Bash: 2022 day 5 (both parts, brute code)

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2022-12-08 16:09:32 +01:00
parent 0c787d9a51
commit d7fa1c4fb5
2 changed files with 210 additions and 1 deletions

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@@ -88,4 +88,82 @@ Elves the message =CMZ=.
/After the rearrangement procedure completes, what crate ends up on top
of each stack?/
To begin, [[file:5/input][get your puzzle input]].
Your puzzle answer was =VQZNJMWTR=.
** --- Part Two ---
As you watch the crane operator expertly rearrange the crates, you
notice the process isn't following your prediction.
Some mud was covering the writing on the side of the crane, and you
quickly wipe it away. The crane isn't a CrateMover 9000 - it's a
/CrateMover 9001/.
The CrateMover 9001 is notable for many new and exciting features: air
conditioning, leather seats, an extra cup holder, and /the ability to
pick up and move multiple crates at once/.
Again considering the example above, the crates begin in the same
configuration:
#+begin_example
[D]
[N] [C]
[Z] [M] [P]
1 2 3
#+end_example
Moving a single crate from stack 2 to stack 1 behaves the same as
before:
#+begin_example
[D]
[N] [C]
[Z] [M] [P]
1 2 3
#+end_example
However, the action of moving three crates from stack 1 to stack 3 means
that those three moved crates /stay in the same order/, resulting in
this new configuration:
#+begin_example
[D]
[N]
[C] [Z]
[M] [P]
1 2 3
#+end_example
Next, as both crates are moved from stack 2 to stack 1, they /retain
their order/ as well:
#+begin_example
[D]
[N]
[C] [Z]
[M] [P]
1 2 3
#+end_example
Finally, a single crate is still moved from stack 1 to stack 2, but now
it's crate =C= that gets moved:
#+begin_example
[D]
[N]
[Z]
[M] [C] [P]
1 2 3
#+end_example
In this example, the CrateMover 9001 has put the crates in a totally
different order: =MCD=.
Before the rearrangement process finishes, update your simulation so
that the Elves know where they should stand to be ready to unload the
final supplies. /After the rearrangement procedure completes, what crate
ends up on top of each stack?/
Your puzzle answer was =NLCDCLVMQ=.
Both parts of this puzzle are complete! They provide two gold stars: **

131
2022/day05/aoc.bash Executable file
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@@ -0,0 +1,131 @@
#!/usr/bin/env bash
#
# aoc.bash: Advent of Code 2022, day 3
#
# 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>
. common.bash
declare -a stacks=() rules=()
printall() {
local i
for (( i = 0; i < ${#stacks[@]}; ++i )); do
printf "stack %d: %s\n" "$i" "${stacks[i]}"
done
#for (( i = 0; i < ${#rules[@]}; ++i )); do
# printf "rule %d: %s\n" "$i" "${rules[i]}"
#done
}
parse() {
local -i _part="$1" _queue li _state=1
local _input
local -a _parse
declare -ig _res=0
li=1
while IFS= read -r _input; do
if [[ -z ${_input} || ${_input:1:1} == "1" ]]; then
printf "\tzobi\n"
_state=2
li=0
continue
fi
case $_state in
1) # stacks description
# get blocks of 4 characters
#echo GGG
for ((i=0, _queue=0; i<${#_input}; i+=4, _queue++)); do
#printf "line=%d queue=%d\n" "$li" "$_queue"
sub=${_input:i:4}
if [[ $sub == " " ]]; then
:
#printf "\t SKIP line %d queue\n" "$li" "$_queue"
else
# printf "\tline %d queue %d CHAR %s\n" "$li" "$_queue" "${sub:1:1}"
stacks[$_queue]+="${sub:1:1}"
fi
done
;;
2) # moves description
#echo HHH
read -ra _parse <<<"$_input"
#printf "RULE: %d\n" ${#_parse[@]}
#printf "rule %d: %d %d %d\n" "$li" "${_parse[1]}" "${_parse[3]}" "${_parse[5]}"
rules[$li]="${_parse[1]} ${_parse[3]} ${_parse[5]}"
esac
((li++))
done
#printall
}
part1() {
local -i _i _j _nb _from
local -a _rule
local _tmp
printall
for (( _i=0; _i<${#rules[@]}; ++_i )); do
# shellcheck disable=2086
read -ra _rule <<< ${rules[$_i]}
(( _nb = _rule[0] ))
((_from = _rule[1] - 1 ))
((_to = _rule[2] - 1 ))
#printf "rule=%d nb=%d from=%d=%s to=%d=%s\n" "$_i" "$_nb" "$_from" "${stacks[$_from]}" "$_to" "${stacks[$_to]}"
#printf "_rule=%d nb=%s from=%s/%s to=%s/%s\n" ${#_rule[@]} "${_rule[0]}" "${_rule[1]}" "${stacks[${_rule[1]}]}" "${_rule[2]}" "${stacks[${_rule[2]}]}"
for ((_j = 0; _j < _nb; ++_j)); do
#printf "moving char %d f=%s t=%s\n" "$_i" "${stacks[$_from]}" "${stacks[$_to]}"
stacks[$_to]="${stacks[$_from]:0:1}${stacks[$_to]}"
stacks[$_from]=${stacks[$_from]:1}
#printf " --> f=%s t=%s\n" "${stacks[$_from]}" "${stacks[$_to]}"
done
printall
done
}
part2() {
local -i _i _nb _from
local -a _rule
local _tmp
#printall
for (( _i=0; _i<${#rules[@]}; ++_i )); do
# shellcheck disable=2086
read -ra _rule <<< ${rules[$_i]}
(( _nb = _rule[0] ))
(( _from = _rule[1] - 1 ))
(( _to = _rule[2] - 1 ))
stacks[$_to]="${stacks[$_from]:0:_nb}${stacks[$_to]}"
stacks[$_from]=${stacks[$_from]:_nb}
#printf "rule=%d nb=%d from=%d=%s to=%d=%s\n" "$_i" "$_nb" "$_from" "${stacks[$_from]}" "$_to" "${stacks[$_to]}"
#printf "_rule=%d nb=%s from=%s/%s to=%s/%s\n" ${#_rule[@]} "${_rule[0]}" "${_rule[1]}" "${stacks[${_rule[1]}]}" "${_rule[2]}" "${stacks[${_rule[2]}]}"
#printall
done
}
solve() {
local -i _i
if (($1 == 1)); then
part1
else
part2
fi
res=""
for ((_i = 0; _i < ${#stacks[@]}; ++_i)); do
res+="${stacks[_i]:0:1}"
done
#echo "res=$res"
}
main "$@"
exit 0