The task of crossing the seven bridges over the Pregola River on a city tour of Königsberg (nowadays known as Kaliningrad) without missing one or walking across one twice [1] is simply captivating.

The Swiss mathematician Leonhard Euler already proved that this was impossible as early as 1736, but the task is still useful as a mathematical brain teaser today because the network of bridges can be converted into a graph (Figure 1) and bombarded with graph-theory axioms and algorithms.

Figure 1: Abstraction of the path over the Königsberg bridges as a graph (Wikipedia). CC BY-SA 3.0, Bogdan Giusca

Euler found that Königsberg's land masses can be represented as nodes in a graph, connected by seven paths – referred to as edges by graph theorists – that represent the connecting bridges (a to g in Figure 2). The task for the Königsberg city guide is therefore to cover all the edges drawn in the graph without passing a segment more than once.

Figure 2: Bridges named a to g between the land connections represented as nodes.

Leonhard Euler recognized that the number of edges leading to a node (he called it "degree" of the node) directly determines whether or not the problem of a repetition-free walk-through can be solved.

Odd or Even

If each node has an even number of edges, a tourist can walk across all the bridges in sequence without ever getting stuck. A similar situation occurs if exactly two of the nodes in the graph have an odd number of access points and the remaining nodes have an even number – in this case, the walker starts walking at the first of the odd nodes and ends the walk at the second odd node. In all other cases, for example also in the present Königsberg bridge problem, where all four nodes exhibit odd degrees, a repetition-free round trip is mathematically impossible.

Brute Force

For fun, you could now tackle the problem with a Python script that marches through the graph and only stops if it cannot continue at a node, because all adjacent segments have already been traversed. If a following check shows that the trip has covered all segments of the graph, the result is the correct solution to the problem. In the case of Königsberg, this is not possible. However, as we'll see later, if you add another connection to the graph, there is a way.

For the script to try all possible path combinations, a recursive depth-first algorithm works through the different bridge combinations. On the journey, it keeps track of the crossed bridges in a dictionary and does not take a route that leads over a bridge that previously has been passed.

If it gets stuck on a node, it resets and tries further combinations of previously chosen directions at branches. It stores the longest path found so far in an object attribute for later reference. If its length later matches the number of all defined bridges, the script has solved the problem.

Inserting the Graph

As a first task, the script in Listing 1 [2] models the graph from Figure 2 in the data structure starting on line 4. Node names ("1", "2", etc.) are used as keys into the dictionary, whose values consist of lists of connected nodes along with a list of path options, for example ["a", "b"] in the first entry, because bridges a and b lead from node 1 to node 2.

01 #!/usr/bin/env python3
02 from bridgewalk import BridgeWalk
03
04 g = { "1" : [["2", ["a", "b"]],
05              ["3", ["d", "e"]],
06              ["4", ["c"]]
07             ],
08       "2" : [["1", ["a", "b"]],
09              ["4", ["f"]]
10             ],
11       "3" : [["1", ["d", "e"]],
12              ["4", ["g"]],
13             ],
14       "4" : [["2", ["f"]],
15              ["3", ["g"]],
16              ["1", ["c"]]
17             ]
18 }
19
20 trail = BridgeWalk(g)
21 trail.explore()
22
23 print(trail.maxpath)
24
25 if len(trail.bridges) == \
26    len(trail.maxpath):
27     print("Solved!")
28 else:
29     print("Impossible!")

This puzzle's definition gets passed to the BridgeWalk class' constructor in line 20; the class itself is defined further down in Listing 2. The explore() method on the resulting object then shimmies through the graph to find the longest path without an overlap.

01 #!/usr/bin/env python3
02 class BridgeWalk(object):
03   def __init__(self, graph):
04     self.graph   = graph
05     self.bridges = {}
06     self.maxpath = []
07
08     for node in self.graph:
09       for fork in self.graph[node]:
10         for bridge in fork[1]:
11           self.bridges[bridge] = 1
12
13   def explore(self):
14     # try different start nodes
15     for node in self.graph:
16       self.scan(node)
17
18   def scan(self, node, path=[], seen={}):
19     # try different connected nodes
20     for fork in self.graph[node]:
21       # try all bridges leading there
22       for bridge in fork[1]:
23         self.bridge_ok(bridge, fork[0],
24           path.copy(), seen.copy())
25
26   def bridge_ok(self, bridge,
27                 node, path, seen):
28     if not bridge in seen:
29       seen[bridge]=1
30       path.append(bridge)
31
32       if len(self.maxpath) < len(path):
33         self.maxpath = path.copy()
34
35       # recurse
36       self.scan(node, path, seen)

At any point in time during the run, the maxpath object attribute contains a list of bridges in the order they were crossed. The print() statement in line 23 in Listing 1 automatically converts the list into a string and outputs it. The bridges attribute maintains a dictionary with the keys for all bridges defined in the graph for easy counting of unique bridges.

If the number of bridges in the graph determined with len() matches the number of bridges on the longest path in maxpath, the script in line 27 reports the successful solution of the puzzle.