BFS, baby. 2 things got me. * S position was not (0,0) in the puzzle input. Such embarrasment to loose time over something so stupid. * I asumed the elevation was fixed to to increase AND DECREASE by 1. From the problem description: > To avoid needing to get out your climbing gear, the elevation of the > destination square can be at most one higher than the elevation of > your current square; that is, if your current elevation is m, you > could step to elevation n, but not to elevation o. _(This also means > that the elevation of the destination square can be much lower than > the elevation of your current square.)_ This means elevation only can _increase_ by one, but _decrease_ with more. So my wrong code: 0 <= abs(n - e) <= 1 got fixed with: n - e <= 1 For pt 2, I reused the loop to find S to collect all "a" positions, and looped them as starting points. Execution time is not the best, but it works.
201 lines
4.9 KiB
Python
201 lines
4.9 KiB
Python
import re
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# Directions/Adjacents for 2D matrices, in the order UP, RIGHT, DOWN, LEFT
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D = [
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(-1, 0),
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(0, 1),
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(1, 0),
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(0, -1),
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]
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Di = [
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(-1, -1),
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(-1, 1),
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(1, -1),
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(1, 1),
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]
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# Directions for 2D matrices, as a dict with keys U, R, D, L
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DD = {
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"U": (-1, 0),
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"R": (0, 1),
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"D": (1, 0),
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"L": (0, -1),
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}
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DDa = {
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"^": (-1, 0),
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">": (0, 1),
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"v": (1, 0),
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"<": (0, -1),
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}
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# Adjacent relative positions including diagonals for 2D matrices, in the order NW, N, NE, W, E, SW, S, SE
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ADJ = [
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(-1, -1),
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(-1, 0),
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(1, -1),
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(0, -1),
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(0, 1),
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(1, 1),
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(1, 0),
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(1, -1),
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]
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def ints(s):
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"""Extract all integers from a string"""
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return [int(n) for n in re.findall(r"\d+", s)]
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def sints(s):
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"""Extract all signed integers from a string"""
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return [int(n) for n in re.findall(r"-?\d+", s)]
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def mhd(a, b):
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"""Calculates the Manhattan distance between 2 positions in the format (y, x) or (x, y)"""
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ar, ac = a
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br, bc = b
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return abs(ar - br) + abs(ac - bc)
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def matrix(d):
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"""Transform a string into an iterable matrix. Returns the matrix, row count and col count"""
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m = [tuple(r) for r in d.split()]
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return m, len(m), len(m[0])
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def mdbg(m):
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"""Print-debug a matrix"""
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for r in m:
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print("".join(r))
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def vdbg(seen, h, w):
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"""Print-debug visited positions of a matrix"""
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for r in range(h):
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print("".join(["#" if (r, c) in seen else " " for c in range(w)]))
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def vvdbg(seen, h, w):
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"""Print-debug visited positions of a matrix, with values"""
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for r in range(h):
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print("".join([seen[(r, c)] if (r, c) in seen else "." for c in range(w)]))
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def cw(y, x):
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"""Flip a (y, x) direction counterwise: U->R, R->D, D->L, L->U.
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>>> cw(-1, 0)
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(0, 1)
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>>> cw(0, 1)
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(1, 0)
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>>> cw(1, 0)
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(0, -1)
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>>> cw(0, -1)
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(-1, 0)
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"""
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return (x, y) if y == 0 else (x, -y)
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def ccw(y, x):
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"""Flip a (y, x) direction counterwise: U->L, L->D, D->R, R->U.
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>>> ccw(-1, 0)
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(0, -1)
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>>> ccw(0, -1)
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(1, 0)
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>>> ccw(1, 0)
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(0, 1)
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>>> ccw(0, 1)
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(-1, 0)
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"""
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return (x, y) if x == 0 else (-x, y)
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def bfs(S, E=None):
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"""BFS algorithm, equal weighted nodes"""
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seen = set()
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q = [(S, 0)]
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g = {} # graph, required to be provided at some point
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while q:
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m, w = q.pop(0)
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if m in seen:
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continue
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seen.add(m)
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# investigate here
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for s in g[m]:
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q.append((s, w + 1))
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# return insights
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def mhd_search(r, c, R=20):
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"""returns all coords that are within R manhattan distance from (r,c)"""
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p = set()
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for d in range(1, R + 1):
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p.add((r, c + d))
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p.add((r, c - d))
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p.add((r + d, c))
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p.add((r - d, c))
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for dd in range(d):
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p.add((r - dd, c - d + dd))
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p.add((r + dd, c - d + dd))
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p.add((r - dd, c - dd + d))
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p.add((r + dd, c - dd + d))
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return p
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def dijkstras(grid, start, target):
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"""
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1. Create an array that holds the distance of each vertex from the starting
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vertex. Initially, set this distance to infinity for all vertices except
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the starting vertex which should be set to 0.
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2. Create a priority queue (heap) and insert the starting vertex with its
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distance of 0.
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3. While there are still vertices left in the priority queue, select the vertex
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with the smallest recorded distance from the starting vertex and visit its
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neighboring vertices.
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4. For each neighboring vertex, check if it is visited already or not. If it
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isn’t visited yet, calculate its tentative distance by adding its weight
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to the smallest distance found so far for its parent/previous node
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(starting vertex in case of first-level vertices).
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5. If this tentative distance is smaller than previously recorded value
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(if any), update it in our ‘distances’ array.
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6. Finally, add this visited vertex with its updated distance to our priority
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queue and repeat step-3 until we have reached our destination or exhausted
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all nodes.
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"""
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import heapq
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target = max(grid)
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seen = set()
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queue = [(start, 0)]
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while queue:
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cost, pos, direction, steps = heapq.heappop(queue)
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y, x = pos
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dy, dx = direction
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if pos == target:
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return cost
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if ((pos, "and stuff")) in seen:
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continue
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seen.add((pos, "and stuff"))
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neighbors = []
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for n in neighbors:
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heapq.heappush(queue, ("stuffs"))
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return -1
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def bk(graph, p, r=set(), x=set()):
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"""Bron-Kerbosch algoritm, no pivot: https://en.wikipedia.org/wiki/Bron%E2%80%93Kerbosch_algorithm"""
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if not p and not x:
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yield r
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while p:
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v = p.pop()
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yield from bk(graph, p & set(graph[v]), r | {v}, x & graph[v])
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x.add(v)
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