The problem defines 3 specific operations for shuffling a deck with parameters. For part 1, it asks us to shuffle a 10,000 card deck once. For part 2, we're asked to shuffle a one trillion card deck a trillion times.
This was the hardest problem this year, and I needed outside help for it. This
was the only problem this year I couldn't solve on my own. Part 1 was solved
easily enough with a deque simulation, but Part 2 would require multiple
terabytes of memory to simulate, and let's not forget about the CPU runtime
required. Clearly a different approach is needed!
I browsed the solutions thread on reddit to see the approaches others took,
then I knew which concepts I needed to research and apply. The key insight
here is that every shuffle operation can be expressed as a linear
transformation of the form y = mx + b, or card = m * index + b operating
in a modulo space. The deck starts with m == 1 and b == 0, such that
card = index, or the 2nd slot holds the card labeled 2.
Each shuffle operation is a linear transformation which modifies the m and
b variables. Playing around with these variables lets you find them; but
for one you will need to use inverse modulo, which might be new to you (it
was new to me).
Since linear transformations can be composed, we can take
a list of 100 shuffle operations and compress it to one y = mx + b equation
that describes the deck. Since we're operating in a modulo space, it's always
safe to set m = m % deck_size or b = b % deck_size if the numbers get too
large.
Finally, for applying the shuffle a trillion times, we can compose the equation with itself to build up one equation that describes shuffling the deck a trillion times. Doubling is especially useful here and gets us to a O(ln x) of building up this equation. I used matrix multiplication here, as that made the most sense to me intuitively. ( 3blue1brown's videos on linear algebra helped me build up this intuition.
Overall, this was a difficult problem, found contentious by the community. Did it even belong in Advent of Code? Where are the data structures? I can see both sides of the argument; it was a bit frustrating in the moment, but I came out on the other side having learned something. It's also amazing just how quickly my final program calculates the result of shuffling a trillion card deck a trillion times; granted it's only looking in one position.
2019 Day 22 on AdventOfCode.com
There isn't much to do while you wait for the droids to repair your ship. At least you're drifting in the right direction. You decide to practice a new card shuffle you've been working on.
Digging through the ship's storage, you find a deck of space cards! Just like any deck of space cards, there are 10007 cards in the deck numbered 0 through 10006. The deck must be new - they're still in factory order, with 0 on the top, then 1, then 2, and so on, all the way through to 10006 on the bottom.
You've been practicing three different techniques that you use while shuffling. Suppose you have a deck of only 10 cards (numbered 0 through 9):
To deal into new stack, create a new stack of cards by dealing the top card of the deck onto the top of the new stack repeatedly until you run out of cards:
Top Bottom
0 1 2 3 4 5 6 7 8 9 Your deck
New stack
1 2 3 4 5 6 7 8 9 Your deck
0 New stack
2 3 4 5 6 7 8 9 Your deck
1 0 New stack
3 4 5 6 7 8 9 Your deck
2 1 0 New stack
Several steps later...
9 Your deck
8 7 6 5 4 3 2 1 0 New stack
Your deck
9 8 7 6 5 4 3 2 1 0 New stack
Finally, pick up the new stack you've just created and use it as the deck for the next technique.
To cut N cards, take the top N cards off the top of the deck and move them as a single unit to the bottom of the deck, retaining their order. For example, to cut 3:
Top Bottom
0 1 2 3 4 5 6 7 8 9 Your deck
3 4 5 6 7 8 9 Your deck
0 1 2 Cut cards
3 4 5 6 7 8 9 Your deck
0 1 2 Cut cards
3 4 5 6 7 8 9 0 1 2 Your deck
You've also been getting pretty good at a version of this technique where N is negative! In that case, cut (the absolute value of) N cards from the bottom of the deck onto the top. For example, to cut -4:
Top Bottom
0 1 2 3 4 5 6 7 8 9 Your deck
0 1 2 3 4 5 Your deck
6 7 8 9 Cut cards
0 1 2 3 4 5 Your deck
6 7 8 9 Cut cards
6 7 8 9 0 1 2 3 4 5 Your deck
To deal with increment N, start by clearing enough space on your table to lay out all of the cards individually in a long line. Deal the top card into the leftmost position. Then, move N positions to the right and deal the next card there. If you would move into a position past the end of the space on your table, wrap around and keep counting from the leftmost card again. Continue this process until you run out of cards.
For example, to deal with increment 3:
0 1 2 3 4 5 6 7 8 9 Your deck
. . . . . . . . . . Space on table
^ Current position
Deal the top card to the current position:
1 2 3 4 5 6 7 8 9 Your deck
0 . . . . . . . . . Space on table
^ Current position
Move the current position right 3:
1 2 3 4 5 6 7 8 9 Your deck
0 . . . . . . . . . Space on table
^ Current position
Deal the top card:
2 3 4 5 6 7 8 9 Your deck
0 . . 1 . . . . . . Space on table
^ Current position
Move right 3 and deal:
3 4 5 6 7 8 9 Your deck
0 . . 1 . . 2 . . . Space on table
^ Current position
Move right 3 and deal:
4 5 6 7 8 9 Your deck
0 . . 1 . . 2 . . 3 Space on table
^ Current position
Move right 3, wrapping around, and deal:
5 6 7 8 9 Your deck
0 . 4 1 . . 2 . . 3 Space on table
^ Current position
And so on:
0 7 4 1 8 5 2 9 6 3 Space on table
Positions on the table which already contain cards are still counted; they're not skipped. Of course, this technique is carefully designed so it will never put two cards in the same position or leave a position empty.
Finally, collect the cards on the table so that the leftmost card ends up at the top of your deck, the card to its right ends up just below the top card, and so on, until the rightmost card ends up at the bottom of the deck.
The complete shuffle process (your puzzle input) consists of applying many of these techniques. Here are some examples that combine techniques; they all start with a factory order deck of 10 cards:
deal with increment 7
deal into new stack
deal into new stack
Result: 0 3 6 9 2 5 8 1 4 7
cut 6
deal with increment 7
deal into new stack
Result: 3 0 7 4 1 8 5 2 9 6
deal with increment 7
deal with increment 9
cut -2
Result: 6 3 0 7 4 1 8 5 2 9
deal into new stack
cut -2
deal with increment 7
cut 8
cut -4
deal with increment 7
cut 3
deal with increment 9
deal with increment 3
cut -1
Result: 9 2 5 8 1 4 7 0 3 6
Positions within the deck count from 0 at the top, then 1 for the card immediately below the top card, and so on to the bottom. (That is, cards start in the position matching their number.)
After shuffling your factory order deck of 10007 cards, what is the position of card 2019?
After a while, you realize your shuffling skill won't improve much more with merely a single deck of cards. You ask every 3D printer on the ship to make you some more cards while you check on the ship repairs. While reviewing the work the droids have finished so far, you think you see Halley's Comet fly past!
When you get back, you discover that the 3D printers have combined their power to create for you a single, giant, brand new, factory order deck of 119315717514047 space cards.
Finally, a deck of cards worthy of shuffling!
You decide to apply your complete shuffle process (your puzzle input) to the deck 101741582076661 times in a row.
You'll need to be careful, though - one wrong move with this many cards and you might overflow your entire ship!
After shuffling your new, giant, factory order deck that many times, what number is on the card that ends up in position 2020?