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refactor of day07 solution
This commit is contained in:
Alex Chao
+191
-73
@@ -1,89 +1,207 @@
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/*
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Intcode struct is defined within this file
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MakePermutations is in the util package as that will likely be reused
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*/
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package main
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import (
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"adventofcode/util"
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"fmt"
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"./intcode"
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"./permutations"
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"log"
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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 to first amp = 0
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// output of each amp is input of next amp
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// final output is from amp #5 / E to thrusters
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// read the input file, modify it to a slice of numbers
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inputFile := util.ReadFile("../input.txt")
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input := []int{3, 8, 1001, 8, 10, 8, 105, 1, 0, 0, 21, 42, 55, 64, 77, 94, 175, 256, 337, 418, 99999, 3, 9, 102, 4, 9, 9, 1001, 9, 5, 9, 102, 2, 9, 9, 101, 3, 9, 9, 4, 9, 99, 3, 9, 102, 2, 9, 9, 101, 5, 9, 9, 4, 9, 99, 3, 9, 1002, 9, 4, 9, 4, 9, 99, 3, 9, 102, 4, 9, 9, 101, 5, 9, 9, 4, 9, 99, 3, 9, 102, 5, 9, 9, 1001, 9, 3, 9, 1002, 9, 5, 9, 4, 9, 99, 3, 9, 1002, 9, 2, 9, 4, 9, 3, 9, 101, 1, 9, 9, 4, 9, 3, 9, 1001, 9, 1, 9, 4, 9, 3, 9, 101, 1, 9, 9, 4, 9, 3, 9, 1002, 9, 2, 9, 4, 9, 3, 9, 101, 1, 9, 9, 4, 9, 3, 9, 101, 2, 9, 9, 4, 9, 3, 9, 1001, 9, 2, 9, 4, 9, 3, 9, 101, 1, 9, 9, 4, 9, 3, 9, 1002, 9, 2, 9, 4, 9, 99, 3, 9, 1002, 9, 2, 9, 4, 9, 3, 9, 1001, 9, 2, 9, 4, 9, 3, 9, 1001, 9, 1, 9, 4, 9, 3, 9, 1002, 9, 2, 9, 4, 9, 3, 9, 1002, 9, 2, 9, 4, 9, 3, 9, 1002, 9, 2, 9, 4, 9, 3, 9, 1002, 9, 2, 9, 4, 9, 3, 9, 101, 2, 9, 9, 4, 9, 3, 9, 1001, 9, 1, 9, 4, 9, 3, 9, 1001, 9, 1, 9, 4, 9, 99, 3, 9, 1002, 9, 2, 9, 4, 9, 3, 9, 102, 2, 9, 9, 4, 9, 3, 9, 101, 2, 9, 9, 4, 9, 3, 9, 101, 1, 9, 9, 4, 9, 3, 9, 1002, 9, 2, 9, 4, 9, 3, 9, 102, 2, 9, 9, 4, 9, 3, 9, 1001, 9, 2, 9, 4, 9, 3, 9, 101, 1, 9, 9, 4, 9, 3, 9, 101, 1, 9, 9, 4, 9, 3, 9, 1002, 9, 2, 9, 4, 9, 99, 3, 9, 101, 2, 9, 9, 4, 9, 3, 9, 102, 2, 9, 9, 4, 9, 3, 9, 1001, 9, 2, 9, 4, 9, 3, 9, 102, 2, 9, 9, 4, 9, 3, 9, 1001, 9, 2, 9, 4, 9, 3, 9, 101, 2, 9, 9, 4, 9, 3, 9, 102, 2, 9, 9, 4, 9, 3, 9, 1002, 9, 2, 9, 4, 9, 3, 9, 101, 1, 9, 9, 4, 9, 3, 9, 1002, 9, 2, 9, 4, 9, 99, 3, 9, 1001, 9, 2, 9, 4, 9, 3, 9, 1002, 9, 2, 9, 4, 9, 3, 9, 102, 2, 9, 9, 4, 9, 3, 9, 102, 2, 9, 9, 4, 9, 3, 9, 1002, 9, 2, 9, 4, 9, 3, 9, 101, 1, 9, 9, 4, 9, 3, 9, 101, 1, 9, 9, 4, 9, 3, 9, 1002, 9, 2, 9, 4, 9, 3, 9, 1002, 9, 2, 9, 4, 9, 3, 9, 1001, 9, 1, 9, 4, 9, 99}
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splitStrings := strings.Split(inputFile, ",")
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// should give 18216
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// input := []int{3, 52, 1001, 52, -5, 52, 3, 53, 1, 52, 56, 54, 1007, 54, 5, 55, 1005, 55, 26, 1001, 54, -5, 54, 1105, 1, 12, 1, 53, 54, 53, 1008, 54, 0, 55, 1001, 55, 1, 55, 2, 53, 55, 53, 4, 53, 1001, 56, -1, 56, 1005, 56, 6, 99, 0, 0, 0, 0, 10}
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inputNumbers := make([]int, len(splitStrings))
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for i, v := range splitStrings {
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fmt.Println(v)
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inputNumbers[i], _ = strconv.Atoi(v)
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}
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fmt.Println(inputNumbers)
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// Make perms via a util function
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perms := util.MakePermutations([]int{5, 6, 7, 8, 9})
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// create all permutations of 5, 6, 7, 8, 9
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perms := permutations.CreatePermutations(5, 9)
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// fmt.Println(perms)
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// iterate over all perms and run through a single pass of the Amps
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// if the final output (from Amp E) is higher, update the highestOutput variable
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highestOutput := 0
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for _, perm := range *perms {
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// initialize 5 computers
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ampA := MakeComputer(inputNumbers, perm[0])
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ampB := MakeComputer(inputNumbers, perm[1])
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ampC := MakeComputer(inputNumbers, perm[2])
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ampD := MakeComputer(inputNumbers, perm[3])
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ampE := MakeComputer(inputNumbers, perm[4])
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// will be the final returned value
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var highestReturn int
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// loop over all of the permutations
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for _, onePerm := range perms {
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// needs to have the right length to copy into
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input1 := make([]int, len(input))
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input2 := make([]int, len(input))
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input3 := make([]int, len(input))
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input4 := make([]int, len(input))
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input5 := make([]int, len(input))
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// make copies of the input, one for each thruster
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copy(input1, input)
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copy(input2, input)
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copy(input3, input)
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copy(input4, input)
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copy(input5, input)
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// fmt.Println(input1)
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// fmt.Println(onePerm)
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// This will be the final thruster signal, and we want to return the maximum one
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var last5Return int // zero nil value!
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// Initial load for the permutation of 5-9
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index1, _, lastOutput1 := intcode.RunDiagnostics(input1, onePerm[0], 0)
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index2, _, lastOutput2 := intcode.RunDiagnostics(input2, onePerm[1], 0)
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index3, _, lastOutput3 := intcode.RunDiagnostics(input3, onePerm[2], 0)
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index4, _, lastOutput4 := intcode.RunDiagnostics(input4, onePerm[3], 0)
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index5, exitCode5, lastOutput5 := intcode.RunDiagnostics(input5, onePerm[4], 0)
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// fmt.Println("INITIAL LOADS DONE"
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// Here's the fix to my problem:
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// I needed to account for the initial load but then also the second input coming from the preceeding Amp thingy
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// But because this is the exact same code as the loop below, I've commented it out
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// index1, _, lastOutput1 = intcode.RunDiagnostics(input1, lastOutput5, index1)
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// index2, _, lastOutput2 = intcode.RunDiagnostics(input2, lastOutput1, index2)
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// index3, _, lastOutput3 = intcode.RunDiagnostics(input3, lastOutput2, index3)
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// index4, _, lastOutput4 = intcode.RunDiagnostics(input4, lastOutput3, index4)
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// index5, _, lastOutput5 = intcode.RunDiagnostics(input5, lastOutput4, index5)
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// fmt.Println("Second input load done")
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for exitCode5 != 99 {
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index1, _, lastOutput1 = intcode.RunDiagnostics(input1, lastOutput5, index1)
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index2, _, lastOutput2 = intcode.RunDiagnostics(input2, lastOutput1, index2)
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index3, _, lastOutput3 = intcode.RunDiagnostics(input3, lastOutput2, index3)
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index4, _, lastOutput4 = intcode.RunDiagnostics(input4, lastOutput3, index4)
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index5, exitCode5, lastOutput5 = intcode.RunDiagnostics(input5, lastOutput4, index5)
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if exitCode5 == 4 {
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last5Return = lastOutput5
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} else if exitCode5 == 99 {
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// fmt.Println("99 exit code!", lastOutput5) // loop will end now
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}
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// fmt.Println("one full pass") // using this to check if multiple inputs are being asked for
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// Iterate while the Amps are still running (i.e. not terminated)
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for ampA.IsRunning && ampB.IsRunning && ampC.IsRunning && ampD.IsRunning && ampE.IsRunning {
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// first input to Amp A is a zero, this is the zero-value of an int!
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ampA.Step(ampE.LastOutput)
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ampB.Step(ampA.LastOutput)
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ampC.Step(ampB.LastOutput)
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ampD.Step(ampC.LastOutput)
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ampE.Step(ampD.LastOutput)
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}
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// fmt.Println(lastOutput5) // should be changing for each perm
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// check if the max thrust input is higher than highest return
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if last5Return > highestReturn {
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highestReturn = last5Return
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if ampE.LastOutput > highestOutput {
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highestOutput = ampE.LastOutput
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}
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}
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fmt.Println("highestReturn is", highestReturn)
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// print highest output found
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fmt.Printf("Highest output is %v\n", highestOutput)
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}
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/*
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Intcode is an OOP approach *************************************************
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MakeComputer is equivalent to the constructor
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Step takes in an input int and updates properties in the computer:
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- InstructionIndex: where to read the next instruction from
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- LastOutput, what the last opcode 4 outputted
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- PuzzleIndex based if the last instruction modified the puzzle at all
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****************************************************************************/
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type Intcode struct {
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PhaseSetting int // initial input: ID or number used to "prime"/setup the comp
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PuzzleInput []int // file/puzzle input parsed into slice of ints
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InstructionIndex int // stores the index where the next instruction is
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LastOutput int // last output from an opcode 4
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IsRunning bool // will be true until a 99 opcode is hit
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}
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// MakeComputer initializes a new comp
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func MakeComputer(PuzzleInput []int, PhaseSetting int) Intcode {
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puzzleInputCopy := make([]int, len(PuzzleInput))
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copy(puzzleInputCopy, PuzzleInput)
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comp := Intcode{
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PhaseSetting,
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puzzleInputCopy,
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0,
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0,
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true,
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}
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// Prime the computer by running its initial phase setting through it
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// This will update the comp's InstructionIndex so it's pointing to the next command
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// will also update the PuzzleInput itself via opcode 3's insert
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// AND will run the computer until it asks for the next input, _comp is now primed_
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comp.Step(PhaseSetting)
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return comp
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}
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// Step will read the next 4 values in the input `sli` and make updates
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// according to the opcodes
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func (comp *Intcode) Step(input int) int {
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// read the instruction, opcode and the indexes where the params point to
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opcode, paramIndexes := comp.GetOpCodeAndIndexes()
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switch opcode {
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case 99: // 99: Terminates program
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fmt.Println("Terminating...")
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comp.IsRunning = false
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return input
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case 1: // 1: Add next two paramIndexes, store in third
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comp.PuzzleInput[paramIndexes[2]] = comp.PuzzleInput[paramIndexes[0]] + comp.PuzzleInput[paramIndexes[1]]
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comp.InstructionIndex += 4
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return comp.Step(input)
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case 2: // 2: Multiply next two and store in third
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comp.PuzzleInput[paramIndexes[2]] = comp.PuzzleInput[paramIndexes[0]] * comp.PuzzleInput[paramIndexes[1]]
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comp.InstructionIndex += 4
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return comp.Step(input)
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case 3: // 3: Takes one input and saves it to position of one parameter
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// check if input has already been used (i.e. input == -1)
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// if it's been used, return the LastOutput
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// NOTE: making a big assumption that -1 will never be an input...
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if input == -1 {
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return comp.LastOutput
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}
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// otherwise use the input, then recurse with a -1 to signal the initial input has been used
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comp.PuzzleInput[paramIndexes[0]] = input
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comp.InstructionIndex += 2
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return comp.Step(-1)
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case 4: // 4: outputs its input value
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// set LastOutput of the computer & log it
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comp.LastOutput = comp.PuzzleInput[paramIndexes[0]]
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// fmt.Printf("Opcode 4 output: %v\n", comp.LastOutput)
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comp.InstructionIndex += 2
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// continue running until terminates or asks for another input
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return comp.Step(input)
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// 5: jump-if-true: if first param != 0, move pointer to second param, else nothing
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case 5:
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if comp.PuzzleInput[paramIndexes[0]] != 0 {
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comp.InstructionIndex = comp.PuzzleInput[paramIndexes[1]]
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} else {
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comp.InstructionIndex += 3
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}
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return comp.Step(input)
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// 6: jump-if-false, if first param == 0 then set instruction pointer to 2nd param, else nothing
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case 6:
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if comp.PuzzleInput[paramIndexes[0]] == 0 {
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comp.InstructionIndex = comp.PuzzleInput[paramIndexes[1]]
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} else {
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comp.InstructionIndex += 3
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}
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return comp.Step(input)
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// 7: less-than, if param1 < param2 then store 1 in postion of 3rd param, else store 0
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case 7:
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if comp.PuzzleInput[paramIndexes[0]] < comp.PuzzleInput[paramIndexes[1]] {
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comp.PuzzleInput[paramIndexes[2]] = 1
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} else {
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comp.PuzzleInput[paramIndexes[2]] = 0
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}
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comp.InstructionIndex += 4
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return comp.Step(input)
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// 8: equals, if param1 == param2 then set position of 3rd param to 1, else store 0
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case 8:
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if comp.PuzzleInput[paramIndexes[0]] == comp.PuzzleInput[paramIndexes[1]] {
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comp.PuzzleInput[paramIndexes[2]] = 1
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} else {
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comp.PuzzleInput[paramIndexes[2]] = 0
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}
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comp.InstructionIndex += 4
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return comp.Step(input)
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default:
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log.Fatal("Error: unknown opcode: ", opcode)
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}
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// this should never be called b/c switch statement will always return
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return -1
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}
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/*
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GetOpCodeAndIndexes will parse the instruction at comp.PuzzleInput[comp.InstructionIndex]
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- opcode will be the left two digits, mod by 100 will get that
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- rest of instructions will be grabbed via mod 10
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- these also have to be parsed for the
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*/
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func (comp *Intcode) GetOpCodeAndIndexes() (int, [3]int) {
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instruction := comp.PuzzleInput[comp.InstructionIndex]
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// opcode is the lowest two digits, so mod by 100
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opcode := instruction % 100
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instruction /= 100
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// assign the indexes that need to be read by reading the parameter modes
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var paramIndexes [3]int
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for i := 1; i <= 3 && comp.InstructionIndex+i < len(comp.PuzzleInput); i++ {
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// grab the mode with a mod, last digit
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mode := instruction % 10
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instruction /= 10
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switch mode {
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case 1: // immediate mode, the index itself
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paramIndexes[i-1] = comp.InstructionIndex + i
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case 0: // position mode, index will be the value at the index
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paramIndexes[i-1] = comp.PuzzleInput[comp.InstructionIndex+i]
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}
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}
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return opcode, paramIndexes
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}
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