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Cake day: September 17th, 2023

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  • Go

    Now I’m behind by 1 day, will try to catch up.

    For part 2 I spent a good while thinking about it, then when I convinced myself my plan could work, struggled a bit with the implementation. But it worked in the end. Basically grid[i][j] is how many different ways you can reach a cell. Start at 1 on the S cell, then propagate the values down and keep adding up the nums when you reach cells through different paths. The answer is the sum of the nums in the last row.

    spoiler
    func part2() {
    	// file, _ := os.Open("sample.txt")
    	file, _ := os.Open("input.txt")
    	defer file.Close()
    	scanner := bufio.NewScanner(file)
    
    	input := [][]rune{}
    
    	for scanner.Scan() {
    		line := []rune(scanner.Text())
    		input = append(input, line)
    	}
    
    	m := len(input)
    	n := len(input[0])
    	grid := make([][]int, m)
    	for i := range m {
    		grid[i] = make([]int, n)
    	}
    
    	for i := range m {
    		for j := range n {
    			c := input[i][j]
    			if i == 0 {
    				if c == 'S' {
    					grid[i][j] = 1
    				}
    				continue
    			}
    			if c == '^' {
    				grid[i][j-1] += grid[i-1][j]
    				grid[i][j+1] += grid[i-1][j]
    			} else {
    				grid[i][j] = grid[i][j] + grid[i-1][j]
    			}
    		}
    	}
    
    	paths := 0
    	for j := range n {
    		paths += grid[m-1][j]
    	}
    
    	fmt.Println(paths)
    }
    

  • Go

    Part 2: Read the whole input in a rune matrix. Scan it column by column, store the numbers as you go, ignoring all spaces, and store the operand when you find it. When you hit an empty column or the end, do the operation and add it to the total.

    spoiler
    func part2() {
    	// file, _ := os.Open("sample.txt")
    	file, _ := os.Open("input.txt")
    	defer file.Close()
    	scanner := bufio.NewScanner(file)
    
    	chars := [][]rune{}
    	for scanner.Scan() {
    		chars = append(chars, []rune(scanner.Text()))
    	}
    
    	m := len(chars)
    	n := len(chars[0])
    	var op rune
    	nums := []int{}
    	total := 0
    
    	for j := range n {
    		current := []rune{}
    		for i := range m {
    			if chars[i][j] == '+' || chars[i][j] == '*' {
    				op = chars[i][j]
    			} else if chars[i][j] != ' ' {
    				current = append(current, chars[i][j])
    			}
    		}
    		if len(current) > 0 {
    			x, _ := strconv.Atoi(string(current))
    			nums = append(nums, x)
    		}
    		if len(current) == 0 || j == n-1 {
    			result := 0
    			if op == '*' {
    				result = 1
    			}
    			for _, x := range nums {
    				if op == '+' {
    					result = result + x
    				} else {
    					result = result * x
    				}
    			}
    			total += result
    			nums = []int{}
    		}
    	}
    
    	fmt.Println(total)
    }
    

  • First I tried to to part 2 with a very poor regex strategy and the performance was abysmal. Switched to plain substrings and boom, instant result.

    Golang

    func part1() {
    	ranges := readInput()
    	invalidSum := 0
    
    	for _, r := range ranges {
    		parts := strings.Split(r, "-")
    		start, _ := strconv.Atoi(parts[0])
    		end, _ := strconv.Atoi(parts[1])
    
    		for num := start; num <= end; num++ {
    			current := strconv.Itoa(num)
    			n := len(current)
    			if n%2 != 0 {
    				continue
    			}
    			left := current[:n/2]
    			right := current[n/2:]
    			if left == right {
    				invalidSum += num
    			}
    		}
    	}
    
    	fmt.Println(invalidSum)
    }
    
    func part2() {
    	ranges := readInput()
    	invalidSum := 0
    
    	for _, r := range ranges {
    		parts := strings.Split(r, "-")
    		start, _ := strconv.Atoi(parts[0])
    		end, _ := strconv.Atoi(parts[1])
    
    		for num := start; num <= end; num++ {
    			current := strconv.Itoa(num)
    			n := len(current)
    
    			for index := 1; index <= n/2; index++ {
    				if n%index != 0 {
    					continue
    				}
    
    				left := 0
    				right := index
    				prefix := current[left:right]
    				isRepeated := true
    				for left < n && right < n {
    					left = right
    					right = right + index
    					next := current[left:right]
    					if next != prefix {
    						isRepeated = false
    						break
    					}
    				}
    
    				if isRepeated {
    					invalidSum += num
    					break
    				}
    			}
    		}
    	}
    
    	fmt.Println(invalidSum)
    }
    

  • Golang

    func part1() {
    	// file, _ := os.Open("sample.txt")
    	file, _ := os.Open("input.txt")
    	defer file.Close()
    
    	scanner := bufio.NewScanner(file)
    	n := 100
    	current := 50
    	pointingAt0 := 0
    
    	for scanner.Scan() {
    		line := scanner.Text()
    		num, _ := strconv.Atoi(line[1:])
    		if line[0] == 'L' {
    			current = ((current-num)%n + n) % n
    		} else {
    			current = (current + num) % n
    		}
    		if current == 0 {
    			pointingAt0++
    		}
    	}
    
    	fmt.Println(pointingAt0)
    }
    
    func part2() {
    	// file, _ := os.Open("sample.txt")
    	file, _ := os.Open("input.txt")
    	defer file.Close()
    
    	scanner := bufio.NewScanner(file)
    	n := 100
    	current := 50
    	pointingAt0 := 0
    
    	for scanner.Scan() {
    		line := scanner.Text()
    		num, _ := strconv.Atoi(line[1:])
    
    		rounds := num / n
    		pointingAt0 += rounds
    		num = num % n
    		new := -1
    
    		if line[0] == 'L' {
    			new = ((current-num)%n + n) % n
    			if current != 0 && (new > current || new == 0) {
    				pointingAt0++
    			}
    		} else {
    			new = (current + num) % n
    			if current != 0 && (new < current || new == 0) {
    				pointingAt0++
    			}
    		}
    
    		current = new
    	}
    
    	fmt.Println(pointingAt0)
    }
    










  • Just pointing out that these 2 are not necessarily contradicting statements. Just because it hasn’t taken any jobs yet, doesn’t mean it won’t take any in the next 10 years. Not that I really want to believe those numbers though. But yeah as others have pointed out, these studies need to be taken with a grain of salt.








  • reboot6675@sopuli.xyztoComic Strips@lemmy.world1/2 off haircuts
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    7 months ago

    When I was a kid I encountered this problem when I wondered what would happen if I half-empty a bottle of soda, re-fill it with water, and repeat. Will it eventually become just water or will there always be some soda left? It boggled my mind for a while, then I forgot about it until I reached university calculus haha