day 10 remove unused fragments

10/a.py

@@ -7,7 +7,7 @@ lines.add(0)
 
 def a(transformers):
     v = 0
-    dv = [3] #Always a 3 v transform at the end
+    dv = [3] #Always a 3v transformer at the end
     while v < max(transformers):
         v_ = min(set(range(v+1,v+4)).intersection(transformers))
         dv.append(v_ - v)
@@ -17,10 +17,6 @@ def a(transformers):
     count = dict(Counter(dv))
     return count[3]*count[1]
 
-# print(lines)
-
-# This is a recursive solution. IT works but it'll take an eternity to go through the whole list of transformers.
-# Instead, we split 
 def b(c, v, transformers):
     for v_ in set(range(v+1,v+4)).intersection(lines):
         # print(v_)
@@ -29,29 +25,20 @@ def b(c, v, transformers):
         else:
             c = b(c, v_, transformers)
     return c
-# print(b(0, 0, lines)) #Does not work for big sets!
-# print(b(0, 1, {1, 2, 3, 4}))
-# print(b(0, 7, {7, 8, 9, 10, 11}))
-
+# print(b(0, 0, lines)) #This is a recursive solution. IT works but it'll take an eternity to go through the whole list of transformers.
+                        #My solution: Split the set by 'gaps' (where there is only one path through to the next number), then multiply together
 def b_optimized(transformers):
     rel = { v: set(range(v+1,v+4)).intersection(transformers) for v in transformers }
     rel.pop(max(transformers))
-    # print(rel)
-    # print({ x if (vf in rel[x]) else None for x in rel }.intersection(transformers))
     gaps = { list(rel[x])[0] if (len(rel[x]) == 1) else None for x in rel }.intersection(transformers)
     gaps.add(list(rel)[0])
     gaps = sorted(gaps)
-    # print(gaps)
-
     idx = 0
     C = 1
     while idx+1 < len(gaps):
         s = sorted(set(range(gaps[idx],gaps[idx+1]+1)).intersection(transformers))
-        # print(s)
         if len(s) > 2:
             C *= b(0, list(s)[0], s)
-            # print(C, b(0, list(s)[0], s))
-
         idx += 1
     return C