82 lines
3.2 KiB
Markdown
82 lines
3.2 KiB
Markdown
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## \-\-- Day 3: Mull It Over \-\--
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\"Our computers are having issues, so I have no idea if we have any
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Chief Historians [in
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stock]{title="There's a spot reserved for Chief Historians between the green toboggans and the red toboggans. They've never actually had any Chief Historians in stock, but it's best to be prepared."}!
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You\'re welcome to check the warehouse, though,\" says the mildly
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flustered shopkeeper at the [North Pole Toboggan Rental
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Shop](/2020/day/2). The Historians head out to take a look.
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The shopkeeper turns to you. \"Any chance you can see why our computers
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are having issues again?\"
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The computer appears to be trying to run a program, but its memory (your
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puzzle input) is *corrupted*. All of the instructions have been jumbled
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up!
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It seems like the goal of the program is just to *multiply some
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numbers*. It does that with instructions like `mul(X,Y)`, where `X` and
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`Y` are each 1-3 digit numbers. For instance, `mul(44,46)` multiplies
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`44` by `46` to get a result of `2024`. Similarly, `mul(123,4)` would
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multiply `123` by `4`.
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However, because the program\'s memory has been corrupted, there are
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also many invalid characters that should be *ignored*, even if they look
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like part of a `mul` instruction. Sequences like `mul(4*`, `mul(6,9!`,
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`?(12,34)`, or `mul ( 2 , 4 )` do *nothing*.
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For example, consider the following section of corrupted memory:
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xmul(2,4)%&mul[3,7]!@^do_not_mul(5,5)+mul(32,64]then(mul(11,8)mul(8,5))
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Only the four highlighted sections are real `mul` instructions. Adding
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up the result of each instruction produces `161`
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(`2*4 + 5*5 + 11*8 + 8*5`).
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Scan the corrupted memory for uncorrupted `mul` instructions. *What do
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you get if you add up all of the results of the multiplications?*
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Your puzzle answer was `174561379`.
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## \-\-- Part Two \-\-- {#part2}
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As you scan through the corrupted memory, you notice that some of the
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conditional statements are also still intact. If you handle some of the
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uncorrupted conditional statements in the program, you might be able to
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get an even more accurate result.
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There are two new instructions you\'ll need to handle:
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- The `do()` instruction *enables* future `mul` instructions.
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- The `don't()` instruction *disables* future `mul` instructions.
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Only the *most recent* `do()` or `don't()` instruction applies. At the
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beginning of the program, `mul` instructions are *enabled*.
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For example:
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xmul(2,4)&mul[3,7]!^don't()_mul(5,5)+mul(32,64](mul(11,8)undo()?mul(8,5))
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This corrupted memory is similar to the example from before, but this
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time the `mul(5,5)` and `mul(11,8)` instructions are *disabled* because
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there is a `don't()` instruction before them. The other `mul`
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instructions function normally, including the one at the end that gets
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re-*enabled* by a `do()` instruction.
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This time, the sum of the results is `48` (`2*4 + 8*5`).
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Handle the new instructions; *what do you get if you add up all of the
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results of just the enabled multiplications?*
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Your puzzle answer was `106921067`.
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Both parts of this puzzle are complete! They provide two gold stars:
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\*\*
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At this point, you should [return to your Advent calendar](/2024) and
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try another puzzle.
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If you still want to see it, you can [get your puzzle
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input](3/input).
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