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72 lines
2.0 KiB
Python
72 lines
2.0 KiB
Python
# Eulerian Path is a path in graph that visits every edge exactly once.
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# Eulerian Circuit is an Eulerian Path which starts and ends on the same
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# vertex.
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# time complexity is O(V+E)
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# space complexity is O(VE)
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# using dfs for finding eulerian path traversal
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def dfs(u, graph, visited_edge, path=None):
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path = (path or []) + [u]
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for v in graph[u]:
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if visited_edge[u][v] is False:
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visited_edge[u][v], visited_edge[v][u] = True, True
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path = dfs(v, graph, visited_edge, path)
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return path
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# for checking in graph has euler path or circuit
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def check_circuit_or_path(graph, max_node):
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odd_degree_nodes = 0
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odd_node = -1
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for i in range(max_node):
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if i not in graph:
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continue
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if len(graph[i]) % 2 == 1:
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odd_degree_nodes += 1
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odd_node = i
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if odd_degree_nodes == 0:
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return 1, odd_node
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if odd_degree_nodes == 2:
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return 2, odd_node
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return 3, odd_node
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def check_euler(graph, max_node):
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visited_edge = [[False for _ in range(max_node + 1)] for _ in range(max_node + 1)]
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check, odd_node = check_circuit_or_path(graph, max_node)
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if check == 3:
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print("graph is not Eulerian")
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print("no path")
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return
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start_node = 1
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if check == 2:
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start_node = odd_node
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print("graph has a Euler path")
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if check == 1:
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print("graph has a Euler cycle")
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path = dfs(start_node, graph, visited_edge)
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print(path)
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def main():
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g1 = {1: [2, 3, 4], 2: [1, 3], 3: [1, 2], 4: [1, 5], 5: [4]}
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g2 = {1: [2, 3, 4, 5], 2: [1, 3], 3: [1, 2], 4: [1, 5], 5: [1, 4]}
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g3 = {1: [2, 3, 4], 2: [1, 3, 4], 3: [1, 2], 4: [1, 2, 5], 5: [4]}
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g4 = {1: [2, 3], 2: [1, 3], 3: [1, 2]}
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g5 = {
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1: [],
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2: [],
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# all degree is zero
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}
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max_node = 10
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check_euler(g1, max_node)
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check_euler(g2, max_node)
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check_euler(g3, max_node)
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check_euler(g4, max_node)
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check_euler(g5, max_node)
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if __name__ == "__main__":
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main()
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