Solved 2024/25 P1
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import sys,time,re,copy
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from pprint import pprint
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sys.path.insert(0, '../../')
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from fred import list2int,get_re,nprint,lprint,loadFile,dijkstra,toGrid,findInGrid,create_graph_from_grid,get_value_in_direction,addTuples
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from fred import list2int,get_re,nprint,lprint,loadFile,dijkstra,toGrid,findInGrid,create_graph_from_grid,get_value_in_direction,addTuples,dprint
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start_time = time.time()
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input_f = 'input'
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@ -24,42 +24,57 @@ def part1():
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path, dist = dijkstra(graph,start,end)
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cheatWalls = []
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for r,row in enumerate(grid):
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for c,pos in enumerate(row):
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if pos == '.':
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if get_value_in_direction(grid,(r,c),'up') == '#' and get_value_in_direction(grid,(r,c),'down') == '#':
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cheatWalls.append(addTuples((r,c),(-1,0)))
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cheatWalls.append(addTuples((r,c),(1,0)))
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if addTuples((r,c),(-1,0)) not in cheatWalls and r != 1 and r != len(grid)-2 and c != 1 and c != len(row)-2:
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cheatWalls.append(addTuples((r,c),(-1,0)))
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if addTuples((r,c),(1,0)) not in cheatWalls and r != 1 and r != len(grid)-2 and c != 1 and c != len(row)-2:
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cheatWalls.append(addTuples((r,c),(1,0)))
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elif get_value_in_direction(grid,(r,c),'left') == '#' and get_value_in_direction(grid,(r,c),'right') == '#':
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cheatWalls.append(addTuples((r,c),(0,-1)))
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cheatWalls.append(addTuples((r,c),(0,1)))
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if addTuples((r,c),(0,-1)) not in cheatWalls and r != 1 and r != len(grid)-2 and c != 1 and c != len(row)-2:
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cheatWalls.append(addTuples((r,c),(0,-1)))
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if addTuples((r,c),(0,1)) not in cheatWalls and r != 1 and r != len(grid)-2 and c != 1 and c != len(row)-2:
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cheatWalls.append(addTuples((r,c),(0,1)))
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if pos == '#':
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if get_value_in_direction(grid,(r,c),'up') == '.' and get_value_in_direction(grid,(r,c),'down') == '.':
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cheatWalls.append((r,c))
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cheatWalls.append((r,c))
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if (r,c) not in cheatWalls and r != 0 and r != len(grid)-1 and c != 0 and c != len(row)-1:
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cheatWalls.append((r,c))
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elif get_value_in_direction(grid,(r,c),'left') == '.' and get_value_in_direction(grid,(r,c),'right') == '.':
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cheatWalls.append((r,c))
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cheatWalls.append((r,c))
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if (r,c) not in cheatWalls and r != 0 and r != len(grid)-1 and c != 0 and c != len(row)-1:
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cheatWalls.append((r,c))
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#print(dist)
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#nprint(grid,positions=path)
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#nprint(grid,positions=cheatWalls)
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results = {}
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startDist = dist
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#print(cheatWalls)
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count = 0
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for c in cheatWalls:
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#print(c)
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newGrid = copy.deepcopy(grid)
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newGrid[c[0]][c[1]] = '.'
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graph = create_graph_from_grid(newGrid,start,end,'#')
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path, dist = dijkstra(graph,start,end)
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#nprint(grid,(c[0],c[1]),sign='o',positions=path)
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#input()
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dist = startDist-dist
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if dist not in results:
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results[dist] = 0
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results[dist] += 1
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pprint(results)
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if dist != 0:
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if dist > 100:
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count += 1
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#if dist not in results:
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# results[dist] = 0
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#results[dist] += 1
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print(count)
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print(results)
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return count
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start_time = time.time()
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print('Part 1:',part1(), '\t\t', round((time.time() - start_time)*1000), 'ms')
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190
2024/25/25.md
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190
2024/25/25.md
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## \-\-- Day 25: Code Chronicle \-\--
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Out of ideas and time, The Historians agree that they should go back to
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check the *Chief Historian\'s office* one last time, just in case he
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went back there without you noticing.
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When you get there, you are surprised to discover that the door to his
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office is *locked*! You can hear someone inside, but knocking
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[yields]{title="function knock() {
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yield no_response;
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}"} no response. The locks on this floor are all fancy, expensive,
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virtual versions of [five-pin tumbler
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locks](https://en.wikipedia.org/wiki/Pin_tumbler_lock),
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so you contact North Pole security to see if they can help open the
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door.
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Unfortunately, they\'ve lost track of which locks are installed and
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which keys go with them, so the best they can do is send over
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*schematics of every lock and every key* for the floor you\'re on (your
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puzzle input).
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The schematics are in a cryptic file format, but they do contain
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manufacturer information, so you look up their support number.
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\"Our Virtual Five-Pin Tumbler product? That\'s our most expensive
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model! *Way* more secure than\--\" You explain that you need to open a
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door and don\'t have a lot of time.
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\"Well, you can\'t know whether a key opens a lock without actually
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trying the key in the lock (due to quantum hidden variables), but you
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*can* rule out some of the key/lock combinations.\"
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\"The virtual system is complicated, but part of it really is a crude
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simulation of a five-pin tumbler lock, mostly for marketing reasons. If
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you look at the schematics, you can figure out whether a key could
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possibly fit in a lock.\"
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He transmits you some example schematics:
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#####
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.####
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.####
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.####
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.#.#.
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.#...
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.....
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#####
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##.##
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.#.##
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...##
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...#.
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...#.
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.....
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.....
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#....
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#....
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#...#
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#.#.#
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#.###
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#####
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.....
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.....
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#.#..
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###..
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###.#
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###.#
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#####
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.....
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.....
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.....
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#....
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#.#..
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#.#.#
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#####
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\"The locks are schematics that have the top row filled (`#`) and the
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bottom row empty (`.`); the keys have the top row empty and the bottom
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row filled. If you look closely, you\'ll see that each schematic is
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actually a set of columns of various heights, either extending downward
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from the top (for locks) or upward from the bottom (for keys).\"
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\"For locks, those are the pins themselves; you can convert the pins in
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schematics to a list of heights, one per column. For keys, the columns
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make up the shape of the key where it aligns with pins; those can also
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be converted to a list of heights.\"
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\"So, you could say the first lock has pin heights `0,5,3,4,3`:\"
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#####
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.####
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.####
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.####
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.#.#.
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.#...
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.....
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\"Or, that the first key has heights `5,0,2,1,3`:\"
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.....
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#....
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#....
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#...#
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#.#.#
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#.###
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#####
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\"These seem like they should fit together; in the first four columns,
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the pins and key don\'t overlap. However, this key *cannot* be for this
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lock: in the rightmost column, the lock\'s pin overlaps with the key,
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which you know because in that column the sum of the lock height and key
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height is more than the available space.\"
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\"So anyway, you can narrow down the keys you\'d need to try by just
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testing each key with each lock, which means you would have to check\...
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wait, you have *how* many locks? But the only installation *that* size
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is at the North\--\" You disconnect the call.
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In this example, converting both locks to pin heights produces:
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0,5,3,4,3
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1,2,0,5,3
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Converting all three keys to heights produces:
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5,0,2,1,3
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4,3,4,0,2
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3,0,2,0,1
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Then, you can try every key with every lock:
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- Lock `0,5,3,4,3` and key `5,0,2,1,3`: *overlap* in the last column.
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- Lock `0,5,3,4,3` and key `4,3,4,0,2`: *overlap* in the second
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column.
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- Lock `0,5,3,4,3` and key `3,0,2,0,1`: all columns *fit*!
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- Lock `1,2,0,5,3` and key `5,0,2,1,3`: *overlap* in the first column.
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- Lock `1,2,0,5,3` and key `4,3,4,0,2`: all columns *fit*!
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- Lock `1,2,0,5,3` and key `3,0,2,0,1`: all columns *fit*!
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So, in this example, the number of unique lock/key pairs that fit
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together without overlapping in any column is `3`.
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Analyze your lock and key schematics. *How many unique lock/key pairs
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fit together without overlapping in any column?*
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Your puzzle answer was `3395`.
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The first half of this puzzle is complete! It provides one gold star: \*
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## \-\-- Part Two \-\-- {#part2}
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You and The Historians crowd into the office, startling the Chief
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Historian awake! The Historians all take turns looking confused until
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one asks where he\'s been for the last few months.
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\"I\'ve been right here, working on this high-priority request from
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Santa! I think the only time I even stepped away was about a month ago
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when I went to grab a cup of coffee\...\"
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Just then, the Chief notices the time. \"Oh no! I\'m going to be late! I
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must have fallen asleep trying to put the finishing touches on this
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*chronicle* Santa requested, but now I don\'t have enough time to go
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visit the last 50 places on my list and complete the chronicle before
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Santa leaves! He said he needed it before tonight\'s sleigh launch.\"
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One of The Historians holds up the list they\'ve been using this whole
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time to keep track of where they\'ve been searching. Next to each place
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you all visited, they checked off that place with a *star*. Other
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Historians hold up their own notes they took on the journey; as The
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Historians, how could they resist writing everything down while visiting
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all those historically significant places?
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The Chief\'s eyes get wide. \"With all this, we might just have enough
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time to finish the chronicle! Santa said he wanted it wrapped up with a
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bow, so I\'ll call down to the wrapping department and\... hey, could
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*you* bring it up to Santa? I\'ll need to be in my seat to watch the
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sleigh launch by then.\"
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You nod, and The Historians quickly work to collect their notes into the
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final set of pages for the chronicle.
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You don\'t have enough stars to finish the chronicle, though. You need
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13 more.
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Although it hasn\'t changed, you can still [get your puzzle
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input](25/input).
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66
2024/25/solution.py
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66
2024/25/solution.py
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#!/bin/python3
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import sys,time,re
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from pprint import pprint
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sys.path.insert(0, '../../')
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from fred import list2int,get_re,nprint,lprint,loadFile
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start_time = time.time()
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input_f = 'input'
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#########################################
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# #
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# Part 1 #
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# #
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#########################################
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def part1():
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instructions = []
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with open(input_f) as file:
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for line in file:
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instructions.append(list(line.rstrip()))
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keys = []
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locks = []
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for i in range(0,len(instructions),8):
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tmp = []
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# lock
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if instructions[i] == list('#####'):
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for l in range(5):
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tmp.append(sum(element == '#' for element in [row[l] for row in instructions[i:i+7]])-1)
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if tmp not in locks:
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locks.append(tmp)
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# key
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elif instructions[i] == list('.....'):
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for l in range(5):
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tmp.append(sum(element == '#' for element in [row[l] for row in instructions[i:i+7]])-1)
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if tmp not in keys:
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keys.append(tmp)
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fits = 0
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works = True
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for k in keys:
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for l in locks:
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for i in range(5):
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if k[i] + l[i] > 5:
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works = False
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if works:
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fits += 1
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works = True
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return fits
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start_time = time.time()
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print('Part 1:',part1(), '\t\t', round((time.time() - start_time)*1000), 'ms')
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#########################################
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# #
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# Part 2 #
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# #
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#########################################
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def part2():
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return
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start_time = time.time()
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print('Part 2:',part2(), '\t\t', round((time.time() - start_time)*1000), 'ms')
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