mirror of
https://github.com/TheAlgorithms/Python.git
synced 2024-10-06 21:59:30 +00:00
0f09530707
```
=============================== warnings summary ===============================
graphs/dijkstra.py:81
/home/travis/build/TheAlgorithms/Python/graphs/dijkstra.py:81: DeprecationWarning: invalid escape sequence \
"""
graphs/dijkstra.py:97
/home/travis/build/TheAlgorithms/Python/graphs/dijkstra.py:97: DeprecationWarning: invalid escape sequence \
"""
-- Docs: https://docs.pytest.org/en/latest/warnings.html
=================== 126 passed, 7 warnings in 19.35 seconds ====================
```
119 lines
3.1 KiB
Python
119 lines
3.1 KiB
Python
"""pseudo-code"""
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"""
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DIJKSTRA(graph G, start vertex s, destination vertex d):
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//all nodes initially unexplored
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1 - let H = min heap data structure, initialized with 0 and s [here 0 indicates
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the distance from start vertex s]
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2 - while H is non-empty:
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3 - remove the first node and cost of H, call it U and cost
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4 - if U has been previously explored:
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5 - go to the while loop, line 2 //Once a node is explored there is no need
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to make it again
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6 - mark U as explored
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7 - if U is d:
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8 - return cost // total cost from start to destination vertex
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9 - for each edge(U, V): c=cost of edge(U,V) // for V in graph[U]
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10 - if V explored:
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11 - go to next V in line 9
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12 - total_cost = cost + c
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13 - add (total_cost,V) to H
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You can think at cost as a distance where Dijkstra finds the shortest distance
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between vertexes s and v in a graph G. The use of a min heap as H guarantees
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that if a vertex has already been explored there will be no other path with
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shortest distance, that happens because heapq.heappop will always return the
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next vertex with the shortest distance, considering that the heap stores not
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only the distance between previous vertex and current vertex but the entire
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distance between each vertex that makes up the path from start vertex to target
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vertex.
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"""
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import heapq
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def dijkstra(graph, start, end):
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"""Return the cost of the shortest path between vertexes start and end.
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>>> dijkstra(G, "E", "C")
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6
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>>> dijkstra(G2, "E", "F")
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3
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>>> dijkstra(G3, "E", "F")
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3
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"""
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heap = [(0, start)] # cost from start node,end node
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visited = set()
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while heap:
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(cost, u) = heapq.heappop(heap)
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if u in visited:
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continue
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visited.add(u)
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if u == end:
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return cost
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for v, c in graph[u]:
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if v in visited:
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continue
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next = cost + c
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heapq.heappush(heap, (next, v))
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return -1
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G = {
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"A": [["B", 2], ["C", 5]],
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"B": [["A", 2], ["D", 3], ["E", 1], ["F", 1]],
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"C": [["A", 5], ["F", 3]],
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"D": [["B", 3]],
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"E": [["B", 4], ["F", 3]],
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"F": [["C", 3], ["E", 3]],
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}
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r"""
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Layout of G2:
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E -- 1 --> B -- 1 --> C -- 1 --> D -- 1 --> F
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\ /\
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\ ||
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----------------- 3 --------------------
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"""
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G2 = {
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"B": [["C", 1]],
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"C": [["D", 1]],
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"D": [["F", 1]],
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"E": [["B", 1], ["F", 3]],
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"F": [],
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}
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r"""
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Layout of G3:
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E -- 1 --> B -- 1 --> C -- 1 --> D -- 1 --> F
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\ /\
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\ ||
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-------- 2 ---------> G ------- 1 ------
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"""
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G3 = {
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"B": [["C", 1]],
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"C": [["D", 1]],
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"D": [["F", 1]],
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"E": [["B", 1], ["G", 2]],
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"F": [],
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"G": [["F", 1]],
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}
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shortDistance = dijkstra(G, "E", "C")
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print(shortDistance) # E -- 3 --> F -- 3 --> C == 6
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shortDistance = dijkstra(G2, "E", "F")
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print(shortDistance) # E -- 3 --> F == 3
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shortDistance = dijkstra(G3, "E", "F")
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print(shortDistance) # E -- 2 --> G -- 1 --> F == 3
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if __name__ == "__main__":
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import doctest
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doctest.testmod()
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