Advent of code day 4

I ranked (((109))) on the global leaderboard for the first star. So close

my solution was so bad that im gonna rewrite it before posting so i dont look like even more of a mouthbreather than ppl already know i am

here was mine. i was off on a goose chase for about 40 minutes finding a off by one error (yes i was missing a + 1 in range())

from y2022.scaffold import *

class Day04(Day):
	def __init__(self):
		super()
	
	@property
	def day(self): return :marseymonke: 4

	def prepare_data(self) -> Any:
		data = self.get_data().splitlines()
		for i in range(0, len(data)):
			data[i] = data[i].split(',')
			data[i][0] = range(int(data[i][0].split('-')[0]), int(data[i][0].split('-')[1]) + 1)
			data[i][1] = range(int(data[i][1].split('-')[0]), int(data[i][1].split('-')[1]) + 1)
		return data

	def a(self):
		data = self.prepare_data()
		cnt = 0
		i = 0
		for x in data:
			if set(x[0]).issuperset(set(x[1])) or set(x[1]).issuperset(set(x[0])):
				cnt += 1
			i += 1
		print(cnt)
		print(len(data))
	
	def b(self):
		data = self.prepare_data()
		cnt = 0
		for x in data:
			if any(y for y in x[0] if y in x[1]): cnt += 1
		print(cnt)

This day was a bit boring but here's my sir, do the needful solution

i would have finished both parts in a couple minutes but ranges excluding the end fucked me up for so goddamn long, i just couldn't figure out what was wrong

with open("day4input.txt") as f:
    pairs = [line for line in f]

# part 1

full_overlaps = 0

for pair in pairs:
    pair = pair.split(",") # [1-2, 2-3] 
    pair = [elf.split("-") for elf in pair] # [(1,2), (2,3)]
    elf1 = set(range(int(pair[0][0]), int(pair[0][1])+1))
    elf2 = set(range(int(pair[1][0]), int(pair[1][1])+1)) 
    if elf1.issubset(elf2) or elf2.issubset(elf1):
        full_overlaps += 1

print(full_overlaps)

# part 2 

overlaps = 0

for pair in pairs:
    pair = pair.split(",") # [1-2, 2-3] 
    pair = [elf.split("-") for elf in pair] # [(1,2), (2,3)]
    elf1 = set(range(int(pair[0][0]), int(pair[0][1])+1))
    elf2 = set(range(int(pair[1][0]), int(pair[1][1])+1)) 
    if elf1.intersection(elf2) != set():
        overlaps += 1

print(overlaps)

Not difficult yet. Misread the problem per usual and sunk 20 minutes into a problem that didnt exist though.

Part 1

subSets = 0

elfAssignments = []

with open('input', 'r') as f:

for line in f:

    cur_pair = line.split(',')

    first_assign = range(int(cur_pair[0].split('-')[0]), int(cur_pair[0].split('-')[1])+1)

    second_assign = range(int(cur_pair[1].split('-')[0]), int(cur_pair[1].split('-')[1])+1)

    l1 = set(list(first_assign))

    l2 = set(list(second_assign))

    if (l1.issubset(l2)) or l2.issubset(l1):

        subSets += 1

print(subSets)

Part 2:

overlaps = 0

elfAssignments = []

with open('input', 'r') as f:

    for line in f:

        cur_pair = line.split(',')

        first_assign = range(int(cur_pair[0].split('-')[0]), int(cur_pair[0].split('-')[1])+1)

        second_assign = range(int(cur_pair[1].split('-')[0]), int(cur_pair[1].split('-')[1])+1)

        l1 = set(list(first_assign))

        l2 = set(list(second_assign))

        if not l1.isdisjoint(l2) or not l2.isdisjoint(l1):

            overlaps += 1

print(overlaps)

#include <iostream>
#include <regex>
#include <string>

struct Range
{
	long low;
	long upper;

	bool inline isContainedIn(const struct Range &larger) const
	{
		return larger.low <= low and larger.upper >= upper;
	}

	bool inline isIntersectionEmptySet(const struct Range &r) const
	{
		return upper < r.low or low > r.upper;
	}
};

// G-d forgive me bc I used C++ Slowpoke regex
std::regex ranges_reg("(\\d+)\\-(\\d+),(\\d+)\\-(\\d+)");

int main()
{
	std::string line;
	unsigned card_pairs = 0;

	while(getline(std::cin, line))
	{
		std::smatch match;
		std::regex_search(line, match, ranges_reg);

		Range elf1 = {std::stol(match.str(1)), std::stol(match.str(2))};
		Range elf2 = {std::stol(match.str(3)), std::stol(match.str(4))};

		// PART A
		//if(elf1.isContainedIn(elf2) or elf2.isContainedIn(elf1))
		//	card_pairs ++;

		// PART B
		if(not elf1.isIntersectionEmptySet(elf2))
			card_pairs ++;
	}

	std::cout << card_pairs << std::endl;
}

woke up early at 1PM today for this.

https://fsdfsd.net/cyberdick/AdventOfCode2022/src/branch/master/AOC2022/Puzzles/Day4.cs

had to think hard and draw a diagram to figure out overlaps(), ngmi

Anyone else doing Rust?

with range and set its easy. Don't care about performance

with open('input04.txt') as file:
    lines = [line.split(',') for line in file]

# a and b
count_a = 0
count_b = 0
for pair in lines:
    lower,upper = pair[0].split('-')
    id_set_1 = set(range(int(lower),int(upper)+1))
    lower,upper = pair[1].split('-')
    id_set_2 = set(range(int(lower),int(upper)+1))
    if id_set_1.issubset(id_set_2) or id_set_1.issuperset(id_set_2): # a
        count_a += 1
    if len(id_set_1.intersection(id_set_2)) > 0: # b
        count_b += 1
print("Part a:", count_a)
print("Part b:", count_b)

r←⊢⎕FIO[49] 'day4raw.txt'
v←⍎¨¨¨('-'≠¨r⊂¨⍨','≠¨r)⊂¨¨r⊂¨⍨','≠¨r ⋄ f←{(1⌷⍵){1-⍨⍺+⍳1+⍵-⍺}2⌷⍵}
d1←{f↑1⌷⍵}¨v ⋄ d2←{f↑2⌷⍵}¨v ⋄ +/d1{+/1∊+/¨(⍺⍷⍵)(⍵⍷⍺)}¨d2 ⋄ ↑+/⍉⊃0≠⍴¨d1∩¨d2

c++

struct Elf {
    int min;
    int max;

    bool inline isContainedOrContains(const Elf& b) const {
        return (min >= b.min and max <= b.max) || (b.min >= min && b.max <= max);
    }

    bool inline intersect(const Elf& b) const {
        return (min <= b.max && b.min <= max);
    }
};



std::tuple<int int=""> solution()
{
    std::cout << "Day 4" << '\n';

    std::ifstream file("input.txt");
    std::string line;

    int p1{ 0 };
    int p2{ 0 };

    while (std::getline(file, line)) {

        int a{ 0 };
        int b{ 0 };
        Elf elf1{};
        Elf elf2{};

        std::string buffer;
        for (char c : line) {
            if (c == '-') {
                a = std::stoi(buffer);
                buffer = "";
            }
            else if (c == ',') {
                b = std::stoi(buffer);
                elf1 = { a, b };
                buffer = "";
            }
            else buffer += c;
        }
        elf2 = { a, std::stoi(buffer)};

        if (elf1.isContainedOrContains(elf2)) ++p1;
        if (elf1.intersect(elf2)) ++p2;
    }

    file.close();

    return { p1, p2 };
}
</int>

import * as fs from 'fs'

console.log(
  fs.readFileSync(process.argv[2], 'utf-8').split('\n').reduce((p, l) => {
      const m = l.match(/\d+/g)?.map(s => Number(s));
      return [ 
        p[0] + (m && (m[0] <= m[2] && m[1] >= m[3] || m[2] <= m[0] && m[3] >= m[1]) ? 1 : 0),
        p[1] + (m && (m[0] <= m[2] && m[1] >= m[2] || m[2] <= m[0] && m[3] >= m[0]) ? 1 : 0)
      ]
    }, [0,0])
);

Part 1:

with open('input.txt') as f:
    lines = f.readlines()
  
total_overlap = 0
any_overlap = 0
  
for line in lines:
    elf1, elf2 = line.strip().split(',')
    elf1_start, elf1_end = elf1.split('-')
    elf2_start, elf2_end = elf2.split('-')
    
    elf1_start = int(elf1_start)
    elf1_end = int(elf1_end)
    elf2_start = int(elf2_start)
    elf2_end = int(elf2_end)
    
    if elf1_start <= elf2_start and elf1_end >= elf2_end:
        total_overlap += 1
    elif elf1_start >= elf2_start and elf1_end <= elf2_end:
        total_overlap += 1
        
    if elf1_start <= elf2_end and elf1_start >= elf2_start:
        any_overlap += 1
    elif elf2_start <= elf1_end and elf2_end >= elf1_start:
        any_overlap += 1


print("There are " + str(total_overlap) + " total overlapping sets")
print("There are " + str(any_overlap) + " overlapping sets")

Solution with no comparisons, only bitflips and abuse of casting

def to_unary(number):
    to_return = 0
    for i in range(int(number)):
        to_return+=2**i
    return to_return

# PART 2 BITWISE
matcher = re.compile(R"([0-9]*)-([0-9]*),([0-9]*)-([0-9]*)")
score = 0
for pair in lines:
    result = matcher.search(pair)
    a_min = to_unary(result.group(1))
    a_max = to_unary(result.group(2))
    b_min = to_unary(result.group(3))
    b_max = to_unary(result.group(4))
    #print(f"A MIN: {a_min:b} A MAX: {a_max:b} B MIN: {b_min:b} B MAX: {b_max:b}")
    
    score+=int(bool(((a_max ^ a_min) | ((a_min+1) >> 1)) & ((b_max ^ b_min) | ((b_min+1) >> 1))))

yes I stole the intersection function, no I don't care

package four

import (
	"bufio"
	"os"
	"strconv"
	"strings"
)

func grabRange(line string) (map[string]int, map[string]int) {
	a := make(map[string]int)
	b := make(map[string]int)

	pair := strings.Split(line, ",")

	left := strings.Split(pair[0], "-")
	right := strings.Split(pair[1], "-")

	a["min"], _ = strconv.Atoi(left[0])
	a["max"], _ = strconv.Atoi(left[1])

	b["min"], _ = strconv.Atoi(right[0])
	b["max"], _ = strconv.Atoi(right[1])

	return a, b
}

func One() int {
	var deliver int

	fp, _ := os.Open("four.txt")
	defer fp.Close()

	rd := bufio.NewScanner(fp)

	for rd.Scan() {
		left, right := grabRange(rd.Text())

		if (left["min"] <= right["min"] && right["max"] <= left["max"]) ||
			(left["min"] >= right["min"] && right["max"] >= left["max"]) {
			deliver++
		}
	}

	return deliver
}

func Two() int {
	var deliver int

	fp, _ := os.Open("four.txt")
	defer fp.Close()

	rd := bufio.NewScanner(fp)

	for rd.Scan() {
		var a, b []int
		left, right := grabRange(rd.Text())

		for i := left["min"]; ; i++ {
			a = append(a, i)

			if i >= left["max"] {
				break
			}
		}

		for i := right["min"]; ; i++ {
			b = append(b, i)

			if i >= right["max"] {
				break
			}
		}

		intersection := func(a, b []int) bool {
			var c []int
			m := make(map[int]bool)

			for _, item := range a {
				m[item] = true
			}

			for _, item := range b {
				if _, ok := m[item]; ok {
					c = append(c, item)
				}
			}
			return (len(c) > 0)
		}

		if intersection(a, b) {
			deliver++
		}

		a = nil
		b = nil
	}

	return deliver

Ban all pythoncels from Advent of Code, we can't trust them

Do they get harder at some point? Since the timing format is very euromisic I'm just aiming to complete every challenge before the next one drops.

import re
foo = []
with open('day4_input.txt', 'r') as inp:
    foo = inp.read().split('\n')
total1, total2 = 0, 0
for f in foo:
    f = [eval(x) for x in re.split(',|-', f)]
    if (f[0] <= f[2] <= f[3] <= f[1]) or (f[2] <= f[0] <= f[1] <= f[3]):
        total1 += 1
    if not (f[2] > f[1] or f[0] > f[3]):
        total2 += 1
print(total1, total2)

had important business to attend to today sorry I'm late boss here's the homework. Bodged together at 11:30pm as usual

file = open("input.txt")
cc = 0
co = 0
for line in file:
    line = line.rstrip()
    h_i = [i for i, x in enumerate(line) if x == '-']
    c_i = [i for i, x in enumerate(line) if x == ',']
    e1_min = int(line[0:h_i[0]])
    e1_max = int(line[h_i[0]+1:c_i[0]])
    e2_min = int(line[c_i[0]+1:h_i[1]])
    e2_max = int(line[h_i[1]+1:len(line)])
    if ((e1_min <= e2_min and e1_max>=e2_max) or (e2_min <= e1_min and e2_max>=e1_max)):
        cc += 1
    if not ((e1_max < e2_min) or (e2_max < e1_min)):
        co += 1
print(cc)
print(co)

I'm never sure if calling the same function with the same args twice in the same line makes it compute twice of if it's optimized away, so added another functional composition layer on top. This way is faster for part 2 than just intersecting ranges. in matlab ()

A = readlines("input.txt","EmptyLineRule","skip")';

comp1 = @(x) ((x(1)>=x(3))&&(x(2)<=x(4)))||((x(1)<=x(3))&&(x(2)>=x(4)));
comp2 = @(x) ((x(1)>=x(3))&&(x(1)<=x(4)))||((x(2)>=x(3))&&(x(2)<=x(4)));
aux = @(u) comp2(u)||comp2([u(3:4) u(1:2)]);

iterfunc = @(u) [comp1(u); aux(u)];
iterfunc2 = @(i) iterfunc(sscanf(i,'%d-%d,%d-%d'));

res = sum(cell2mat(arrayfun(iterfunc2,A,'UniformOutput',false)),2);

fprintf("Part 1 : %d\n",res(1))
fprintf("Part 2 : %d\n",res(2))

with open("./input4.txt") as infile:
    ct = 0
    for line in map(str.strip, infile):
        a, b = line.split(',')
        a_min, a_max = map(int, a.split("-"))
        b_min, b_max = map(int, b.split("-"))
        a_set = set(range(a_min, a_max +1))
        b_set = set(range(b_min, b_max +1))
        u = a_set & b_set

        # part one
        # if len(u) in {len(a_set), len(b_set)}:
        #     ct += 1
        
        # part two
        # if u:
        #     ct += 1
    print(ct)

Frick me I keep forgetting to do these each day

People always talk about how poorly written and unrealistic female characters are but even in real life women are poorly written