reboot6675

joined 2 years ago
[–] reboot6675@sopuli.xyz 2 points 4 months ago

Unfortunately I've struggled with this myself :/ (moving to a new country didn't help either)

[–] reboot6675@sopuli.xyz 6 points 4 months ago (2 children)

I'd say go for it if you get the chance. Playing with other people is a game changer in your music journey. Then it's not only focusing on your playing but also on what the others are playing, following the tempo, the dynamics...

[–] reboot6675@sopuli.xyz 4 points 4 months ago (1 children)

Related question, what CalDAV server are you using? Been looking for something lightweight

[–] reboot6675@sopuli.xyz 3 points 4 months ago (9 children)

Wasn't Spotify Swedish tho? But yeah

[–] reboot6675@sopuli.xyz 2 points 4 months ago

Agree with Zero Mission. Second would be Dread and then Super. Haven't played Fusion tho, long overdue...

[–] reboot6675@sopuli.xyz 10 points 4 months ago

Usually artists go on tour after releasing a new album, to promote said album. So of course they're going to play new stuff.

The ideal for me is a good balance. Some new songs mixed with the classics.

[–] reboot6675@sopuli.xyz 1 points 7 months ago (1 children)

Go

Using a map to store u|v relations. Part 2 sorting with a custom compare function worked very nicely

spoiler

func main() {
	file, _ := os.Open("input.txt")
	defer file.Close()
	scanner := bufio.NewScanner(file)

	mapPages := make(map[string][]string)
	rulesSection := true
	middleSumOk := 0
	middleSumNotOk := 0

	for scanner.Scan() {
		line := scanner.Text()
		if line == "" {
			rulesSection = false
			continue
		}

		if rulesSection {
			parts := strings.Split(line, "|")
			u, v := parts[0], parts[1]
			mapPages[u] = append(mapPages[u], v)
		} else {
			update := strings.Split(line, ",")
			isOk := true

			for i := 1; i < len(update); i++ {
				u, v := update[i-1], update[i]
				if !slices.Contains(mapPages[u], v) {
					isOk = false
					break
				}
			}

			middlePos := len(update) / 2
			if isOk {
				middlePage, _ := strconv.Atoi(update[middlePos])
				middleSumOk += middlePage
			} else {
				slices.SortFunc(update, func(u, v string) int {
					if slices.Contains(mapPages[u], v) {
						return -1
					} else if slices.Contains(mapPages[v], u) {
						return 1
					}
					return 0
				})
				middlePage, _ := strconv.Atoi(update[middlePos])
				middleSumNotOk += middlePage
			}
		}
	}

	fmt.Println("Part 1:", middleSumOk)
	fmt.Println("Part 2:", middleSumNotOk)
}

[–] reboot6675@sopuli.xyz 1 points 7 months ago

Same, I initially also thought a|b and a|c implies a|c. However when I drew the graph of the example on paper, I suspected that all relations will be given, and coded it with that assumption, that turned out to be correct

[–] reboot6675@sopuli.xyz 2 points 7 months ago

I thought of Ernest & Celestine

[–] reboot6675@sopuli.xyz 2 points 7 months ago

Agree, I filled like half notebook with this game

[–] reboot6675@sopuli.xyz 1 points 7 months ago

Go

Just a bunch of ifs and bounds checking. Part 2 was actually simpler.

Code

func part1(W [][]rune) {
	m := len(W)
	n := len(W[0])
	xmasCount := 0

	for i := 0; i < m; i++ {
		for j := 0; j < n; j++ {
			if W[i][j] != 'X' {
				continue
			}
			if j < n-3 && W[i][j+1] == 'M' && W[i][j+2] == 'A' && W[i][j+3] == 'S' {
				// Horizontal left to right
				xmasCount++
			}
			if j >= 3 && W[i][j-1] == 'M' && W[i][j-2] == 'A' && W[i][j-3] == 'S' {
				// Horizontal right to left
				xmasCount++
			}
			if i < m-3 && W[i+1][j] == 'M' && W[i+2][j] == 'A' && W[i+3][j] == 'S' {
				// Vertical up to down
				xmasCount++
			}
			if i >= 3 && W[i-1][j] == 'M' && W[i-2][j] == 'A' && W[i-3][j] == 'S' {
				// Vertical down to up
				xmasCount++
			}
			if j < n-3 && i < m-3 && W[i+1][j+1] == 'M' && W[i+2][j+2] == 'A' && W[i+3][j+3] == 'S' {
				// Diagonal left to right and up to down
				xmasCount++
			}
			if j >= 3 && i < m-3 && W[i+1][j-1] == 'M' && W[i+2][j-2] == 'A' && W[i+3][j-3] == 'S' {
				// Diagonal right to left and up to down
				xmasCount++
			}
			if j < n-3 && i >= 3 && W[i-1][j+1] == 'M' && W[i-2][j+2] == 'A' && W[i-3][j+3] == 'S' {
				// Diagonal left to right and down to up
				xmasCount++
			}
			if j >= 3 && i >= 3 && W[i-1][j-1] == 'M' && W[i-2][j-2] == 'A' && W[i-3][j-3] == 'S' {
				// Diagonal right to left and down to up
				xmasCount++
			}
		}
	}

	fmt.Println(xmasCount)
}

func part2(W [][]rune) {
	m := len(W)
	n := len(W[0])
	xmasCount := 0

	for i := 0; i <= m-3; i++ {
		for j := 0; j <= n-3; j++ {
			if W[i+1][j+1] != 'A' {
				continue
			}
			if W[i][j] == 'M' && W[i][j+2] == 'M' && W[i+2][j] == 'S' && W[i+2][j+2] == 'S' {
				xmasCount++
			} else if W[i][j] == 'M' && W[i][j+2] == 'S' && W[i+2][j] == 'M' && W[i+2][j+2] == 'S' {
				xmasCount++
			} else if W[i][j] == 'S' && W[i][j+2] == 'S' && W[i+2][j] == 'M' && W[i+2][j+2] == 'M' {
				xmasCount++
			} else if W[i][j] == 'S' && W[i][j+2] == 'M' && W[i+2][j] == 'S' && W[i+2][j+2] == 'M' {
				xmasCount++
			}
		}
	}

	fmt.Println(xmasCount)
}

func main() {
	file, _ := os.Open("input.txt")
	defer file.Close()
	scanner := bufio.NewScanner(file)

	var W [][]rune
	for scanner.Scan() {
		line := scanner.Text()
		W = append(W, []rune(line))
	}

	part1(W)
	part2(W)
}

[–] reboot6675@sopuli.xyz 1 points 7 months ago

Honestly this is soo much better, I'm not proud of my code at all haha. Thanks for sharing, definitely adding that | to my bag of tricks

 

Context: this is a legit screenshot I took on my workplace around 1.5 years ago. Hopefully it's been patched by now? Completely ridiculous behavior

 

Had it saved on my hard drive for years and recently re-discovered it. Seems to be from 2011 (!!)

Google is getting worse by the year, but it seems even back then some people were already seeing where things were going.

(Credits to the original author whoever it might be)

 

I've been following this account for a few years on Twitter and it's awesome. I was out of the loop since I stopped using Twitter, but now I found that they launched a blog and also started to be active in other social media. Thought you might appreciate it like I do!

 

Hello. I sometimes receive images in HEIC format and I use an app to convert them to JPG. However I realized that this app uploads the images and makes the conversion in the cloud, so I stopped using it.

Does anyone know of an app that does this conversion offline?

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