Day 8

day08.py

@@ -0,0 +1,104 @@
+#!/usr/bin/env python
+
+# https://adventofcode.com/2025/day/8
+
+import math
+
+f = open("day08input.txt", "r")
+#f = open("testinput.txt", "r")
+jbox = []
+for line in f.readlines():
+  line = line.strip().split(",")
+  jbox.append(list(map(int, line)))
+
+
+def print2d(field):
+  for line in field:
+    print("\t".join(["{:6.2f}".format(el) if el else "None" for el in line]))
+
+
+def get_distance(a: list[int], b: list[int]) -> float:
+  assert len(a) == len(b)
+  return math.sqrt(sum([(a[i] - b[i]) ** 2 for i in range(len(a))]))
+
+
+
+def solve(jbox: list[list[int]], connections: int) -> int:
+  if connections >= 0:
+    print(f"Solving part 1: Adding {connections} connections")
+  else:
+    print(f"Solving part 2: Adding connections until all boxes are on the same circuit")
+
+  l = len(jbox)
+  
+  # calculate all distances
+  distance = [[None] * l] * l
+  for p in range(l):
+    distance[p] = [get_distance(jbox[p], jbox[q]) if q > p else None for q in range(l)]
+
+  # start with one-box circuits
+  circuits = [[i] for i in range(l)]
+
+  # connect junction boxes to form circuits
+  conn_count = 0
+  while True:
+    print(f"{conn_count=}")
+
+    # find current shortest distance
+    min_pq = None
+    for p in range(l - 1):
+      line = [distance[p][q] for q in range(p + 1, l)]
+      min_q = distance[p].index(min(list(filter(lambda x: x, line))))
+      # check if distance wasn't used before
+      if min_q:
+        # check if distance is shorter than the shortest distance found so far
+        if not min_pq or distance[p][min_q] < distance[min_pq[0]][min_pq[1]]:
+          min_pq = (p, min_q)
+    assert min_pq is not None
+
+    # find the circuits the two boxes with the shortest distance to each other belong to
+    circuit_p, circuit_q = None, None
+    for i in range(len(circuits)):
+      if min_pq[0] in circuits[i]:
+        circuit_p = i
+      if min_pq[1] in circuits[i]:
+        circuit_q = i
+      if circuit_p and circuit_q:
+        break
+
+    # make sure both boxes are on any circuits 
+    assert circuit_p is not None and circuit_q is not None
+
+    # make new connection
+    if circuit_p == circuit_q:
+      print(f"Boxes {min_pq[0]} and {min_pq[1]} are on the same circuit -> there are still {len(circuits)} circuits")
+    else:
+      print(f"Boxes {min_pq[0]} and {min_pq[1]} are on different circuits -> combining circuits {circuit_p} and {circuit_q} -> there are now {len(circuits) - 1} circuits")
+      if circuit_p < circuit_q:
+        circuits[circuit_p] += circuits.pop(circuit_q)
+      else:
+        circuits[circuit_q] += circuits.pop(circuit_p)
+      assert sum([len(c) for c in circuits]) == l
+
+    # set distance to None so it won't be considered going forward 
+    distance[min_pq[0]][min_pq[1]] = None
+
+    # increment counter
+    conn_count += 1
+    
+    # check if while loop is finished or not 
+    if connections >= 0:  
+      # break condition for part 1: break if number of connection is reached
+      if conn_count == connections:
+        print("Finished solving part 1! Returning product of the highest three circuit lengths …")
+        circuits.sort(reverse=True, key=len)
+        return len(circuits[0]) * len(circuits[1]) * len(circuits[2])
+    else: 
+      # break condition for part 2: break if there is only one circuit
+      if len(circuits) == 1:
+        print("Finished solving part 2! Returning product of the final two boxes' X coordinates …")
+        return jbox[min_pq[0]][0] * jbox[min_pq[1]][0]
+
+#print(solve(jbox, 1000))
+print(solve(jbox, -1))
+