x = (x - b)^-1\n",
" \n",
" \n",
"def convergents(partial_denoms):\n",
" h, hprev = 1, 0\n",
" k, kprev = 0, 1\n",
" for b in partial_denoms:\n",
" h, hprev = b * h + hprev, h\n",
" k, kprev = b * k + kprev, k\n",
" yield h/k"
]
},
{
"cell_type": "markdown",
"id": "d5cbf5eb",
"metadata": {},
"source": [
"Then we can just iterate over each convergent to see if its numerator and denominator are a solution to the Pell equation."
]
},
{
"cell_type": "code",
"execution_count": 5,
"id": "5d95125c", "id": "5d95125c",
"metadata": {}, "metadata": {},
"outputs": [], "outputs": [],
"source": [ "source": [
"def pell_fundamental_solution(d):\n", "def pell_fundamental_solution(d):\n",
" partial_denoms = continued_fraction_sqrt(d)\n", " for f in continued_fraction(sqrt(d)).convergents():\n",
" for f in convergents(partial_denoms):\n",
" x, y = f.as_integer_ratio()\n", " x, y = f.as_integer_ratio()\n",
" if x^2 - d*y^2 == 1:\n", " if x^2 - d*y^2 == 1:\n",
" return (x, y)" " return (x, y)"
@ -157,7 +126,7 @@
}, },
{ {
"cell_type": "code", "cell_type": "code",
"execution_count": 6, "execution_count": 5,
"id": "03d7ba9d", "id": "03d7ba9d",
"metadata": {}, "metadata": {},
"outputs": [ "outputs": [
@ -167,7 +136,7 @@
"661" "661"
] ]
}, },
"execution_count": 6, "execution_count": 5,
"metadata": {}, "metadata": {},
"output_type": "execute_result" "output_type": "execute_result"
} }