Solved 2015/09 P1+P2 and added TSP (short/long) function and dijkstras to helper file

2015/09/9.md

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+## \-\-- Day 9: All in a Single Night \-\--
+
+Every year, Santa manages to deliver all of his presents in a single
+night.
+
+This year, however, he has some [new
+locations]{title="Bonus points if you recognize all of the locations."}
+to visit; his elves have provided him the distances between every pair
+of locations. He can start and end at any two (different) locations he
+wants, but he must visit each location exactly once. What is the
+*shortest distance* he can travel to achieve this?
+
+For example, given the following distances:
+
+    London to Dublin = 464
+    London to Belfast = 518
+    Dublin to Belfast = 141
+
+The possible routes are therefore:
+
+    Dublin -> London -> Belfast = 982
+    London -> Dublin -> Belfast = 605
+    London -> Belfast -> Dublin = 659
+    Dublin -> Belfast -> London = 659
+    Belfast -> Dublin -> London = 605
+    Belfast -> London -> Dublin = 982
+
+The shortest of these is `London -> Dublin -> Belfast = 605`, and so the
+answer is `605` in this example.
+
+What is the distance of the shortest route?
+
+Your puzzle answer was `117`.
+
+## \-\-- Part Two \-\-- {#part2}
+
+The next year, just to show off, Santa decides to take the route with
+the *longest distance* instead.
+
+He can still start and end at any two (different) locations he wants,
+and he still must visit each location exactly once.
+
+For example, given the distances above, the longest route would be `982`
+via (for example) `Dublin -> London -> Belfast`.
+
+What is the distance of the longest route?
+
+Your puzzle answer was `909`.
+
+Both parts of this puzzle are complete! They provide two gold stars:
+\*\*
+
+At this point, you should [return to your Advent calendar](/2015) and
+try another puzzle.
+
+If you still want to see it, you can [get your puzzle
+input](9/input).
+

2015/09/solution.py

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+#!/bin/python3
+import sys,time,re,json
+from pprint import pprint
+
+sys.path.insert(0, '../../')
+from fred import list2int,get_re,nprint,lprint,loadFile,dprint,TSP
+start_time = time.time()
+
+input_f = 'input'
+
+part = 2
+#########################################
+#                                       #
+#              Part 1                   #
+#                                       #
+#########################################
+
+def constructGraph(input_f):
+    """
+        The graph of cities should look something like this
+
+        graph = {
+            "node": [("dist1", distance1), ("dist", distance2)],
+        }
+    """
+    inst = []
+    graph = {}
+    with open(input_f) as file:
+        for line in file:
+            inst = line.rstrip().split(' = ')
+            inst = inst[0].split(' to ') + [int(inst[1])]
+            if inst[0] not in graph:
+                graph[inst[0]] = []
+            if inst[1] not in graph:
+               graph[inst[1]] = []
+            graph[inst[0]].append((inst[1],inst[2]))
+            graph[inst[1]].append((inst[0],inst[2]))
+            
+    return graph
+    
+if part == 1:
+    graph = constructGraph(input_f)
+    shortest = float('inf')
+    routes = []
+    cities = set(graph.keys()) 
+    for neighbors in graph.values():
+        for neighbor, _ in neighbors:
+            cities.add(neighbor) 
+    for c in cities:
+        print(TSP(graph, c))
+        shortest = min(shortest,TSP(graph, c)[1])
+        
+    print(shortest)
+
+#########################################
+#                                       #
+#              Part 2                   #
+#                                       #
+#########################################
+if part == 2:
+    graph = constructGraph(input_f)
+    #print(graph)
+    longest = 0
+    routes = []
+    cities = set(graph.keys()) 
+    for neighbors in graph.values():
+        for neighbor, _ in neighbors:
+            cities.add(neighbor) 
+    for c in cities:
+        longest = max(longest,TSP(graph, c,'longest')[1])
+        
+    print(longest)
+
+print("--- %s seconds ---" % (time.time() - start_time))