4 files changed, 223 insertions, 0 deletions
diff --git a/src/22/input b/src/22/input
new file mode 100644
index 0000000..9b643c6
--- /dev/null
+++ b/src/22/input
@@ -0,0 +1,2 @@
+depth: 4848
+target: 15,700
diff --git a/src/22/part1 b/src/22/part1
new file mode 100644
index 0000000..bb29305
--- /dev/null
+++ b/src/22/part1
@@ -0,0 +1,106 @@
+--- Day 22: Mode Maze ---
+
+This is it, your final stop: the year -483. It's snowing and dark outside; the only light you can
+see is coming from a small cottage in the distance. You make your way there and knock on the door.
+
+A portly man with a large, white beard answers the door and invites you inside. For someone living
+near the North Pole in -483, he must not get many visitors, but he doesn't act surprised to see you.
+Instead, he offers you some milk and cookies.
+
+After talking for a while, he asks a favor of you. His friend hasn't come back in a few hours, and
+he's not sure where he is.  Scanning the region briefly, you discover one life signal in a cave
+system nearby; his friend must have taken shelter there.  The man asks if you can go there to
+retrieve his friend.
+
+The cave is divided into square [1m[97mregions[0m which are either dominantly [1m[97mrocky[0m, [1m[97mnarrow[0m, or
+[1m[97mwet[0m (called its [1m[97mtype[0m). Each region occupies exactly one [1m[97mcoordinate[0m in X,Y format where X and Y are
+integers and zero or greater. (Adjacent regions can be the same type.)
+
+The scan (your puzzle input) is not very detailed: it only reveals the [1m[97mdepth[0m of the cave system and
+the [1m[97mcoordinates of the target[0m. However, it does not reveal the type of each region. The mouth of the
+cave is at 0,0.
+
+The man explains that due to the unusual geology in the area, there is a method to determine any
+region's type based on its [1m[97merosion level[0m. The erosion level of a region can be determined from its
+[1m[97mgeologic index[0m. The geologic index can be determined using the first rule that applies from the list
+below:
+
+
+ - The region at 0,0 (the mouth of the cave) has a geologic index of 0.
+
+ - The region at the coordinates of the target has a geologic index of 0.
+
+ - If the region's Y coordinate is 0, the geologic index is its X coordinate times 16807.
+
+ - If the region's X coordinate is 0, the geologic index is its Y coordinate times 48271.
+
+ - Otherwise, the region's geologic index is the result of multiplying the erosion
+[1m[97mlevels[0m of the regions at X-1,Y and X,Y-1.
+
+
+A region's [1m[97merosion level[0m is its [1m[97mgeologic index[0m plus the cave system's [1m[97mdepth[0m, all modulo 20183. Then:
+
+
+ - If the [1m[97merosion level modulo 3[0m is 0, the region's type is [1m[97mrocky[0m.
+
+ - If the [1m[97merosion level modulo 3[0m is 1, the region's type is [1m[97mwet[0m.
+
+ - If the [1m[97merosion level modulo 3[0m is 2, the region's type is [1m[97mnarrow[0m.
+
+
+For example, suppose the cave system's depth is 510 and the target's coordinates are 10,10. Using %
+to represent the modulo operator, the cavern would look as follows:
+
+
+ - At 0,0, the geologic index is 0. The erosion level is (0 + 510) % 20183 = 510. The type is 510 %
+3 = 0, [1m[97mrocky[0m.
+
+ - At 1,0, because the Y coordinate is 0, the geologic index is 1 * 16807 = 16807. The erosion level
+is (16807 + 510) % 20183 = 17317. The type is 17317 % 3 = 1, [1m[97mwet[0m.
+
+ - At 0,1, because the X coordinate is 0, the geologic index is  1 * 48271 = 48271. The erosion
+level is (48271 + 510) % 20183 = 8415. The type is 8415 % 3 = 0, [1m[97mrocky[0m.
+
+ - At 1,1, neither coordinate is 0 and it is not the coordinate of the target, so the geologic index
+is the erosion level of 0,1 (8415) times the erosion level of 1,0 (17317), 8415 * 17317 = 145722555.
+The erosion level is (145722555 + 510) % 20183 = 1805. The type is 1805 % 3 = 2,
+[1m[97mnarrow[0m.
+
+ - At 10,10, because they are the target's coordinates, the geologic index is 0. The erosion level
+is (0 + 510) % 20183 = 510. The type is 510 % 3 = 0, [1m[97mrocky[0m.
+
+
+Drawing this same cave system with rocky as ., wet as =, narrow as |, the mouth as M, the target as
+T, with 0,0 in the top-left corner, X increasing to the right, and Y increasing downward, the
+top-left corner of the map looks like this:
+
+[1m[97mM[0m=.|=.|.|=.|=|=.
+.|=|=|||..|.=...
+.==|....||=..|==
+=.|....|.==.|==.
+=|..==...=.|==..
+=||.=.=||=|=..|=
+|.=.===|||..=..|
+|..==||=.|==|===
+.=..===..=|.|||.
+.======|||=|=.|=
+.===|=|===[1m[97mT[0m===||
+=|||...|==..|=.|
+=.=|=.=..=.||==|
+||=|=...|==.=|==
+|=.=||===.|||===
+||.|==.|.|.||=||
+
+Before you go in, you should determine the [1m[97mrisk level[0m of the area. For the rectangle that has a
+top-left corner of region 0,0 and a bottom-right corner of the region containing the target, add up
+the risk level of each individual region: 0 for rocky regions, 1 for wet regions, and 2 for narrow
+regions.
+
+In the cave system above, because the mouth is at 0,0 and the target is at 10,10, adding up the risk
+level of all regions with an X coordinate from 0 to 10 and a Y coordinate from 0 to 10, this total
+is [1m[97m114[0m.
+
+[1m[97mWhat is the total risk level for the smallest rectangle that includes 0,0 and the target's
+coordinates?[0m
+
+
diff --git a/src/22/part2 b/src/22/part2
new file mode 100644
index 0000000..fbcfdd7
--- /dev/null
+++ b/src/22/part2
@@ -0,0 +1,22 @@
+--- Part Two ---
+
+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, [1m[97mfactory order[0m deck of [1m[97m119315717514047 space cards[0m.
+
+Finally, a deck of cards worthy of shuffling!
+
+You decide to apply your complete shuffle process (your puzzle input) to the deck
+[1m[97m101741582076661 times in a row[0m.
+
+You'll need to be careful, though - one wrong move with this many cards and you might
+[1m[97moverflow[0m your entire ship!
+
+After shuffling your new, giant, [1m[97mfactory order[0m deck that many times, [1m[97mwhat number is on the card that
+ends up in position 2020?[0m
+
+
diff --git a/src/22/solve.py b/src/22/solve.py
new file mode 100644
index 0000000..375ae23
--- /dev/null
+++ b/src/22/solve.py
@@ -0,0 +1,93 @@
+import sys
+sys.path.append("../common")
+import aoc
+import networkx
+
+adjacent = [[-1, 0], [0, -1], [1, 0], [0, 1]]
+
+# create symbol definitions
+symbols = [".", "=", "|"]
+rocky, wet, narrow = 0, 1, 2
+torch, gear, neither = 0, 1, 2
+
+tooloptions = dict()
+tooloptions[rocky] = (torch, gear)
+tooloptions[wet] = (gear, neither)
+tooloptions[narrow] = (torch, neither)
+
+# parse input file
+def parse_input(si):
+    s = si.split("\n")
+    depth = int(s[0].split(": ")[1])
+    target = [int(v) for v in s[1].split(": ")[1].split(",")]
+    return depth, target
+
+# get erosion level from geological index
+def get_erosion_level(n):
+    return (n + depth) % 20183
+
+# generate map of erosion types
+def gen_map():
+    grid = [[0 for x in range(mrange[0])] for y in range(mrange[1])]
+
+    # generate values on x side
+    grid[0] = [get_erosion_level(x * 16807) for x in range(mrange[0])] 
+
+    # generate values on y side
+    for y in range(mrange[1]):
+        grid[y][0] = get_erosion_level(y * 48271)
+
+    # fill in other positions
+    for y in range(1, mrange[1]):
+        for x in range(1, mrange[0]):
+            grid[y][x] = get_erosion_level(grid[y-1][x] * grid[y][x-1])
+
+    # mod 3 for env type
+    grid = [[grid[y][x] % 3 for x in range(mrange[0])] for y in range(mrange[1])]
+
+    # start position is rocky
+    grid[target[1]][target[0]] = 0
+
+    return grid
+
+# define constants from input file
+depth, target = parse_input(aoc.data)
+mrange = (target[0] + 100, target[1] + 100) # 100 block padding for potential hook-paths
+
+vmap = gen_map()
+
+def risk_level(width, height):
+    risk = 0
+    for y in range(height + 1):
+        for x in range(width + 1):
+            risk += vmap[y][x]
+    return risk
+
+def adj(p):
+    return [[p[0] + di[0], p[1] + di[1]] for di in adjacent]
+
+def dijkstra():
+    graph = networkx.Graph()
+    for y in range(mrange[1]):
+        for x in range(mrange[0]):
+            tools = tooloptions[vmap[y][x]]
+            # always takes 7 minutes to switch, 2 z-options for each tool
+            graph.add_edge((x, y, tools[0]), (x, y, tools[1]), weight = 7)
+            for (nx, ny) in adj((x, y)):
+                if 0 <= nx < mrange[0] and 0 <= ny < mrange[1]:
+                    ntools = tooloptions[vmap[ny][nx]]
+                    # if tool is usable in both environments
+                    for tool in [t for t in tools if t in ntools]:
+                         # then it only takes 1 minute
+                        graph.add_edge((x, y, tool), (nx, ny, tool), weight = 1)
+
+    return networkx.dijkstra_path_length(graph,
+        (0, 0, torch), (target[0], target[1], torch))
+
+def solve1(args):
+    return risk_level(target[0], target[1])
+
+def solve2(args):
+    return dijkstra()
+
+aoc.run(solve1, solve2, sols=[11359, 976])