@papierabreisser.bsky.social
No bell so loud, no clothing so bright, no helmet so hard to even out bad cycling infrastructure!
"Immerhin".
Und zum Glück sind keine Menschen an Bord, denn wir wissen ja alle, die größte Gefahr geht für Insassen von PKW aus, im Gegensatz zu irgendwelchen Leuten ohne Blechhülle auf der Straße.
Kritik gibt es erst hinter der Bezahlschranke im #SPON.
Und zum Glück sind keine Menschen an Bord, denn wir wissen ja alle, die größte Gefahr geht für Insassen von PKW aus, im Gegensatz zu irgendwelchen Leuten ohne Blechhülle auf der Straße.
Kritik gibt es erst hinter der Bezahlschranke im #SPON.
December 16, 2025 at 12:04 PM
"Immerhin".
Und zum Glück sind keine Menschen an Bord, denn wir wissen ja alle, die größte Gefahr geht für Insassen von PKW aus, im Gegensatz zu irgendwelchen Leuten ohne Blechhülle auf der Straße.
Kritik gibt es erst hinter der Bezahlschranke im #SPON.
Und zum Glück sind keine Menschen an Bord, denn wir wissen ja alle, die größte Gefahr geht für Insassen von PKW aus, im Gegensatz zu irgendwelchen Leuten ohne Blechhülle auf der Straße.
Kritik gibt es erst hinter der Bezahlschranke im #SPON.
One last time this year: a flowless solution to #AdventOfCode day 12.
It is cheating a bit, as it does not flip and rotate but rather check the total space occupied by each packet. Will not work on the sample!
Thanks, @was.tl! #AOC is fun each year and I learn something new each time. 🎄🎁
It is cheating a bit, as it does not flip and rotate but rather check the total space occupied by each packet. Will not work on the sample!
Thanks, @was.tl! #AOC is fun each year and I learn something new each time. 🎄🎁
![// My solution for day 12 to this challenge: https://www.reddit.com/r/adventofcode/comments/1p2ral9/flowless_challenge_2025/
// Purely functional and branchless programming
// Just paste this code into the developer console on https://adventofcode.com/2025/day/12/input (try F12 or CTRL+I in your browser)
// The number is the solution to the first part of the challenge.
// There is no second part to this days challenge, just have 23 stars total
const parseInput = input => input
.trim()
.split('\n\n')
.reduce(([boxes, regions], block) => {
const lines = block.split('\n')
const hasX = +(lines[0].search('x') != -1)
boxes = [[...boxes, lines.slice(1).join('').split('').filter(c => c == '#').length], boxes][hasX]
regions = [regions, lines.reduce((reg, line) => [...reg, line.match(/\d+/g)?.map(Number)], regions)][hasX]
return [boxes, regions]
}, [[], []])
const part1 = input => {
const [boxes, regions] = parseInput(input)
return regions.reduce((total, region) => {
const area = region[0] * region[1]
const needed = region.slice(2).reduce((acc, cur, i) => acc + cur * boxes[i], 0)
return total + +(area >= needed)
}, 0)
}
console.time('Advent of Code Day 12 both parts flowless')
console.log(part1(document.body.innerText.trim()))
console.timeEnd('Advent of Code Day 12 both parts flowless')](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:imdouboftoohbcdk4tvnlez7/bafkreieu72jxdxemwbzkckxtqu57ie4bh4dt4czsattngwocjfkzg5gcmy@jpeg)
December 12, 2025 at 7:40 AM
One last time this year: a flowless solution to #AdventOfCode day 12.
It is cheating a bit, as it does not flip and rotate but rather check the total space occupied by each packet. Will not work on the sample!
Thanks, @was.tl! #AOC is fun each year and I learn something new each time. 🎄🎁
It is cheating a bit, as it does not flip and rotate but rather check the total space occupied by each packet. Will not work on the sample!
Thanks, @was.tl! #AOC is fun each year and I learn something new each time. 🎄🎁
After my hiatus yesterday I am now back in business with my #AOC #AdventOfCode day 9 solution to the flowless challenge.
But it is slooooooow! Took about 40 Minutes to process my real input.
My non-flowless solution with the same algo runs in 7 seconds.
But it is slooooooow! Took about 40 Minutes to process my real input.
My non-flowless solution with the same algo runs in 7 seconds.
December 9, 2025 at 6:24 PM
After my hiatus yesterday I am now back in business with my #AOC #AdventOfCode day 9 solution to the flowless challenge.
But it is slooooooow! Took about 40 Minutes to process my real input.
My non-flowless solution with the same algo runs in 7 seconds.
But it is slooooooow! Took about 40 Minutes to process my real input.
My non-flowless solution with the same algo runs in 7 seconds.
This is my solution to the #AOC #AdventOfCode day 7 flowless challenge, this time handling 4 cases in a recursion with memoization.
It's astonishing how readable these still are. And how fast they perform despite changing vars each time, especially in the recursion.
It's astonishing how readable these still are. And how fast they perform despite changing vars each time, especially in the recursion.
![// My solution for day 7 to this challenge: https://www.reddit.com/r/adventofcode/comments/1p2ral9/flowless_challenge_2025/
// Purely functional and branchless programming
// Just paste this code into the developer console on https://adventofcode.com/2025/day/7/input (try F12 or CTRL+I in your browser)
// The two numbers are the results for the first and second part for your personal input
const steps = count => [...(new Array(count)).keys()]
const memo = new Map()
const solve = (grid, timelines, row, col) => {
const key = `${row},${col}`
const val = memo.get(key)
|| (row == grid.length - 1) * (timelines + 1)
|| (grid[row][col] == '.') * solve(grid, timelines, row + 1, col)
|| solve(grid, timelines, row + 1, col - 1) + solve(grid, timelines, row + 1, col + 1)
memo.set(key, val)
return val
}
const part1 = input => steps(input.length)
.slice(1)
.reduce((total, row) => steps(input[row].length)
.reduce((t, col) => {
const above = +('S|'.includes(input[row - 1][col]))
const splitter = +(input[row][col] == '^')
input[row][col] = [input[row][col], '|'][above * !splitter]
input[row][col - 1] = [input[row][col - 1], '|'][above * splitter]
input[row][col + 1] = [input[row][col + 1], '|'][above * splitter]
return above * splitter + t
}, total), 0)
const part2 = input => solve(input, 0, 1, input[0].indexOf('S'))
const input = document.body.innerText.trim().split('\n').map(l => [...l])
console.time('Advent of Code day 7 flowless challenge both parts')
console.log([part1(input.map(r => r.map(c => c))), part2(input)])
console.timeEnd('Advent of Code day 7 flowless challenge both parts')](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:imdouboftoohbcdk4tvnlez7/bafkreiafc2k7xwkxi763k55zywdlkvg74fumy3m7ancmlwguhz7rqdvava@jpeg)
December 7, 2025 at 1:07 PM
This is my solution to the #AOC #AdventOfCode day 7 flowless challenge, this time handling 4 cases in a recursion with memoization.
It's astonishing how readable these still are. And how fast they perform despite changing vars each time, especially in the recursion.
It's astonishing how readable these still are. And how fast they perform despite changing vars each time, especially in the recursion.
My solution to the #AOC #AdventOfCode day 6 flowless challenge.
Had to refactor my transpose utility function to branchless style. Part 2 has a dirty re-use of the line argument just to skip creating another variable. I need to get better at consistent naming schemes...
Had to refactor my transpose utility function to branchless style. Part 2 has a dirty re-use of the line argument just to skip creating another variable. I need to get better at consistent naming schemes...
![// My solution for day 6 to this challenge: https://www.reddit.com/r/adventofcode/comments/1p2ral9/flowless_challenge_2025/
// Purely functional and branchless programming
// Just paste this code into the developer console on https://adventofcode.com/2025/day/6/input (try F12 or CTRL+I in your browser)
// The two numbers are the results for the first and second part for your personal input
const transpose = grid => grid.reduce((cols, row, r, arr) =>
[ cols, row.map((_, c) => arr.map(l => l[c])) ][+(r == 0)] , [])
const part1 = input => {
input = input.trim().split('\n').map(l => {
l = l.trim().replace(/ +/g, ' ').split(' ')
return [l, l.map(Number)][+(l[0].match(/\d+/) != null)]
})
const operations = input.pop()
return transpose(input)
.map((vals, i) => [
vals.reduce((acc, cur) => acc * cur, 1),
vals.reduce((acc, cur) => acc + cur, 0)
][+(operations[i] == '+')])
.reduce((acc, cur) => acc + cur, 0)
}
const part2 = input => {
input = transpose(input.trim().split('\n').map(l => l.split('')))
input = input.map(l => l.map(c => [c, ' '][+(typeof c === 'undefined')])).reverse()
return input.reduce(([total, block], line) => {
const op = line.pop()
const val = Number(line.join(''))
block = [[...block, val], block][+(val == 0)]
line = [
0,
block.reduce((a, v) => a + v, 0),
block.reduce((a, v) => a * v, 1)
][' +*'.indexOf(op)]
block = [ block, [], [] ][' +*'.indexOf(op)]
return [total + line, block]
}, [0, []])[0]
}
const db = document.body.innerText
console.time('Advent of Code day 6 flowless challenge both parts')
console.log([part1(db), part2(db)])
console.timeEnd('Advent of Code day 6 flowless challenge both parts')](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:imdouboftoohbcdk4tvnlez7/bafkreiadevscuyattbzdg6t4c4kc7izctttx3hielcqtioi2adlc33elre@jpeg)
December 6, 2025 at 11:45 AM
My solution to the #AOC #AdventOfCode day 6 flowless challenge.
Had to refactor my transpose utility function to branchless style. Part 2 has a dirty re-use of the line argument just to skip creating another variable. I need to get better at consistent naming schemes...
Had to refactor my transpose utility function to branchless style. Part 2 has a dirty re-use of the line argument just to skip creating another variable. I need to get better at consistent naming schemes...
Managed to solve #AOC #AdventOfCode 2025 day 5 under the restrictions of the flowless challenge again!
![// My solution for day 5 to this challenge: https://www.reddit.com/r/adventofcode/comments/1p2ral9/flowless_challenge_2025/
// Purely functional and branchless programming
// Just paste this code into the developer console on https://adventofcode.com/2025/day/5/input (try F12 or CTRL+I in your browser)
// The two numbers are the results for the first and second part for your personal input
const parseInput = input => {
const [ranges, ing] = input.trim().split('\n\n')
return [
ranges.split('\n').map(l => l.split('-').map(Number)),
ing.split('\n').map(Number)
]
}
const part1 = ([ranges, ings]) => ings.reduce((acc, ing) => acc + ranges.some(([l, h]) => ing >= l && ing <= h), 0)
const part2 = ([ranges]) => {
ranges.sort((a, b) => a[0] - b[0])
return ranges
.reduce((comp, [l, h]) => {
const found = +(comp.some(([s, e], i) => {
const contained = +(s <= l && l < e && e > h)
const extend = +(s <= l && l <= e && e <= h)
comp[i][1] = [e, h][extend]
return contained || extend
}))
return [[...comp, [l, h]], comp][found]
}, [ranges[0]])
.reduce((acc, [l, h]) => acc + h - l + 1, 0)
}
const db = parseInput(document.body.innerText)
console.time('Advent of Code day 5 flowless challenge both parts')
console.log([part1(db), part2(db)])
console.timeEnd('Advent of Code day 5 flowless challenge both parts')](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:imdouboftoohbcdk4tvnlez7/bafkreifbf5tjvhf622se5oxk7vn7mt3yqnufbnswoea3a4ayjq5s4gtwuq@jpeg)
December 5, 2025 at 3:42 PM
Managed to solve #AOC #AdventOfCode 2025 day 5 under the restrictions of the flowless challenge again!
Totally rad solution to the #AOC #AdventOfCode day 4 flowless challenge, featuring a recursive function handling 3 different cases with a single return statement 🤓
Even had to refactor my grid functions because they used a ternary operator that is forbidden in flowless.
Even had to refactor my grid functions because they used a ternary operator that is forbidden in flowless.
![// My solution for day 4 to this challenge: https://www.reddit.com/r/adventofcode/comments/1p2ral9/flowless_challenge_2025/
// Purely functional and branchless programming
// Just paste this code into the developer console on https://adventofcode.com/2025/day/4/input (try F12 or CTRL+I in your browser)
// The two numbers are the results for the first and second part for your personal input
const eightWayDeltas = [[-1, -1], [-1, 0], [-1, 1], [0, -1], [0, 1], [1, -1], [1, 0], [1, 1]]
const gridCells = grid => grid.reduce((cells, line, row) => [...cells, ...line.map((value, col) => { return {row, col, value} })], [])
const validCoordForGrid = (row, col, grid) => row >= 0 && row < grid.length && col >= 0 && col < grid[row].length
const getSurroundingGridCoords = (grid, row, col, deltas) =>
deltas.flatMap(([r, c]) => Array(+(validCoordForGrid(row + r, col + c, grid))).fill([row + r, col + c]));
const getSurrounding = (grid, row, col, deltas) =>
getSurroundingGridCoords(grid, row, col, deltas).reduce((tiles, [r, c]) => [...tiles, { tile: grid[r][c], row: r, col: c }], [])
const solve = (grid, total, isPart1) => {
let changed = gridCells(grid)
.filter(f => f.value == '@')
.reduce((acc, {row, col}) => {
const removable = +(getSurrounding(grid, row, col, eightWayDeltas).filter(f => f.tile == '@').length < 4)
grid[row][col] = '@.'[removable * !isPart1]
return acc + removable
}, 0)
return ((changed == 0) * total) || (isPart1 * changed) || solve(grid, total + changed, isPart1)
}
const input = document.body.innerText.trim().split('\n').map(l => l.split(''))
console.time('Advent of Code 2025 day 4 both parts')
console.log([
solve(input.map(r => r.map(c => c)), 0, 1),
solve(input.map(r => r.map(c => c)), 0, 0)
])
console.timeEnd('Advent of Code 2025 day 4 both parts')](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:imdouboftoohbcdk4tvnlez7/bafkreihonmgnflcpthcdfb6g7f3qohdwm7hmbmbl2ymuantbkeifx5kzsm@jpeg)
December 4, 2025 at 4:14 PM
Totally rad solution to the #AOC #AdventOfCode day 4 flowless challenge, featuring a recursive function handling 3 different cases with a single return statement 🤓
Even had to refactor my grid functions because they used a ternary operator that is forbidden in flowless.
Even had to refactor my grid functions because they used a ternary operator that is forbidden in flowless.
![// My solution for day 3 to this challenge: https://www.reddit.com/r/adventofcode/comments/1p2ral9/flowless_challenge_2025/
// Purely functional and branchless programming
// Just paste this code into the developer console on https://adventofcode.com/2025/day/3/input (try F12 or CTRL+I in your browser)
// The two numbers are the results for the first and second part for your personal input
const best = (digits, row) => Number(
(new Array(digits))
.fill(0)
.reduce(([res, pos, i]) => {
const r = res + Math.max(...row.slice(pos + 1, row.length - digits + 1 + i))
return [ r, pos + row.slice(pos + 1).indexOf(~~r[r.length - 1]) + 1, i + 1 ]
}, ['', -1, 0])[0]
)
console.log(
document.body.innerText
.trim()
.split('\n')
.map(l => l.split('').map(Number))
.reduce(([p1, p2], row) => [p1 + best(2, row), p2 + best(12, row)], [0, 0])
)](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:imdouboftoohbcdk4tvnlez7/bafkreifpvlmoej2khbaionsyiqgqao2hgwec522k54kix4berseolqfc2y@jpeg)
December 3, 2025 at 3:39 PM
Upping the ante for #AdventOfCode by providing a solution for the flowless challenge on day 2:
![// My solution for day 2 to this challenge: https://www.reddit.com/r/adventofcode/comments/1p2ral9/flowless_challenge_2025/
// Purely functional and branchless programming
// Just paste this code into the developer console on https://adventofcode.com/2025/day/2/input (try F12 or CTRL+I in your browser)
// The two numbers are the results for the first and second part for your personal input
console.log(
document.body.innerText
.split(',')
.map(range => range.split('-').map(Number))
.reduce(([p1, p2], [start, end]) =>
(new Array(end - start))
.fill(0)
.reduce(([a, b, v]) => [
a + /^(\d+)\1$/.test(v) * v,
b + /^(\d+)\1+$/.test(v) * v,
v + 1
], [p1, p2, start])
, [0, 0])
.slice(0, 2)
)](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:imdouboftoohbcdk4tvnlez7/bafkreihf6usrdsnnqxx7pz6qomjfzrtxkdvhncyldflvay3fp72ny4k7ti@jpeg)
December 2, 2025 at 7:51 AM
Upping the ante for #AdventOfCode by providing a solution for the flowless challenge on day 2:
Upping the ante for #AdventOfCode by providing a solution for the flowless challenge on day 1:
![// My solution to this challenge: https://www.reddit.com/r/adventofcode/comments/1p2ral9/flowless_challenge_2025/
// Purely functional and branchless programming
// Just paste this code into the developer console on https://adventofcode.com/2025/day/1/input (try F12 or CTRL+I in your browser)
// The two numbers are the results for the first and second part for your personal input
console.log(
document.body.innerText
.split('\n')
.map(l => [(l[0] == 'R') * 2 - 1, ~~l.slice(1)])
.reduce(([p1, p2, pos], [dir, val]) => {
[pos, p2] = (new Array(val))
.fill(0)
.reduce(([p, z, d]) => {
p = (p + d + 100) % 100
z += p == 0
return [p, z, d]
}, [pos, p2, dir])
p1 += pos == 0
return [p1, p2, pos]
}, [0, 0, 50])
.slice(0, 2)
)](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:imdouboftoohbcdk4tvnlez7/bafkreid3qlej3f4yj6mwn3e4xsok65vco3vhu2an7yjlbyyvdmmbwhufna@jpeg)
December 2, 2025 at 7:50 AM
Upping the ante for #AdventOfCode by providing a solution for the flowless challenge on day 1:
Möchte mir mal jemand kurz erklären, in wie weit Apple Geräte bislang als Maßstab für bezahlbar galten?
Muss ich hier wirklich irgendwen an die Apple Tax erinnern? Oder den ach so beliebten Monitorständer für fast 1000 Euro?
www.spiegel.de/wirtschaft/u...
Muss ich hier wirklich irgendwen an die Apple Tax erinnern? Oder den ach so beliebten Monitorständer für fast 1000 Euro?
www.spiegel.de/wirtschaft/u...
(S+) Apple und die Zölle: Wie Trump die iPhone-Erfolgsgeschichte ausradieren könnte
Der Aufstieg von Apple gilt als beispielloser Erfolg. Doch Trumps Zölle könnten iPhones bald unbezahlbar machen. Konzernchef Tim Cook versucht, das Schlimmste zu verhindern.
www.spiegel.de
April 15, 2025 at 3:36 PM
Möchte mir mal jemand kurz erklären, in wie weit Apple Geräte bislang als Maßstab für bezahlbar galten?
Muss ich hier wirklich irgendwen an die Apple Tax erinnern? Oder den ach so beliebten Monitorständer für fast 1000 Euro?
www.spiegel.de/wirtschaft/u...
Muss ich hier wirklich irgendwen an die Apple Tax erinnern? Oder den ach so beliebten Monitorständer für fast 1000 Euro?
www.spiegel.de/wirtschaft/u...
Wählen ist ein wenig wie Vorsorgeuntersuchungen.
Der reine Tätigkeit an sich ist ganz OK, eine winzige Unannehmlichkeit.
Aber direkt danach diese Ungewissheit, wie schlimm das Ergebnis wohl ausfallen wird, das ist echt beschissen.
Durchhalten, hier kommt das Ergebnis zum Glück recht zügig!
Der reine Tätigkeit an sich ist ganz OK, eine winzige Unannehmlichkeit.
Aber direkt danach diese Ungewissheit, wie schlimm das Ergebnis wohl ausfallen wird, das ist echt beschissen.
Durchhalten, hier kommt das Ergebnis zum Glück recht zügig!
February 23, 2025 at 12:12 PM
Wählen ist ein wenig wie Vorsorgeuntersuchungen.
Der reine Tätigkeit an sich ist ganz OK, eine winzige Unannehmlichkeit.
Aber direkt danach diese Ungewissheit, wie schlimm das Ergebnis wohl ausfallen wird, das ist echt beschissen.
Durchhalten, hier kommt das Ergebnis zum Glück recht zügig!
Der reine Tätigkeit an sich ist ganz OK, eine winzige Unannehmlichkeit.
Aber direkt danach diese Ungewissheit, wie schlimm das Ergebnis wohl ausfallen wird, das ist echt beschissen.
Durchhalten, hier kommt das Ergebnis zum Glück recht zügig!
Sieht das für Dich wie eine Steigerung der Erfahrungen aus der Demo-Tour nach Lützerath aus, @katjadiehl.bsky.social ?
www.grundrechtekomitee.de/details/pres...
www.grundrechtekomitee.de/details/pres...
January 13, 2025 at 3:26 PM
Sieht das für Dich wie eine Steigerung der Erfahrungen aus der Demo-Tour nach Lützerath aus, @katjadiehl.bsky.social ?
www.grundrechtekomitee.de/details/pres...
www.grundrechtekomitee.de/details/pres...
Lasst uns Musk bei jedem neuen Amoklauf in Amerika an seinen "incompetent fool" Tweet erinnern! "Should resign immediately!"
Und natürlich bei jedem weiteren Todesopfer durch einen ach so intelligenten Tesla.
www.spiegel.de/panorama/jus...
Und natürlich bei jedem weiteren Todesopfer durch einen ach so intelligenten Tesla.
www.spiegel.de/panorama/jus...
Anschlag in Magdeburg: Elon Musk fordert Olaf Scholz zum sofortigen Rücktritt auf
Die genauen Hintergründe zur Tat auf dem Magdeburger Weihnachtsmarkt sind noch unklar, doch Elon Musk nutzt sie umgehend für einen Frontalangriff auf den Kanzler. Dieser sei ein »unfähiger Idiot« und ...
www.spiegel.de
December 21, 2024 at 4:54 AM
Lasst uns Musk bei jedem neuen Amoklauf in Amerika an seinen "incompetent fool" Tweet erinnern! "Should resign immediately!"
Und natürlich bei jedem weiteren Todesopfer durch einen ach so intelligenten Tesla.
www.spiegel.de/panorama/jus...
Und natürlich bei jedem weiteren Todesopfer durch einen ach so intelligenten Tesla.
www.spiegel.de/panorama/jus...