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Added additional solutions to problem 31.

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This commit is contained in:
Felix Martin
2018-10-17 23:20:33 -04:00
parent c34ebd6181
commit 1444f7cc7e
2 changed files with 320 additions and 71 deletions
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194
ipython/EulerProblem031.ipynb
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@@ -29,9 +29,7 @@
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"metadata": {},
"outputs": [
{
"name": "stdout",
@@ -51,59 +49,58 @@
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"execution_count": 8,
"metadata": {},
"outputs": [],
"source": [
"from math import ceil\n",
"\n",
"r = 200\n",
"c = 0\n",
"def calculate():\n",
" r = 200\n",
" c = 0\n",
"\n",
"if r % 200 == 0:\n",
" c += 1\n",
"for p200 in range(ceil(r / 200)):\n",
" r200 = r - p200 * 200\n",
" if r200 % 100 == 0:\n",
" if r % 200 == 0:\n",
" c += 1\n",
" for p100 in range(ceil(r200 / 100)):\n",
" r100 = r200 - p100 * 100\n",
" if r100 % 50 == 0:\n",
" for p200 in range(ceil(r / 200)):\n",
" r200 = r - p200 * 200\n",
" if r200 % 100 == 0:\n",
" c += 1\n",
" for p50 in range(ceil(r100 / 50)):\n",
" r50 = r100 - p50 * 50\n",
" if r50 % 20 == 0:\n",
" for p100 in range(ceil(r200 / 100)):\n",
" r100 = r200 - p100 * 100\n",
" if r100 % 50 == 0:\n",
" c += 1\n",
" for p20 in range(ceil(r50 / 20)):\n",
" r20 = r50 - p20 * 20\n",
" if r20 <= 0:\n",
" break\n",
" if r20 % 10 == 0:\n",
" for p50 in range(ceil(r100 / 50)):\n",
" r50 = r100 - p50 * 50\n",
" if r50 % 20 == 0:\n",
" c += 1\n",
" for p10 in range(ceil(r20 / 10)):\n",
" r10 = r20 - p10 * 10\n",
" if r10 % 5 == 0:\n",
" for p20 in range(ceil(r50 / 20)):\n",
" r20 = r50 - p20 * 20\n",
" if r20 <= 0:\n",
" break\n",
" if r20 % 10 == 0:\n",
" c += 1\n",
" for p5 in range(ceil(r10 / 5)):\n",
" r5 = r10 - p5 * 5\n",
" if r5 % 2 == 0:\n",
" for p10 in range(ceil(r20 / 10)):\n",
" r10 = r20 - p10 * 10\n",
" if r10 % 5 == 0:\n",
" c += 1\n",
" for p2 in range(ceil(r5 / 2)):\n",
" r2 = r5 - p2 * 2\n",
" if r2 % 1 == 0:\n",
" for p5 in range(ceil(r10 / 5)):\n",
" r5 = r10 - p5 * 5\n",
" if r5 % 2 == 0:\n",
" c += 1\n",
" for p2 in range(ceil(r5 / 2)):\n",
" r2 = r5 - p2 * 2\n",
" if r2 % 1 == 0:\n",
" c += 1\n",
"\n",
"s = c"
" return c\n",
"\n",
"s = calculate()"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"execution_count": 9,
"metadata": {},
"outputs": [
{
"name": "stdout",
@@ -117,13 +114,128 @@
"print(s)\n",
"assert(s == 73682)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"I came up with a more concise solution that can be implemented in Scheme. I would like to compare the performance of the solutions."
]
},
{
"cell_type": "code",
"execution_count": 34,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"73682\n"
]
}
],
"source": [
"def count_change(change, coins):\n",
" if not coins:\n",
" return 0\n",
" count = 0\n",
" coin, coins = coins[0], coins[1:]\n",
" for i in range(change // coin + 1):\n",
" if change - i * coin == 0:\n",
" count += 1\n",
" continue\n",
" if change - i * coin < 0:\n",
" break\n",
" count += count_change(change - i * coin, coins)\n",
" return count\n",
"\n",
"def calculate_iterative():\n",
" return count_change(200, [200, 100, 50, 20, 10, 5, 2, 1])\n",
"\n",
"s = calculate_iterative()\n",
"print(s)"
]
},
{
"cell_type": "code",
"execution_count": 35,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"0.16300111500095227\n",
"11.429953110000497\n"
]
}
],
"source": [
"import timeit\n",
"print(timeit.timeit(calculate, number=10))\n",
"print(timeit.timeit(calculate_iterative, number=10))"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Turns out I was way smarter than I thought when I have developped calculate. It checks the remainder and by doing that saves a ton of iterations."
]
},
{
"cell_type": "code",
"execution_count": 36,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"73682\n",
"0.34602016500139143\n"
]
}
],
"source": [
"def count_change(change, coins):\n",
" count = 0\n",
" coin, coins = coins[0], coins[1:]\n",
"\n",
" if change % coin == 0:\n",
" count += 1\n",
" \n",
" if not coins:\n",
" return count\n",
" \n",
" for i in range(ceil(change / coin)):\n",
" count += count_change(change - i * coin, coins)\n",
" return count\n",
"\n",
"def calculate_iterative():\n",
" return count_change(200, [200, 100, 50, 20, 10, 5, 2, 1])\n",
"\n",
"s = calculate_iterative()\n",
"print(s)\n",
"print(timeit.timeit(calculate_iterative, number=10))"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": []
}
],
"metadata": {
"completion_date": "Fri, 25 Aug 2017, 13:02",
"kernelspec": {
"display_name": "Python 3",
"language": "python",
"language": "python3.6",
"name": "python3"
},
"language_info": {
@@ -136,7 +248,7 @@
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython3",
"version": "3.5.4"
"version": "3.6.5"
},
"tags": [
"recursion",