2024-11-24 19:21:15 +01:00
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## \-\-- Day 13: Packet Scanners \-\--
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You need to cross a vast *firewall*. The firewall consists of several
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layers, each with a *security scanner* that moves back and forth across
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the layer. To succeed, you must not be detected by a scanner.
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By studying the firewall briefly, you are able to record (in your puzzle
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input) the *depth* of each layer and the *range* of the scanning area
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for the scanner within it, written as `depth: range`. Each layer has a
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thickness of exactly `1`. A layer at depth `0` begins immediately inside
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the firewall; a layer at depth `1` would start immediately after that.
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For example, suppose you\'ve recorded the following:
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0: 3
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1: 2
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4: 4
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6: 4
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This means that there is a layer immediately inside the firewall (with
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range `3`), a second layer immediately after that (with range `2`), a
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third layer which begins at depth `4` (with range `4`), and a fourth
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layer which begins at depth 6 (also with range `4`). Visually, it might
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look like this:
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0 1 2 3 4 5 6
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[ ] [ ] ... ... [ ] ... [ ]
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[ ] [ ] [ ] [ ]
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[ ] [ ] [ ]
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[ ] [ ]
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Within each layer, a security scanner moves back and forth within its
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range. Each security scanner starts at the top and moves down until it
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reaches the bottom, then moves up until it reaches the top, and repeats.
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A security scanner takes *one picosecond* to move one step. Drawing
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scanners as `S`, the first few picoseconds look like this:
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Picosecond 0:
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0 1 2 3 4 5 6
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[S] [S] ... ... [S] ... [S]
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[ ] [ ] [ ] [ ]
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[ ] [ ] [ ]
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[ ] [ ]
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Picosecond 1:
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0 1 2 3 4 5 6
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[ ] [ ] ... ... [ ] ... [ ]
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[S] [S] [S] [S]
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[ ] [ ] [ ]
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[ ] [ ]
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Picosecond 2:
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0 1 2 3 4 5 6
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[ ] [S] ... ... [ ] ... [ ]
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[ ] [ ] [ ] [ ]
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[S] [S] [S]
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[ ] [ ]
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Picosecond 3:
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0 1 2 3 4 5 6
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[ ] [ ] ... ... [ ] ... [ ]
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[S] [S] [ ] [ ]
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[ ] [ ] [ ]
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[S] [S]
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Your plan is to hitch a ride on a packet about to move through the
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firewall. The packet will travel along the top of each layer, and it
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moves at *one layer per picosecond*. Each picosecond, the packet moves
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one layer forward (its first move takes it into layer 0), and then the
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scanners move one step. If there is a scanner at the top of the layer
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*as your packet enters it*, you are *caught*. (If a scanner moves into
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the top of its layer while you are there, you are *not* caught: it
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doesn\'t have time to notice you before you leave.) If you were to do
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this in the configuration above, marking your current position with
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parentheses, your passage through the firewall would look like this:
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Initial state:
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0 1 2 3 4 5 6
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[S] [S] ... ... [S] ... [S]
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[ ] [ ] [ ] [ ]
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[ ] [ ] [ ]
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[ ] [ ]
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Picosecond 0:
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0 1 2 3 4 5 6
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(S) [S] ... ... [S] ... [S]
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[ ] [ ] [ ] [ ]
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[ ] [ ] [ ]
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[ ] [ ]
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0 1 2 3 4 5 6
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( ) [ ] ... ... [ ] ... [ ]
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[S] [S] [S] [S]
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[ ] [ ] [ ]
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[ ] [ ]
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Picosecond 1:
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0 1 2 3 4 5 6
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[ ] ( ) ... ... [ ] ... [ ]
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[S] [S] [S] [S]
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[ ] [ ] [ ]
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[ ] [ ]
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0 1 2 3 4 5 6
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[ ] (S) ... ... [ ] ... [ ]
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[ ] [ ] [ ] [ ]
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[S] [S] [S]
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[ ] [ ]
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Picosecond 2:
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0 1 2 3 4 5 6
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[ ] [S] (.) ... [ ] ... [ ]
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[ ] [ ] [ ] [ ]
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[S] [S] [S]
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[ ] [ ]
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0 1 2 3 4 5 6
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[ ] [ ] (.) ... [ ] ... [ ]
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[S] [S] [ ] [ ]
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[ ] [ ] [ ]
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[S] [S]
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Picosecond 3:
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0 1 2 3 4 5 6
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[ ] [ ] ... (.) [ ] ... [ ]
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[S] [S] [ ] [ ]
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[ ] [ ] [ ]
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[S] [S]
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0 1 2 3 4 5 6
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[S] [S] ... (.) [ ] ... [ ]
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[ ] [ ] [ ] [ ]
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[ ] [S] [S]
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[ ] [ ]
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Picosecond 4:
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0 1 2 3 4 5 6
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[S] [S] ... ... ( ) ... [ ]
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[ ] [ ] [ ] [ ]
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[ ] [S] [S]
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[ ] [ ]
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0 1 2 3 4 5 6
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[ ] [ ] ... ... ( ) ... [ ]
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[S] [S] [S] [S]
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[ ] [ ] [ ]
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[ ] [ ]
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Picosecond 5:
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0 1 2 3 4 5 6
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[ ] [ ] ... ... [ ] (.) [ ]
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[S] [S] [S] [S]
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[ ] [ ] [ ]
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[ ] [ ]
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0 1 2 3 4 5 6
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[ ] [S] ... ... [S] (.) [S]
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[ ] [ ] [ ] [ ]
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[S] [ ] [ ]
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[ ] [ ]
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Picosecond 6:
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0 1 2 3 4 5 6
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[ ] [S] ... ... [S] ... (S)
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[ ] [ ] [ ] [ ]
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[S] [ ] [ ]
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[ ] [ ]
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0 1 2 3 4 5 6
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[ ] [ ] ... ... [ ] ... ( )
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[S] [S] [S] [S]
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[ ] [ ] [ ]
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[ ] [ ]
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In this situation, you are *caught* in layers `0` and `6`, because your
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packet entered the layer when its scanner was at the top when you
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entered it. You are *not* caught in layer `1`, since the scanner moved
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into the top of the layer once you were already there.
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The *severity* of getting caught on a layer is equal to its *depth*
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multiplied by its *range*. (Ignore layers in which you do not get
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caught.) The severity of the whole trip is the sum of these values. In
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the example above, the trip severity is `0*3 + 6*4 = 24`.
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Given the details of the firewall you\'ve recorded, if you leave
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immediately, *what is the severity of your whole trip*?
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2024-11-25 12:27:11 +01:00
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Your puzzle answer was `1844`.
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## \-\-- Part Two \-\-- {#part2}
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Now, you need to pass through the firewall without being caught - easier
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said than done.
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You can\'t control the [speed of the
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packet]{title="Seriously, what network stack doesn't let you adjust the speed of light?"},
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but you can *delay* it any number of picoseconds. For each picosecond
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you delay the packet before beginning your trip, all security scanners
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move one step. You\'re not in the firewall during this time; you don\'t
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enter layer `0` until you stop delaying the packet.
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In the example above, if you delay `10` picoseconds (picoseconds `0` -
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`9`), you won\'t get caught:
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State after delaying:
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0 1 2 3 4 5 6
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[ ] [S] ... ... [ ] ... [ ]
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[ ] [ ] [ ] [ ]
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[S] [S] [S]
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[ ] [ ]
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Picosecond 10:
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0 1 2 3 4 5 6
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( ) [S] ... ... [ ] ... [ ]
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[ ] [ ] [ ] [ ]
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[S] [S] [S]
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[ ] [ ]
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0 1 2 3 4 5 6
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( ) [ ] ... ... [ ] ... [ ]
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[S] [S] [S] [S]
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[ ] [ ] [ ]
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[ ] [ ]
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Picosecond 11:
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0 1 2 3 4 5 6
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[ ] ( ) ... ... [ ] ... [ ]
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[S] [S] [S] [S]
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[ ] [ ] [ ]
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[ ] [ ]
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0 1 2 3 4 5 6
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[S] (S) ... ... [S] ... [S]
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[ ] [ ] [ ] [ ]
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[ ] [ ] [ ]
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[ ] [ ]
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Picosecond 12:
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0 1 2 3 4 5 6
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[S] [S] (.) ... [S] ... [S]
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[ ] [ ] [ ] [ ]
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[ ] [ ] [ ]
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[ ] [ ]
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0 1 2 3 4 5 6
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[ ] [ ] (.) ... [ ] ... [ ]
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[S] [S] [S] [S]
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[ ] [ ] [ ]
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[ ] [ ]
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Picosecond 13:
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0 1 2 3 4 5 6
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[ ] [ ] ... (.) [ ] ... [ ]
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[S] [S] [S] [S]
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[ ] [ ] [ ]
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[ ] [ ]
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0 1 2 3 4 5 6
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[ ] [S] ... (.) [ ] ... [ ]
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[ ] [ ] [ ] [ ]
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[S] [S] [S]
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[ ] [ ]
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Picosecond 14:
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0 1 2 3 4 5 6
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[ ] [S] ... ... ( ) ... [ ]
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[ ] [ ] [ ] [ ]
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[S] [S] [S]
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[ ] [ ]
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0 1 2 3 4 5 6
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[ ] [ ] ... ... ( ) ... [ ]
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[S] [S] [ ] [ ]
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[ ] [ ] [ ]
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[S] [S]
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Picosecond 15:
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0 1 2 3 4 5 6
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[ ] [ ] ... ... [ ] (.) [ ]
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[S] [S] [ ] [ ]
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[ ] [ ] [ ]
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[S] [S]
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0 1 2 3 4 5 6
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[S] [S] ... ... [ ] (.) [ ]
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[ ] [ ] [ ] [ ]
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[ ] [S] [S]
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[ ] [ ]
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Picosecond 16:
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0 1 2 3 4 5 6
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[S] [S] ... ... [ ] ... ( )
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[ ] [ ] [ ] [ ]
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[ ] [S] [S]
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[ ] [ ]
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0 1 2 3 4 5 6
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[ ] [ ] ... ... [ ] ... ( )
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[S] [S] [S] [S]
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[ ] [ ] [ ]
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[ ] [ ]
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Because all smaller delays would get you caught, the fewest number of
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picoseconds you would need to delay to get through safely is `10`.
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*What is the fewest number of picoseconds* that you need to delay the
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packet to pass through the firewall without being caught?
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2024-11-24 19:21:15 +01:00
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2024-11-25 14:57:38 +01:00
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Your puzzle answer was `3897604`.
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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](/2017) and
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try another puzzle.
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2024-11-25 12:27:11 +01:00
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2024-11-25 14:57:38 +01:00
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If you still want to see it, you can [get your puzzle
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2024-11-25 12:27:11 +01:00
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input](13/input).
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