Transformation dictionary - attempt 1

.gitignore

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+/__pycache__

transform_dict.py

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+# Distance between two words
+def distance(a, b):
+    return sum([int(ord(ac) == ord(bc)) for ac, bc in zip(list(a), list(b))])
+
+
+# Create a map for each word with the list of other words in the set
+# which are the given distance from the key word
+def dist_map(words, dist):
+    dist_map = {}
+    for w in words:
+        dist_map[w] = [y for y in words if distance(w, y) == dist]
+
+    return dist_map
+
+
+# Shortest path? Technically we only need to find if there *is* a path
+# so there might be a simpler way to do this.
+
+# For shortest path, should use something like A* with the edge weight
+# being the relative difference in distance from current word to the
+# target word
+
+# But I don't know A* actually, so I'm going to just try a DFS that
+# prioritizes in order of the edge weight, hopefully that's close
+# I think the real thing might use the sum of all the edge weights from
+# the word on the other side of each edge, when calculating the weights
+# of that edge. I also think they might say 'cost' instead of 'weight' --
+# either way we want to minimize that value for the shortest path.
+
+# So lets find any path at all
+def has_path(start, end, dist, dmap, visited):
+    if dist < 2:
+        return True
+
+    visited.add(start)
+    words = dmap[start]
+    dists = [(w, distance(w, end)) for w in words]
+    for w, d in sorted(dists, key=lambda t: t[1] - dist):
+        if w in visited:
+            continue
+        if has_path(w, end, d, dmap, visited):
+            return True
+    return False
+
+
+def is_transformable(start, end, d):
+    dmap = dist_map(d + [start, end], 1)
+    return has_path(start, end, distance(start, end), dmap, set())