Day 23: LAN Party

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FAQ

  • eco_game@discuss.tchncs.de
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    6 hours ago

    Kotlin

    Part1 was pretty simple, just check all neighbors of a node for overlap, then filter out triples which don’t have nodes beginning with ‘t’.

    For part2, I seem to have picked a completely different strategy to everyone else. I was a bit lost, then just boldly assumed, that if I take overlap of all triples with 1 equal node, I might be able to find the answer that way. To my surprise, this worked for my input. I’d be very curious to know if I just got lucky or if the puzzle is designed to work with this approach.

    The full code is also on GitHub.

    Solution
    class Day23 : Puzzle {
    
        private val connections = ArrayList<Pair<String, String>>()
    
        private val tripleCache = HashSet<Triple<String, String, String>>()
    
        override fun readFile() {
            val input = readInputFromFile("src/main/resources/day23.txt")
            for (line in input.lines()) {
                val parts = line.split("-")
                connections.add(Pair(parts[0], parts[1]))
            }
        }
    
        override fun solvePartOne(): String {
            val triples = getConnectionTriples(connections)
            tripleCache.addAll(triples) // for part 2
            val res = triples.count { it.first.startsWith("t") || it.second.startsWith("t") || it.third.startsWith("t") }
            return res.toString()
        }
    
        private fun getConnectionTriples(connectionList: List<Pair<String, String>>): List<Triple<String, String, String>> {
            val triples = ArrayList<Triple<String, String, String>>()
            for (connection in connectionList) {
                val connectionListTemp = getAllConnections(connection.first, connectionList)
                for (i in connectionListTemp.indices) {
                    for (j in i + 1 until connectionListTemp.size) {
                        val con1 = connectionListTemp[i]
                        val con2 = connectionListTemp[j]
                        if (Pair(con1, con2) in connectionList || Pair(con2, con1) in connectionList) {
                            val tripleList = mutableListOf(connection.first, con1, con2)
                            tripleList.sort()
                            triples.add(Triple(tripleList[0], tripleList[1], tripleList[2]))
                        }
                    }
                }
            }
            return triples.distinct()
        }
    
        private fun getAllConnections(connection: String, connectionList: List<Pair<String, String>>): List<String> {
            val res = HashSet<String>()
            for (entry in connectionList) {
                when (connection) {
                    entry.first -> res.add(entry.second)
                    entry.second -> res.add(entry.first)
                }
            }
            return res.toList()
        }
    
        override fun solvePartTwo(): String {
            val pools = getPools(connections)
            println(pools)
            val res = pools.maxByOrNull { it.size }!!
            return res.joinToString(",")
        }
    
        // will get all pools with a minimum size of 4
        // this method makes some naive assumptions, but works for the example and my puzzle input
        private fun getPools(connectionList: List<Pair<String, String>>): List<List<String>> {
            val pools = ArrayList<List<String>>()
            val triples = tripleCache
            val nodes = connectionList.map { listOf(it.first, it.second) }.flatten().toHashSet()
    
            for (node in nodes) {
                val contenders = triples.filter { it.first == node || it.second == node || it.third == node }
                if (contenders.size < 2) continue // expect the minimum result to be 4, for efficiency
    
                // if *all* nodes within *all* triples are interconnected, add to pool
                // this may not work for all inputs!
                val contenderList = contenders.map { listOf(it.first, it.second, it.third) }.flatten().distinct()
                if (checkAllConnections(contenderList, connectionList)) {
                    pools.add(contenderList.sorted())
                }
            }
    
            return pools.distinct()
        }
    
        private fun checkAllConnections(pool: List<String>, connectionList: List<Pair<String, String>>): Boolean {
            for (i in pool.indices) {
                for (j in i + 1 until pool.size) {
                    val con1 = pool[i]
                    val con2 = pool[j]
                    if (Pair(con1, con2) !in connectionList && Pair(con2, con1) !in connectionList) {
                        return false
                    }
                }
            }
            return true
        }
    }
    
    • lwhjp@lemmy.sdf.org
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      6 hours ago

      That’s a fun approach. The largest totally connected group will of course contain overlapping triples, so I think you’re effectively doing the same thing as checking a node at a time, just more efficiently.