Hamiltonian Circuits

Learning Outcomes

  • Identify whether a graph has a Hamiltonian circuit or path
  • Find the optimal Hamiltonian circuit for a graph using the brute force algorithm, the nearest neighbor algorithm, and the sorted edges algorithm
  • Identify a connected graph that is a spanning tree
  • Use Kruskal’s algorithm to form a spanning tree, and a minimum cost spanning tree

Hamiltonian Circuits and the Traveling Salesman Problem

In the last section, we considered optimizing a walking route for a postal carrier. How is this different than the requirements of a package delivery driver? While the postal carrier needed to walk down every street (edge) to deliver the mail, the package delivery driver instead needs to visit every one of a set of delivery locations. Instead of looking for a circuit that covers every edge once, the package deliverer is interested in a circuit that visits every vertex once.

putting all your study skills together

This section contains a bit of everything you’ve encountered so far: new vocabulary, new usages of familiar words, multi-step processes, and new ideas. Allow yourself ample time to work through the examples and problems multiple times so that you can develop a good understanding.

Hamiltonian Circuits and Paths

A Hamiltonian circuit is a circuit that visits every vertex once with no repeats. Being a circuit, it must start and end at the same vertex. A Hamiltonian path also visits every vertex once with no repeats, but does not have to start and end at the same vertex.

Hamiltonian circuits are named for William Rowan Hamilton who studied them in the 1800’s.

Example

One Hamiltonian circuit is shown on the graph below. There are several other Hamiltonian circuits possible on this graph. Notice that the circuit only has to visit every vertex once; it does not need to use every edge.

This circuit could be notated by the sequence of vertices visited, starting and ending at the same vertex: ABFGCDHMLKJEA. Notice that the same circuit could be written in reverse order, or starting and ending at a different vertex.

Rectangular graph with 12 vertices labeled a through M (without I)

Unlike with Euler circuits, there is no nice theorem that allows us to instantly determine whether or not a Hamiltonian circuit exists for all graphs.[1]

Example

Does a Hamiltonian path or circuit exist on the graph below?

Arrow shaped graph with 5 vertices labeled A- E, Edge from C to E is not part of the arrow shape. A and C are connected by two edges.

We can see that once we travel to vertex E there is no way to leave without returning to C, so there is no possibility of a Hamiltonian circuit. If we start at vertex E we can find several Hamiltonian paths, such as ECDAB and ECABD

Try It

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.

Watch this video to see the examples above worked out.

This problem is called the Traveling salesman problem (TSP) because the question can be framed like this: Suppose a salesman needs to give sales pitches in four cities. He looks up the airfares between each city, and puts the costs in a graph. In what order should he 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. The first option that might come to mind is to just try all different possible circuits.

Star Shaped graph with 5 vertices named home (Seattle), Dallas, Atlanta, Chicago, LA and the following dollar amounts between the cities: home and dallas - $12, home and atlanta - $14, home and LA $70, LA and chicago - $100, chicago and atlanta - $75, atlanta and dallas - $85, dallas and chicago - $16, LA and atlanta - $170, chicago and dallas - $16

question can be framed like this: Suppose a salesman needs to give sales pitches in four cities. He looks up the airfares between each city, and puts the costs in a graph. In what order should he 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. The first option that might come to mind is to just try all different possible circuits.

Brute Force Algorithm (a.k.a. exhaustive search)

1.     List all possible Hamiltonian circuits

2.     Find the length of each circuit by adding the edge weights

3.     Select the circuit with minimal total weight.

 

Example

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

triangular graph with 4 vertices and 6 edges. There is one vertex in the center of the triangle.

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

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

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, ABDCA, is the optimal circuit.

 

 

Watch these examples worked again in the following video.

Try It

 

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 thereare [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.

Number of Possible Circuits

For N vertices in a complete graph, there will be [latex](n-1)!=(n-1)(n-2)(n-3)\dots{3}\cdot{2}\cdot{1}[/latex] routes. Half of these are duplicates in reverse order, so there are [latex]\frac{(n-1)!}{2}[/latex] unique circuits.

The exclamation symbol, !, is read “factorial” and is shorthand for the product shown.

Example

How many circuits would a complete graph with 8 vertices have?

A complete graph with 8 vertices would have = 5040 possible Hamiltonian circuits. Half of the circuits are duplicates of other circuits but in reverse order, leaving 2520 unique routes.

While this is a lot, it doesn’t seem unreasonably huge. But consider what happens as the number of cities increase:

Cities Unique Hamiltonian Circuits
9 8!/2 = 20,160
10 9!/2 = 181,440
11 10!/2 = 1,814,400
15 14!/2 = 43,589,145,600
20 19!/2 = 60,822,550,204,416,000

Watch these examples worked again in the following video.

 

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 20 cities! Certainly Brute Force is not an efficient algorithm.

Nearest Neighbor Algorithm (NNA)

1.     Select a starting point.

2.     Move to the nearest unvisited vertex (the edge with smallest weight).

3.     Repeat until the circuit is complete.

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.

Example

Consider our earlier graph, shown to the right.

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

From D, the nearest neighbor is C, with a weight of 8.

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

From B we return to A with a weight of 4.

triangular graph with 4 vertices and 6 edges. There is one vertex in the center of the triangle.

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

Watch the example worked out in the following video.

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 AD forced us to use the very expensive edge BC later.

Example

Consider again our salesman. Starting in Seattle, the nearest neighbor (cheapest flight) is to LA, at a cost of $70. From there:

LA to Chicago: $100

Chicago to Atlanta: $75

Atlanta to Dallas: $85

Dallas to Seattle: $120

Total cost: $450

 

In this case, nearest neighbor did find the optimal circuit.

Watch this example worked out again in this video.

 

Going back to our first example, 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.

Example

We will revisit the graph from Example 17.

triangular graph with 4 vertices and 6 edges. There is one vertex in the center of the triangle.

Starting at vertex A resulted in a circuit with weight 26.

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

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

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

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

 

Try It

The table below shows the time, in milliseconds, it takes to send a packet of data between computers on a network. If data needed to be sent in sequence to each computer, then notification needed to come back to the original computer, we would be solving the TSP. The computers are labeled A-F for convenience.

A B C D E F
A — 44 34 12 40 41
B 44 — 31 43 24 50
C 34 31 — 20 39 27
D 12 43 20 — 11 17
E 40 24 39 11 — 42
F 41 50 27 17 42 —

a.     Find the circuit generated by the NNA starting at vertex B.

b.     Find the circuit generated by the RNNA.

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.

Example

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

triangular graph with 4 vertices and 6 edges. There is one vertex in the center of the triangle.

The cheapest edge is AD, with a cost of 1. We highlight that edge to mark it selected.

The next shortest edge is AC, with a weight of 2, so we highlight that edge.

triangular graph with 4 vertices and 6 edges. There is one vertex in the center of the triangle. The edge between A and D is highlighted

triangular graph with 4 vertices and 6 edges. There is one vertex in the center of the triangle. Edges between A and D and A and C are highlighted.

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

triangular graph with 4 vertices and 6 edges. There is one vertex in the center of the triangle. Edges between A and D and A and C are highlighted. Edge between A and B is highlighted.

BAD

triangular graph with 4 vertices and 6 edges. There is one vertex in the center of the triangle. Edges between A and D and A and C are highlighted.

BAD

triangular graph with 4 vertices and 6 edges. There is one vertex in the center of the triangle. Edges between A and D and A and C are highlighted.

OK

We then add the last edge to complete the circuit: ACBDA with weight 25.

Notice that the algorithm did not produce the optimal circuit in this case; the optimal circuit is ACDBA with weight 23.

Notice that the algorithm did not produce the optimal circuit in this case; the optimal circuit is ACDBA with weight 23.

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.

Example

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

To see the entire table, scroll to the right

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 47, to Salem. Following that idea, our circuit will be:

Portland to Salem                    47

Salem to Corvallis                   40

Corvallis to Eugene                 47

Eugene to Newport                 91

Newport to Seaside                117

Seaside to Astoria                   17

Astoria to Bend                      255

Bend to Ashland                     200

Ashland to Crater Lake           108

Crater Lake to Portland          344

Total trip length:                     1266 miles

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 3.

We start adding the shortest edges:

Seaside to Astoria       17 miles

Corvallis to Salem       40 miles

Portland to Salem        47 miles

Corvallis to Eugene     47 miles

ring of dots with city names in problem as labels. edges between Seaside and Astoria, Eugene and Corvallis, Salem and Corvallis, and Salem and Portland.

The graph after adding these edges is shown to the right.   The next shortest edge is from Corvallis to Newport at 52 miles, but adding that edge would give Corvallis degree 3.

 

Continuing on, we can skip over any edge pair that contains Salem or Corvallis, since they both already have degree 2.

Portland to Seaside                 78 miles

Eugene to Newport                 91 miles

Portland to Astoria                 (reject – closes circuit)

Ashland to Crater Lk  108 miles

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 2.

 

Newport to Astoria                (reject – closes circuit)

Newport to Bend                    180 miles

Bend to Ashland                     200 miles

 

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

Crater Lk to Astoria   433 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: 1241 miles.

While better than the NNA route, neither algorithm produced the optimal route. The following route can make the tour in 1069 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.

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

Try It

Find the circuit produced by the Sorted Edges algorithm using the graph below.

graph with 6 vertices and 14 edges. between B and E is 13, between E and G is 45, between G and F is 19, between F and C is 37, between c and A is 33 between A and B is 11. Between B and C is 25, between B and F is 23, between E and A is 14, between E and F is 15. Between G and B is 13, and between G and C is 36. Between G and A is 27.

https://www.myopenmath.com/multiembedq.php?id=6565&theme=oea&iframe_resize_id=mom1

 Spanning Trees

A company requires reliable internet and phone connectivity between their five offices (named A, B, C, D, and E for simplicity) in New York, so they decide to lease dedicated lines from the phone company. The phone company will charge for each link made. The costs, in thousands of dollars per year, are shown in the graph.

graph with 5 vertices and 11 edges. between A and B is $4, between B and C is $10, between C and D is $7, between D and E is $13, between E and B is $6, between E and C is $11, between A and D is $6, between A and C is $8. Between B and E is $6, between B and D is $14.

In this case, we don’t need to find a circuit, or even a specific path; all we need to do is make sure we can make a call from any office to any other. In other words, we need to be sure there is a path from any vertex to any other vertex.

Spanning Tree

A spanning tree is a connected graph using all vertices in which there are no circuits.

In other words, there is a path from any vertex to any other vertex, but no circuits.

 

Some examples of spanning trees are shown below. Notice there are no circuits in the trees, and it is fine to have vertices with degree higher than two.

Five triangular graphs where there are no Euler circuits, but every vertex has a path to every other vertex.

Usually we have a starting graph to work from, like in the phone example above. In this case, we form our spanning tree by finding a subgraph – a new graph formed using all the vertices but only some of the edges from the original graph. No edges will be created where they didn’t already exist.

Of course, any random spanning tree isn’t really what we want. We want the minimum cost spanning tree (MCST).

Minimum Cost Spanning Tree (MCST)

The minimum cost spanning tree is the spanning tree with the smallest total edge weight.

A nearest neighbor style approach doesn’t make as much sense here since we don’t need a circuit, so instead we will take an approach similar to sorted edges.

Kruskal’s Algorithm

  1. Select the cheapest unused edge in the graph.
  2. Repeat step 1, adding the cheapest unused edge, unless:
  3. adding the edge would create a circuit

Repeat until a spanning tree is formed

 

Example

Using our phone line graph from above, begin adding edges:

AB      $4        OK

AE       $5        OK

BE       $6        reject – closes circuit ABEA

DC      $7        OK

AC      $8        OK

A graph with five vertices labeled A through E. A to B is $4, A to E is $5, B to E is $6, D to C is $7, A to C is $8. AB, AC, AE, and DC are shown in red.

At this point we stop – every vertex is now connected, so we have formed a spanning tree with cost $24 thousand a year.

Remarkably, Kruskal’s algorithm is both optimal and efficient; we are guaranteed to always produce the optimal MCST.

Example

The power company needs to lay updated distribution lines connecting the ten Oregon cities below to the power grid. How can they minimize the amount of new line to lay?

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

Using Kruskal’s algorithm, we add edges from cheapest to most expensive, rejecting any that close a circuit. We stop when the graph is connected.

Seaside to Astoria                   17 milesCorvallis to Salem                   40 miles

Portland to Salem                    47 miles

Corvallis to Eugene                 47 miles

Corvallis to Newport              52 miles

Salem to Eugene           reject – closes circuit

Portland to Seaside                 78 miles

The graph up to this point is shown below.

Continuing,

Newport to Salem                   reject

Corvallis to Portland               reject

Eugene to Newport                 reject

Portland to Astoria                 reject

Ashland to Crater Lk              108 miles

Eugene to Portland                  reject

Newport to Portland              reject

Newport to Seaside                reject

Salem to Seaside                      reject

Bend to Eugene                       128 miles

Bend to Salem                         reject

Astoria to Newport                reject

Salem to Astoria                     reject

Corvallis to Seaside                 reject

Portland to Bend                     reject

Astoria to Corvallis                reject

Eugene to Ashland                  178 miles

 

This connects the graph. The total length of cable to lay would be 695 miles.

Watch the example above worked out in the following video, without a table.

Now we present the same example, with a table in the following video.

Try It

Find a minimum cost spanning tree on the graph below using Kruskal’s algorithm.

graph with 6 vertices and 14 edges. between B and E is 13, between E and G is 45, between G and F is 19, between F and C is 37, between c and A is 33 between A and B is 11. Between B and C is 25, between B and F is 23, between E and A is 14, between E and F is 15. Between G and B is 13, and between G and C is 36. Between G and A is 27.

https://www.myopenmath.com/multiembedq.php?id=6581&theme=oea&iframe_resize_id=mom5

 

 

[1] There are some theorems that can be used in specific circumstances, such as Dirac’s theorem, which says that a Hamiltonian circuit must exist on a graph with n vertices if each vertex has degree n/2 or greater.

Time to Practice!

Complete the homework assigned by your teacher before moving on to the next section.