init day 9

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2022-12-12 20:47:45 +01:00
parent 31a255a9ac
commit dfe2207e8e
7 changed files with 2538 additions and 0 deletions

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# AOC daily Makefile - GNU make only.
#
# Copyright (C) 2021-2022 Bruno Raoult ("br")
# Licensed under the GNU General Public License v3.0 or later.
# Some rights reserved. See COPYING.
#
# You should have received a copy of the GNU General Public License along with this
# program. If not, see <https://www.gnu.org/licenses/gpl-3.0-standalone.html>.
#
# SPDX-License-Identifier: GPL-3.0-or-later <https://spdx.org/licenses/GPL-3.0-or-later.html>
#
INPUT := input/input.txt
SHELL := /bin/bash
CC := gcc
BEAR := bear
CCLSFILE:= compile_commands.json
LIB := aoc_$(shell uname -m)
INCDIR := ../include
LIBDIR := ../lib
LDFLAGS := -L$(LIBDIR)
#LDLIB := -l$(LIB) -lm
LDLIB := -l$(LIB)
export LD_LIBRARY_PATH = $(LIBDIR)
CFLAGS += -std=gnu11
CFLAGS += -O2
CFLAGS += -g
# for gprof
# CFLAGS += -pg
CFLAGS += -Wall
CFLAGS += -Wextra
CFLAGS += -march=native
# Next one may be useful for valgrind (some invalid instructions)
# CFLAGS += -mno-tbm
CFLAGS += -Wmissing-declarations
CFLAGS += -Wno-unused-result
CFLAGS += -DDEBUG_DEBUG # activate general debug (debug.c)
CFLAGS += -DDEBUG_POOL # memory pools management
VALGRIND := valgrind
VALGRINDFLAGS := --leak-check=full --show-leak-kinds=all --track-origins=yes \
--sigill-diagnostics=yes --quiet --show-error-list=yes
TIME := \time -f "\ttime: %E real, %U user, %S sys\n\tcontext-switch:\t%c+%w, page-faults: %F+%R\n"
export PATH := .:$(PATH)
.PHONY: clean all compile assembly memcheck memcheck1 memcheck2 part1 part2 ccls bear org
all: README.org ccls part1 part2
memcheck: memcheck1 memcheck2
memcheck1: aoc-c
@$(VALGRIND) $(VALGRINDFLAGS) aoc-c -p 1 < $(INPUT)
memcheck2: aoc-c
@$(VALGRIND) $(VALGRINDFLAGS) aoc-c -p 2 < $(INPUT)
@#@valgrind -s --track-origins=yes aoc-c -p 2 < $(INPUT)
compile: aoc-c
cpp: aoc-c.i
assembly: aoc-c.s
part1: aoc-c
@$(TIME) aoc.bash -p 1 < $(INPUT) 2>&1
@$(TIME) aoc-c -p 1 < $(INPUT)
part2: aoc-c
@$(TIME) aoc.bash -p 2 < $(INPUT) 2>&1
@$(TIME) aoc-c -p 2 < $(INPUT)
ccls: $(CCLSFILE)
clean:
@rm -f aoc-c core* vgcore* gmon.out aoc-c.s aoc-c.i README.html compile_commands.json
aoc-c: aoc-c.c common.c
@echo compiling $<
$(CC) $(CFLAGS) $(LDFLAGS) -I $(INCDIR) $^ $(LDLIB) -o $@
# generate pre-processed file (.i) and assembler (.s)
%.i: %.c
@echo generating $@
@$(CC) -E $(CFLAGS) -I $(INCDIR) $< -o $@
%.s: %.c
@echo generating $@
@$(CC) -S -fverbose-asm $(CFLAGS) -I $(INCDIR) $< -o $@
# generate README.org from README.html (must cleanup !)
org: README.org
%.org: %.html
@echo generating $@. Cleanup before commit !
@pandoc $< -o $@
# generate compile_commands.json
$(CCLSFILE): aoc-c.c Makefile
$(BEAR) -- make clean compile
bear: clean
@touch .ccls-root
@$(BEAR) -- make compile

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** --- Day 9: Rope Bridge ---
This rope bridge creaks as you walk along it. You aren't sure how old it
is, or whether it can even support your weight.
It seems to support the Elves just fine, though. The bridge spans a
gorge which was carved out by the massive river far below you.
You step carefully; as you do, the ropes stretch and twist. You decide
to distract yourself by modeling rope physics; maybe you can even figure
out where /not/ to step.
Consider a rope with a knot at each end; these knots mark the /head/ and
the /tail/ of the rope. If the head moves far enough away from the tail,
the tail is pulled toward the head.
Due to nebulous reasoning involving
[[https://en.wikipedia.org/wiki/Planck_units#Planck_length][Planck
lengths]], you should be able to model the positions of the knots on a
two-dimensional grid. Then, by following a hypothetical /series of
motions/ (your puzzle input) for the head, you can determine how the
tail will move.
Due to the aforementioned Planck lengths, the rope must be quite short;
in fact, the head (=H=) and tail (=T=) must /always be touching/
(diagonally adjacent and even overlapping both count as touching):
#+begin_example
....
.TH.
....
....
.H..
..T.
....
...
.H. (H covers T)
...
#+end_example
If the head is ever two steps directly up, down, left, or right from the
tail, the tail must also move one step in that direction so it remains
close enough:
#+begin_example
..... ..... .....
.TH.. -> .T.H. -> ..TH.
..... ..... .....
... ... ...
.T. .T. ...
.H. -> ... -> .T.
... .H. .H.
... ... ...
#+end_example
Otherwise, if the head and tail aren't touching and aren't in the same
row or column, the tail always moves one step diagonally to keep up:
#+begin_example
..... ..... .....
..... ..H.. ..H..
..H.. -> ..... -> ..T..
.T... .T... .....
..... ..... .....
..... ..... .....
..... ..... .....
..H.. -> ...H. -> ..TH.
.T... .T... .....
..... ..... .....
#+end_example
You just need to work out where the tail goes as the head follows a
series of motions. Assume the head and the tail both start at the same
position, overlapping.
For example:
#+begin_example
R 4
U 4
L 3
D 1
R 4
D 1
L 5
R 2
#+end_example
This series of motions moves the head /right/ four steps, then /up/ four
steps, then /left/ three steps, then /down/ one step, and so on. After
each step, you'll need to update the position of the tail if the step
means the head is no longer adjacent to the tail. Visually, these
motions occur as follows (=s= marks the starting position as a reference
point):
#+begin_example
== Initial State ==
......
......
......
......
H..... (H covers T, s)
== R 4 ==
......
......
......
......
TH.... (T covers s)
......
......
......
......
sTH...
......
......
......
......
s.TH..
......
......
......
......
s..TH.
== U 4 ==
......
......
......
....H.
s..T..
......
......
....H.
....T.
s.....
......
....H.
....T.
......
s.....
....H.
....T.
......
......
s.....
== L 3 ==
...H..
....T.
......
......
s.....
..HT..
......
......
......
s.....
.HT...
......
......
......
s.....
== D 1 ==
..T...
.H....
......
......
s.....
== R 4 ==
..T...
..H...
......
......
s.....
..T...
...H..
......
......
s.....
......
...TH.
......
......
s.....
......
....TH
......
......
s.....
== D 1 ==
......
....T.
.....H
......
s.....
== L 5 ==
......
....T.
....H.
......
s.....
......
....T.
...H..
......
s.....
......
......
..HT..
......
s.....
......
......
.HT...
......
s.....
......
......
HT....
......
s.....
== R 2 ==
......
......
.H.... (H covers T)
......
s.....
......
......
.TH...
......
s.....
#+end_example
After simulating the rope, you can count up all of the positions the
/tail visited at least once/. In this diagram, =s= again marks the
starting position (which the tail also visited) and =#= marks other
positions the tail visited:
#+begin_example
..##..
...##.
.####.
....#.
s###..
#+end_example
So, there are =13= positions the tail visited at least once.
Simulate your complete hypothetical series of motions. /How many
positions does the tail of the rope visit at least once?/

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/* aoc.c: Advent of Code 2022
*
* Copyright (C) 2022 Bruno Raoult ("br")
* Licensed under the GNU General Public License v3.0 or later.
* Some rights reserved. See COPYING.
*
* You should have received a copy of the GNU General Public License along with this
* program. If not, see <https://www.gnu.org/licenses/gpl-3.0-standalone.html>.
*
* SPDX-License-Identifier: GPL-3.0-or-later <https://spdx.org/licenses/GPL-3.0-or-later.html>
*/
#ifndef _AOC_H_
#define _AOC_H_
extern int parseargs(int ac, char**av);
#endif /* _AOC_H_ */

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#!/usr/bin/env bash
#
# common.bash: Advent of Code 2022, common bash functions
#
# Copyright (C) 2022 Bruno Raoult ("br")
# Licensed under the GNU General Public License v3.0 or later.
# Some rights reserved. See COPYING.
#
# You should have received a copy of the GNU General Public License along with this
# program. If not, see <https://www.gnu.org/licenses/gpl-3.0-standalone.html>.
#
# SPDX-License-Identifier: GPL-3.0-or-later <https://spdx.org/licenses/GPL-3.0-or-later.html>
# shellcheck disable=2034
export cmdname=${0##*/}
export debug=0
export res
export LANG=C
shopt -s extglob
set -o noglob
usage() {
printf "usage: %s [-d DEBUG] [-p PART]\n" "$cmdname"
exit 1
}
checkargs() {
local part=1
while getopts p:d: todo; do
case "$todo" in
d)
if [[ "$OPTARG" =~ ^[[:digit:]+]$ ]]; then
debug="$OPTARG"
else
printf "%s: illegal [%s] debug level.\n" "$CMD" "$OPTARG"
exit 1
fi
;;
p)
if [[ "$OPTARG" =~ ^[12]$ ]]; then
part="$OPTARG"
else
printf "%s: illegal [%s] part.\n" "$CMD" "$OPTARG"
exit 1
fi
;;
*)
usage
;;
esac
done
# Now check remaining argument (backup directory)
shift $((OPTIND - 1))
(( $# > 1 )) && usage
return "$part"
}
main() {
local -i part
checkargs "$@"
part=$?
parse "$part"
solve "$part"
printf "%s: res=%s\n" "$cmdname" "$res"
}

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/* common.c: Advent of Code 2022, common functions
*
* Copyright (C) 2022 Bruno Raoult ("br")
* Licensed under the GNU General Public License v3.0 or later.
* Some rights reserved. See COPYING.
*
* You should have received a copy of the GNU General Public License along with this
* program. If not, see <https://www.gnu.org/licenses/gpl-3.0-standalone.html>.
*
* SPDX-License-Identifier: GPL-3.0-or-later <https://spdx.org/licenses/GPL-3.0-or-later.html>
*/
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include "aoc.h"
#include "debug.h"
static int usage(char *prg)
{
fprintf(stderr, "Usage: %s [-d debug_level] [-p part] [-i input]\n", prg);
return 1;
}
int parseargs(int ac, char **av)
{
int opt, part = 1;
while ((opt = getopt(ac, av, "d:p:")) != -1) {
switch (opt) {
case 'd':
debug_level_set(atoi(optarg));
break;
case 'p': /* 1 or 2 */
part = atoi(optarg);
if (part < 1 || part > 2)
return usage(*av);
break;
case 'i':
default:
return usage(*av);
}
}
if (optind < ac)
return usage(*av);
return part;
}

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R 4
U 4
L 3
D 1
R 4
D 1
L 5
R 2

2000
2022/day09/input/input.txt Normal file

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