Solve 2025 day 7 pt 1-2

For part 1, BFS is used since there is no need to
visit each splitter more than once to count visited
splitters.

For part 2, the code initially removed the visited
check. This failed miserably, losing momentum aound
Y=54-56.

A recursive function with memoization solves it much
faster.

2025-python/output/day_07.py

@@ -0,0 +1,50 @@
+import functools
+from output import grid
+
+
+def solve(data):
+    G = grid(data, o="^")
+    p1 = set()
+    p2 = 0
+    H = len(data.split())
+    W = len(data.split()[0])
+    S = (0, data.split()[0].index("S"))
+    Q = [S]
+    while Q:
+        y, x = Q.pop(0)
+        if y == H:
+            continue
+        if (y, x) in p1:
+            continue
+        if (y, x) in G:
+            Q.append((y, x - 1))
+            Q.append((y, x + 1))
+            p1.add((y, x))
+        else:
+            Q.append((y + 1, x))
+    p1 = len(p1)
+
+    @functools.cache
+    def _timelines(p):
+        y, x = p
+        if not 0 <= y < H or not 0 <= x < W:
+            return 1
+        if p in G:
+            return _timelines((y, x - 1)) + _timelines((y, x + 1))
+        return _timelines((y + 1, x))
+
+    p2 = _timelines(S)
+    return p1, p2
+
+
+if __name__ == "__main__":
+    with open("./input/07.txt", "r") as f:
+        inp = f.read().strip()
+
+    p1, p2 = solve(inp)
+
+    print(p1)
+    print(p2)
+
+    assert p1 == 1539
+    assert p2 == 6479180385864