Added Dijkstra's shortest path problem

Author: Farai Mugaviri

searches/dijkstra.py

@@ -0,0 +1,85 @@
+"""Dijkstra's shortest path algorithm.
+
+This module provides a simple Dijkstra implementation that works on a
+graph represented as an adjacency mapping: {node: [(neighbor, weight), ...], ...}.
+
+Functions:
+- dijkstra(graph, source) -> (dist, prev)
+- shortest_path(prev, target) -> list
+
+Doctests include a small example graph.
+"""
+from __future__ import annotations
+
+import heapq
+from typing import Dict, Iterable, List, Tuple, Any
+
+
+def dijkstra(graph: Dict[Any, Iterable[Tuple[Any, float]]], source: Any) -> Tuple[Dict[Any, float], Dict[Any, Any]]:
+    """Compute shortest path distances from source to all reachable nodes.
+
+    Args:
+        graph: adjacency mapping where graph[u] yields (v, weight) pairs.
+        source: start node.
+
+    Returns:
+        (dist, prev)
+        - dist: mapping node -> distance (float). Unreachable nodes are absent.
+        - prev: mapping node -> predecessor on shortest path (or None for source).
+
+    Example:
+    >>> graph = {
+    ...     'A': [('B', 1), ('C', 4)],
+    ...     'B': [('C', 2), ('D', 5)],
+    ...     'C': [('D', 1)],
+    ...     'D': []
+    ... }
+    >>> dist, prev = dijkstra(graph, 'A')
+    >>> dist['D']
+    4
+    >>> shortest_path(prev, 'D')
+    ['A', 'B', 'C', 'D']
+    """
+    dist: Dict[Any, float] = {}
+    prev: Dict[Any, Any] = {}
+    pq: List[Tuple[float, Any]] = []  # (distance, node)
+
+    heapq.heappush(pq, (0.0, source))
+    dist[source] = 0.0
+    prev[source] = None
+
+    while pq:
+        d, u = heapq.heappop(pq)
+        # Skip stale entries
+        if d != dist.get(u, float('inf')):
+            continue
+        for v, w in graph.get(u, []):
+            nd = d + float(w)
+            if nd < dist.get(v, float('inf')):
+                dist[v] = nd
+                prev[v] = u
+                heapq.heappush(pq, (nd, v))
+
+    return dist, prev
+
+
+def shortest_path(prev: Dict[Any, Any], target: Any) -> List[Any]:
+    """Reconstruct path from source to target using predecessor map.
+
+    If target is not in `prev`, returns an empty list.
+    """
+    if target not in prev:
+        return []
+    path: List[Any] = []
+    cur = target
+    while cur is not None:
+        path.append(cur)
+        cur = prev.get(cur)
+    path.reverse()
+    return path
+
+
+if __name__ == "__main__":
+    import doctest
+
+    doctest.testmod()