Day 8 - Part 1 & 2

cmd/main.go

@@ -9,6 +9,7 @@ import (
 	day04 "github.com/doniacld/adventofcode/puzzles/2020/04"
 	day05 "github.com/doniacld/adventofcode/puzzles/2020/05"
 	day06 "github.com/doniacld/adventofcode/puzzles/2020/06"
+	day08 "github.com/doniacld/adventofcode/puzzles/2020/08"
 	"github.com/doniacld/adventofcode/puzzles/solver"
 	"log"
 )
@@ -53,6 +54,8 @@ func main() {
 		s = day05.New("./puzzles/2020/05/input.txt")
 	case 6:
 		s = day06.New("./puzzles/2020/06/input.txt")
+	case 8:
+		s = day08.New("./puzzles/2020/08/input.txt")
 	}
 
 	out, err := s.Solve()

puzzles/2020/06/input.txt

@@ -278,9 +278,9 @@ omvnxidwzhc
 mzkilcfhudtex
 qrhcdyxvwmzi
 
-i
-i
-i
+currentIdx
+currentIdx
+currentIdx
 
 mrspfekayvo
 fmrysvpkoa
@@ -626,9 +626,9 @@ yjzciltueoshvbkfmdgnwr
 jbswehmkiyvgdnzurotlfc
 xgqrhwsozelnyvudmftbkacji
 
-i
+currentIdx
 cyeti
-i
+currentIdx
 
 gahjt
 thbljy
@@ -924,11 +924,11 @@ x
 rc
 fcr
 
-i
-i
+currentIdx
+currentIdx
 ronsi
-i
-i
+currentIdx
+currentIdx
 
 e
 g
@@ -1534,7 +1534,7 @@ eqmbovwcakngzhpr
 rhwzokanpvqbegmc
 
 w
-i
+currentIdx
 kuen
 vp
 p
@@ -1915,7 +1915,7 @@ oqzlbvtpgca
 bqoeg
 qugbreo
 
-i
+currentIdx
 fd
 mxy
 ebgy

puzzles/2020/08/README.md

@@ -0,0 +1,89 @@
+# Day 8: Handheld Halting
+
+## Part One
+
+Your flight to the major airline hub reaches cruising altitude without incident. While you consider checking the in-flight menu for one of those drinks that come with a little umbrella, you are interrupted by the kid sitting next to you.
+
+Their handheld game console won't turn on! They ask if you can take a look.
+
+You narrow the problem down to a strange infinite loop in the boot code (your puzzle input) of the device. You should be able to fix it, but first you need to be able to run the code in isolation.
+
+The boot code is represented as a text file with one instruction per line of text. Each instruction consists of an Operation (acc, jmp, or nop) and an argument (a signed number like +4 or -20).
+
+    acc increases or decreases a single global value called the accumulator by the value given in the argument. For example, acc +7 would increase the accumulator by 7. The accumulator starts at 0. After an acc instruction, the instruction immediately below it is executed next.
+    jmp jumps to a new instruction relative to itself. The next instruction to execute is found using the argument as an offset from the jmp instruction; for example, jmp +2 would skip the next instruction, jmp +1 would continue to the instruction immediately below it, and jmp -20 would cause the instruction 20 lines above to be executed next.
+    nop stands for No OPeration - it does nothing. The instruction immediately below it is executed next.
+
+For example, consider the following program:
+
+    nop +0
+    acc +1
+    jmp +4
+    acc +3
+    jmp -3
+    acc -99
+    acc +1
+    jmp -4
+    acc +6
+
+These instructions are visited in this order:
+
+    nop +0  | 1
+    acc +1  | 2, 8(!)
+    jmp +4  | 3
+    acc +3  | 6
+    jmp -3  | 7
+    acc -99 |
+    acc +1  | 4
+    jmp -4  | 5
+    acc +6  |
+
+First, the nop +0 does nothing. Then, the accumulator is increased from 0 to 1 (acc +1) and jmp +4 sets the next instruction to the other acc +1 near the bottom. After it increases the accumulator from 1 to 2, jmp -4 executes, setting the next instruction to the only acc +3. It sets the accumulator to 5, and jmp -3 causes the program to continue back at the first acc +1.
+
+This is an infinite loop: with this sequence of jumps, the program will run forever. The moment the program tries to run any instruction a second time, you know it will never terminate.
+
+Immediately before the program would run an instruction a second time, the value in the accumulator is 5.
+
+Run your copy of the boot code. Immediately before any instruction is executed a second time, what value is in the accumulator?
+
+Your puzzle answer was 2014.
+
+## Part Two
+
+After some careful analysis, you believe that exactly one instruction is corrupted.
+
+Somewhere in the program, either a jmp is supposed to be a nop, or a nop is supposed to be a jmp. (No acc instructions were harmed in the corruption of this boot code.)
+
+The program is supposed to terminate by attempting to execute an instruction immediately after the last instruction in the file. By changing exactly one jmp or nop, you can repair the boot code and make it terminate correctly.
+
+For example, consider the same program from above:
+
+    nop +0
+    acc +1
+    jmp +4
+    acc +3
+    jmp -3
+    acc -99
+    acc +1
+    jmp -4
+    acc +6
+
+If you change the first instruction from nop +0 to jmp +0, it would create a single-instruction infinite loop, never leaving that instruction. If you change almost any of the jmp instructions, the program will still eventually find another jmp instruction and loop forever.
+
+However, if you change the second-to-last instruction (from jmp -4 to nop -4), the program terminates! The instructions are visited in this order:
+
+    nop +0  | 1
+    acc +1  | 2
+    jmp +4  | 3
+    acc +3  |
+    jmp -3  |
+    acc -99 |
+    acc +1  | 4
+    nop -4  | 5
+    acc +6  | 6
+
+After the last instruction (acc +6), the program terminates by attempting to run the instruction below the last instruction in the file. With this change, after the program terminates, the accumulator contains the value 8 (acc +1, acc +1, acc +6).
+
+Fix the program so that it terminates normally by changing exactly one jmp (to nop) or nop (to jmp). What is the value of the accumulator after the program terminates?
+
+Your puzzle answer was 2251.

puzzles/2020/08/day08.go

@@ -0,0 +1,37 @@
+package day08
+
+import (
+	"fmt"
+
+	"github.com/doniacld/adventofcode/internal/filereader"
+	"github.com/doniacld/adventofcode/puzzles/2020/08/game"
+	"github.com/doniacld/adventofcode/puzzles/solver"
+)
+
+func New(fileName string) solver.Solver {
+	return day08{fileName: fileName}
+}
+
+type day08 struct {
+	fileName string
+}
+
+func (d day08) Solve() (string, error) {
+
+	ops := game.NewOperations(0)
+	err := filereader.ReadAndExtract(d.fileName, func(line string) error {
+		err := ops.ParseLine(line)
+		return err
+	})
+	if err != nil {
+		return "", err
+	}
+	// store the initial ops before any modification by the Part one
+	part2 := ops.DuplicateOps()
+
+	_, accCounter := ops.ComputeCorruptedAcc()
+	accCounterFixed := part2.ComputeFixedAcc()
+	out := fmt.Sprintf("acc value: %d, after fix acc value: %d", accCounter, accCounterFixed)
+
+	return out, nil
+}

puzzles/2020/08/game/corrupted.go

@@ -0,0 +1,38 @@
+package game
+
+// ComputeCorruptedAcc computes the accumulator in the corrupted game
+// ends when an Operation is seen twice
+func (ops Operations) ComputeCorruptedAcc() (idx int, accCounter int) {
+
+	o := ops[idx]
+	for ; !o.seen; o = ops[idx] {
+		switch o.name {
+		case nopOp:
+			idx = ops.nop(idx)
+		case accOp:
+			idx, accCounter = ops.acc(idx, accCounter)
+		case jmpOp:
+			idx = ops.jump(idx)
+		}
+	}
+	return idx, accCounter
+}
+
+// nop for no Operation does nothing, go to the next instruction
+func (ops Operations) nop(idx int) int {
+	ops[idx].seen = true
+	return (idx + 1) % len(ops)
+}
+
+// acc for accumulator, increase the current value of the acc counter
+// and go to the next instruction
+func (ops Operations) acc(idx int, acc int) (int, int) {
+	ops[idx].seen = true
+	return (idx + 1) % len(ops), ops[idx].value + acc
+}
+
+// jump add the offset to the current index and go to this instruction
+func (ops Operations) jump(idx int) int {
+	ops[idx].seen = true
+	return (idx + ops[idx].value) % len(ops)
+}

puzzles/2020/08/game/corrupted_test.go

@@ -0,0 +1,32 @@
+package game
+
+import (
+	"testing"
+
+	"github.com/doniacld/adventofcode/internal/filereader"
+	"github.com/stretchr/testify/assert"
+)
+
+func TestOperations_ComputeOperations(t *testing.T) {
+	tt := []struct {
+		description string
+		input       string
+		output      int
+	}{
+		{"nominal case", "./resources/input-test.txt", 5},
+	}
+
+	for _, tc := range tt {
+		t.Run(tc.description, func(t *testing.T) {
+			var ops Operations
+			err := filereader.ReadAndExtract(tc.input, func(line string) error {
+				err := ops.ParseLine(line)
+				return err
+			})
+			assert.NoError(t, err)
+			idx, acc := ops.ComputeCorruptedAcc()
+			assert.Equal(t, tc.output, acc)
+			assert.Equal(t, 1, idx)
+		})
+	}
+}

puzzles/2020/08/game/fixed.go

@@ -0,0 +1,104 @@
+package game
+
+// ComputeFixedAcc returns the accumulator value
+// game ends at the last instruction
+// a jmp is replaced by a nop or the other way around
+func (ops Operations) ComputeFixedAcc() int {
+	for i := 0; i < len(ops)-1; i++ {
+
+		o := ops[i]
+		tmpOps := ops.DuplicateOps()
+
+		// switch on the nop or jmp to check if the replace is successful
+		if o.value > 1 {
+			continue
+		}
+
+		switch o.name {
+		case nopOp:
+			tmpOps.replace(i, jmpOp)
+		case jmpOp:
+			tmpOps.replace(i, nopOp)
+		}
+
+		// if the game ends at the last instruction, return accCounter
+		v, accCounter := tmpOps.computeToTheLastOp()
+		if v {
+			return accCounter
+		}
+	}
+	// should not happened
+	return 0
+}
+
+func (ops Operations) DuplicateOps() Operations {
+	// store the tmp ops for modification
+	tmpOps := NewOperations(len(ops))
+	copy(tmpOps, ops)
+	return tmpOps
+}
+
+// replace the current instruction by the new wanted one
+func (ops Operations) replace(idx int, newOp string) {
+	ops[idx].name = newOp
+}
+
+// computeToTheLastOp calculates the accumulator counter depending on operations
+// returns false if the  operation is already seen or the index is not in the range anymore
+func (ops Operations) computeToTheLastOp() (bool, int) {
+	currentIdx, nextIdx, accCounter := 0, 0, 0
+
+	o := ops[currentIdx]
+	for ; nextIdx != len(ops)-1; o = ops[currentIdx] {
+		switch o.name {
+		case nopOp:
+			nextIdx = ops.nops(currentIdx)
+			break
+		case accOp:
+			nextIdx, accCounter = ops.accs(currentIdx, accCounter)
+			break
+		case jmpOp:
+			nextIdx = ops.jumps(currentIdx)
+			break
+		}
+
+		// operation already seen, this is not normal
+		if ops[nextIdx].seen {
+			return false, accCounter
+		}
+
+		// nextIdx not in the range anymore
+		if nextIdx < 0 || nextIdx > len(ops)-1 {
+			return false, accCounter
+		}
+
+		// let's continue, Operation is seen and save the index
+		ops[nextIdx].seen = true
+		currentIdx = nextIdx
+	}
+
+	// compute the last value
+	if ops[nextIdx].name == accOp {
+		_, accCounter = ops.accs(nextIdx, accCounter)
+	}
+	return true, accCounter
+}
+
+// nop for no Operation does nothing, go to the next instruction
+func (ops Operations) nops(idx int) int {
+	return idx + 1
+}
+
+// acc for accumulator, increase the current value of the acc counter
+// and go to the next instruction
+func (ops Operations) accs(idx int, acc int) (int, int) {
+	if idx == len(ops)-1 {
+		return idx, ops[idx].value + acc
+	}
+	return idx + 1, ops[idx].value + acc
+}
+
+// jump add the offset to the current index and go to this instruction
+func (ops Operations) jumps(idx int) int {
+	return idx + ops[idx].value
+}