With Hamiltonian circuits, our focus will not be on existence, but on the question of optimization; given a graph where the edges have weights, can we find the optimal Hamiltonian circuit; the one with lowest total weight.

This problem is called the Traveling salesperson problem (TSP) because the question can be framed like this: Suppose a salesperson needs to give sales pitches in four cities. They looks up the airfares between each city, and put the costs in a graph. The salesperson needs to figure out what order they should travel to visit each city once then return home with the lowest cost?

To answer this question of how to find the lowest cost Hamiltonian circuit, we will consider some possible approaches.

Brute Force Algorithm

The first option that might come to mind is to just try all different possible circuits. We call this the brute force algorithm.

Apply the Brute force algorithm to find the minimum cost Hamiltonian circuit on the graph below.

Show Solution

To apply the Brute force algorithm, we list all possible Hamiltonian circuits and calculate their weight:

Circuit	Weight
[latex]ABCDA[/latex]	[latex]4+13+8+1 = 26[/latex]
[latex]ABDCA[/latex]	[latex]4+9+8+2 = 23[/latex]
[latex]ACBDA[/latex]	[latex]2+13+9+1 = 25[/latex]

 

Note: These are the unique circuits on this graph. All other possible circuits are the reverse of the listed ones or start at a different vertex, but result in the same weights. From this we can see that the second circuit, [latex]ABDCA[/latex], is the optimal circuit.

Watch this example worked again in the following video.

You can view the transcript for “TSP by brute force” here (opens in new window).

The Brute force algorithm is optimal; it will always produce the Hamiltonian circuit with minimum weight. Is it efficient? To answer that question, we need to consider how many Hamiltonian circuits a graph could have. For simplicity, let’s look at the worst-case possibility, where every vertex is connected to every other vertex. This is called a complete graph.

Suppose we had a complete graph with five vertices like the air travel graph above. From Seattle there are four cities we can visit first. From each of those, there are three choices. From each of those cities, there are two possible cities to visit next. There is then only one choice for the last city before returning home. This can be shown visually:

Counting the number of routes, we can see there are [latex]4\cdot{3}\cdot{2}\cdot{1}[/latex] routes. For six cities there would be [latex]5\cdot{4}\cdot{3}\cdot{2}\cdot{1}[/latex] routes.  

As you can see the number of circuits is growing extremely quickly. If a computer looked at one billion circuits a second, it would still take almost two years to examine all the possible circuits with only [latex]20[/latex] cities! Certainly brute force is not an efficient algorithm. Let’s look at another option.

Nearest Neighbor Algorithm

Unfortunately, no one has yet found an efficient and optimal algorithm to solve the TSP, and it is very unlikely anyone ever will. Since it is not practical to use brute force to solve the problem, we turn instead to heuristic algorithms; efficient algorithms that give approximate solutions. In other words, heuristic algorithms are fast, but may or may not produce the optimal circuit.

 

Consider again our salesperson.

Starting at Seattle, the nearest neighbor (cheapest flight) is to LA, at a cost of [latex]$70[/latex]. From there:

• LA to Chicago: [latex]$100[/latex]
• Chicago to Atlanta: [latex]$75[/latex]
• Atlanta to Dallas: [latex]$85[/latex]
• Dallas to Seattle: [latex]$120[/latex]
• Total cost: [latex]$450[/latex]

In this case, nearest neighbor found the optimal circuit.

What if the NNA does not pick the most optimal edge?

Consider our earlier graph, shown to the right.

Starting at vertex [latex]A[/latex], the nearest neighbor is vertex [latex]D[/latex] with a weight of [latex]1[/latex].

From [latex]D[/latex], the nearest neighbor is [latex]C[/latex], with a weight of [latex]8[/latex].

From [latex]C[/latex], our only option is to move to vertex [latex]B[/latex], the only unvisited vertex, with a cost of [latex]13[/latex].

From [latex]B[/latex] we return to [latex]A[/latex] with a weight of [latex]4[/latex].

The resulting circuit is [latex]ADCBA[/latex] with a total weight of [latex]1+8+13+4 = 26[/latex].

We ended up finding the worst circuit in the graph! What happened? Unfortunately, while it is very easy to implement, the NNA is a greedy algorithm, meaning it only looks at the immediate decision without considering the consequences in the future. In this case, following the edge [latex]AD[/latex] forced us to use the very expensive edge [latex]BC[/latex] later. How could we improve the outcome? One option would be to redo the nearest neighbor algorithm with a different starting point to see if the result changed. Since nearest neighbor is so fast, doing it several times isn’t a big deal.

Starting at vertex [latex]A[/latex] resulted in a circuit with weight [latex]26[/latex], as seen above. Lets repeat the NNA for the other vertices.

Starting at vertex [latex]B[/latex], the nearest neighbor circuit is [latex]BADCB[/latex] with a weight of [latex]4+1+8+13 = 26[/latex]. This is the same circuit we found starting at vertex [latex]A[/latex]. No better.

Starting at vertex [latex]C[/latex], the nearest neighbor circuit is [latex]CADBC[/latex] with a weight of [latex]2+1+9+13 = 25[/latex]. Better!

Starting at vertex [latex]D[/latex], the nearest neighbor circuit is [latex]DACBA[/latex]. Notice that this is actually the same circuit we found starting at [latex]C[/latex], just written with a different starting vertex.

The repeat nearest neighbor algorithm (RNNA) was able to produce a slightly better circuit with a weight of [latex]25[/latex], but still not the optimal circuit in this case. Notice that even though we found the circuit by starting at vertex [latex]C[/latex], we could still write the circuit starting at [latex]A[/latex]: [latex]ADBCA[/latex] or [latex]ACBDA[/latex].

While certainly better than the basic NNA, unfortunately, the RNNA is still greedy and will produce very bad results for some graphs. As an alternative, our next approach will step back and look at the “big picture” – it will select first the edges that are shortest, and then fill in the gaps.

Sorted Edges Algorithm

Using the four vertex graph from earlier, we can use the Sorted Edges algorithm.

 

The cheapest edge is [latex]AD[/latex], with a cost of [latex]1[/latex]. We highlight that edge to mark it selected. The next shortest edge is [latex]AC[/latex], with a weight of [latex]2[/latex], so we highlight that edge.

 

 

For the third edge, we’d like to add [latex]AB[/latex], but that would give vertex [latex]A[/latex] degree [latex]3[/latex], which is not allowed in a Hamiltonian circuit. The next shortest edge is [latex]CD[/latex], but that edge would create a circuit [latex]ACDA[/latex] that does not include vertex [latex]B[/latex], so we reject that edge. The next shortest edge is [latex]BD[/latex], so we add that edge to the graph.

Bad

Bad

Okay

We then add the last edge to complete the circuit: [latex]ACBDA[/latex] with weight [latex]25[/latex]. Notice that the algorithm did not produce the optimal circuit in this case; the optimal circuit is [latex]ACDBA[/latex] with weight [latex]23[/latex].

 

While the Sorted Edge algorithm overcomes some of the shortcomings of NNA, it is still only a heuristic algorithm, and does not guarantee the optimal circuit.

Your teacher’s band, Derivative Work, is doing a bar tour in Oregon. The driving distances are shown below. Plan an efficient route for your teacher to visit all the cities and return to the starting location. Use NNA starting at Portland, and then use Sorted Edges.

	Ashland	Astoria	Bend	Corvallis	Crater Lake	Eugene	Newport	Portland	Salem	Seaside
Ashland	–	[latex]374[/latex]	[latex]200[/latex]	[latex]223[/latex]	[latex]108[/latex]	[latex]178[/latex]	[latex]252[/latex]	[latex]285[/latex]	[latex]240[/latex]	[latex]356[/latex]
Astoria	[latex]374[/latex]	–	[latex]255[/latex]	[latex]166[/latex]	[latex]433[/latex]	[latex]199[/latex]	[latex]135[/latex]	[latex]95[/latex]	[latex]136[/latex]	[latex]17[/latex]
Bend	[latex]200[/latex]	[latex]255[/latex]	–	[latex]128[/latex]	[latex]277[/latex]	[latex]128[/latex]	[latex]180[/latex]	[latex]160[/latex]	[latex]131[/latex]	[latex]247[/latex]
Corvallis	[latex]223[/latex]	[latex]166[/latex]	[latex]128[/latex]	–	[latex]430[/latex]	[latex]47[/latex]	[latex]52[/latex]	[latex]84[/latex]	[latex]40[/latex]	[latex]155[/latex]
Crater Lake	[latex]108[/latex]	[latex]433[/latex]	[latex]277[/latex]	[latex]430[/latex]	–	[latex]453[/latex]	[latex]478[/latex]	[latex]344[/latex]	[latex]389[/latex]	[latex]423[/latex]
Eugene	[latex]178[/latex]	[latex]199[/latex]	[latex]128[/latex]	[latex]47[/latex]	[latex]453[/latex]	–	[latex]91[/latex]	[latex]110[/latex]	[latex]64[/latex]	[latex]181[/latex]
Newport	[latex]252[/latex]	[latex]135[/latex]	[latex]180[/latex]	[latex]52[/latex]	[latex]478[/latex]	[latex]91[/latex]	–	[latex]114[/latex]	[latex]83[/latex]	[latex]117[/latex]
Portland	[latex]285[/latex]	[latex]95[/latex]	[latex]160[/latex]	[latex]84[/latex]	[latex]344[/latex]	[latex]110[/latex]	[latex]114[/latex]	–	[latex]47[/latex]	[latex]78[/latex]
Salem	[latex]240[/latex]	[latex]136[/latex]	[latex]131[/latex]	[latex]40[/latex]	[latex]389[/latex]	[latex]64[/latex]	[latex]83[/latex]	[latex]47[/latex]	–	[latex]118[/latex]
Seaside	[latex]356[/latex]	[latex]17[/latex]	[latex]247[/latex]	[latex]155[/latex]	[latex]423[/latex]	[latex]181[/latex]	[latex]117[/latex]	[latex]78[/latex]	[latex]118[/latex]	–

Show Solution

Using NNA with a large number of cities, you might find it helpful to mark off the cities as they’re visited to keep from accidently visiting them again. Looking in the row for Portland, the smallest distance is [latex]47[/latex], to Salem.

Following that idea, our circuit will be:

[latex]\begin{array} {ll} \text{Portland to Salem} & 47 \\ \text{Salem to Corvallis} & 40 \\ \text{Corvallis to Eugene} & 47 \\ \text{Eugene to Newport} & 91 \\ \text{Newport to Seaside} & 117 \\ \text{Seaside to Astoria} & 17 \\ \text{Astoria to Bend} & 255 \\ \text{Bend to Ashland} & 200 \\ \text{Ashland to Crater Lake} & 108 \\ \text{Crater Lake to Portland} & 344 \\ \text{Total trip length: } & 1266\text{ miles} \end{array}[/latex]

Using Sorted Edges, you might find it helpful to draw an empty graph, perhaps by drawing vertices in a circular pattern. Adding edges to the graph as you select them will help you visualize any circuits or vertices with degree [latex]3[/latex].

 

We start adding the shortest edges:

[latex]\begin{array} {ll} \text{Seaside to Astoria} & 17\text{ miles} \\ \text{Corvallis to Salem} & 40\text{ miles} \\ \text{Portland to Salem} & 47\text{ miles} \\ \text{Corvallis to Eugene} & 47\text{ miles} \end{array}[/latex]

The graph after adding these edges is shown to the right.  The next shortest edge is from Corvallis to Newport at [latex]52[/latex] miles, but adding that edge would give Corvallis degree [latex]3[/latex]. Continuing on, we can skip over any edge pair that contains Salem or Corvallis, since they both already have degree [latex]2[/latex].

[latex]\begin{array} {ll} \text{Portland to Seaside} & 78\text{ miles} \\ \text{Eugene to Newport} & 91\text{ miles} \\ \text{Portland to Astoria} & \text{(reject – closes circuit)} \\ \text{Ashland to Crater Lk} & 108\text{ miles} \end{array}[/latex]

The graph after adding these edges is shown to the right. At this point, we can skip over any edge pair that contains Salem, Seaside, Eugene, Portland, or Corvallis since they already have degree [latex]2[/latex].  

[latex]\begin{array} {ll} \text{Newport to Astoria} & \text{(reject – closes circuit)} \\ \text{Newport to Bend} & 180\text{ miles} \\ \text{Bend to Ashland} & 200\text{ miles} \end{array}[/latex]

At this point the only way to complete the circuit is to add:

Crater Lk to Astoria   [latex]433[/latex] miles.

 The final circuit, written to start at Portland, is:

Portland, Salem, Corvallis, Eugene, Newport, Bend, Ashland, Crater Lake, Astoria, Seaside, Portland.

 Total trip length: [latex]1241[/latex] miles.

While better than the NNA route, neither algorithm produced the optimal route. The following route can make the tour in [latex]1069[/latex] miles:

Portland, Astoria, Seaside, Newport, Corvallis, Eugene, Ashland, Crater Lake, Bend, Salem, Portland

 

Watch the example of nearest neighbor algorithm for traveling from city to city using a table worked out in the video below.

You can view the transcript for “Nearest Neighbor from a table” here (opens in new window).

In the next video we use the same table, but use sorted edges to plan the trip.

You can view the transcript for “Sorted Edges from a table” here (opens in new window).