555 lines
15 KiB
Plaintext
555 lines
15 KiB
Plaintext
{
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"cells": [
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"# Advent of Code 2017\n",
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"\n",
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"## December 2nd\n",
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"\n",
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"For each row, determine the difference between the largest value and the smallest value; the checksum is the sum of all of these differences.\n",
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"\n",
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"For example, given the following spreadsheet:\n",
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"\n",
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" 5 1 9 5\n",
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" 7 5 3\n",
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" 2 4 6 8\n",
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"\n",
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"* The first row's largest and smallest values are 9 and 1, and their difference is 8.\n",
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"* The second row's largest and smallest values are 7 and 3, and their difference is 4.\n",
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"* The third row's difference is 6.\n",
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"\n",
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"In this example, the spreadsheet's checksum would be 8 + 4 + 6 = 18."
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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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"metadata": {},
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"outputs": [],
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"source": [
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"from notebook_preamble import J, V, define"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"I'll assume the input is a Joy sequence of sequences of integers.\n",
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"\n",
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" [[5 1 9 5]\n",
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" [7 5 3]\n",
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" [2 4 6 8]]\n",
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"\n",
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"So, obviously, the initial form will be a `step` function:\n",
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"\n",
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" AoC2017.2 == 0 swap [F +] step"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"This function `F` must get the `max` and `min` of a row of numbers and subtract. We can define a helper function `maxmin` which does this:"
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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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"metadata": {},
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"outputs": [],
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"source": [
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"define('maxmin == [max] [min] cleave')"
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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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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"3 1\n"
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]
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}
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],
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"source": [
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"J('[1 2 3] maxmin')"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Then `F` just does that then subtracts the min from the max:\n",
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"\n",
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" F == maxmin -\n",
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"\n",
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"So:"
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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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"metadata": {},
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"outputs": [],
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"source": [
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"define('AoC2017.2 == [maxmin - +] step_zero')"
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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": 5,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"18\n"
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]
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}
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],
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"source": [
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"J('''\n",
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"\n",
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"[[5 1 9 5]\n",
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" [7 5 3]\n",
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" [2 4 6 8]] AoC2017.2\n",
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"\n",
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"''')"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"...find the only two numbers in each row where one evenly divides the other - that is, where the result of the division operation is a whole number. They would like you to find those numbers on each line, divide them, and add up each line's result.\n",
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"\n",
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"For example, given the following spreadsheet:\n",
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"\n",
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" 5 9 2 8\n",
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" 9 4 7 3\n",
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" 3 8 6 5\n",
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"\n",
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"* In the first row, the only two numbers that evenly divide are 8 and 2; the result of this division is 4.\n",
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"* In the second row, the two numbers are 9 and 3; the result is 3.\n",
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"* In the third row, the result is 2.\n",
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"\n",
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"In this example, the sum of the results would be 4 + 3 + 2 = 9.\n",
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"\n",
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"What is the sum of each row's result in your puzzle input?"
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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": 6,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"[9 8 5 2]\n"
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]
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}
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],
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"source": [
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"J('[5 9 2 8] sort reverse')"
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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": 7,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"[8 5 2] [9 divmod] [8 5 2]\n"
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]
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}
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],
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"source": [
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"J('[9 8 5 2] uncons [swap [divmod] cons] dupdip')"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"\n",
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" [9 8 5 2] uncons [swap [divmod] cons F] dupdip G\n",
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" [8 5 2] [9 divmod] F [8 5 2] G\n",
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"\n"
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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": 8,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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" . [8 5 2] [9 divmod] [uncons swap] dip dup [i not] dip\n",
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" [8 5 2] . [9 divmod] [uncons swap] dip dup [i not] dip\n",
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" [8 5 2] [9 divmod] . [uncons swap] dip dup [i not] dip\n",
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" [8 5 2] [9 divmod] [uncons swap] . dip dup [i not] dip\n",
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" [8 5 2] . uncons swap [9 divmod] dup [i not] dip\n",
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" 8 [5 2] . swap [9 divmod] dup [i not] dip\n",
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" [5 2] 8 . [9 divmod] dup [i not] dip\n",
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" [5 2] 8 [9 divmod] . dup [i not] dip\n",
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" [5 2] 8 [9 divmod] [9 divmod] . [i not] dip\n",
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"[5 2] 8 [9 divmod] [9 divmod] [i not] . dip\n",
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" [5 2] 8 [9 divmod] . i not [9 divmod]\n",
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" [5 2] 8 . 9 divmod not [9 divmod]\n",
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" [5 2] 8 9 . divmod not [9 divmod]\n",
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" [5 2] 1 1 . not [9 divmod]\n",
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" [5 2] 1 False . [9 divmod]\n",
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" [5 2] 1 False [9 divmod] . \n"
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]
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}
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],
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"source": [
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"V('[8 5 2] [9 divmod] [uncons swap] dip dup [i not] dip')"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Tricky\n",
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"\n",
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"Let's think.\n",
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"\n",
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"Given a *sorted* sequence (from highest to lowest) we want to \n",
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"* for head, tail in sequence\n",
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" * for term in tail:\n",
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" * check if the head % term == 0\n",
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" * if so compute head / term and terminate loop\n",
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" * else continue"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"### So we want a `loop` I think\n",
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"\n",
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" [a b c d] True [Q] loop\n",
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" [a b c d] Q [Q] loop\n",
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"\n",
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"`Q` should either leave the result and False, or the `rest` and True.\n",
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"\n",
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" [a b c d] Q\n",
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" -----------------\n",
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" result 0\n",
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"\n",
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" [a b c d] Q\n",
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" -----------------\n",
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" [b c d] 1"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"This suggests that `Q` should start with:\n",
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"\n",
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" [a b c d] uncons dup roll<\n",
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" [b c d] [b c d] a\n",
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"\n",
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"Now we just have to `pop` it if we don't need it.\n",
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"\n",
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" [b c d] [b c d] a [P] [T] [cons] app2 popdd [E] primrec\n",
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" [b c d] [b c d] [a P] [a T] [E] primrec\n",
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"\n",
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"-------------------\n",
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"\n",
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" w/ Q == [% not] [T] [F] primrec\n",
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"\n",
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" [a b c d] uncons\n",
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" a [b c d] tuck\n",
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" [b c d] a [b c d] uncons\n",
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" [b c d] a b [c d] roll>\n",
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" [b c d] [c d] a b Q\n",
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" [b c d] [c d] a b [% not] [T] [F] primrec\n",
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"\n",
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" [b c d] [c d] a b T\n",
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" [b c d] [c d] a b / roll> popop 0\n",
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"\n",
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" [b c d] [c d] a b F Q\n",
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" [b c d] [c d] a b pop swap uncons ... Q\n",
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" [b c d] [c d] a swap uncons ... Q\n",
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" [b c d] a [c d] uncons ... Q\n",
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" [b c d] a c [d] roll> Q\n",
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" [b c d] [d] a c Q\n",
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"\n",
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" Q == [% not] [/ roll> popop 0] [pop swap uncons roll>] primrec\n",
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" \n",
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" uncons tuck uncons roll> Q"
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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": 9,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"[8 5 3 2] [9 swap] [9 % not]\n"
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]
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}
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],
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"source": [
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"J('[8 5 3 2] 9 [swap] [% not] [cons] app2 popdd')"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"-------------------\n",
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"\n",
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" [a b c d] uncons\n",
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" a [b c d] tuck\n",
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" [b c d] a [b c d] [not] [popop 1] [Q] ifte\n",
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"\n",
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" [b c d] a [] popop 1\n",
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" [b c d] 1\n",
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"\n",
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" [b c d] a [b c d] Q \n",
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"\n",
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"\n",
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" a [...] Q\n",
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" ---------------\n",
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" result 0\n",
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"\n",
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" a [...] Q\n",
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" ---------------\n",
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" 1\n",
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"\n",
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"\n",
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" w/ Q == [first % not] [first / 0] [rest [not] [popop 1]] [ifte]\n",
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"\n",
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"\n",
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"\n",
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" a [b c d] [first % not] [first / 0] [rest [not] [popop 1]] [ifte]\n",
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" a [b c d] first % not\n",
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" a b % not\n",
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" a%b not\n",
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" bool(a%b)\n",
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"\n",
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" a [b c d] [first % not] [first / 0] [rest [not] [popop 1]] [ifte]\n",
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" a [b c d] first / 0\n",
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" a b / 0\n",
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" a/b 0\n",
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"\n",
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" a [b c d] [first % not] [first / 0] [rest [not] [popop 1]] [ifte]\n",
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" a [b c d] rest [not] [popop 1] [Q] ifte\n",
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" a [c d] [not] [popop 1] [Q] ifte\n",
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" a [c d] [not] [popop 1] [Q] ifte\n",
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"\n",
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" a [c d] [not] [popop 1] [Q] ifte\n",
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" a [c d] not\n",
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"\n",
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" a [] popop 1\n",
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" 1\n",
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"\n",
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" a [c d] Q\n",
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"\n",
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"\n",
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" uncons tuck [first % not] [first / 0] [rest [not] [popop 1]] [ifte]\n",
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" \n",
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" \n"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"### I finally sat down with a piece of paper and blocked it out.\n",
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"\n",
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"First, I made a function `G` that expects a number and a sequence of candidates and return the result or zero:\n",
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"\n",
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" n [...] G\n",
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" ---------------\n",
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" result\n",
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"\n",
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" n [...] G\n",
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" ---------------\n",
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" 0\n",
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"\n",
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"It's a recursive function that conditionally executes the recursive part of its recursive branch\n",
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"\n",
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" [Pg] [E] [R1 [Pi] [T]] [ifte] genrec\n",
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"\n",
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"The recursive branch is the else-part of the inner `ifte`:\n",
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"\n",
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" G == [Pg] [E] [R1 [Pi] [T]] [ifte] genrec\n",
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" == [Pg] [E] [R1 [Pi] [T] [G] ifte] ifte\n",
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"\n",
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"But this is in hindsight. Going forward I derived:\n",
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"\n",
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" G == [first % not]\n",
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" [first /]\n",
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" [rest [not] [popop 0]]\n",
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" [ifte] genrec\n",
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"\n",
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"The predicate detects if the `n` can be evenly divided by the `first` item in the list. If so, the then-part returns the result. Otherwise, we have:\n",
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"\n",
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" n [m ...] rest [not] [popop 0] [G] ifte\n",
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" n [...] [not] [popop 0] [G] ifte\n",
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"\n",
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"This `ifte` guards against empty sequences and returns zero in that case, otherwise it executes `G`."
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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": 10,
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"metadata": {},
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"outputs": [],
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"source": [
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"define('G == [first % not] [first /] [rest [not] [popop 0]] [ifte] genrec')"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Now we need a word that uses `G` on each (head, tail) pair of a sequence until it finds a (non-zero) result. It's going to be designed to work on a stack that has some candidate `n`, a sequence of possible divisors, and a result that is zero to signal to continue (a non-zero value implies that it is the discovered result):\n",
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"\n",
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" n [...] p find-result\n",
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" ---------------------------\n",
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" result\n",
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"\n",
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"It applies `G` using `nullary` because if it fails with one candidate it needs the list to get the next one (the list is otherwise consumed by `G`.)\n",
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"\n",
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" find-result == [0 >] [roll> popop] [roll< popop uncons [G] nullary] primrec\n",
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"\n",
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" n [...] p [0 >] [roll> popop] [roll< popop uncons [G] nullary] primrec\n",
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"\n",
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"The base-case is trivial, return the (non-zero) result. The recursive branch...\n",
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"\n",
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" n [...] p roll< popop uncons [G] nullary find-result\n",
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" [...] p n popop uncons [G] nullary find-result\n",
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" [...] uncons [G] nullary find-result\n",
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" m [..] [G] nullary find-result\n",
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" m [..] p find-result\n",
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"\n",
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"The puzzle states that the input is well-formed, meaning that we can expect a result before the row sequence empties and so do not need to guard the `uncons`."
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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": 11,
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"metadata": {},
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"outputs": [],
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"source": [
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"define('find-result == [0 >] [roll> popop] [roll< popop uncons [G] nullary] primrec')"
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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": 14,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"3.0\n"
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]
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}
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],
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"source": [
|
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"J('[11 9 8 7 3 2] 0 tuck find-result')"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
|
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"In order to get the thing started, we need to `sort` the list in descending order, then prime the `find-result` function with a dummy candidate value and zero (\"continue\") flag."
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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": 12,
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"metadata": {},
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"outputs": [],
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"source": [
|
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"define('prep-row == sort reverse 0 tuck')"
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]
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},
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|
{
|
|
"cell_type": "markdown",
|
|
"metadata": {},
|
|
"source": [
|
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"Now we can define our program."
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]
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},
|
|
{
|
|
"cell_type": "code",
|
|
"execution_count": 13,
|
|
"metadata": {},
|
|
"outputs": [],
|
|
"source": [
|
|
"define('AoC20017.2.extra == [prep-row find-result +] step_zero')"
|
|
]
|
|
},
|
|
{
|
|
"cell_type": "code",
|
|
"execution_count": 15,
|
|
"metadata": {},
|
|
"outputs": [
|
|
{
|
|
"name": "stdout",
|
|
"output_type": "stream",
|
|
"text": [
|
|
"9.0\n"
|
|
]
|
|
}
|
|
],
|
|
"source": [
|
|
"J('''\n",
|
|
"\n",
|
|
"[[5 9 2 8]\n",
|
|
" [9 4 7 3]\n",
|
|
" [3 8 6 5]] AoC20017.2.extra\n",
|
|
"\n",
|
|
"''')"
|
|
]
|
|
}
|
|
],
|
|
"metadata": {
|
|
"kernelspec": {
|
|
"display_name": "Python 2",
|
|
"language": "python",
|
|
"name": "python2"
|
|
},
|
|
"language_info": {
|
|
"codemirror_mode": {
|
|
"name": "ipython",
|
|
"version": 2
|
|
},
|
|
"file_extension": ".py",
|
|
"mimetype": "text/x-python",
|
|
"name": "python",
|
|
"nbconvert_exporter": "python",
|
|
"pygments_lexer": "ipython2",
|
|
"version": "2.7.13"
|
|
}
|
|
},
|
|
"nbformat": 4,
|
|
"nbformat_minor": 2
|
|
}
|