In this challenge I got to learn a few things. First and foremost, Golang does not have a built-in "set" datatype like Python does, which makes things interesting. However, a set type can be implemented with a little know-how. Specifically through the use of maps.
Generally, the idea is you create a map of a datatype that you want to compare against and that map essentially becomes the set. So:
type Point struct {
X, Y int
}
mySet := make(map[Point]bool)Now this essentially creates an un-ordered set with unique values. So you would add to it like so:
mySet[Point{50, 100}] = true
mySet[Point{25, 50}] = true
// etc...This way if you add another point to the set, you are simply overriding the point.
Setting a datatype of bool is a length of 1 bit. Why use 1 bit when you can use...no bits? Dr. Evil pinky thing
var exists = struct{}{} // sets a composite literal struct object which is 0 bytes in length
myStructSet := make(map[Point]struct{})
// add to the set
myStructSet[Point{50, 100}] = exists
myStructSet[Point{25, 50}] = existsYou can use the same technique from the day 14 challenge with interfaces to determine if an object exists:
if _, ok := myStructSet[Point{50,100}]; ok {
fmt.Println("This exists!")
}Because the second value printed from setting an object to the struct is a boolean value that states it was able to retrieve the data. This technique was learned from David Kaya!
I used this concept a bit in practice, but when I got to part 2 of this challenge my thoughts went towards trying to check every single possible point on the provided map and checking if it was the distress signal. This was the wrong way to go about doing it of course, and with the help of 0xdf's youtube video I learned a better way of handling it, which I did myself.
...hey, I'm not here to figure out the algorithm by myself, I'm here to force myself to learn Go.
You feel the ground rumble again as the distress signal leads you to a large network of subterranean tunnels. You don't have time to search them all, but you don't need to: your pack contains a set of deployable sensors that you imagine were originally built to locate lost Elves.
The sensors aren't very powerful, but that's okay; your handheld device indicates that you're close enough to the source of the distress signal to use them. You pull the emergency sensor system out of your pack, hit the big button on top, and the sensors zoom off down the tunnels.
Once a sensor finds a spot it thinks will give it a good reading, it attaches itself to a hard surface and begins monitoring for the nearest signal source beacon. Sensors and beacons always exist at integer coordinates. Each sensor knows its own position and can determine the position of a beacon precisely; however, sensors can only lock on to the one beacon closest to the sensor as measured by the Manhattan distance. (There is never a tie where two beacons are the same distance to a sensor.)
It doesn't take long for the sensors to report back their positions and closest beacons (your puzzle input). For example:
Sensor at x=2, y=18: closest beacon is at x=-2, y=15
Sensor at x=9, y=16: closest beacon is at x=10, y=16
Sensor at x=13, y=2: closest beacon is at x=15, y=3
Sensor at x=12, y=14: closest beacon is at x=10, y=16
Sensor at x=10, y=20: closest beacon is at x=10, y=16
Sensor at x=14, y=17: closest beacon is at x=10, y=16
Sensor at x=8, y=7: closest beacon is at x=2, y=10
Sensor at x=2, y=0: closest beacon is at x=2, y=10
Sensor at x=0, y=11: closest beacon is at x=2, y=10
Sensor at x=20, y=14: closest beacon is at x=25, y=17
Sensor at x=17, y=20: closest beacon is at x=21, y=22
Sensor at x=16, y=7: closest beacon is at x=15, y=3
Sensor at x=14, y=3: closest beacon is at x=15, y=3
Sensor at x=20, y=1: closest beacon is at x=15, y=3
So, consider the sensor at 2,18; the closest beacon to it is at -2,15. For the sensor at 9,16, the closest beacon to it is at 10,16.
Drawing sensors as S and beacons as B, the above arrangement of sensors and beacons looks like this:
1 1 2 2
0 5 0 5 0 5
0 ....S.......................
1 ......................S.....
2 ...............S............
3 ................SB..........
4 ............................
5 ............................
6 ............................
7 ..........S.......S.........
8 ............................
9 ............................
10 ....B.......................
11 ..S.........................
12 ............................
13 ............................
14 ..............S.......S.....
15 B...........................
16 ...........SB...............
17 ................S..........B
18 ....S.......................
19 ............................
20 ............S......S........
21 ............................
22 .......................B....
This isn't necessarily a comprehensive map of all beacons in the area, though. Because each sensor only identifies its closest beacon, if a sensor detects a beacon, you know there are no other beacons that close or closer to that sensor. There could still be beacons that just happen to not be the closest beacon to any sensor. Consider the sensor at 8,7:
1 1 2 2
0 5 0 5 0 5
-2 ..........#.................
-1 .........###................
0 ....S...#####...............
1 .......#######........S.....
2 ......#########S............
3 .....###########SB..........
4 ....#############...........
5 ...###############..........
6 ..#################.........
7 .#########S#######S#........
8 ..#################.........
9 ...###############..........
10 ....B############...........
11 ..S..###########............
12 ......#########.............
13 .......#######..............
14 ........#####.S.......S.....
15 B........###................
16 ..........#SB...............
17 ................S..........B
18 ....S.......................
19 ............................
20 ............S......S........
21 ............................
22 .......................B....
This sensor's closest beacon is at 2,10, and so you know there are no beacons that close or closer (in any positions marked #).
None of the detected beacons seem to be producing the distress signal, so you'll need to work out where the distress beacon is by working out where it isn't. For now, keep things simple by counting the positions where a beacon cannot possibly be along just a single row.
So, suppose you have an arrangement of beacons and sensors like in the example above and, just in the row where y=10, you'd like to count the number of positions a beacon cannot possibly exist. The coverage from all sensors near that row looks like this:
1 1 2 2
0 5 0 5 0 5
9 ...#########################...
10 ..####B######################..
11 .###S#############.###########.
In this example, in the row where y=10, there are 26 positions where a beacon cannot be present.
Consult the report from the sensors you just deployed. In the row where y=2000000, how many positions cannot contain a beacon?
Your puzzle answer was 5403290.
Your handheld device indicates that the distress signal is coming from a beacon nearby. The distress beacon is not detected by any sensor, but the distress beacon must have x and y coordinates each no lower than 0 and no larger than 4000000.
To isolate the distress beacon's signal, you need to determine its tuning frequency, which can be found by multiplying its x coordinate by 4000000 and then adding its y coordinate.
In the example above, the search space is smaller: instead, the x and y coordinates can each be at most 20. With this reduced search area, there is only a single position that could have a beacon: x=14, y=11. The tuning frequency for this distress beacon is 56000011.
Find the only possible position for the distress beacon. What is its tuning frequency?
Your puzzle answer was 10291582906626.
Both parts of this puzzle are complete! They provide two gold stars: **
At this point, you should return to your Advent calendar and try another puzzle.
If you still want to see it, you can get your puzzle input.