Uiua

Part 1 was fun, though I had quite some frustration with getting the loop work the way I wanted it to.
For part 2 I didn’t actually need to modify that much, just pass another function to deal with the list of tachyons: part 1, remove duplicates, part 2, don’t remove and count at the end. Easy. Or so I thought.
I realized something was wrong when it just wouldn’t stop running and the fan got louder.

Today I started over from scratch and decided to solve both parts at once since I basically got the counts for each row anyways, so I just had to change the way the splitting was handled.
There was one occasion where I got an error that the resulting array would be too big (~4GB) but at least I got a warning instead of seeing RAM usage spike suddenly :P

Run with example input

$ .......S.......
$ ...............
$ .......^.......
$ ...............
$ ......^.^......
$ ...............
$ .....^.^.^.....
$ ...............
$ ....^.^...^....
$ ...............
$ ...^.^...^.^...
$ ...............
$ ..^...^.....^..
$ ...............
$ .^.^.^.^.^...^.
$ ...............
# &fras "input-7.txt" ◌
⊜∘⊸≠@\n
⊃(=@S⊡₀)(⊏⊚>0⊸≡/+=@^↘₂)
ShootSplit ← (
  ⤚(=2+>0)
  ⧻⊸⊚
  ⊙(⟜₂(
      ⟜⊏⊚⊙⟜[⧻]
      ⍚(⊂⊃-₁+₁)
      /+≡⌟(⬚0˜↯×◇°⊚)
    )
    +⍜⊏(˜↯0⧻)⊚
  )
)

Solve ← (
  0
  ⍥(⊙(+|ShootSplit|⊃⊡₀↘₁))◡⋅⋅⧻
  ⊟⊙⊙◌⊙/+
)

Solve
≡(&p $"Part _: _") 1_2

And for completeness sake:

Prep ← ⊃(⊚=@S)⊜∘⊸≠@\n

Splits! ← (
  ˙⍤>0◡⋅⧻
  ⤚(=⧻⤙˜⨂)
  ∩⌟▽⊸¬
  ⊸⧻
  ⊙(♭≡[⊃-₁+₁]
    ^⊂)
)

Shoot! ← (
  0
  ⍢(⊙(+
    | ⍣Splits!^⊸⋅0
    | ⊚=@^°⊂
    )
  | >0⋅⋅⧻)
)

PartOne ← (
  &fras "input-7.txt"
  Prep
  ⊙⋅◌Shoot!◴
)

PartTwo ← (
  &fras "input-7.txt"
  Prep
  ⧻⋅⊙◌Shoot!∘
)

Uiua

Heavily copied ahem, inspired by @[email protected]’s solution :)

Parse     ← °⊂ ▽⊸≡/↥≠@.⊜∘⊸≠@\n
Flow      ← +⊃(/+≡↻1_¯1¤×|×⊙¬)
Propagate ← ˜∧(⊸˜Flow)
Part₁     ← /+♭↧ ⊸Propagate
Part₂     ← /+⊣Propagate ⊙(˜⊂0)

$ .......S.......
$ ...............
$ .......^.......
$ ...............
$ ......^.^......
$ ...............
$ .....^.^.^.....
$ ...............
$ ....^.^...^....
$ ...............
$ ...^.^...^.^...
$ ...............
$ ..^...^.....^..
$ ...............
$ .^.^.^.^.^...^.
$ ...............
&fras "7.txt"

⊃(Part₁|Part₂) Parse

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.

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)
}

Janet

I hate dynamic programming I hate dynamic programming I hate dynamic programming I hate dynamic programming I hate dynamic programming I hate dynamic programming I hate dynamic programming I hate dynamic programming I hate dynamic programming I hate dynamic programming

• part1
• part2

Uiua

A bit late getting to this, but happy with this solution, despite it being nothing more than just building/summing all paths. As usual, you can drop your own file onto that solution.

D ← ".......S.......\n...............\n.......^.......\n...............\n......^.^......\n...............\n.....^.^.^.....\n...............\n....^.^...^....\n...............\n...^.^...^.^...\n...............\n..^...^.....^..\n...............\n.^.^.^.^.^...^.\n..............."
# D ← &fras"AOC2025day07.txt" # <- Uncomment and drop file here.

Parse ← ⊃↘↙1≠@.⊜∘⊸≠@\n
Flow  ← ⊃(+⊃↻₁↻₋₁×|׬)⊙⊸⊣
P₁    ← /+>0♭⬚0×⟜∧(˜⊂↥Flow)
P₂    ← /+⊣∧(˜⊂+Flow)
⊃P₁ P₂ Parse D

QIR (Quantum Intermediate Representation)

Nah, just kidding - I used Rust. The only tricky part seemed to be finding time on a Sunday to get it coded!

Part 1 I swept down with a bool array for beams and part 2 I swept up with a score array and summed when it split (joined).

struct Teleporter(String);

impl Teleporter {
    fn new(s: String) -> Self {
        Self(s)
    }

    fn start_pos(line: &str) -> Result<usize> {
        line.find('S').ok_or_eyre("Start not found")
    }

    fn splits(&self) -> Result<usize> {
        let mut input = self.0.lines();
        let start_line = input.next().ok_or_eyre("No start line")?;
        let start = Self::start_pos(start_line)?;

        let mut beams = vec![false; start_line.len()];
        beams[start] = true;

        let mut splits = 0;
        for line in input {
            for (i, ch) in line.bytes().enumerate() {
                if beams[i] && ch == b'^' {
                    splits += 1;
                    beams[i] = false;
                    beams[i - 1] = true;
                    beams[i + 1] = true;
                }
            }
        }
        Ok(splits)
    }

    fn timelines(&self) -> Result<usize> {
        let mut input = self.0.lines();
        let start_line = input.next().ok_or_eyre("No start line")?;
        let start = Self::start_pos(start_line)?;

        let mut num_paths = vec![1; start_line.len()];
        for line in input.rev() {
            for (i, c) in line.bytes().enumerate() {
                if c == b'^' || c == b'S' {
                    num_paths[i] = num_paths[i - 1] + num_paths[i + 1];
                }
            }
        }
        Ok(num_paths[start])
    }
}

Rust

Dynamic programming? Nah, just keep track of the number of overlapping beams and part 2 becomes no different to part 1.

View on github

use std::collections::VecDeque;

fn parse_input(input: &str) -> (Vec<Vec<bool>>, (usize, usize)) {
    let splits = input
        .lines()
        .map(|l| l.chars().map(|c| c == '^').collect())
        .collect();
    // Assume start is on first row
    let start = (input.chars().position(|c| c == 'S').unwrap(), 0);
    (splits, start)
}

fn solve(input: String) {
    let (splits, start) = parse_input(&input);
    let mut nsplits = 0u32;
    let mut timelines = 1u64;
    let mut frontier = VecDeque::from([(start, 1)]);
    while let Some((pos, multiplicity)) = frontier.pop_front() {
        let (x, y) = (pos.0, pos.1 + 1);
        if y == splits.len() {
            // Falls out of bottom
            continue;
        }
        if splits[y][x] {
            nsplits += 1;
            timelines += multiplicity;
            if let Some((b, m2)) = frontier.back_mut()
                && *b == (x - 1, y)
            {
                *m2 += multiplicity;
            } else {
                frontier.push_back(((x - 1, y), multiplicity));
            }
            frontier.push_back(((x + 1, y), multiplicity));
        } else if let Some((b, m2)) = frontier.back_mut()
            && *b == (x, y)
        {
            *m2 += multiplicity;
        } else {
            frontier.push_back(((x, y), multiplicity));
        }
    }
    println!("Part 1: {nsplits}");
    println!("Part 2: {timelines}");
}

fn main() -> std::io::Result<()> {
    let (input, _) = util::get_input("day7.txt")?;
    solve(input);
    Ok(())
}

I went for recursion (and a simple graph node class) but my favourite solution is transfer matrices one that I have seen some people do

from pathlib import Path
import numpy as np
import itertools as it

cwd = Path(__file__).parent


def parse_input(file_path):
  with file_path.open("r") as fp:
    data = list(map(str.strip, fp.readlines()))

  return np.array([list(row) for row in data], dtype=object)


class Node:

  def __init__(self, symbol, row, col):
    self.symbol = symbol
    self.loc = tuple([row, col])
    self.parents = []
    self.npaths_to_start = -1

  @property
  def is_connected(self):
    return self.symbol=='S' or len(self.parents)>0

  def connect(self, node_dict, s0, s1):

    if self.loc[0]==0:
      return

    top = node_dict[self.loc[0]-1, self.loc[1]]

    if top.symbol in ['.','S'] and top.is_connected:
      self.parents.append(top)

    if self.symbol=='^' and top.is_connected:
      left =  node_dict[self.loc[0], self.loc[1]-1]
      right =  node_dict[self.loc[0], self.loc[1]+1]

      left.parents.append(self)
      right.parents.append(self)


def count_paths_to_start(node0, node1):
  '''
  node0 should always be the start node else
  result is meaningless
  '''

  if node0 in node1.parents:
    return 1
  else:
    npaths = 0
    for p in node1.parents:
      if p.npaths_to_start != -1:
        npaths += p.npaths_to_start
      else:
        p.npaths_to_start = count_paths_to_start(node0, p)
        npaths += p.npaths_to_start

    return npaths


def solve_problem(file_name, quantum=False):

  grid = parse_input(Path(cwd, file_name))
  s0,s1 = grid.shape

  node_dict = {(i,j):Node(grid[i,j], i, j) for i,j in
               it.product(range(s0), range(s1))}
  [node.connect(node_dict, s0, s1) for node in node_dict.values()]

  if not quantum:
    return len([x for x in node_dict.values() if x.symbol == '^' and
                x.is_connected>0])
  else:
    start_ind = [(0, j) for j in range(s1) if node_dict[0,j].symbol=='S'][0]

    return\
      sum([count_paths_to_start(node_dict[start_ind], node_dict[s0-1, j]) for
           j in range(s1) if node_dict[s0-1,j].is_connected])



if __name__ == "__main__":

  assert solve_problem("test_input") == 21
  assert solve_problem("input") == 1633
  assert solve_problem("test_input", True) == 40
  assert solve_problem("input", True) == 34339203133559

Rust

Part 1 was quite straightforward: just keep track of a frontier of every beam, and keep following it, adding splitter locations to a set. Easy.

I had to give some thought to Part 2 and ended up doing dynamic programming: for every position, keep track of how many timelines go through it, solving recursively where necesary.

use std::collections::{HashMap, HashSet, VecDeque};

use crate::{grid::{Coordinate, Direction, Grid}, solver::DaySolver};

pub struct Day07Solver;

impl DaySolver for Day07Solver {
    fn part1(&self, input: String) -> String {
        let grid = Grid::from_grid_string(
            &input,
            |c| match c {
                'S' => Location::Start,
                '^' => Location::Splitter,
                '.' => Location::Empty,
                _ => panic!("Invalid location type"),
            }
        );

        let mut reached_splitters: HashSet<Coordinate> = HashSet::new();

        let mut beam_coordinates: VecDeque<Coordinate> = grid.iter()
            .filter_map(|(c, l)| match l {
                Location::Start => Some(c),
                _ => None,
            })
            .collect();

        while let Some(next) = beam_coordinates.pop_front() {
            if let Some(next_coord) = grid.neighbor_in_direction(next, Direction::Down) {
                match grid[next_coord] {
                    Location::Start => { panic!("Encountered a second start!"); },
                    Location::Splitter => {
                        if !reached_splitters.contains(&next_coord) {
                            reached_splitters.insert(next_coord);
                            [
                                grid.neighbor_in_direction(next_coord, Direction::Left),
                                grid.neighbor_in_direction(next_coord, Direction::Right),
                            ]
                                .into_iter()
                                .flatten()
                                .for_each(|c| beam_coordinates.push_back(c));
                        }
                    },
                    Location::Empty => { beam_coordinates.push_back(next_coord); }
                }
            }
        }

        reached_splitters
            .len()
            .to_string()
    }

    fn part2(&self, input: String) -> String {
        let grid = Grid::from_grid_string(
            &input,
            |c| match c {
                'S' => Location::Start,
                '^' => Location::Splitter,
                '.' => Location::Empty,
                _ => panic!("Invalid location type"),
            }
        );

        let start_coord = grid.iter().find(|(_, l)| **l == Location::Start).unwrap().0;

        let mut num_timelines: HashMap<Coordinate, usize> = HashMap::new();
        count_timelines(&mut num_timelines, &grid, 1, start_coord);

        num_timelines[&start_coord].to_string()
    }
}

fn count_timelines(
    num_timelines: &mut HashMap<Coordinate, usize>,
    grid: &Grid<Location>,
    timelines_so_far: usize,
    coordinate: Coordinate,
) {
    if num_timelines.contains_key(&coordinate) {
        return
    }

    match grid[coordinate] {
        Location::Splitter => {
            let neighbors: Vec<Coordinate> = [
                grid.neighbor_in_direction(coordinate, Direction::Left),
                grid.neighbor_in_direction(coordinate, Direction::Right),
            ]
                .into_iter()
                .flatten()
                .collect();

            neighbors.iter()
                .for_each(|next| {
                    count_timelines(num_timelines, grid, timelines_so_far, *next);
                });

            let timelines_from_here = neighbors.iter().map(|c| num_timelines[c]).sum();
            num_timelines.insert(coordinate, timelines_from_here);
        },
        _ => {
            let timelines_from_here = if let Some(next) = grid.neighbor_in_direction(coordinate, Direction::Down) {
                count_timelines(num_timelines, grid, timelines_so_far, next);
                num_timelines[&next]
            } else {
                timelines_so_far
            };

            num_timelines.insert(coordinate, timelines_from_here);
        },
    };
}

#[derive(PartialEq, Eq, Copy, Clone)]
enum Location {
    Start,
    Splitter,
    Empty,
}

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn part1() {
        let input = include_str!("../../inputs/test/07");

        let solver = Day07Solver {};
        assert_eq!("21", solver.part1(input.to_string()));
    }


    #[test]
    fn part2() {
        let input = include_str!("../../inputs/test/07");

        let solver = Day07Solver {};
        assert_eq!("40", solver.part2(input.to_string()));
    }
}

My choice of representation for part 1 ended up being fortunate, as all I needed to change for part 2 was three setf in propagate that became incf. This was also another good use case for multiple return values, as it’s the beam vector you want to reduce on, but for part 1 the split count needs to come along for the ride.

(defun parse-line (line)
  (let ((result (make-array (length line) :initial-element 0)))
    (loop for i from 0 to (1- (length line))
          if (member (char line i) '(#\^ #\S))
            do (setf (aref result i) 1))
    result))

(defun read-inputs (filename)
  (let ((input-lines (uiop:read-file-lines filename)))
    (mapcar #'parse-line input-lines)))

(defun propagate (in-beams splitters)
  (let ((out-beams (make-array (length in-beams) :initial-element 0))
        (splits 0))
    (loop for i from 0 to (1- (length in-beams))
          if (not (zerop (aref in-beams i)))
            do (if (not (zerop (aref splitters i)))
                   (progn (incf (aref out-beams (1- i)) (aref in-beams i))
                          (incf (aref out-beams (1+ i)) (aref in-beams i))
                          (incf splits))
                   (incf (aref out-beams i) (aref in-beams i))))
    (values out-beams splits)))

(defun propagate-all (initial-beam-vector splitter-map)
  (let* ((total-splits 0)
         (final-beam-vector
           (reduce #'(lambda (in-beams splitters)
                       (multiple-value-bind (out-beams this-splits)
                           (propagate in-beams splitters)
                         (incf total-splits this-splits)
                         out-beams))
                   splitter-map :initial-value initial-beam-vector)))
    (values final-beam-vector total-splits)))

(defun main-1 (filename)
  (let ((grid (read-inputs filename)))
    (multiple-value-bind (final-beam-vector total-splits)
        (propagate-all (car grid) (cdr grid))
      total-splits)))

(defun main-2 (filename)
  (let ((grid (read-inputs filename)))
    (multiple-value-bind (final-beam-vector total-splits)
        (propagate-all (car grid) (cdr grid))
      (reduce #'+ (coerce final-beam-vector 'list)))))

Haskell

import Control.Arrow
import Control.Monad
import Control.Monad.Writer.Strict
import Data.Array
import Data.Functor
import Data.List
import Data.Maybe
import Data.Monoid

parse content = (elemIndices 'S' x, filter (not . null) $ elemIndices '^' <$> xs)
  where
    (x : xs) = lines content

split :: [Int] -> Int -> Writer (Sum Int) [Int]
split splitters beam
    | beam `elem` splitters = tell 1 $> [pred beam, succ beam]
    | otherwise = pure [beam]

part1 = getSum . execWriter . uncurry process
  where
    process start =
        foldl'
            (\beams splitters -> nub . concat <$> (beams >>= mapM (split splitters)))
            (pure start)

part2 :: ([Int], [[Int]]) -> Int
part2 (start, splitterList) = go (head start, 0)
  where
    go (i, j)
        | j >= depth = 1
        | hasSplitter i j = r ! (pred i, succ j) + r ! (succ i, succ j)
        | otherwise = r ! (i, succ j)

    r = listArray bounds [go (i, j) | (i, j) <- range bounds]
    bounds = ((0, 0), (width, depth))

    hasSplitter i j = j < length splitterList && i `elem` splitterList !! j

    depth = length splitterList
    width = succ . maximum $ concat splitterList

main = getContents >>= print . (part1 &&& part2) . parse

Kotlin

Tried recursive for part1, didn’t work. LUCKILY for once I was smart and committed anyway, as it was the right solution for part2! I was losing my mind for a bit though as I had originally written my method with Integers…

class Day07 : Puzzle {

    val grid = mutableListOf<MutableList<Char>>()
    val partTwoCache = mutableMapOf<Pair<Int, Int>, Long>()

    override fun readFile() {
        val input = readInputFromFile(2025, 7, false)
        for (line in input.lines().filter { it.isNotBlank() }) {
            grid.add(line.toCharArray().toMutableList())
        }
    }

    override fun solvePartOne(): String {
        grid[1][grid[0].indexOf('S')] = '|'

        var splits = 0
        for (r in 1..<grid.size - 1) {
            for (c in 0..<grid[r].size) {
                if (grid[r][c] == '|') {
                    if (grid[r+1][c] == '.') {
                        grid[r+1][c] = '|'
                    } else if (grid[r+1][c] == '^') {
                        grid[r+1][c-1] = '|'
                        grid[r+1][c+1] = '|'
                        splits++
                    }
                }
            }
        }
        return splits.toString()
    }

    override fun solvePartTwo(): String {
        val start = grid[0].indexOf('S')
        return (1 + processBeamPartTwo(1, start)).toString() // don't forget to count the original timeline
    }

    private fun processBeamPartTwo(row: Int, column: Int): Long {
        if (partTwoCache.contains(Pair(row, column))) {
            return partTwoCache[Pair(row, column)]!!
        }

        if (row == grid.size) return 0L
        if (column == grid[row].size || column < 0) return 0L

        val out = if (grid[row][column] == '^') { // splitter
            1L + processBeamPartTwo(row, column - 1) + processBeamPartTwo(row, column + 1)
        } else {
            processBeamPartTwo(row + 1, column)
        }
        partTwoCache[Pair(row, column)] = out
        return out
    }
}

full code on Codeberg

Haskell

That was a fun little problem.

import Data.Map qualified as Map  
import Data.Set qualified as Set  
import Data.Tuple (swap)  

readInput s =  
  Map.fromDistinctAscList  
    [((i, j), c) | (i, l) <- zip [0 ..] $ lines s, (j, c) <- zip [0 ..] l]  

beamPaths input = scanl step (Map.singleton startX 1) [startY .. endY]  
  where  
    Just (startY, startX) = lookup 'S' $ map swap $ Map.assocs input  
    Just ((endY, _), _) = Map.lookupMax input  
    step beams y =  
      Map.unionsWith (+) $  
        [ if input Map.!? (y + 1, j) == Just '^'  
            then Map.fromList [(j - 1, n), (j + 1, n)]  
            else Map.singleton j n  
          | (j, n) <- Map.assocs beams  
        ]  

part1 = sum . map Set.size . (zipWith (Set.\\) <*> tail) . map Map.keysSet . beamPaths  

part2 = sum . last . beamPaths  

main = do  
  input <- readInput <$> readFile "input07"  
  print $ part1 input  
  print $ part2 input  

C

Accidentally solved part 2 first but had the foresight to save the code. Otherwise my solution looks similar to what other people are doing, just with more code 😅

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <inttypes.h>
#include <assert.h>

#define GW 148
#define GH 74

static char g[GH][GW];
static uint64_t dp[2][GW];
static int w,h;

/* recursively traces beam for part 1, marking visited splitters with 'X' */
static uint64_t
solve_p1(int x, int y)
{
	if (x<0 || x>=w || y<0 || y>=h || g[y][x] == 'X')
		return 0;
	else if (g[y][x] != '^')
		return solve_p1(x, y+1);
	else {
		g[y][x] = 'X';
		return 1 + solve_p1(x-1, y) + solve_p1(x+1, y);
	}
}

/* DP for part 2 */
static uint64_t
solve_p2(void)
{
	int x,y, odd;

	for (y=h-1; y>=0; y--)
	for (x=1; x<w-1; x++) {
		/* only two rows relevant at a time, so don't store any more */
		odd = y&1;

		if (g[y][x] == 'S') {
			printf("\n");
			return dp[!odd][x] + 1;
		}

		dp[odd][x] = g[y][x] == '^' || g[y][x] == 'X'
		    ? dp[!odd][x-1] + dp[!odd][x+1] + 1
		    : dp[!odd][x];
	}

	return 0;
}

int
main()
{
	int x,y, sx,sy;

	for (h=0; ; ) {
		/* one bottom row of padding */
		assert(h < GH-1);
		/* input already side padded, plus we have \n\0 */
		if (!fgets(g[h], GW, stdin))
			break;
		/* skip empty rows */
		for (x=0; g[h][x]; x++)
			if (g[h][x] == 'S' || g[h][x] == '^')
				{ h++; break; }
	}

	w = strlen(g[0])-1; /* strip \n */

	for (y=0; y<h; y++)
	for (x=0; x<w; x++)
		if (g[y][x] == 'S')
			{ sx=x; sy=y; break; }

	printf("07: %"PRIu64" %"PRIu64"\n", solve_p1(sx,sy), solve_p2());
}

Python

Not too difficult, especially since there are no edge cases (particles leaving the grid, adjacent splitters).

My initial solution mutated the input for the simulation which I cleaned up after by creating an array that would record the number of particle paths at every column location + some other optimizations. I chose to implement both parts in the same function because they share the majority of the logic.

def solve(data: str):
    grid = data.splitlines()
    m, n = len(grid), len(grid[0])

    # find the first particle
    particle_paths = [0] * n  # count of particle paths that will reach this column
    for j in range(n):
        if grid[0][j] == 'S':
            particle_paths[j] = 1
            break
    
    # count the number of splits for part 1
    splits = 0

    # simulate the particle moving down the grid
    #   optimization 1: we can start from the 3rd row (index 2) because that's where the first splitter is
    #   optimization 2: we can skip alternating rows because every other row is empty
    for i in range(2, m, 2):
        # particle paths per column after this row is processed
        next_particle_paths = [0] * n

        for j in range(n):
            if particle_paths[j] == 0:
                # skip if there are no particle paths coming from above in this column
                continue
            
            if grid[i][j] == '.':
                # no splitter here, the number of paths in this column remains the same
                # make sure to use += to account for neighboring splitters dumping additional paths into this column
                next_particle_paths[j] += particle_paths[j]
            else:
                # splitter activated here, any particle arriving here can end up in the left or right column
                # this can be simulated by adding the number of paths to the columns on either side
                splits += 1
                next_particle_paths[j-1] += particle_paths[j]
                next_particle_paths[j+1] += particle_paths[j]
        
        # update vars for next iteration
        particle_paths = next_particle_paths

    # return both 
    #   the number of splits AND 
    #   the count of timelines a particle would create
    return splits, sum(particle_paths)

sample = """.......S.......
...............
.......^.......
...............
......^.^......
...............
.....^.^.^.....
...............
....^.^...^....
...............
...^.^...^.^...
...............
..^...^.....^..
...............
.^.^.^.^.^...^.
..............."""
assert solve(sample) == (21, 40)