Solved 2024/18 P1
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## \-\-- Day 16: Reindeer Maze \-\--
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It\'s time again for the [Reindeer Olympics](/2015/day/14)! This year,
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the big event is the *Reindeer Maze*, where the Reindeer compete for the
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*[lowest
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score]{title="I would say it's like Reindeer Golf, but knowing Reindeer, it's almost certainly nothing like Reindeer Golf."}*.
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You and The Historians arrive to search for the Chief right as the event
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is about to start. It wouldn\'t hurt to watch a little, right?
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The Reindeer start on the Start Tile (marked `S`) facing *East* and need
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to reach the End Tile (marked `E`). They can move forward one tile at a
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time (increasing their score by `1` point), but never into a wall (`#`).
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They can also rotate clockwise or counterclockwise 90 degrees at a time
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(increasing their score by `1000` points).
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To figure out the best place to sit, you start by grabbing a map (your
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puzzle input) from a nearby kiosk. For example:
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###############
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#.......#....E#
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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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#S..#.....#...#
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###############
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There are many paths through this maze, but taking any of the best paths
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would incur a score of only `7036`. This can be achieved by taking a
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total of `36` steps forward and turning 90 degrees a total of `7` times:
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###############
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#.......#....E#
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#.#.###.#.###^#
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#.....#.#...#^#
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#.###.#####.#^#
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#.#.#.......#^#
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#.#.#####.###^#
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#..>>>>>>>>v#^#
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###^#.#####v#^#
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#>>^#.....#v#^#
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#^#.#.###.#v#^#
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#^....#...#v#^#
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#^###.#.#.#v#^#
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#S..#.....#>>^#
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###############
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Here\'s a second example:
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#################
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#...#...#...#..E#
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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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#S#.............#
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#################
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In this maze, the best paths cost `11048` points; following one such
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path would look like this:
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#################
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#...#...#...#..E#
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#.#.#.#.#.#.#.#^#
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#.#.#.#...#...#^#
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#.#.#.#.###.#.#^#
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#>>v#.#.#.....#^#
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#^#v#.#.#.#####^#
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#^#v..#.#.#>>>>^#
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#^#v#####.#^###.#
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#^#v#..>>>>^#...#
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#^#v###^#####.###
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#^#v#>>^#.....#.#
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#^#v#^#####.###.#
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#^#v#^........#.#
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#^#v#^#########.#
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#S#>>^..........#
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#################
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Note that the path shown above includes one 90 degree turn as the very
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first move, rotating the Reindeer from facing East to facing North.
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Analyze your map carefully. *What is the lowest score a Reindeer could
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possibly get?*
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Your puzzle answer was `108504`.
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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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Now that you know what the best paths look like, you can figure out the
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best spot to sit.
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Every non-wall tile (`S`, `.`, or `E`) is equipped with places to sit
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along the edges of the tile. While determining which of these tiles
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would be the best spot to sit depends on a whole bunch of factors (how
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comfortable the seats are, how far away the bathrooms are, whether
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there\'s a pillar blocking your view, etc.), the most important factor
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is *whether the tile is on one of the best paths through the maze*. If
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you sit somewhere else, you\'d miss all the action!
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So, you\'ll need to determine which tiles are part of *any* best path
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through the maze, including the `S` and `E` tiles.
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In the first example, there are `45` tiles (marked `O`) that are part of
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at least one of the various best paths through the maze:
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###############
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#.......#....O#
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#.#.###.#.###O#
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#.....#.#...#O#
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#.###.#####.#O#
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#.#.#.......#O#
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#.#.#####.###O#
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#..OOOOOOOOO#O#
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###O#O#####O#O#
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#OOO#O....#O#O#
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#O#O#O###.#O#O#
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#OOOOO#...#O#O#
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#O###.#.#.#O#O#
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#O..#.....#OOO#
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###############
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In the second example, there are `64` tiles that are part of at least
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one of the best paths:
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#################
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#...#...#...#..O#
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#.#.#.#.#.#.#.#O#
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#.#.#.#...#...#O#
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#.#.#.#.###.#.#O#
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#OOO#.#.#.....#O#
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#O#O#.#.#.#####O#
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#O#O..#.#.#OOOOO#
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#O#O#####.#O###O#
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#O#O#..OOOOO#OOO#
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#O#O###O#####O###
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#O#O#OOO#..OOO#.#
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#O#O#O#####O###.#
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#O#O#OOOOOOO..#.#
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#O#O#O#########.#
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#O#OOO..........#
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#################
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Analyze your map further. *How many tiles are part of at least one of
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the best paths through the maze?*
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Answer:
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Although it hasn\'t changed, you can still [get your puzzle
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input](16/input).
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@@ -0,0 +1,87 @@
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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,dijkstra,toGrid,dfs,bfs,findInGrid,get_value_in_direction,addTuples,subTuples,create_graph_from_grid,manhattan_distance
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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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grid = toGrid(input_f)
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start = findInGrid(grid,'S')
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end = findInGrid(grid,'E')
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print(start,end)
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def a_star(graph, start, end, heuristic):
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import heapq
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# Priority queue stores (f_score, current_node, current_direction)
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priority_queue = [(0, start, None)]
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g_scores = {node: float('inf') for node in graph} # Cost from start to each node
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g_scores[start] = 0
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f_scores = {node: float('inf') for node in graph}
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f_scores[start] = heuristic(start, end)
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previous_nodes = {node: None for node in graph}
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while priority_queue:
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current_f_score, current_node, previous_direction = heapq.heappop(priority_queue)
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if current_node == end:
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break
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for neighbor, weight in graph[current_node]:
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current_direction = (
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neighbor[0] - current_node[0], # Row direction (-1, 0, 1)
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neighbor[1] - current_node[1] # Col direction (-1, 0, 1)
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)
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turn_cost = 1000 if previous_direction and current_direction != previous_direction else 0
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tentative_g_score = g_scores[current_node] + weight + turn_cost
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if tentative_g_score < g_scores[neighbor]:
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g_scores[neighbor] = tentative_g_score
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f_scores[neighbor] = tentative_g_score + heuristic(neighbor, end)
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previous_nodes[neighbor] = current_node
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heapq.heappush(priority_queue, (f_scores[neighbor], neighbor, current_direction))
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path = []
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while end is not None:
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path.append(end)
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end = previous_nodes[end]
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path.reverse()
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return path, g_scores[path[-1]]
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graph = create_graph_from_grid(grid,start,end,'#')
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path, score = a_star(graph,start,end,manhattan_distance)
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#print(path)
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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.
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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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