Day 17 part 1 , bash version
This commit is contained in:
@@ -20,7 +20,7 @@ ex1:
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@#$(TIME) ex1-c 2020 < $(INPUT) 2>&1
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ex2:
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@$(TIME) ex2.bash < $(INPUT) 2>&1
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@#$(TIME) ex2.bash < $(INPUT) 2>&1
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@#$(TIME) ex1-c 30000000 < $(INPUT) 2>&1
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clean:
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4
day17/OUTPUT
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4
day17/OUTPUT
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@@ -0,0 +1,4 @@
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ex1.bash : res=263
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time: 1:24.06 real, 60.13 user, 30.29 sys
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context-switch: 5369+166373, page-faults: 0+7362621
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257
day17/README
257
day17/README
@@ -144,7 +144,262 @@ After the full six-cycle boot process completes, 112 cubes are left in the activ
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Starting with your given initial configuration, simulate six cycles. How many cubes are left in the active state after the sixth cycle?
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To begin, get your puzzle input.
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Your puzzle answer was 263.
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The first half of this puzzle is complete! It provides one gold star: *
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--- Part Two ---
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For some reason, your simulated results don't match what the experimental energy source engineers expected. Apparently, the pocket dimension actually has four spatial dimensions, not three.
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The pocket dimension contains an infinite 4-dimensional grid. At every integer 4-dimensional coordinate (x,y,z,w), there exists a single cube (really, a hypercube) which is still either active or inactive.
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Each cube only ever considers its neighbors: any of the 80 other cubes where any of their coordinates differ by at most 1. For example, given the cube at x=1,y=2,z=3,w=4, its neighbors include the cube at x=2,y=2,z=3,w=3, the cube at x=0,y=2,z=3,w=4, and so on.
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The initial state of the pocket dimension still consists of a small flat region of cubes. Furthermore, the same rules for cycle updating still apply: during each cycle, consider the number of active neighbors of each cube.
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For example, consider the same initial state as in the example above. Even though the pocket dimension is 4-dimensional, this initial state represents a small 2-dimensional slice of it. (In particular, this initial state defines a 3x3x1x1 region of the 4-dimensional space.)
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Simulating a few cycles from this initial state produces the following configurations, where the result of each cycle is shown layer-by-layer at each given z and w coordinate:
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Before any cycles:
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z=0, w=0
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.#.
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..#
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###
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After 1 cycle:
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z=-1, w=-1
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#..
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..#
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.#.
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z=0, w=-1
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#..
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..#
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.#.
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z=1, w=-1
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#..
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..#
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.#.
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z=-1, w=0
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#..
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..#
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.#.
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z=0, w=0
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#.#
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.##
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.#.
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z=1, w=0
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#..
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..#
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.#.
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z=-1, w=1
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#..
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..#
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.#.
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z=0, w=1
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#..
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..#
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.#.
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z=1, w=1
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#..
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..#
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.#.
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After 2 cycles:
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z=-2, w=-2
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.....
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.....
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..#..
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.....
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.....
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z=-1, w=-2
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.....
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.....
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.....
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.....
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.....
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z=0, w=-2
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###..
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##.##
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#...#
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.#..#
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.###.
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z=1, w=-2
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.....
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.....
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.....
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.....
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.....
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z=2, w=-2
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.....
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.....
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..#..
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.....
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.....
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z=-2, w=-1
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.....
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.....
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.....
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.....
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.....
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z=-1, w=-1
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.....
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.....
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.....
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.....
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.....
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z=0, w=-1
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.....
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.....
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.....
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.....
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.....
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z=1, w=-1
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.....
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.....
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.....
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.....
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.....
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z=2, w=-1
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.....
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.....
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.....
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.....
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.....
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z=-2, w=0
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###..
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##.##
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#...#
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.#..#
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.###.
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z=-1, w=0
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.....
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.....
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.....
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.....
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.....
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z=0, w=0
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.....
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.....
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.....
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.....
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.....
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z=1, w=0
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.....
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.....
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.....
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.....
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.....
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z=2, w=0
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###..
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##.##
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#...#
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.#..#
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.###.
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z=-2, w=1
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.....
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.....
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.....
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.....
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.....
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z=-1, w=1
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.....
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.....
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.....
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.....
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.....
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z=0, w=1
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.....
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.....
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.....
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.....
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.....
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z=1, w=1
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.....
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.....
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.....
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.....
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.....
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z=2, w=1
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.....
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.....
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.....
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.....
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.....
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z=-2, w=2
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.....
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.....
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..#..
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.....
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.....
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z=-1, w=2
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.....
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.....
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.....
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.....
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.....
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z=0, w=2
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###..
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##.##
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#...#
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.#..#
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.###.
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z=1, w=2
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.....
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.....
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.....
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.....
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.....
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z=2, w=2
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.....
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.....
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..#..
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.....
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.....
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After the full six-cycle boot process completes, 848 cubes are left in the active state.
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Starting with your given initial configuration, simulate six cycles in a 4-dimensional space. How many cubes are left in the active state after the sixth cycle?
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Answer:
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Although it hasn't changed, you can still get your puzzle input.
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You can also [Share] this puzzle.
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109
day17/ex1.bash
Executable file
109
day17/ex1.bash
Executable file
@@ -0,0 +1,109 @@
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#!/bin/bash
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#
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# ex1.bash: Advent2020 game, day 17/game 1.
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CMD=${0##*/}
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#shopt -s extglob
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declare -A life=()
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declare -i x=-1 y=-1 z=-1 res=0
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declare -i maxx maxy maxz
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function print_life() {
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local -i x=0 y=0 z=0 foundx foundy
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for ((z=0; z<maxz; ++z)); do
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foundy=1
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for ((y=0; y<maxy; ++y)); do
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foundx=1
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for ((x=0; x<maxx; ++x)); do
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if [[ -v life[$x-$y-$z] ]]; then
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#printf "%d-%d-%d:" $x $y $z
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printf "%c" ${life["$x-$y-$z"]}
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foundx=1
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else
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printf "%c" "."
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fi
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done
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((foundx==1)) && foundy=1 && printf "\n"
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done
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((foundy==1)) && printf "z=%d\n\n" "$z"
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done
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}
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function count_neighbors () {
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local -i x=$1 y=$2 z=$3 x1 y1 z1 count=0
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local str=""
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for ((x1=x-1; x1<x+2; ++x1)); do
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for ((y1=y-1; y1<y+2; ++y1)); do
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for ((z1=z-1; z1<z+2; ++z1)); do
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if ((x1!=x || y1!=y || z1!=z)); then
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if [[ ${life[$x1-$y1-$z1]} == "#" ]]; then
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((count++))
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str="$str $x1-$y1-$z1"
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fi
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fi
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done
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done
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done
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#((count)) && printf "neighbours (%d, %d, %d)=%s\n" "$x" "$y" "$z" "$str" >&2
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echo "$count"
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}
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function run_cycle () {
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local -i x y z count=0
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local -A lifetmp=()
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for ((x=0; x<maxx; ++x)); do
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for ((y=0; y<maxy; ++y)); do
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for ((z=0; z<maxz; ++z)); do
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count=$(count_neighbors "$x" "$y" "$z")
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if [[ ${life[$x-$y-$z]} == "#" ]]; then
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((count!=2 && count!=3)) && lifetmp[$x-$y-$z]="."
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else
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((count==3)) && lifetmp[$x-$y-$z]="#"
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fi
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done
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done
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done
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for k in "${!lifetmp[@]}"; do
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life[$k]=${lifetmp[$k]}
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done
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}
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function count_active () {
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local k
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local -i count=0
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for k in "${!life[@]}"; do
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if [[ ${life[$k]} == "#" ]]; then
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((count++))
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fi
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done
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echo "$count"
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}
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x=6; y=6; z=6
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while read -r line; do
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if ((y==6)); then # 1st line
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((maxx=6+${#line}+6+1))
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((maxy=maxx))
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((maxz=maxx))
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fi
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for ((j=0; j<${#line}; ++j)); do
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((curx=x+j))
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life[$curx-$y-$z]=${line:j:1}
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done
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((y++))
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done
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for ((i=0; i<6; ++i)); do
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run_cycle
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done
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res=$(count_active)
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printf "%s : res=%d\n" "$CMD" "$res"
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exit 0
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