Published in Programming - 8 mins to read

The problems are slowly beginning to get harder, although I did find these three still very achievable. With a little bit of practice I am finding my approach to solving them is improving very fast, and I am better at mentally modelling how the code should look, as well as improving my understanding of Go - I feel like I am finally making progress grokking addresses and pointers thanks to these challenges.

## Day 4

The challenge for day four was to take a set of called numbers and a set of bingo cards, and calculate which bingo card would win first, and then which would win last. The “score” for that card was produced by multiplying the last number that had been called before the card won by the sum of all the uncalled numbers remaining on the card. Overall this one was pretty easy, once I got past a small hiccough with my logic in the `iswinner()` method.

``````Example input:
7,4,9,5,11,17,23,2,0,14,21,24,10,16,13,6,15,25,12,22,18,20,8,19,3,26,1

22 13 17 11  0
8  2 23  4 24
21  9 14 16  7
6 10  3 18  5
1 12 20 15 19

3 15  0  2 22
9 18 13 17  5
19  8  7 25 23
20 11 10 24  4
14 21 16 12  6

14 21 17 24  4
10 16 15  9 19
18  8 23 26 20
22 11 13  6  5
2  0 12  3  7
``````
``````type bingo struct {
values int
called bool
turnsToWin int
lastNumberCalled int
unmarkedSum int
}

func Four(input []string) (int, int) {
bingos := []bingo{}

m := input

// ignore the first two lines as these are
// the numbers called and whitespace and
// then increment by six to account for more
// blank lines
for i := 2; i < len(input); i += 6 {
bingoStringo := strings.Join(input[i:i+5], "")
// probably my favourite line of code I've ever written
bingos = append(bingos, stringoToBingo(bingoStringo, m))
}

fastestBingo := bingos
slowestBingo := bingos

for _, b := range bingos {
if b.turnsToWin < fastestBingo.turnsToWin {
fastestBingo = b
}
if b.turnsToWin > slowestBingo.turnsToWin {
slowestBingo = b
}
}

return fastestBingo.lastNumberCalled * fastestBingo.unmarkedSum,
slowestBingo.lastNumberCalled * slowestBingo.unmarkedSum
}

func (b bingo) isWinner() bool {
for i := 0; i < 5; i++ {
if (
// horizontal
b.called[5*i] &&
b.called[5*i+1] &&
b.called[5*i+2] &&
b.called[5*i+3] &&
b.called[5*i+4]
) || (
// vertical
b.called[i] &&
b.called[i+5] &&
b.called[i+10] &&
b.called[i+15] &&
b.called[i+20] ) {
return true
}
}

return false
}

func stringoToBingo(s, m string) bingo {
b := bingo{}
vals := strings.Fields(s)
for i := 0; i < 25; i++ {
b.values[i], _ = strconv.Atoi(vals[i])
}
b.numberOfTurnsToWin(m)
return b
}

func (b *bingo) numberOfTurnsToWin(nums string) {
// parse nums
// for each one, mark as seen && check winner
// if winner, mark last number
// calculate everything else

// kinda gross but kinda funny
// I later wrote a helper method
// that you'll see later for this
fakeJson := "[" + nums + "]"
var moves []int
_ = json.Unmarshal([]byte(fakeJson), &moves)

// for every move
for i := 0; i < len(moves); i++ {
// for every value
for j := 0; j < len(b.values); j++ {
if moves[i] == b.values[j] {
b.called[j] = true
if b.isWinner() {
b.lastNumberCalled = moves[i]
b.turnsToWin = i
b.sum()
return
}
}
}
}
}

func (b *bingo) sum() {
sum := 0

for i := 0; i < 25; i++ {
if !b.called[i] {
sum += b.values[i]
}
}

b.unmarkedSum = sum
}
``````

## Day 5

Today’s challenge was to take a series of coordinates in the form `x1,y1 -> x2,y2` and find every point on the grid where two or more of the lines intersected. For part one diagonal lines were ignored, for part two they weren’t. Again after not quite getting my diagonal logic correct out of the gate, this one was still fairly easy.

``````Example input:
0,9 -> 5,9
8,0 -> 0,8
9,4 -> 3,4
2,2 -> 2,1
7,0 -> 7,4
6,4 -> 2,0
0,9 -> 2,9
3,4 -> 1,4
0,0 -> 8,8
5,5 -> 8,2
``````

I started a `utils` package for this one, which for now only has one function in:

``````func ExtractInts(s string) []int {
var ints []int

r, _ := regexp.Compile(`\d+`)
matches := r.FindAllString(s, -1)

for _, match := range matches {
i, _ := strconv.Atoi(match)
ints = append(ints, i)
}

return ints
}

``````
``````func Five(input []string) (int, int) {
var ventLines []ventLine

for _, vl := range input {
ventLines = append(ventLines, stringToVentLine(vl))
}

return five(ventLines, false), five(ventLines, true)
}

type ventLine struct {
x1, x2, y1, y2 int
}

func stringToVentLine(s string) ventLine {
ints := utils.ExtractInts(s)

return ventLine{
x1: ints,
x2: ints,
y1: ints,
y2: ints,
}
}

func five(vls []ventLine, diagonals bool) int {
// Doing it like this felt pretty gross
// I could've iterated over the list of coordinates
// to find the highest values and then make slices
// from that, but that felt pretty gross too
grid := int{}

for _, v := range vls {
v.calculateVents(&grid, diagonals)
}

count := 0

for _, x := range grid {
for _, y := range x {
if y > 1 { count++ }
}
}

return count
}

func (v ventLine) calculateVents(
grid *int,
diagonals bool) {
if v.x1 != v.x2 && v.y1 != v.y2 {
if diagonals { v.diagonal(grid) }
} else {
v.straight(grid)
}
}

func (v ventLine) straight(grid *int) {
var xl, xu, yl, yu int

if v.x2 < v.x1 {
xl = v.x2
xu = v.x1
} else {
xl = v.x1
xu = v.x2
}

if v.y2 < v.y1 {
yl = v.y2
yu = v.y1
} else {
yl = v.y1
yu = v.y2
}

for x := xl; x <= xu; x++ {
for y := yl; y <= yu; y++ {
grid[x][y]++
}
}
}

// there is definitely a more readable, more concise
// way of doing this
func (v ventLine) diagonal(grid *int) {
// x and y both ascend
if v.x1 < v.x2 && v.y1 < v.y2 {
for i := 0; i <= v.x2 - v.x1; i++ {
grid[v.x1+i][v.y1+i]++
}
}

// x ascends y descends
if v.x1 < v.x2 && v.y1 > v.y2 {
for i := 0; i <= v.x2 - v.x1; i++ {
grid[v.x1+i][v.y1-i]++
}
}

// x descends y ascends
if v.x1 > v.x2 && v.y1 < v.y2 {
for i := 0; i <= v.x1 - v.x2; i++ {
grid[v.x1-i][v.y1+i]++
}
}

// x and y both descend
if v.x1 > v.x2 && v.y1 > v.y2 {
for i := 0; i <= v.x1 - v.x2; i++ {
grid[v.x1-i][v.y1-i]++
}
}
}
``````

## Day 6

The challenge for day 6 was to model the population of a group of fish, assuming that each fish spawns a new fish every 7 days and that if a fish is new it takes 9 days to spawn a new fish instead. The first part required the population to be calculated after 80 days, the second after 256. Admittedly with this one I fell into old habits, and started writing code without adequately planning it, and so the code that may have found me the answer for part one was nowhere near efficient enough to find an answer for part two unless I was running some kind of crazy hardware, which obviously I’m not… I included it here for posterity, given I have refactored most of the other code posted here to some extent.

``````Example input:
3,4,3,1,2
``````
``````// This... is not good. My brain is bad in the morning.
func Six(input []string) (int, int) {
var lf []lanternfish
for _, days := range utils.ExtractInts(input) {
lf = append(lf, newFish(days))
}

for i := 0; i < 80; i++ {
for _, fish := range lf {
fish.processDay(&lf)
}
}

return len(lf), 0
}

type lanternfish struct {
daysUntilSpawn int
}

func newFish(dus int) lanternfish {
return lanternfish{ daysUntilSpawn: dus }
}

func (lf *lanternfish) processDay(fishList *[]lanternfish) {
*lf = newFish(lf.daysUntilSpawn - 1)
if lf.daysUntilSpawn < 0 {
lf.daysUntilSpawn = 6
*fishList = append(*fishList, newFish(8))
}
}
``````
``````// This is both easier to implement
// and easier to read afterwards...
// I regret not thinking about it first.
func Six(input []string) (int, int) {
initialFish := utils.ExtractInts(input)

return fishPop(initialFish, 80), fishPop(initialFish, 256)
}

func fishPop(in []int, days int) int {
lf := map[int]int{
0: 0,
1: 0,
2: 0,
3: 0,
4: 0,
5: 0,
6: 0,
7: 0,
8: 0,
}

for _, timer := range in {
lf[timer]++
}

for i := 0; i < days; i++ {
t0 := lf
t1 := lf
t2 := lf
t3 := lf
t4 := lf
t5 := lf
t6 := lf
t7 := lf
t8 := lf
lf = t0
lf = t8
lf = t7 + t0
lf = t6
lf = t5
lf = t4
lf = t3
lf = t2
lf = t1
}

return lf + lf + lf + lf +
lf + lf + lf + lf + lf
}
``````

See other posts in the Advent of Code series