day 8 - lucky lucky kind inputs

2023/day08/main.go

@@ -0,0 +1,145 @@
+package main
+
+import (
+	_ "embed"
+	"flag"
+	"fmt"
+	"strings"
+
+	"github.com/alexchao26/advent-of-code-go/mathy"
+	"github.com/alexchao26/advent-of-code-go/util"
+)
+
+//go:embed input.txt
+var input string
+
+func init() {
+	// do this in init (not main) so test file has same input
+	input = strings.TrimRight(input, "\n")
+	if len(input) == 0 {
+		panic("empty input.txt file")
+	}
+}
+
+func main() {
+	var part int
+	flag.IntVar(&part, "part", 1, "part 1 or 2")
+	flag.Parse()
+	fmt.Println("Running part", part)
+
+	if part == 1 {
+		ans := part1(input)
+		util.CopyToClipboard(fmt.Sprintf("%v", ans))
+		fmt.Println("Output:", ans)
+	} else {
+		ans := part2(input)
+		util.CopyToClipboard(fmt.Sprintf("%v", ans))
+		fmt.Println("Output:", ans)
+	}
+}
+
+func part1(input string) int {
+	steps, graph := parseInput(input)
+
+	location := "AAA"
+	numOfSteps := 0
+	for location != "ZZZ" {
+		dir := steps[numOfSteps%len(steps)]
+
+		if dir == "L" {
+			location = graph[location][0]
+		} else {
+			location = graph[location][1]
+		}
+
+		numOfSteps++
+	}
+
+	return numOfSteps
+}
+
+func part2(input string) int {
+	steps, graph := parseInput(input)
+
+	locations := []string{}
+	for loc := range graph {
+		if loc[2:3] == "A" {
+			locations = append(locations, loc)
+		}
+	}
+
+	/**
+	brute force doesn't work... need to figure out cycle times of each starting location
+	but they won't cycle just based on number of steps because of the weird L-R randomness
+
+	so we can only rely on the "full cycle" of all steps before it loops
+
+	- there are six starting locations
+
+	NOTE: BIG assumptions based on KIND inputs
+	- assume that the Z-end locations will sync EXACTLY at the end of a cycle of steps
+	- after further analyzing logs of the end of each cycle, the entry point VERY kindly deposits us
+	  	at the very start of a cycle that will eventually end in a Z-end location
+		AAA -> MLM -> ... -> XKZ -> MLM -> ... -> XKZ -> MLM -> ... -> XKZ -> MLM
+		and this holds true for all six locations in my input
+		Therefore the cycles are not offset by a particular number of steps at the start to get to the cycle
+		such as START --> LOC1 --> LOC2 --> Start -> A
+											  ^		 |
+											  |		 v
+											  D <--  C
+		this makes the maths fairly straight forward with just having to find the LCM (least common multiple)
+		of all the cycle periods because that is when they will all sync up and land on a Z
+	*/
+
+	numOfSteps := 0
+
+	locationCyclePeriods := []int{}
+	for cycle := 0; len(locations) > 0; cycle++ {
+		for _, dir := range steps {
+			for i, loc := range locations {
+				if dir == "L" {
+					locations[i] = graph[loc][0]
+				} else {
+					locations[i] = graph[loc][1]
+				}
+			}
+			numOfSteps++
+		}
+
+		// if any location is at a z-end at the end of a cycle, record the cycle time
+		// to do the final maths at the end
+		newLocations := []string{}
+		for _, loc := range locations {
+			if loc[2:3] == "Z" {
+				locationCyclePeriods = append(locationCyclePeriods, numOfSteps)
+			} else {
+				newLocations = append(newLocations, loc)
+			}
+		}
+		locations = newLocations
+	}
+
+	// combine all into an LCM (helper function added to mathy package)
+	lcm := locationCyclePeriods[0]
+	for i := 1; i < len(locationCyclePeriods); i++ {
+		lcm = mathy.LeastCommonMultiple(lcm, locationCyclePeriods[i])
+	}
+
+	return lcm
+}
+
+func parseInput(input string) (steps []string, graph map[string][]string) {
+	graph = map[string][]string{}
+
+	parts := strings.Split(input, "\n\n")
+	steps = strings.Split(parts[0], "")
+
+	for _, line := range strings.Split(parts[1], "\n") {
+		graph[line[0:3]] = []string{
+			line[7:10],
+			line[12:15],
+		}
+	}
+
+	return steps, graph
+}