Keep top level comments as only solutions, if you want to say something other than a solution put it in a new post. (replies to comments can be whatever)
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This turned out to be reasonably easy in Uiua, though this solution relies on macros which maybe slow it down.
(edit: removing one macro sped it up quite a bit)
(edit2: Letting Uiua build up an n-dimensional array turned out to be the solution, though sadly my mind only works in 3 dimensions. Now runs against the live data in around 0.3 seconds.)
Thanks to your solution I learned more about how to use reduce :D
My solution did work for the example input but not for the actual one. When I went here and saw this tiny code block and you saying
This turned out to be reasonably easy
I was quite taken aback. And it's so much better performance-wise too :D
(well, until part 2 comes along in my case. Whatever this black magic is you used there is too high for my fried brain atm)
Haha, sorry about that, it does seem quite smug :-) I went into it expecting it to be a nightmare of boxes and dimensions, but finding it something I could deal with was a massive relief. Of course once I had a working solution I reversed it back into a multi-dimensional nightmare. That's where the performance gains came from: about 10x speedup from letting Uiua build up as many dimensions as it needed before doing a final deshaping.
I enjoyed reading a different approach to this, and thanks for reminding me that ⋕$"__" exists, that's a great idiom to have up your sleeve.
Let me know if there's any bits of my solution that you'd like me to talk you through.
def parse_input(path):
with path.open("r") as fp:
lines = fp.read().splitlines()
roots = [int(line.split(':')[0]) for line in lines]
node_lists = [[int(x) for x in line.split(':')[1][1:].split(' ')] for line in lines]
return roots, node_lists
def construct_tree(root, nodes, include_concat):
levels = [[] for _ in range(len(nodes)+1)]
levels[0] = [(str(root), "")]
# level nodes are tuples of the form (val, operation) where both are str
# val can be numerical or empty string
# operation can be *, +, || or empty string
for indl, level in enumerate(levels[1:], start=1):
node = nodes[indl-1]
for elem in levels[indl-1]:
if elem[0]=='':
continue
if elem[0][-len(str(node)):] == str(node) and include_concat:
levels[indl].append((elem[0][:-len(str(node))], "||"))
if (a:=int(elem[0]))%(b:=int(node))==0:
levels[indl].append((str(int(a/b)), '*'))
if (a:=int(elem[0])) - (b:=int(node))>0:
levels[indl].append((str(a - b), "+"))
return levels[-1]
def solve_problem(file_name, include_concat):
roots, node_lists = parse_input(Path(cwd, file_name))
valid_roots = []
for root, nodes in zip(roots, node_lists):
top = construct_tree(root, nodes[::-1], include_concat)
if any((x[0]=='1' and x[1]=='*') or (x[0]=='0' and x[1]=='+') or
(x[0]=='' and x[1]=='||') for x in top):
valid_roots.append(root)
return sum(valid_roots)
I asked ChatGPT to explain your code and mentioned you said it was a binary search. idk why, but it output a matter of fact response that claims you were wrong. lmao, I still don't understand how your code works
This code doesn’t perform a classic binary search. Instead, it uses each input node to generate new possible states or “branches,” forming a tree of transformations. At each level, it tries up to three operations on the current value (remove digits, divide, subtract). These expansions create multiple paths, and the code checks which paths end in a successful condition. While the author may have described it as a “binary search,” it’s more accurately a state-space search over a tree of possible outcomes, not a binary search over a sorted data structure.
I understand it now! I took your solution and made it faster. it is now like 36 milliseconds faster for me. which is interesting because if you look at the code. I dont process the entire list of integers. I sometimes stop prematurely before the next level, clear it, and add the root. I don't know why but it just works for my input and the test input.
code
def main(input_data):
input_data = input_data.replace('\r', '')
parsed_data = {int(line[0]): [int(i) for i in line[1].split()[::-1]] for line in [l.split(': ') for l in input_data.splitlines()]}
part1 = 0
part2 = 0
for item in parsed_data.items():
root, num_array = item
part_1_branches = [set() for _ in range(len(num_array)+1)]
part_2_branches = [set() for _ in range(len(num_array)+1)]
part_1_branches[0].add(root)
part_2_branches[0].add(root)
for level,i in enumerate(num_array):
if len(part_1_branches[level]) == 0 and len(part_2_branches[level]) == 0:
break
for branch in part_1_branches[level]:
if branch == i:
part_1_branches[level+1] = set() # clear next level to prevent adding root again
part1 += root
break
if branch % i == 0:
part_1_branches[level+1].add(branch//i)
if branch - i > 0:
part_1_branches[level+1].add(branch-i)
for branch in part_2_branches[level]:
if branch == i or str(branch) == str(i):
part_2_branches[level+1] = set() # clear next level to prevent adding root again
part2 += root
break
if branch % i == 0:
part_2_branches[level+1].add(branch//i)
if branch - i > 0:
part_2_branches[level+1].add(branch-i)
if str(i) == str(branch)[-len(str(i)):]:
part_2_branches[level+1].add(int(str(branch)[:-len(str(i))]))
print("Part 1:", part1, "\nPart 2:", part2)
return [part1, part2]
if __name__ == "__main__":
with open('input', 'r') as f:
main(f.read())
however what I notice is that the parse_input causes it to be the reason why it is slower by 20+ milliseconds. I find that even if I edited your solution like so to be slightly faster, it is still 10 milliseconds slower than mine:
code
def parse_input():
with open('input',"r") as fp:
lines = fp.read().splitlines()
roots = [int(line.split(':')[0]) for line in lines]
node_lists = [[int(x) for x in line.split(': ')[1].split(' ')] for line in lines]
return roots, node_lists
def construct_tree(root, nodes):
levels = [[] for _ in range(len(nodes)+1)]
levels[0] = [(root, "")]
# level nodes are tuples of the form (val, operation) where both are str
# val can be numerical or empty string
# operation can be *, +, || or empty string
for indl, level in enumerate(levels[1:], start=1):
node = nodes[indl-1]
for elem in levels[indl-1]:
if elem[0]=='':
continue
if (a:=elem[0])%(b:=node)==0:
levels[indl].append((a/b, '*'))
if (a:=elem[0]) - (b:=node)>0:
levels[indl].append((a - b, "+"))
return levels[-1]
def construct_tree_concat(root, nodes):
levels = [[] for _ in range(len(nodes)+1)]
levels[0] = [(str(root), "")]
# level nodes are tuples of the form (val, operation) where both are str
# val can be numerical or empty string
# operation can be *, +, || or empty string
for indl, level in enumerate(levels[1:], start=1):
node = nodes[indl-1]
for elem in levels[indl-1]:
if elem[0]=='':
continue
if elem[0][-len(str(node)):] == str(node):
levels[indl].append((elem[0][:-len(str(node))], "||"))
if (a:=int(elem[0]))%(b:=int(node))==0:
levels[indl].append((str(int(a/b)), '*'))
if (a:=int(elem[0])) - (b:=int(node))>0:
levels[indl].append((str(a - b), "+"))
return levels[-1]
def solve_problem():
roots, node_lists = parse_input()
valid_roots_part1 = []
valid_roots_part2 = []
for root, nodes in zip(roots, node_lists):
top = construct_tree(root, nodes[::-1])
if any((x[0]==1 and x[1]=='*') or (x[0]==0 and x[1]=='+') for x in top):
valid_roots_part1.append(root)
top = construct_tree_concat(root, nodes[::-1])
if any((x[0]=='1' and x[1]=='*') or (x[0]=='0' and x[1]=='+') or (x[0]=='' and x[1]=='||') for x in top):
valid_roots_part2.append(root)
return sum(valid_roots_part1),sum(valid_roots_part2)
if __name__ == "__main__":
print(solve_problem())
Wow I got thrashed by chatgpt. Strictly speaking that is correct, it is more akin to Tree Search. But even then not strictly because in tree search you are searching through a list of objects that is known, you build a tree out of it and based on some conditions eliminate half of the remaining tree each time. Here I have some state space (as chatgpt claims!) and again based on applying certain conditions, I eliminate some portion of the search space successively (so I dont have to evaluate that part of the tree anymore). To me both are very similar in spirit as both methods avoid evaluating some function on all the possible inputs and successively chops off a fraction of the search space. To be more correct I will atleast replace it with tree search though, thanks. And thanks for taking a close look at my solution and improving it.
Credits to @[email protected] for the approach of using reduce and also how to split the input by multiple characters.
I can happily say that I learned quite a bit today, even though the first part made me frustrated enough that I went searching for other approaches ^^
Part two just needed a simple modification. Changing how the input is parsed and passed to the adapted function took longer than changing the function itself actually.
I wanted to this the way yo did, by repeatedly applying functions, but I didn't dare to because I like to mess up and spend some minutes debugging signatures, may I ask what your IDE setup is for the LSP-Hints with Haskell?
Setting up on my PC was a little bit of a pain because it needed matching ghc and ghcide versions, so I hadn't bothered doing it on my Laptop yet.
I use neovim with haskell-tools.nvim plugin. For ghc, haskell-language-server and others I use nix which, among other benefits makes my development environment reproducible and all haskellPackages are built on the same version so there are no missmatches.
But, as much as I love nix, there are probably easier ways to setup your environment.
Ah, well, I have a bit of a weird setup. GHC is 9.8.4, built from git. I'm using HLS version 2.9.0.1 (again built from git) under Emacs with the LSP and flycheck packages. There are probably much easier ways of getting it to work :)
Takes ~5.3s on my machine to get both outputs. Not sure how to optimize it any further other than running the math in threads? Took me longer than it should have to realize a lot of unnecessary math could be cut if the running total becomes greater than the target while doing the math. Also very happy to see that none of the inputs caused the recursive function to hit Python's max stack depth.
Code
import argparse
import os
class Calibrations:
def __init__(self, target, operators) -> None:
self.operators = operators
self.target = target
self.target_found = False
def do_math(self, numbers, operation) -> int:
if operation == "+":
return numbers[0] + numbers[1]
elif operation == "*":
return numbers[0] * numbers[1]
elif operation == "||":
return int(str(numbers[0]) + str(numbers[1]))
def all_options(self, numbers, last) -> int:
if len(numbers) < 1:
return last
for j in self.operators:
# If we found our target already, abort
# If the last value is greater than the target, abort
if self.target_found or last > self.target:
return
total = self.all_options(
numbers[1:], self.do_math((last, numbers[0]), j)
)
if total == self.target:
self.target_found = True
def process_line(self, line) -> int:
numbers = [int(x) for x in line.split(":")[1].strip().split()]
self.all_options(numbers[1:], numbers[0])
if self.target_found:
return self.target
return 0
def main() -> None:
path = os.path.dirname(os.path.abspath(__file__))
parser = argparse.ArgumentParser(description="Bridge Repair")
parser.add_argument("filename", help="Path to the input file")
args = parser.parse_args()
sum_of_valid = [0, 0]
with open(f"{path}/{args.filename}", "r") as f:
for line in f:
line = line.strip()
target = int(line.split(":")[0])
for idx, ops in enumerate([["+", "*"], ["+", "*", "||"]]):
c = Calibrations(target, ops)
found = c.process_line(line)
sum_of_valid[idx] += found
if found:
break
for i in range(0, 2):
part = i + 1
print(
"The sum of valid calibrations for Part "
+ f"{part} is {sum(sum_of_valid[:part])}"
)
if __name__ == "__main__":
main()
If you havent already done so, doing it in the form of "tree search", the code completes in the blink of an eye (though on a high end cpu 11th Gen Intel(R) Core(TM) i7-11800H @ 2.30GHz). posted the code below
Wasted too much time on part 2 trying to make combinations iterator (it was very slow). In the end solved both parts with 3^n and toTernary.
Runtime: 1.5s
func digits(n: int): int =
result = 1; var n = n
while (n = n div 10; n) > 0: inc result
func concat(a: var int, b: int) =
a = a * (10 ^ b.digits) + b
func toTernary(n: int, len: int): seq[int] =
result = newSeq[int](len)
if n == 0: return
var n = n
for i in 0..<len:
result[i] = n mod 3
n = n div 3
proc solve(input: string): AOCSolution[int, int] =
for line in input.splitLines():
let parts = line.split({':',' '})
let res = parts[0].parseInt
var values: seq[int]
for i in 2..parts.high:
values.add parts[i].parseInt
let opsCount = values.len - 1
var solvable = (p1: false, p2: false)
for s in 0 ..< 3^opsCount:
var sum = values[0]
let ternary = s.toTernary(opsCount)
for i, c in ternary:
case c
of 0: sum *= values[i+1]
of 1: sum += values[i+1]
of 2: sum.concat values[i+1]
else: raiseAssert"!!"
if sum == res:
if ternary.count(2) == 0:
solvable.p1 = true
solvable.p2 = true
if solvable == (true, true): break
if solvable.p1: result.part1 += res
if solvable.p2: result.part2 += res
Today was pretty easy one but for some reason I spent like 20 minutes overthinking part 2 when all it needed was one new else if. I initially through the concatenation operator would take precedence even tho it clearly says "All operators are still evaluated left-to-right" in the instructions..
I'm sure there are optimizations to do but using parallelStreams it only takes around 300ms total on my machine so there's no point really
The code
import java.io.IOException;
import java.nio.charset.StandardCharsets;
import java.nio.file.Files;
import java.nio.file.Path;
import java.util.Arrays;
import java.util.List;
public class Day7 {
public static void main(final String[] _args) throws IOException {
final List<Equation> equations = Files.readAllLines(Path.of("2024\\07\\input.txt"), StandardCharsets.UTF_8).stream()
.map(line -> line.split(":\\s"))
.map(line -> new Equation(
Long.parseLong(line[0]),
Arrays.stream(line[1].split("\\s"))
.map(Integer::parseInt)
.toArray(Integer[]::new)
)
).toList();
final char[] part1Operators = {'+', '*'};
System.out.println("Part 1: " + equations.parallelStream()
.mapToLong(equation -> getResultIfPossible(equation, part1Operators))
.sum()
);
final char[] part2Operators = {'+', '*', '|'};
System.out.println("Part 2: " + equations.parallelStream()
.mapToLong(equation -> getResultIfPossible(equation, part2Operators))
.sum()
);
}
private static Long getResultIfPossible(final Equation equation, final char[] operators) {
final var permutations = Math.pow(operators.length, equation.values.length - 1);
for (int i = 0; i < permutations; i++) {
long result = equation.values[0];
int count = i;
for (int j = 0; j < equation.values.length - 1; j++) {
// If the result is already larger than the expected one, we can short circuit here to save some time
if (result > equation.result) {
break;
}
final char operator = operators[count % operators.length];
count /= operators.length;
if (operator == '+') { result += equation.values[j + 1]; }
else if (operator == '*') { result *= equation.values[j + 1]; }
else if (operator == '|') { result = Long.parseLong(String.valueOf(result) + String.valueOf(equation.values[j + 1])); }
else {
throw new RuntimeException("Unsupported operator " + operator);
}
}
if (result == equation.result) {
return result;
}
}
return 0L;
}
private static record Equation(long result, Integer[] values) {}
}
Could probably go much faster if I kept track of calculations to not repeat, but 4 seconds for part 2 on my old laptop is good enough for me. Also, not really a big change from part 1 to part 2.
Part 1 and 2
(defstruct calibration result inputs)
(defun p1-process-line (line)
(let ((parts (str:words line)))
(make-calibration :result (parse-integer (car parts) :junk-allowed t)
:inputs (mapcar #'parse-integer (cdr parts)))))
(defun apply-opperators (c opps)
(let ((accum (car (calibration-inputs c))))
(loop for o in opps
for v in (cdr (calibration-inputs c))
until (> accum (calibration-result c))
if (eql o 'ADD)
do (setf accum (+ accum v))
else if (eql o 'MUL)
do (setf accum (* accum v))
else
do (setf accum (+ v (* accum (expt 10 (1+ (floor (log v 10)))))))
finally (return accum)
)))
(defun generate-operators (item-count)
(labels ((g-rec (c results)
(if (< c 1)
results
(g-rec (1- c) (loop for r in results
collect (cons 'ADD r)
collect (cons 'MUL r))))))
(g-rec (1- item-count) '((ADD) (MUL)))))
(defun generate-ops-hash (c gen-ops)
(let ((h (make-hash-table)))
(dotimes (x c)
(setf (gethash (+ 2 x) h) (funcall gen-ops (+ 1 x))))
h))
(defun validate-calibration (c ops-h)
(let ((r (calibration-result c))
(ops (gethash (length (calibration-inputs c)) ops-h)))
(loop for o in ops
for v = (apply-opperators c o)
when (= v r)
return t)))
(defun run-p1 (file)
(let ((calibrations (read-file file #'p1-process-line))
(ops (generate-ops-hash 13 #'generate-operators)))
(loop for c in calibrations
when (validate-calibration c ops)
sum (calibration-result c))))
(defun generate-operators-p2 (item-count)
(labels ((g-rec (c results)
(if (< c 1)
results
(g-rec (1- c) (loop for r in results
collect (cons 'ADD r)
collect (cons 'MUL r)
collect (cons 'CAT r))))))
(g-rec (1- item-count) '((ADD) (MUL) (CAT)))))
(defun run-p2 (file)
(let ((calibrations (read-file file #'p1-process-line))
(ops (generate-ops-hash 13 #'generate-operators-p2)))
(loop for c in calibrations
when (validate-calibration c ops)
sum (calibration-result c))))
45s on my machine for first shot, trying to break my will to brute force 😅. I'll try improving on it in a bit after I smoke another bowl and grab another drink.
solution
import itertools
import re
import aoc
def ltr(e):
r = int(e[0])
for i in range(1, len(e), 2):
o = e[i]
n = int(e[i + 1])
if o == '+':
r += n
elif o == '*':
r *= n
elif o == '||':
r = int(f"{r}{n}")
return r
def pl(l, os):
d = [int(x) for x in re.findall(r'\d+', l)]
t, ns = d[0], d[1:]
ops = list(itertools.product(os, repeat=len(ns) - 1))
for o in ops:
e = str(ns[0])
for i, op in enumerate(o):
e += f" {op} {ns[i + 1]}"
r = ltr(e.split())
if r == t:
return r
return 0
def one():
lines = aoc.get_lines(7)
acc = 0
for l in lines:
acc += pl(l, ['+', '*'])
print(acc)
def two():
lines = aoc.get_lines(7)
acc = 0
for l in lines:
acc += pl(l, ['+', '*', '||'])
print(acc)
one()
two()
It's not a long lived project, it's a puzzle, and once solved never needs to run again. My objective here is to get the correct answer, not win a style contest.
Can you provide a link to your solution? I'd like to check it out.
I wrote my own iterator because I'm a big dummy. Also brute forced (~8s). Might be worth adding a cache to skip all the questions that have been computed / done.
Solution
import { AdventOfCodeSolutionFunction } from "./solutions";
function MakeCombination<T>(choices: Array<T>, state: Array<number>): Array<T> {
return state.map((v) => choices[v]);
}
function MakeStateArray(length: number) {
const newArray = [];
while (length-- > 0)
newArray.push(0);
return newArray;
}
function IncrementState(state: Array<number>, max: number): [state: Array<number>, overflow: boolean] {
state[0]++;
for (let index = 0; index < state.length; index++) {
if (state[index] == max) {
state[index] = 0;
if (index + 1 == state.length)
return [state, true];
state[index + 1]++;
}
}
return [state, false];
}
function GenerateCombinations<T>(choices: Array<T>, length: number): Array<Array<T>> {
const states = MakeStateArray(length);
const combinations: Array<Array<T>> = [];
let done = false
while (!done) {
combinations.push(MakeCombination(choices, states));
done = IncrementState(states, choices.length)[1];
}
return combinations;
}
enum Op {
MUL = "*",
ADD = "+",
CON = "|",
}
function ApplyOp(a: number, b: number, op: Op): number {
switch (op) {
case Op.MUL:
return a * b;
case Op.ADD:
return a + b;
case Op.CON:
return Number(`${a}${b}`);
}
}
function ApplyOperatorsToNumbers(numbers: Array<number>, ops: Array<Op>): number {
let prev = ApplyOp(numbers[0], numbers[1], ops[0]);
for (let index = 2; index < numbers.length; index++) {
prev = ApplyOp(prev, numbers[index], ops[index - 1])
}
return prev;
}
export const solution_7: AdventOfCodeSolutionFunction = (input) => {
const numbers = // [{target: 123, numbers: [1, 2, 3, ...]}, ...]
input.split("\n")
.map(
(v) => v.trim()
.split(":")
.map(v => v.trim().split(" ").map(v => Number(v)))
).map(
(v) => {
return { target: v[0][0], numbers: v[1] }
}
);
let part_1 = 0;
let part_2 = 0;
for (let index = 0; index < numbers.length; index++) {
const target = numbers[index].target;
const numbs = numbers[index].numbers;
// GenerateCombinations(["+", "*"], 2) => [["+", "+"], ["+", "*"], ["*", "+"], ["*", "*"]]
const combinations = GenerateCombinations([Op.ADD, Op.MUL], numbs.length - 1);
// part 1 calculations
for (let combinationIndex = 0; combinationIndex < combinations.length; combinationIndex++) {
const combination = combinations[combinationIndex];
const result = ApplyOperatorsToNumbers(numbs, combination);
if (result == target) {
part_1 += result;
break;
}
}
const combinations2 = GenerateCombinations([Op.ADD, Op.MUL, Op.CON], numbs.length - 1);
// part 2 calculations
for (let combinationIndex = 0; combinationIndex < combinations2.length; combinationIndex++) {
const combination = combinations2[combinationIndex];
const result = ApplyOperatorsToNumbers(numbs, combination);
if (result == target) {
part_2 += result;
break;
}
}
}
return {
part_1,
part_2,
}
}
Made a couple of attempts to munge the input data into some kind of binary search tree, lost some time to that, then threw my hands into the air and did a more naïve sort-of breadth-first search instead. Which turned out to be better for part 2 anyway.
Also, maths. Runs in just over a hundred milliseconds when using AsParallel, around half a second without.
C#
List<(long, int[])> data = new List<(long, int[])>();
public void Input(IEnumerable<string> lines)
{
foreach (var line in lines)
{
var parts = line.Split(':', StringSplitOptions.TrimEntries);
data.Add((long.Parse(parts.First()), parts.Last().Split(' ').Select(int.Parse).ToArray()));
}
}
public void Part1()
{
var correct = data.Where(kv => CalcPart(kv.Item1, kv.Item2)).Select(kv => kv.Item1).Sum();
Console.WriteLine($"Correct: {correct}");
}
public void Part2()
{
var correct = data.AsParallel().Where(kv => CalcPart2(kv.Item1, kv.Item2)).Select(kv => kv.Item1).Sum();
Console.WriteLine($"Correct: {correct}");
}
public bool CalcPart(long res, Span<int> num, long carried = 0)
{
var next = num[0];
if (num.Length == 1)
return res == carried + next || res == carried * next;
return CalcPart(res, num.Slice(1), carried + next) || CalcPart(res, num.Slice(1), carried * next);
}
public bool CalcPart2(long res, Span<int> num, long carried = 0)
{
var next = num[0];
// Get the 10 logarithm for the next number, expand the carried value by 10^<next 10log + 1>, add the two together
// For 123 || 45
// 45 ⇒ 10log(45) + 1 == 2
// 123 * 10^2 + 45 == 12345
long combined = carried * (long)Math.Pow(10, Math.Floor(Math.Log10(next) + 1)) + next;
if (num.Length == 1)
return res == carried + next || res == carried * next || res == combined;
return CalcPart2(res, num.Slice(1), carried + next) || CalcPart2(res, num.Slice(1), carried * next) || CalcPart2(res, num.Slice(1), combined);
}
I'm using the library Combinatorics in dotnet, which I've used in the past, generate in this case every duplicating possibility of the operations. I the only optimization that I did was to use a function to concatenate numbers without converting to strings, but that didn't actually help much.
I have parser helpers that use ReadOnlySpans over strings to prevent unnecessary allocations. However, here I'm adding to a C# mutable list and then converting to an FSharp (linked) list, which this language is more familiar with. Not optimal, but runtime was pretty good.
I'm not terribly good with F#, but I think I did ok for this challenge.
F#
// in another file:
let concatenateLong (a:Int64) (b:Int64) : Int64 =
let rec countDigits (n:int64) =
if n = 0 then 0
else 1 + countDigits (n / (int64 10))
let bDigits = if b = 0 then 1 else countDigits b
let multiplier = pown 10 bDigits |> int64
a * multiplier + b
// challenge file
type Operation = {Total:Int64; Inputs:Int64 list }
let parse (s:ReadOnlySpan<char>) : Operation =
let sep = s.IndexOf(':')
let total = Int64.Parse(s.Slice(0, sep))
let inputs = System.Collections.Generic.List<Int64>()
let right:ReadOnlySpan<char> = s.Slice(sep + 1).Trim()
// because the Split function on a span returns a SpanSplitEnumerator, which is a ref-struct and can only live on the stack,
// I can't use the F# list syntax here
for range in right.Split(" ") do
inputs.Add(Int64.Parse(sliceRange right range))
{Total = total; Inputs = List.ofSeq(inputs) }
let part1Ops = [(+); (*)]
let execute ops input =
input
|> PSeq.choose (fun op ->
let total = op.Total
let inputs = op.Inputs
let variations = Variations(ops, inputs.Length - 1, GenerateOption.WithRepetition)
variations
|> Seq.tryFind (fun v ->
let calcTotal = (inputs[0], inputs[1..], List.ofSeq(v)) |||> List.fold2 (fun acc n f -> f acc n)
calcTotal = total
)
|> Option.map(fun _ -> total)
)
|> PSeq.fold (fun acc n -> acc + n) 0L
let part1 input =
(read input parse)
|> execute part1Ops
let part2Ops = [(+); (*); concatenateLong]
let part2 input = (read input parse) |> execute part2Ops
The Gen0 garbage collection looks absurd, but Gen0 is generally considered "free".
Method
Mean
Error
StdDev
Gen0
Gen1
Allocated
Part1
19.20 ms
0.372 ms
0.545 ms
17843.7500
156.2500
106.55 MB
Part2
17.94 ms
0.355 ms
0.878 ms
17843.7500
156.2500
106.55 MB
V2 - concatenate numbers did little for the runtime, but did help with Gen1 garbage,
but not the overall allocation.
Took quite some time to debug but in the end I think it's a nice solution using base 2 and 3 numbers counting up to check all operator combinations.
Code
function readInput(inputFile::String)::Vector{Vector{Int}}
f = open(inputFile,"r")
lines::Vector{String} = readlines(f)
close(f)
equations::Vector{Vector{Int}} = []
function getValues(line::String)
return map(sp->parse(Int,sp),(sp=split(line," ");sp[1]=sp[1][1:end-1];sp))
end
map(l->push!(equations,getValues(l)),lines)
return equations
end
function checkEq(eq::Vector{Int},withConCat::Bool)::Bool
function calcEq(eq::Vector{Int},operators::Vector{Int},withConCat::Bool)::Int
res::Int = eq[2]
for (i,op) in enumerate(operators)
if op == 0 #+
res += eq[i+2]
elseif op ==1 #*
res *= eq[i+2]
else #op==2 ||
res = parse(Int,string(res)*string(eq[i+2]))
end
end
return res
end
opInt::Int = 0
operators = Vector{Int}(undef,length(eq)-2)
while opInt < (withConCat ? 3^(length(eq)-2) : 2^(length(eq)-2))
withConCat==true ? operators=digits(opInt,base=3,pad=length(eq)-2) : operators=digits(opInt,base=2,pad=length(eq)-2)
#calcEq(eq,operators,withConCat)==eq[1] ? (return true) : opInt -= 1
calcEq(eq,operators,withConCat)==eq[1] ? (return true) : opInt += 1
end
return false
end
function calcTotCalRes(equations::Vector{Vector{Int}},withConCat::Bool)::Int
totCalRes::Int = 0
for e in equations
checkEq(e,withConCat) ? totCalRes+=e[1] : nothing
end
return totCalRes
end
@info "Part 1"
println("result: $(calcTotCalRes(readInput("day07Input"),false))")
@info "Part 2"
println("result: $(calcTotCalRes(readInput("day07Input"),true))")
I finally got around to doing day 7. I try the brute force method (takes several seconds), but I'm particularly proud of my sequence generator for operation permutations.
The Collection#rotate method is in the file Utils.kt, which can be found in my repo.
Solution
import kotlin.collections.any
import kotlin.math.pow
fun main() {
fun part1(input: List<String>): Long {
val operations = setOf(CalibrationOperation.Plus, CalibrationOperation.Multiply)
return generalizedSolution(input, operations)
}
fun part2(input: List<String>): Long {
val operations = setOf(CalibrationOperation.Plus, CalibrationOperation.Multiply, CalibrationOperation.Concat)
return generalizedSolution(input, operations)
}
val testInput = readInput("Day07_test")
check(part1(testInput) == 3749L)
check(part2(testInput) == 11387L)
val input = readInput("Day07")
part1(input).println()
part2(input).println()
}
fun parseInputDay7(input: List<String>) = input.map {
val calibrationResultAndInput = it.split(':')
calibrationResultAndInput[0].toLong() to calibrationResultAndInput[1].split(' ').filter { it != "" }.map { it.toLong() }
}
fun generalizedSolution(input: List<String>, operations: Set<CalibrationOperation>): Long {
val parsedInput = parseInputDay7(input)
val operationsPermutations = CalibrationOperation.operationPermutationSequence(*operations.toTypedArray()).take(calculatePermutationsNeeded(parsedInput, operations)).toList()
return sumOfPossibleCalibrationEquations(parsedInput, operationsPermutations)
}
fun calculatePermutationsNeeded(parsedInput: List<Pair<Long, List<Long>>>, operations: Set<CalibrationOperation>): Int {
val highestNumberOfOperations = parsedInput.maxOf { it.second.size - 1 }
return (1..highestNumberOfOperations).sumOf { operations.size.toDouble().pow(it).toInt() }
}
fun sumOfPossibleCalibrationEquations(parsedInput: List<Pair<Long, List<Long>>>, operationPermutationCollection: Collection<OperationPermutation>): Long {
val permutationsGrouped = operationPermutationCollection.groupBy { it.size }
return parsedInput.sumOf { (equationResult, equationInput) ->
if (permutationsGrouped[equationInput.size - 1]!!.any { operations ->
equationResult == equationInput.drop(1)
.foldIndexed(equationInput[0]) { index, acc, lng -> operations[index](acc, lng) }
}) equationResult else 0
}
}
typealias OperationPermutation = List<CalibrationOperation>
sealed class CalibrationOperation(val operation: (Long, Long) -> Long) {
operator fun invoke(a: Long, b: Long) = operation(a, b)
object Plus : CalibrationOperation({ a: Long, b: Long -> a + b })
object Multiply : CalibrationOperation({ a: Long, b: Long -> a * b })
object Concat : CalibrationOperation({ a: Long, b: Long -> "$a$b".toLong() })
companion object {
fun operationPermutationSequence(vararg operations: CalibrationOperation) = sequence<OperationPermutation> {
val cache = mutableListOf<OperationPermutation>()
val calculateCacheRange = { currentLength: Int ->
val sectionSize = operations.size.toDouble().pow(currentLength - 1).toInt()
val sectionStart = (1 until currentLength - 1).sumOf { operations.size.toDouble().pow(it).toInt() }
sectionStart..(sectionStart + sectionSize - 1)
}
// Populate the cache with initial values for permutation length 1.
operations.forEach { operation -> yield(listOf(operation).also { cache.add(it) }) }
var currentLength = 2
var offset = 0
var cacheRange = calculateCacheRange(currentLength)
var rotatingOperations = operations.toList()
yieldAll(
generateSequence {
if (cacheRange.count() == offset) {
rotatingOperations = rotatingOperations.rotated(1)
if (rotatingOperations.first() == operations.first()) {
currentLength++
}
offset = 0
cacheRange = calculateCacheRange(currentLength)
}
val cacheSlice = cache.slice(cacheRange)
return@generateSequence (cacheSlice[offset] + rotatingOperations.first()).also {
cache += it
offset++
}
}
)
}
}
}