this post was submitted on 18 Dec 2024
17 points (94.7% liked)

Advent Of Code

1011 readers
1 users here now

An unofficial home for the advent of code community on programming.dev!

Advent of Code is an annual Advent calendar of small programming puzzles for a variety of skill sets and skill levels that can be solved in any programming language you like.

AoC 2024

Solution Threads

M T W T F S S
1
2 3 4 5 6 7 8
9 10 11 12 13 14 15
16 17 18 19 20 21 22
23 24 25

Rules/Guidelines

Relevant Communities

Relevant Links

Credits

Icon base by Lorc under CC BY 3.0 with modifications to add a gradient

console.log('Hello World')

founded 2 years ago
MODERATORS
[email protected]
[email protected]
[email protected]
17
Is it possible to not brute force day 7? (lemmy.ml)
submitted 2 months ago by [email protected] to c/[email protected]
11 comments fedilink hide all child comments
 

I thought about it for 15 mins, but couldn't think of any mathematical tricks. I thought of lots of minor tricks, like comparing the number to the result and not adding any more multiplications if it's over, things that would cut 10%-20% here and there, but nothing which fundamentally changes big O running time.

For reference, here's my solution for part 2 in smalltalk. I just generated every possible permutation and tested it. Part 1 is similar, mainly I just used bit magic to avoid generating permutations.

(even if you haven't used it, smalltalk is fairly readable, everything is left to right, except in parens)

day7p2: in
	| input | 
	
	input := in lines collect: [ :l | (l splitOn: '\:|\s' asRegex) reject: #isEmpty thenCollect: #asInteger ].
	
	^ (input select: [ :line |
		(#(1 2 3) permutationsWithRepetitionsOfSize: line size - 2) 
			anySatisfy: [ :num | (self d7addmulcat: line ops: num) = (line at: 1)]
	]) sum: #first.

d7addmulcat: nums ops: ops
	| final |
	
	final := nums at: 2.
	ops withIndexDo: [ :op :i |
		op = 1 ifTrue: [ final := final * (nums at: i + 2) ].
		op = 2 ifTrue: [ final := final + (nums at: i + 2) ].
		op = 3 ifTrue: [ final := (final asString, (nums at: i+2) asString) asInteger ]
	].

	^ final
you are viewing a single comment's thread
view the rest of the comments
[โ€“] [email protected] 3 points 2 months ago (5 children)

Probably the easiest optimization (which admittedly I didn't think of myself) is to work backwards: you can eliminate multiplication and concatenation early if you start with the answer and check terms from the right.

[โ€“] Acters 3 points 1 week ago

Call it optimization and people will assume it is magic. Instead I call this a simple algebra challenge.(With part two having that quirky concatenation operation)

It is like solving for x. Let's take an example:

3267: 81 40 27

If you change it to equations:

81 (+ or *) 40 (+ or *) 27 = 3267
Which effectively is:
81 (+ or *) 40 = x (+ or *) 27 = 3267

So what is the x that either add or multiply would need to create 3267? Or solve for x for only two equations.
x + 27 = 3267 -> x = 3267 - 27 = 3240
x * 27 = 3267 -> x = 3267 / 27 = 121

(Special note since multiplying and adding will never make a float then a division that will have a remainder(or float) is impossible for x, we can easily remove multiplying as a possible choice)

Now with our example we go plug in x:
81 (+ or *) 40 = (3240 or 121) (+ or *) 27 = 3267

So now we see if either 40 or 81 can divide or subtract from x to get the other number. Since it is easier to just program to use the next number in the list, we will use 40.

81 + 40 = 3240 -> x = 3240 - 40 = 3200 != 81
81 * 40 = 3240 -> x = 3240 / 40 = 81
81 + 40 = 121 -> x = 121 - 40 = 81
81 * 40 = 121 -> x = 121 / 40 = 3.025 != 81

This particular example from the website has two solutions.

For the concatenation operation, you simply check if the number ends with the number in the list. If not then a concatenation cannot occur, and if true remove the number from the end and continue in the list.

Call it pruning tree paths but it is simply algebraic.

load more comments (4 replies)