Solved 2024/18 P1

2024/16/16.md

@@ -0,0 +1,170 @@
+## \-\-- Day 16: Reindeer Maze \-\--
+
+It\'s time again for the [Reindeer Olympics](/2015/day/14)! This year,
+the big event is the *Reindeer Maze*, where the Reindeer compete for the
+*[lowest
+score]{title="I would say it's like Reindeer Golf, but knowing Reindeer, it's almost certainly nothing like Reindeer Golf."}*.
+
+You and The Historians arrive to search for the Chief right as the event
+is about to start. It wouldn\'t hurt to watch a little, right?
+
+The Reindeer start on the Start Tile (marked `S`) facing *East* and need
+to reach the End Tile (marked `E`). They can move forward one tile at a
+time (increasing their score by `1` point), but never into a wall (`#`).
+They can also rotate clockwise or counterclockwise 90 degrees at a time
+(increasing their score by `1000` points).
+
+To figure out the best place to sit, you start by grabbing a map (your
+puzzle input) from a nearby kiosk. For example:
+
+    ###############
+    #.......#....E#
+    #.#.###.#.###.#
+    #.....#.#...#.#
+    #.###.#####.#.#
+    #.#.#.......#.#
+    #.#.#####.###.#
+    #...........#.#
+    ###.#.#####.#.#
+    #...#.....#.#.#
+    #.#.#.###.#.#.#
+    #.....#...#.#.#
+    #.###.#.#.#.#.#
+    #S..#.....#...#
+    ###############
+
+There are many paths through this maze, but taking any of the best paths
+would incur a score of only `7036`. This can be achieved by taking a
+total of `36` steps forward and turning 90 degrees a total of `7` times:
+
+    ###############
+    #.......#....E#
+    #.#.###.#.###^#
+    #.....#.#...#^#
+    #.###.#####.#^#
+    #.#.#.......#^#
+    #.#.#####.###^#
+    #..>>>>>>>>v#^#
+    ###^#.#####v#^#
+    #>>^#.....#v#^#
+    #^#.#.###.#v#^#
+    #^....#...#v#^#
+    #^###.#.#.#v#^#
+    #S..#.....#>>^#
+    ###############
+
+Here\'s a second example:
+
+    #################
+    #...#...#...#..E#
+    #.#.#.#.#.#.#.#.#
+    #.#.#.#...#...#.#
+    #.#.#.#.###.#.#.#
+    #...#.#.#.....#.#
+    #.#.#.#.#.#####.#
+    #.#...#.#.#.....#
+    #.#.#####.#.###.#
+    #.#.#.......#...#
+    #.#.###.#####.###
+    #.#.#...#.....#.#
+    #.#.#.#####.###.#
+    #.#.#.........#.#
+    #.#.#.#########.#
+    #S#.............#
+    #################
+
+In this maze, the best paths cost `11048` points; following one such
+path would look like this:
+
+    #################
+    #...#...#...#..E#
+    #.#.#.#.#.#.#.#^#
+    #.#.#.#...#...#^#
+    #.#.#.#.###.#.#^#
+    #>>v#.#.#.....#^#
+    #^#v#.#.#.#####^#
+    #^#v..#.#.#>>>>^#
+    #^#v#####.#^###.#
+    #^#v#..>>>>^#...#
+    #^#v###^#####.###
+    #^#v#>>^#.....#.#
+    #^#v#^#####.###.#
+    #^#v#^........#.#
+    #^#v#^#########.#
+    #S#>>^..........#
+    #################
+
+Note that the path shown above includes one 90 degree turn as the very
+first move, rotating the Reindeer from facing East to facing North.
+
+Analyze your map carefully. *What is the lowest score a Reindeer could
+possibly get?*
+
+Your puzzle answer was `108504`.
+
+The first half of this puzzle is complete! It provides one gold star: \*
+
+## \-\-- Part Two \-\-- {#part2}
+
+Now that you know what the best paths look like, you can figure out the
+best spot to sit.
+
+Every non-wall tile (`S`, `.`, or `E`) is equipped with places to sit
+along the edges of the tile. While determining which of these tiles
+would be the best spot to sit depends on a whole bunch of factors (how
+comfortable the seats are, how far away the bathrooms are, whether
+there\'s a pillar blocking your view, etc.), the most important factor
+is *whether the tile is on one of the best paths through the maze*. If
+you sit somewhere else, you\'d miss all the action!
+
+So, you\'ll need to determine which tiles are part of *any* best path
+through the maze, including the `S` and `E` tiles.
+
+In the first example, there are `45` tiles (marked `O`) that are part of
+at least one of the various best paths through the maze:
+
+    ###############
+    #.......#....O#
+    #.#.###.#.###O#
+    #.....#.#...#O#
+    #.###.#####.#O#
+    #.#.#.......#O#
+    #.#.#####.###O#
+    #..OOOOOOOOO#O#
+    ###O#O#####O#O#
+    #OOO#O....#O#O#
+    #O#O#O###.#O#O#
+    #OOOOO#...#O#O#
+    #O###.#.#.#O#O#
+    #O..#.....#OOO#
+    ###############
+
+In the second example, there are `64` tiles that are part of at least
+one of the best paths:
+
+    #################
+    #...#...#...#..O#
+    #.#.#.#.#.#.#.#O#
+    #.#.#.#...#...#O#
+    #.#.#.#.###.#.#O#
+    #OOO#.#.#.....#O#
+    #O#O#.#.#.#####O#
+    #O#O..#.#.#OOOOO#
+    #O#O#####.#O###O#
+    #O#O#..OOOOO#OOO#
+    #O#O###O#####O###
+    #O#O#OOO#..OOO#.#
+    #O#O#O#####O###.#
+    #O#O#OOOOOOO..#.#
+    #O#O#O#########.#
+    #O#OOO..........#
+    #################
+
+Analyze your map further. *How many tiles are part of at least one of
+the best paths through the maze?*
+
+Answer:
+
+Although it hasn\'t changed, you can still [get your puzzle
+input](16/input).
+

2024/16/solution.py

@@ -0,0 +1,87 @@
+#!/bin/python3
+import sys,time,re
+from pprint import pprint
+sys.path.insert(0, '../../')
+from fred import list2int,get_re,nprint,lprint,loadFile,dijkstra,toGrid,dfs,bfs,findInGrid,get_value_in_direction,addTuples,subTuples,create_graph_from_grid,manhattan_distance
+start_time = time.time()
+
+input_f = 'input'
+
+#########################################
+#                                       #
+#              Part 1                   #
+#                                       #
+#########################################
+
+def part1():
+    grid = toGrid(input_f)
+    
+    start = findInGrid(grid,'S')
+    end = findInGrid(grid,'E')
+    
+    print(start,end)
+    
+    def a_star(graph, start, end, heuristic):
+    
+        import heapq
+
+        # Priority queue stores (f_score, current_node, current_direction)
+        priority_queue = [(0, start, None)] 
+        g_scores = {node: float('inf') for node in graph}  # Cost from start to each node
+        g_scores[start] = 0
+
+        f_scores = {node: float('inf') for node in graph} 
+        f_scores[start] = heuristic(start, end)
+
+        previous_nodes = {node: None for node in graph}
+
+        while priority_queue:
+            current_f_score, current_node, previous_direction = heapq.heappop(priority_queue)
+
+            if current_node == end:
+                break
+
+            for neighbor, weight in graph[current_node]:
+                current_direction = (
+                    neighbor[0] - current_node[0],  # Row direction (-1, 0, 1)
+                    neighbor[1] - current_node[1]   # Col direction (-1, 0, 1)
+                )
+
+                turn_cost = 1000 if previous_direction and current_direction != previous_direction else 0
+                tentative_g_score = g_scores[current_node] + weight + turn_cost
+
+                if tentative_g_score < g_scores[neighbor]:
+                    g_scores[neighbor] = tentative_g_score
+                    f_scores[neighbor] = tentative_g_score + heuristic(neighbor, end)
+                    previous_nodes[neighbor] = current_node
+                    heapq.heappush(priority_queue, (f_scores[neighbor], neighbor, current_direction))
+
+        path = []
+        while end is not None:
+            path.append(end)
+            end = previous_nodes[end]
+        path.reverse()
+
+        return path, g_scores[path[-1]]
+
+    
+    graph = create_graph_from_grid(grid,start,end,'#')
+    
+    path, score = a_star(graph,start,end,manhattan_distance)
+    #print(path)
+    return score #Instead of implementing edge cases, just try with turning (adding 1000) to the result, if that's wrong, just don't add it. I got it right by not turning right away.
+    
+start_time = time.time()
+print('Part 1:',part1(), '\t\t', round((time.time() - start_time)*1000), 'ms')
+
+
+#########################################
+#                                       #
+#              Part 2                   #
+#                                       #
+#########################################
+def part2():
+    return
+    
+start_time = time.time()
+print('Part 2:',part2(), '\t\t', round((time.time() - start_time)*1000), 'ms')

2024/18/18.md

@@ -0,0 +1,144 @@
+## \-\-- Day 18: RAM Run \-\--
+
+You and The Historians look a lot more pixelated than you remember.
+You\'re [inside a computer](/2017/day/2) at the North Pole!
+
+Just as you\'re about to check out your surroundings, a program runs up
+to you. \"This region of memory isn\'t safe! The User misunderstood what
+a [pushdown
+automaton](https://en.wikipedia.org/wiki/Pushdown_automaton)
+is and their algorithm is pushing whole *bytes* down on top of us!
+Run!\"
+
+The algorithm is fast - it\'s going to cause a byte to fall into your
+memory space once every
+[nanosecond](https://www.youtube.com/watch?v=9eyFDBPk4Yw)!
+Fortunately, you\'re *faster*, and by quickly scanning the algorithm,
+you create a *list of which bytes will fall* (your puzzle input) in the
+order they\'ll land in your memory space.
+
+Your memory space is a two-dimensional grid with coordinates that range
+from `0` to `70` both horizontally and vertically. However, for the sake
+of example, suppose you\'re on a smaller grid with coordinates that
+range from `0` to `6` and the following list of incoming byte positions:
+
+    5,4
+    4,2
+    4,5
+    3,0
+    2,1
+    6,3
+    2,4
+    1,5
+    0,6
+    3,3
+    2,6
+    5,1
+    1,2
+    5,5
+    2,5
+    6,5
+    1,4
+    0,4
+    6,4
+    1,1
+    6,1
+    1,0
+    0,5
+    1,6
+    2,0
+
+Each byte position is given as an `X,Y` coordinate, where `X` is the
+distance from the left edge of your memory space and `Y` is the distance
+from the top edge of your memory space.
+
+You and The Historians are currently in the top left corner of the
+memory space (at `0,0`) and need to reach the exit in the bottom right
+corner (at `70,70` in your memory space, but at `6,6` in this example).
+You\'ll need to simulate the falling bytes to plan out where it will be
+safe to run; for now, simulate just the first few bytes falling into
+your memory space.
+
+As bytes fall into your memory space, they make that coordinate
+*corrupted*. Corrupted memory coordinates cannot be entered by you or
+The Historians, so you\'ll need to plan your route carefully. You also
+cannot leave the boundaries of the memory space; your only hope is to
+reach the exit.
+
+In the above example, if you were to draw the memory space after the
+first `12` bytes have fallen (using `.` for safe and `#` for corrupted),
+it would look like this:
+
+    ...#...
+    ..#..#.
+    ....#..
+    ...#..#
+    ..#..#.
+    .#..#..
+    #.#....
+
+You can take steps up, down, left, or right. After just 12 bytes have
+corrupted locations in your memory space, the shortest path from the top
+left corner to the exit would take `22` steps. Here (marked with `O`) is
+one such path:
+
+    OO.#OOO
+    .O#OO#O
+    .OOO#OO
+    ...#OO#
+    ..#OO#.
+    .#.O#..
+    #.#OOOO
+
+Simulate the first kilobyte (`1024` bytes) falling onto your memory
+space. Afterward, *what is the minimum number of steps needed to reach
+the exit?*
+
+Your puzzle answer was `294`.
+
+The first half of this puzzle is complete! It provides one gold star: \*
+
+## \-\-- Part Two \-\-- {#part2}
+
+The Historians aren\'t as used to moving around in this pixelated
+universe as you are. You\'re afraid they\'re not going to be fast enough
+to make it to the exit before the path is completely blocked.
+
+To determine how fast everyone needs to go, you need to determine *the
+first byte that will cut off the path to the exit*.
+
+In the above example, after the byte at `1,1` falls, there is still a
+path to the exit:
+
+    O..#OOO
+    O##OO#O
+    O#OO#OO
+    OOO#OO#
+    ###OO##
+    .##O###
+    #.#OOOO
+
+However, after adding the very next byte (at `6,1`), there is no longer
+a path to the exit:
+
+    ...#...
+    .##..##
+    .#..#..
+    ...#..#
+    ###..##
+    .##.###
+    #.#....
+
+So, in this example, the coordinates of the first byte that prevents the
+exit from being reachable are `6,1`.
+
+Simulate more of the bytes that are about to corrupt your memory space.
+*What are the coordinates of the first byte that will prevent the exit
+from being reachable from your starting position?* (Provide the answer
+as two integers separated by a comma with no other characters.)
+
+Answer:
+
+Although it hasn\'t changed, you can still [get your puzzle
+input](18/input).
+

2024/18/solution.py

@@ -0,0 +1,76 @@
+#!/bin/python3
+import sys,time,re
+from pprint import pprint
+sys.path.insert(0, '../../')
+from fred import list2int,get_re,nprint,lprint,loadFile,bfs,get_value_in_direction,addTuples,grid_valid
+start_time = time.time()
+
+input_f = 'input'
+
+#########################################
+#                                       #
+#              Part 1                   #
+#                                       #
+#########################################
+def part1():
+    instructions = []
+    grid = []
+    w = 70
+    h = 70
+    end = (h,w)
+    start = (0,0)
+    with open(input_f) as file:
+        for line in file:
+            l = list2int(line.rstrip().split(','))
+            instructions.append((l[0],l[1]))
+    #print(instructions)
+    
+    grid = [[ '.' for x in range(0,w+1)] for y in range(0,h+1)]
+    
+    for i in range(1024):
+        x = instructions[i]
+        grid[x[0]][x[1]] = '#'
+    
+
+    
+    def is_goal(node):
+        #print(node)
+        
+        return True if node == end else False
+    
+    def get_neighbors(node):
+        directions = ['up','down','left','right']
+        offsets = {
+            'up': (-1, 0),
+            'down': (1, 0),
+            'left': (0, -1),
+            'right': (0, 1),
+        }
+        neighbors = []
+
+        # Loop through all the directions
+        for d in directions:
+            tmp = addTuples(offsets[d],node)
+            if get_value_in_direction(grid,node,d) != '#' and grid_valid(tmp[0],tmp[1],grid):
+                neighbors.append((tmp[0],tmp[1]))
+        # Return the list of valid neighbors
+        return neighbors
+    
+    goal_nodes, path = bfs((0,0),is_goal,get_neighbors)
+    print(goal_nodes)
+    return len(path[goal_nodes[0]])-1
+    
+start_time = time.time()
+print('Part 1:',part1(), '\t\t', round((time.time() - start_time)*1000), 'ms')
+
+
+#########################################
+#                                       #
+#              Part 2                   #
+#                                       #
+#########################################
+def part2():
+    return
+    
+start_time = time.time()
+print('Part 2:',part2(), '\t\t', round((time.time() - start_time)*1000), 'ms')

fred.py

@@ -2,6 +2,17 @@ import sys,re,heapq
 from itertools import permutations
 from termcolor import colored
 
+def findInGrid(grid:list,x:str) -> set:
+    """Find the location of a character in a grid.
+
+
+    """
+    
+    for r,row in enumerate(grid):
+        for c, char in enumerate(row):
+            if char == x:
+                return (r,c)
+
 def loadFile(input_f):
     """
     Loads a file and returns its lines as a list.
@@ -179,7 +190,6 @@ def dprint(graph: dict):
     except Exception as e:
         raise RuntimeError(f"An error occurred while printing the dictionary: {e}")
 
-
 def lprint(x: str, log: bool):
     """
     Prints a string if logging is enabled.
@@ -271,7 +281,10 @@ def nprint(grid, cur: set = None, sign: str = None, positions:list = None):
     for idx, i in enumerate(grid):
         for jdx, j in enumerate(i):
             if jdx == 0:
-                prepend = str(idx)+' '
+                if idx < 10:
+                    prepend = ' '+str(idx)+' '
+                else:
+                    prepend = ''+str(idx)+' '
             else: 
                 prepend = ''
             
@@ -286,7 +299,7 @@ def nprint(grid, cur: set = None, sign: str = None, positions:list = None):
             else:
                 if positions is not None:
                     if (idx,jdx) in positions:
-                        print(prepend+colored(grid[idx][jdx],'red'),end=' ')
+                        print(prepend+colored(grid[idx][jdx],'green'),end=' ')
                     else:
                         print(prepend+grid[idx][jdx], end=' ')
                 else:
@@ -296,8 +309,26 @@ def nprint(grid, cur: set = None, sign: str = None, positions:list = None):
     
     for r in range(len(grid[0])):
         if r == 0:
+            print('   0',end='')
+        
+        else:
+            if r%2 == 0:
+                print(r,end='')
+            else:
                 print('   ',end='')
-        print(r,end=' ')
+    print()
+    for r in range(len(grid[0])):
+        if r == 0:
+            print('    ',end='')
+        
+        else:
+            if r%2 != 0:
+                print(r,end='')
+            else:
+                if r != 9:
+                    print(' ',end='')
+                else:
+                    print('',end='')
     print()
 
 def list2int(x):
@@ -647,6 +678,9 @@ def dfs(grid:list, pos:set) -> list:
     dfs(pos[0], pos[1])
     return list(visited)
 
+def manhattan_distance(node, goal):
+    return abs(node[0] - goal[0]) + abs(node[1] - goal[1])
+
 # Should probably be added to the regular dfs.
 def flood_fill(cells, pos):
     """  
@@ -678,3 +712,82 @@ def flood_fill(cells, pos):
 
     dfs(pos)
     return list(visited)
+
+def create_graph_from_grid(grid, start, end,wall):
+    """
+    Converts a grid into a graph representation for use in pathfinding algorithms.
+
+    Parameters:
+    - grid: A 2D list representing the grid, where '#' is a wall, and '.' is a path.
+    - start: The coordinates of the start node (row, col).
+    - end: The coordinates of the end node (row, col).
+    - wall: The string of the wall node.
+    
+    Returns:
+    - graph: A dictionary where each key is a node (row, col), and the value is a list of tuples.
+             Each tuple represents (neighbor, weight).
+    """
+    rows, cols = len(grid), len(grid[0])
+    graph = {}
+
+    # Helper to get neighbors
+    def neighbors(r, c):
+        for dr, dc in [(-1, 0), (1, 0), (0, -1), (0, 1)]:  # Up, Down, Left, Right
+            nr, nc = r + dr, c + dc
+            if 0 <= nr < rows and 0 <= nc < cols and grid[nr][nc] != '#':
+                yield (nr, nc)
+
+    for r in range(rows):
+        for c in range(cols):
+            if grid[r][c] != '#':  # Only process valid cells
+                graph[(r, c)] = [(neighbor, 1) for neighbor in neighbors(r, c)]  # Weight is 1
+
+    return graph
+
+def a_star(graph, start, end, heuristic):
+    """
+    A* Algorithm to find the shortest path between two nodes in a graph.
+
+    Parameters:
+    - graph: A dictionary where each key is a node, and the value is a list of tuples.
+             Each tuple represents (neighbor, distance).
+    - start: The starting node (row, col).
+    - end: The destination node (row, col).
+    - heuristic: A function that estimates the cost from a node to the end.
+
+    Returns:
+    - path: A list of nodes that represent the shortest path from start to end.
+    - total_cost: The total cost of the shortest path.
+    """
+    priority_queue = [(0, start)]  # (f_score, current_node)
+    g_scores = {node: float('inf') for node in graph}  # Cost from start to each node
+    g_scores[start] = 0
+
+    f_scores = {node: float('inf') for node in graph}  # Estimated total cost (g + h)
+    f_scores[start] = heuristic(start, end)
+
+    previous_nodes = {node: None for node in graph}
+
+    while priority_queue:
+        current_f_score, current_node = heapq.heappop(priority_queue)
+
+        if current_node == end:
+            break
+
+        for neighbor, weight in graph[current_node]:
+            tentative_g_score = g_scores[current_node] + weight
+
+            if tentative_g_score < g_scores[neighbor]:
+                g_scores[neighbor] = tentative_g_score
+                f_scores[neighbor] = tentative_g_score + heuristic(neighbor, end)
+                previous_nodes[neighbor] = current_node
+                heapq.heappush(priority_queue, (f_scores[neighbor], neighbor))
+
+    # Reconstruct the path
+    path = []
+    while end is not None:
+        path.append(end)
+        end = previous_nodes[end]
+    path.reverse()
+
+    return path, g_scores[path[-1]]