2018 Day 16 Complete
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270
2018/day16/day16.go
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270
2018/day16/day16.go
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@ -0,0 +1,270 @@
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package main
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import (
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"bufio"
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"errors"
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"fmt"
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"log"
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"os"
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"strconv"
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"strings"
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)
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func main() {
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input := stdinToStringSlice()
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pt2Idx := part1(input)
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part2(input[pt2Idx:])
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}
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var probs map[string]int
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func part1(input []string) int {
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probs = make(map[string]int)
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var messySamples int
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var ret int
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for i := 0; i < len(input); i += 4 {
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if !strings.HasPrefix(input[i], "Before: ") {
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// We're done with the part 1 input
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ret = i
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break
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}
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aft := NewState()
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aft.loadState(input[i+2])
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if cmd, err := aft.parseCommand(input[i+1]); err == nil {
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origCd := cmd[0]
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var numMatches int
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for j := 0; j < 16; j++ {
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bef := NewState()
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bef.loadState(input[i])
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cmd[0] = j
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bef.run(cmd)
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if bef.registerString() == aft.registerString() {
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numMatches++
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key := fmt.Sprintf("%d:%d", origCd, j)
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probs[key]++
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}
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}
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if numMatches > 2 {
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messySamples++
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}
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}
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}
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fmt.Println("= Part 1 =")
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fmt.Println(messySamples)
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return ret + 2
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}
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var opMap [16]int
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func part2(input []string) {
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for i := 0; i < 16; i++ {
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opMap[i] = -1
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}
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fmt.Println("= Part 2 =")
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m1, m2 := -1, -1
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for {
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m1, m2, probs = findOnlyChance(probs)
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if m1 == -1 {
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break
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}
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opMap[m1] = m2
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}
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vpc := NewState()
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for i := range input {
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if cmd, err := vpc.parseCommand(input[i]); err == nil {
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cmd[0] = opMap[cmd[0]]
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vpc.run(cmd)
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}
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}
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fmt.Println(vpc.reg[0])
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}
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func findOnlyChance(of map[string]int) (int, int, map[string]int) {
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m1, m2 := -1, -1
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for i := 0; i < 16; i++ {
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var num int
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for k := range of {
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codes := strings.Split(k, ":")
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if Atoi(codes[0]) == i {
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num++
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if num > 1 {
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m2 = -1
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break
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}
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m2 = Atoi(codes[1])
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}
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}
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if num == 1 {
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m1 = i
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break
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}
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if num > 1 {
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continue
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}
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}
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if m1 == -1 {
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return -1, -1, of
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}
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// If we're here, we found something
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// Remove other possibilities for m1: and :m2 and return
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for k := range of {
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a1 := strconv.Itoa(m1)
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a2 := strconv.Itoa(m2)
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if strings.HasPrefix(k, a1+":") ||
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strings.HasSuffix(k, ":"+a2) {
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delete(of, k)
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}
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}
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return m1, m2, of
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}
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type State struct {
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reg [4]int
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OpCodes map[int]func(a, b, c int)
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}
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func NewState() *State {
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s := &State{}
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s.OpCodes = make(map[int]func(a, b, c int))
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s.OpCodes[0] = s.addr
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s.OpCodes[1] = s.addi
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s.OpCodes[2] = s.mulr
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s.OpCodes[3] = s.muli
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s.OpCodes[4] = s.banr
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s.OpCodes[5] = s.bani
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s.OpCodes[6] = s.borr
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s.OpCodes[7] = s.bori
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s.OpCodes[8] = s.setr
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s.OpCodes[9] = s.seti
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s.OpCodes[10] = s.gtir
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s.OpCodes[11] = s.gtri
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s.OpCodes[12] = s.gtrr
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s.OpCodes[13] = s.eqir
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s.OpCodes[14] = s.eqri
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s.OpCodes[15] = s.eqrr
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return s
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}
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func (s *State) registerString() string {
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return fmt.Sprint(s.reg)
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}
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func (s *State) loadState(inp string) {
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s.reg = [4]int{0, 0, 0, 0}
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inp = strings.Trim(inp, "ABeftor: []")
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pts := strings.Split(inp, ",")
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for i := range pts {
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s.reg[i] = Atoi(strings.TrimSpace(pts[i]))
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}
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}
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func (s *State) parseCommand(inp string) ([]int, error) {
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var ret []int
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pts := strings.Split(inp, " ")
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for i := range pts {
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ret = append(ret, Atoi(pts[i]))
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}
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if len(ret) != 4 {
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return ret, errors.New("Invalid Command: " + inp)
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}
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return ret, nil
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}
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func (s *State) run(cmd []int) error {
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if fn, ok := s.OpCodes[cmd[0]]; !ok {
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return errors.New(fmt.Sprintf("Invalid OpCode: %d", cmd[0]))
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} else {
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fn(cmd[1], cmd[2], cmd[3])
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}
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return nil
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}
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func (s *State) addr(r1, r2, r3 int) {
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s.reg[r3] = s.reg[r1] + s.reg[r2]
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}
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func (s *State) addi(r1, v2, r3 int) {
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s.reg[r3] = s.reg[r1] + v2
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}
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func (s *State) mulr(r1, r2, r3 int) {
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s.reg[r3] = s.reg[r1] * s.reg[r2]
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}
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func (s *State) muli(r1, v2, r3 int) {
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s.reg[r3] = s.reg[r1] * v2
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}
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func (s *State) banr(r1, r2, r3 int) {
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s.reg[r3] = s.reg[r1] & s.reg[r2]
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}
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func (s *State) bani(r1, v2, r3 int) {
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s.reg[r3] = s.reg[r1] & v2
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}
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func (s *State) borr(r1, r2, r3 int) {
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s.reg[r3] = s.reg[r1] | s.reg[r2]
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}
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func (s *State) bori(r1, v2, r3 int) {
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s.reg[r3] = s.reg[r1] | v2
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}
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func (s *State) setr(r1, r2, r3 int) {
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s.reg[r3] = s.reg[r1]
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}
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func (s *State) seti(v1, v2, r3 int) {
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s.reg[r3] = v1
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}
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func (s *State) gtir(v1, r2, r3 int) {
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s.reg[r3] = boolToInt(v1 > s.reg[r2])
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}
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func (s *State) gtri(r1, v2, r3 int) {
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s.reg[r3] = boolToInt(s.reg[r1] > v2)
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}
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func (s *State) gtrr(r1, r2, r3 int) {
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s.reg[r3] = boolToInt(s.reg[r1] > s.reg[r2])
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}
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func (s *State) eqir(v1, r2, r3 int) {
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s.reg[r3] = boolToInt(v1 == s.reg[r2])
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}
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func (s *State) eqri(r1, v2, r3 int) {
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s.reg[r3] = boolToInt(s.reg[r1] == v2)
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}
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func (s *State) eqrr(r1, r2, r3 int) {
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s.reg[r3] = boolToInt(s.reg[r1] == s.reg[r2])
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}
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func boolToInt(v bool) int {
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if v {
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return 1
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}
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return 0
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}
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func stdinToStringSlice() []string {
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var input []string
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scanner := bufio.NewScanner(os.Stdin)
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for scanner.Scan() {
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input = append(input, scanner.Text())
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}
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return input
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}
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func Atoi(i string) int {
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var ret int
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var err error
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if ret, err = strconv.Atoi(i); err != nil {
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log.Fatal("Invalid Atoi: " + i)
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}
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return ret
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}
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4064
2018/day16/input
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4064
2018/day16/input
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File diff suppressed because it is too large
Load Diff
187
2018/day16/problem
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187
2018/day16/problem
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Advent of Code
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--- Day 16: Chronal Classification ---
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As you see the Elves defend their hot chocolate
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successfully, you go back to falling through time. This is
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going to become a problem.
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If you're ever going to return to your own time, you need
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to understand how this device on your wrist works. You
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have a little while before you reach your next
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destination, and with a bit of trial and error, you manage
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to pull up a programming manual on the device's tiny
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screen.
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According to the manual, the device has four registers
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(numbered 0 through 3) that can be manipulated by
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instructions containing one of 16 opcodes. The registers
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start with the value 0.
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Every instruction consists of four values: an opcode, two
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inputs (named A and B), and an output (named C), in that
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order. The opcode specifies the behavior of the
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instruction and how the inputs are interpreted. The
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output, C, is always treated as a register.
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In the opcode descriptions below, if something says "value
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A", it means to take the number given as A literally.
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(This is also called an "immediate" value.) If something
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says "register A", it means to use the number given as A
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to read from (or write to) the register with that number.
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So, if the opcode addi adds register A and value B,
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storing the result in register C, and the instruction addi
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0 7 3 is encountered, it would add 7 to the value
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contained by register 0 and store the sum in register 3,
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never modifying registers 0, 1, or 2 in the process.
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Many opcodes are similar except for how they interpret
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their arguments. The opcodes fall into seven general
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categories:
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Addition:
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* addr (add register) stores into register C the result
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of adding register A and register B.
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* addi (add immediate) stores into register C the result
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of adding register A and value B.
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Multiplication:
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* mulr (multiply register) stores into register C the
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result of multiplying register A and register B.
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* muli (multiply immediate) stores into register C the
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result of multiplying register A and value B.
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Bitwise AND:
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* banr (bitwise AND register) stores into register C the
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result of the bitwise AND of register A and register
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B.
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* bani (bitwise AND immediate) stores into register C
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the result of the bitwise AND of register A and value
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B.
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Bitwise OR:
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* borr (bitwise OR register) stores into register C the
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result of the bitwise OR of register A and register B.
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* bori (bitwise OR immediate) stores into register C the
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result of the bitwise OR of register A and value B.
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Assignment:
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* setr (set register) copies the contents of register A
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into register C. (Input B is ignored.)
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* seti (set immediate) stores value A into register C.
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(Input B is ignored.)
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Greater-than testing:
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* gtir (greater-than immediate/register) sets register C
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to 1 if value A is greater than register B. Otherwise,
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register C is set to 0.
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* gtri (greater-than register/immediate) sets register C
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to 1 if register A is greater than value B. Otherwise,
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register C is set to 0.
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* gtrr (greater-than register/register) sets register C
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to 1 if register A is greater than register B.
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Otherwise, register C is set to 0.
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Equality testing:
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* eqir (equal immediate/register) sets register C to 1
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if value A is equal to register B. Otherwise, register
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C is set to 0.
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* eqri (equal register/immediate) sets register C to 1
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if register A is equal to value B. Otherwise, register
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C is set to 0.
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* eqrr (equal register/register) sets register C to 1 if
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register A is equal to register B. Otherwise, register
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C is set to 0.
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Unfortunately, while the manual gives the name of each
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opcode, it doesn't seem to indicate the number. However,
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you can monitor the CPU to see the contents of the
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registers before and after instructions are executed to
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try to work them out. Each opcode has a number from 0
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through 15, but the manual doesn't say which is which. For
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example, suppose you capture the following sample:
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Before: [3, 2, 1, 1]
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9 2 1 2
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After: [3, 2, 2, 1]
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This sample shows the effect of the instruction 9 2 1 2 on
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the registers. Before the instruction is executed,
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register 0 has value 3, register 1 has value 2, and
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registers 2 and 3 have value 1. After the instruction is
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executed, register 2's value becomes 2.
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The instruction itself, 9 2 1 2, means that opcode 9 was
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executed with A=2, B=1, and C=2. Opcode 9 could be any of
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the 16 opcodes listed above, but only three of them behave
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in a way that would cause the result shown in the sample:
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* Opcode 9 could be mulr: register 2 (which has a value
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of 1) times register 1 (which has a value of 2)
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produces 2, which matches the value stored in the
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output register, register 2.
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* Opcode 9 could be addi: register 2 (which has a value
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of 1) plus value 1 produces 2, which matches the value
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stored in the output register, register 2.
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* Opcode 9 could be seti: value 2 matches the value
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stored in the output register, register 2; the number
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given for B is irrelevant.
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None of the other opcodes produce the result captured in
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the sample. Because of this, the sample above behaves like
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three opcodes.
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You collect many of these samples (the first section of
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your puzzle input). The manual also includes a small test
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program (the second section of your puzzle input) - you
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can ignore it for now.
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Ignoring the opcode numbers, how many samples in your
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puzzle input behave like three or more opcodes?
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Your puzzle answer was 547.
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--- Part Two ---
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Using the samples you collected, work out the number of
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each opcode and execute the test program (the second
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section of your puzzle input).
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What value is contained in register 0 after executing the
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test program?
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Your puzzle answer was 582.
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Both parts of this puzzle are complete! They provide two
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gold stars: **
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References
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Visible links
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. https://adventofcode.com/
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. https://adventofcode.com/2018/about
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. https://adventofcode.com/2018/events
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. https://adventofcode.com/2018/settings
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. https://adventofcode.com/2018/auth/logout
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. Advent of Code Supporter
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https://adventofcode.com/2018/support
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. https://adventofcode.com/2018
|
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. https://adventofcode.com/2018
|
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. https://adventofcode.com/2018/support
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. https://adventofcode.com/2018/sponsors
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. https://adventofcode.com/2018/leaderboard
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. https://adventofcode.com/2018/stats
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. https://adventofcode.com/2018/sponsors
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. https://en.wikipedia.org/wiki/Hardware_register
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. https://en.wikipedia.org/wiki/Instruction_set_architecture#Instructions
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. https://en.wikipedia.org/wiki/Bitwise_AND
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. https://en.wikipedia.org/wiki/Bitwise_OR
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. https://adventofcode.com/2018
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. https://adventofcode.com/2018/day/16/input
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3
2018/day16/testinput
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3
2018/day16/testinput
Normal file
@ -0,0 +1,3 @@
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Before: [3, 2, 1, 1]
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9 2 1 2
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After: [3, 2, 2, 1]
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