# Chapter 6 Hamilton Circuits ```Chapter 6 Traveling Salesman Problem
ESSENTIAL QUESTIONS:
Section 6.1: How does Hamilton’s Circuits and Paths compare to Euler’s?
Section 6.2: What is a complete graph?
Section 6.3: What do the Traveling Salesman Problems (TSPs) use weighted graphs?
Section 6.4: What are simple strategies for solving TSPs?
Section 6.5: How does the Brute-Force and Nearest –Neighbor Algorithms solve TSPs?
Section 6.6: What is an approximate algorithm?
Section 6.7: How does the Repetitive Nearest – Neighbor Algorithm improve your ability to solve TSPs?
Section 6.8: How does the Cheapest-Link Algorithm differ from the Nearest-Neighbor Algorithm?
WORD WALL:
Brute-Force Algorithm
Complete Graph
Weight
Hamilton Circuit
Hamilton Path
Inefficient Algorithm
Nearest-Neighbor Algorithm
Repetitive Nearest-Neighbor Algorithm
Traveling Salesman Problem (TSP)
Weighted Graph
Extra Review/ Possible Homework:
6.1 = pp. 221 - 2 # 2, 4, 6, 7, 10, 11
6.2 = p 223 # 25 – 28
6.3 = pp. 222 – 3 # 13 – 16
6.4/6.5 = pp. 223 – 225 # 29 - 31, 33, 35
6.7 = p. 225 #37 – 39, 41
6.8 = pp. 225 - 6 #43-45, 47
•
Section 6.1: Hamilton Circuits and Hamilton Paths
HAMILTON CIRCUITS AND PATHS
• HAMILTON = Touch each VERTEX exactly
one time (enter/ leave once).
EULER CIRCUITS AND PATHS
EULER = touch each EDGE of a graph
exactly one time.
•
•
By Definition:
must touch every vertex “can touch
more than once depending on degree”
By Definition:
Doesn’t require touching every edge
HAMILTON Circuits/Paths VERSUS EULER Circuits/Paths
For each of the following graphs, use our definitions of Hamilton and Euler to determine if circuits and
paths of each type are possible.
Graph 1
Graph 2
Graph 3
Graph 4
Graph 5
Graph 6
EULER
PATH
NO
YES
NO
NO
YES
NO
EULER
CIRCUIT
YES
NO
NO
YES
NO
NO
HAMILTON
PATH
YES
YES
YES
YES
NO
YES
HAMILTON
CIRCUIT
YES
NO
YES
NO
NO
NO
What do you notice about the relationship between a graph having an Euler Circuit or Path and having a
Hamilton Circuit of path?
There is no relationship
IN GENERAL, THERE ARE NO THEOREMS TO DETERMINE IF A GRAPH HAS A HAMILTON PATH
OR CIRCUIT.
#1: FOR HAMILTON CIRCUITS/ PATHS, VERTICES OF DEGREE 1 OR 2 ARE
VERY HELPFUL BECAUSE THEY REPRESENT REQUIRED EDGES TO REACH
THAT VERTEX.
#2: EXAMPLE: Hamilton Circuit = A, B, C, D, A
SHIFT/SAME Hamilton Circuits:
Follows the SAME rotation order for a circuit, but
with a different start.
B, C, D, A, B
C, D, A, B, C
D, A, B, C, D
A
B
D
C
DISTINCT Hamilton Circuits:
A different order of vertices (Left to Right) from
the same starting vertex. (reverse order counts)
A, D, C, B, A (reverse order)
A, D, B, C, A AND reverse = A, C, B, D, A
A, B, D, C, A AND reverse = A, C, D, B, A
EXAMPLE #1: List all possible Hamiltonian circuits in the following graph
A
B
1. A, B, C, D, E, F, G, A
2. A, B, E, D, C, F, G, A
3. A, F, C, D, C, B, G, A
4. A, F, E, D, C, B, G, A
REVERSE EACH ONE OF THESE 4 CIRCUITS
C
D
G
F
E
What about the Hamiltonian circuits that start at B, C, D, etc.?
Well, since such a circuit must pass through all of the vertices it doesn’t matter which vertex we start with.
Related to the idea of shifts of circuits.
EXAMPLE #2:
(a) Find a Hamiltonian path that begins at A and ends at E.
(b) Find a Hamiltonian circuit that starts at A and ends with the pair of vertices E, A.
(c) Find a Hamiltonian path that begins at F and ends at G.
A
D
E
F
B
G
(a) A, F, B, C, G, D, E
(b) A, F, B, C, G, D, E, A
(c) F, B, C, E, A, D, G
F, A, E, B, C, D, G
C
Example 3: The following graph has no Hamiltonian circuits or paths--why not?
4 bridges exist = can’t have a circuit; No path possible because
multiple vertices of degree 1
EXAMPLE #4:
(a) Find all Hamiltonian circuits that starts at A.
(b) Find all Hamiltonian circuits that starts at D.
(c) Explain why your answers in a and b must have the same
number of circuits.
(a) A, C, F, D, E, B, A
A, D, F, E, B, C, A
A, D, E, F, C, B, A (REVERSE ORDER as well)
(b) Re-order answers in part a
(c) Re-ordering concept
A
D
F
C
E
B
HOMEWORK: pp. 221 - 222 # 2, 4, 6, 7, 10, 11
Section 6.2: Complete Graphs
EXAMPLE PUT 6 OR 8 DOTS ON WHITEBOARD AND START CONNECTING THEM AND ASK
STUDENTS TO DESCRIBE HOW I AM MAKING THE GRAPH: “ONE DOT IS CONNECTED TO
EVERY OTHER DOT” TO START DEFINITION
COMPLETE GRAPH: A graph in which every pair of distinct vertices is joined by exactly one edge.
o Notation: KN = a complete graph of N vertices
EXAMPLES OF COMPLETE GRAPHS for 3, 4, and 5 vertices:
K3
K4
K5
Use the definition of a complete graph to answer the following questions:
1. Does a complete graph have to be connected?
2. Is a complete graph allowed to have loops?
YES; “every pair of vertices connected by an edge”
NO; “Distinct (Different) Vertices only”
3. Is a complete graph allowed to have multiple edges? 4. Can different vertices have different degree?
NO; exactly “one edge”
NO; each vertex must be connected all remaining
vertices
What is the DEGREE of each VERTEX for N vertices?
K3: Degree = 2
K4: Degree = 3
K5: Degree = 4
1 Less than the # of vertices
Degree = N - 1
How many EDGES are in KN (complete graph of N vertices)?
What is the relationship between edges and degrees?
EDGES
TOTAL DEGREE
K3: EDGES = 3
6 = 3*2
Euler’s Sum of Degree Theorem:
(total # of degrees) = 2 * (# of edges)
N(N-1) = 2 * (# of edges)
K4: EDGES = 6
12 = 4*3
K5: EDGES = 10
20 = 5*4
K6: EDGES = 15
30 = 6*5

N ( N  1)
 number of edges in KN
2
Where have you seen this formula before? Chapter 1: the number of pairwise comparisons.
DEGREE and VERTICES are consecutive numbers.
Complete Graph
KN
SUMMARY OF COMPLETE GRAPH INFORMATION
Number of Vertices
Degree of Each Vertex
N
N–1
Connected Graph, No Loops, No Multiple Edges
Number of Edges
N ( N  1)

2
K3= Complete Graph of 4 Vertices
1) How many Hamiltonian circuits does it
have? A
2
B
1
A, B, C, A
2
A, C, B, A
K4 = Complete Graph of 4 Vertices
1) How many Hamiltonian circuits does it
have?
1 A, B, C, D, A
6
A
B
C
K3
C
K4
D
2
3
4
5
6
A, B, D, C, A
A, C, B, D, A
A, C, D, B, A
A, D, B, C, A
A, D, C, B, A
If we were to answer the same questions for K5 we would find the following:
1) How many Hamiltonian circuits does it have?
(Do you notice any possible relation in how we answered this question for K3 and K4 that
(5 - 1) (5 - 2) ( 5 - 3) (5 - 4) = (4)(3)(2)(1) = 24 (see page 202 for exact circuits)
How many DISTINCT HAMILTON CIRCUITS are in KN?
Complete Graph
K3:
K4:
K5:
K6:
Vertices
3
4
5
6
Hamilton Circuits
2
6
24
120
(N – 1)! = (Degree) != Distinct Hamilton Circuits in KN.
Calculator Command Reminder: [MATH] – [PRB] – [4:!]
INDEPENDENT PRACTICE: Apply the Properties of Complete Graphs
(1) How many edges are there in K20? 20(19)/2 = 190 edges
(2) How many distinct Hamilton Circuits exist in K10? (10 – 1)! = 9! = 362880
(3) How many vertices are in the K50 graph? N = 50
(4) What is the degree of every vertex in K200? Degree = 200-1 = 199
(5) How many distinct Hamilton Circuits exist in K22?
(22 – 1)! = 21! Or 5.109094217E19
(6) How many edges are there in K32? Edges = (32)(32 – 1)/2 = 496 edges
(7) What is the degree of every vertex in K15? Degree = 15 – 1 = 14
(8) If KN has 120 distinct Hamiltonian circuits then what is N?
N=6
Trial and Error: Divide 120 by consecutive numbers starting with 2
Add one to last number used = Number of vertices
(9) If KN has 55 edges then what is N?
N = 11
55 = (N)(N-1)/2;
HINT: square root trick of DOUBLE EDGES round up = N and round down = N -1
(10) If the number of edges of K12 is x and the number of edges of K13 is y, what is the
values of y - x?
(13)(12)/2 - (12)(11)/ 2 = (12/2) (13-11) = 12
(11) For each of the following cases find the value of N.
a. KN has 5040 distinct Hamilton Circuits. 5040 = 7!; N = 7 + 1 = 8
b. KN has 66 edges. 66 = 12*11/2; N = 12
c. KN has 80,200 edges. 80,200 = 401*400/2; N = 401
(12) If KN has 362,880 distinct Hamilton Circuits, then… 362,880 = 6!; N = 7
a. How many vertices are in the KN graph? 7 VERTICES
b. What is the degree of each vertex are in the KN graph? 7 -1 = 6
c. How many edges are in the KN graph?7*6/2 = 21 edges
Section 6.3: Traveling Salesman Problems
WEIGHTED GRAPH: Any graph whose edges have numbers associated or assigned to them
(NOT DRAWN TO SCALE ALWAYS)
o Weights: number for each edge
o Examples of Weights: distance, time, cost…
Complete Weighted Graph: Complete graph that has weights associated with it
TRAVELING SALESMAN PROBLEMS:
From home, the traveling salesman must visit different destinations to sell goods and return home. This is
essentially finding a Hamilton Circuit.
GOAL: Try to find the cheapest, most efficient, optimal route that is a Hamilton circuit for a given graph (find
the least total weight)
Example – The Simpsons: Homer Simpson he has to run errands at the Kwik-E-Mart, the Retirement
Home and Moe`s. Assuming that he wants to begin and end his day at home find the route that will allow
him to get back to napping as soon as possible.
A
20
The numbers will represent the
number of blocks between each
destination.
D
8
When we place values like this
on the edges of a graph we
refer to them as the weights of
the edges.
15
17
12
20
C
B
What if Homer preferred going to see Grandpa first to get it out of the way?
A–B–C–D–A
A-B-D-C- A
12 + 20 + 17 + 8 = 57 BLOCKS
12+15+17+20 = 64 BLOCKS
What if Homer preferred going to the KWIK-E-MART last?
A-B-D-C- A
A–D–B–C–A
12+15+17+20 = 64 BLOCKS
8 +15 + 20 + 20 = 63 BLOCKS
Traveling Salesman Problem Steps:
(1) Create graph models
(2) Apply weights to these models
(3) Find optimal Hamilton Circuits. (Not Always a practical statement)
REAL-LIFE EXAMPLES:
1) Routing School Buses
2) Package Deliveries: UPS or FedEx Delivery Truck
3) Running Errands around Town
5) Grocery Shopping: Planning which aisles you need to go down
HOMEWORK: pp. 222 – 223 # 13 – 16
Section 6.4 and 6.5: Simple Algorithms for Solving Traveling Salesman Problems
Strategy #1: BRUTE FORCE (Exhaustive Search)
Circuit
A, B, C, D, A
Mirror Image
A, D, C, B, A
A, B, D, C, A
A, C, D, B, A
A, C, B, D, A
A, D, B, C, A
Weight
13 + 9 + 14 + 15 = 51
13 + 16 + 14 + 12 = 55
15
A
13 12
B
12 + 9 + 16 + 15 = 52
D
16
14
9
BRUTE-FORCE ALGORITHM
Step 1: Make a list of ALL distinct Hamilton circuits of the graph from ONE starting vertex
Step 2: Calculate its TOTAL WEIGHT for each circuit.
Step 3: Choose an OPTIMAL circuit. (There is always more than one optimal circuit because of mirrorimages)
PRO: Guarantee Optimal Hamilton Circuit
CON: INEFFICIENT ALGORITHM
# steps needed to carry out algorithm grows disproportionately with the size of population (vertices)
ie TAKES A LOT OF TIME the more vertices/ edges there are
STRATEGY #2: NEAREST – NEIGHBOR ALGORITHM
From the start vertex, go to the vertex that is the shortest to get to. From each new vertex go to the next new
city that is the shortest to get to. When there are no more new vertices to go to, go to the starting vertex.
A (12) C (9) B (16) D (15) A = 52
NEAREST-NEIGHBOR ALGORITHM
Start: Start at the designated starting vertex (home). If there is no designated starting vertex, pick any vertex.
Step 1: From the starting vertex go to its NEAREST NEIGHBOR (vertex that is the smallest weight away)
Middle Steps: From each vertex go to its nearest neighbor that hasn’t already been visited. (smallest
remaining weight of possible vertices) Keep doing this until all the vertices have been visited.
Last Step: From the last neighbor (vertex) RETURN to the starting vertex (home).
PRO: EFFICIENT ALGORITHM
Amount of computational effort required to implement it grows in some reasonable proportion with the size of
population (vertices)
ie WORKS FAST/ LESS TIME CONSUMING TO COMPLETE METHOD
CON: Not an optimal Algorithm; Answers won’t always be the OPTIMAL, but relatively close to being the
best possible way
C
A weighted Graph may also be represented as table. Vertices along the first
row and column. Weights on the interior of table.
A
B
C
D
D
A
8
A
AB = 12 AC = 20 AD = 8
B
BA = 12
BC = 20 BD = 15
C
CA = 20 CB = 20
D
DA = 8
CD = 17
12
20
15
17
DB = 15 DC = 17
B
20
Observations:
1) Diagonal is blank because NO LOOPS exist.
2) Upper Right Triangle and Lower Left Triangle are Identical Numbers
a. Represent Same Edges, but are in different order of starting and ending vertices
Draw the following complete weighted graph described by table:
A
B
C
D
E
A B
3
3
4 5
5 7
6 9
C D E
4 5 6
5 7 9
6 8
6
1
8 1
FIND THE NEAREST NEIGHBOR SOLUTION STARTING AT D:
D (1) E (6) A (3) B (5) C (6) D = 21
C
The following Complete Weighted Graph can also be represented by the table:
A
A
B
C
D
E
F
G
75
50
28
35
15
22
30
60
80
65
50
40
48
35
28
20
30
29
40
32
B 75
C 50
30
D 28
60
40
E 35
80
48
20
F 15
65
35
30
40
G 22
50
28
29
32
40
60
50
C
13
B
30
C
13
75
A
65
A
28
28
48
B
35
22
35
50
29
DD
32
30
20
15
G
G
13
80
E
E
40
FF
Does BRUTE FORCE seem practical here?
NO; (7-1)!= 720 POSSIBLE HAMILTON CIRCUITS
Start at Vertex A – Find the NEAREST NEIGHBOR SOLUTION:
(Use table or graph)
A – (15) – F – (13) – G – (28) – C – (30) – B – (60) – D – (20) - E – (35) - A =
201
HOMEWORK: pp. 223 – 225 # 29 - 31, 33, 35
Section 6.7: Repetitive Nearest-Neighbor Algorithm
The Nearest Neighbor Algorithm is an EFFICIENT but NOT always
ACCURATE algorithm.
 Is there a way we can improve it because we like its efficiency?
Compare from different starting spots
The REPETITIVE Nearest-Neighbor Algorithm:
 Perform the Nearest-Neighbor Algorithm for every vertex of the
graph as the starting vertex.
 Compare circuits and CHOOSE the circuit with least weight of all
the options.
o If needed, Re-order (shift) that Hamilton Circuit to have the
DESIGNATED START vertex first.
o Re-ordered (shifted) circuits have the same weight but change
its starting vertex
EXP #1: Perform the nearest neighbor algorithm for start of A.
A
9
D
10
15
A: A, B, D, C, A = 9 + 13 + 11+ 15 = 48
B: B, A, D, C, B = 9 + 10 + 11 + 17 = 47
13
11
C: C, D, A, B, C = 11 + 10 + 9 + 17 = 47
B
17
C
D: D, A, B, C, D = 10 + 9 + 17 + 11 = 47
Final Answer: A,B,C,D,A or A,D,C,B,A = 47
EXAMPLE #2: Label the graph and complete the Repetitive nearest neighbor
algorithm for a start of C.
A
E
B
ACEDBA = \$773
C
BCAEDB = \$722
D
A
A
B
C
D
E
\$185
\$119
\$152
\$133
B
C
\$185 \$119
\$121
\$121
\$150 \$174
\$200 \$120
CAEDBC = \$722 (Final)
D
E
\$152 \$133
\$150 \$200
\$174 \$120
\$199
\$199
DBCAED = \$722
EXAMPLE #3: Perform repetitive nearest neighbor algorithm for a start of B.
A
B
C
D
E
F
G
A
B
C
75
50
28
35
15
22
75 50
30
30
60 40
80 48
65 35
50 28
D
E
F
G
28 35 15 22
60 80 65 50
40 48 35 28
20 30 29
20
40 32
30 40
13
29 32 13
B, D, E, A, F, G, C, B
B, C, G, F, A, E, D, B
201 = 5+13+28+30+60+20+35
214 =30+28+13+15+28+20+80
214 =28+13+15+28+20+80+30
221=20+35+15+13+28+50+60
214= 20+28+15+13+28+30+80
229 =15+22+29+20+48+30+65
204=13+15+28+20+48+30+50
HOMEWORK: p. 225 #37 – 39, 41
GOAL: Piece together a Hamilton circuit by individual edges or “LINKS”
of graph trying to choose the smallest or “cheapest” weights first.
The Cheapest-Link Algorithm for N Vertices:
 Step #1: Pick the edge with the smallest weight first. Mark the
edge (or otherwise note that you have chosen it).
o Smallest weight ANYWHERE on the graph
 Step #2: Pick the next ‘cheapest’ edge anywhere in the graph. Mark
or note it.
 Step #3, …, N-1:Continue picking the ‘cheapest’ available edge in
the graph and as long as
(a) it does not close a circuit and
(b) it does not result in three edges coming out of a single vertex.
(DEGREE = 2)
(c) Ties can be broken at random and don’t break the two rules
 HINT: Draw out the edges chosen helps to double check what
is and is not available.
 Step #N: Connect the last two vertices to close the circuit
(you are forced to choose this edge so no need for “cheapest”)
REORDER CIRCUIT TO THE NECESSARY STARTING
VERTEX (N Vertices = N edges chosen for circuit)
B
30
EXAMPLE #1: Start at G
C
C
40
48
60
50
35
B
75
A
65
A
28
28
22
35
50
29
32
DD 30
20
15
G
80
E
E
40
FF
13
G
When table only, draw links to visualize circuit.
A
B
C
D
E
F
G
A B
75
75
50 30
28 60
35 80
15 65
22 50
C D E F
50 28 35 15
30 60 80 65
40 48 35
40
20 30
48 20
40
35 30 40
28 29 32 13
START IT AT G: G
G
22
50
28
29
32
13
– F – A – D – E – C – B – G = 13 + 15 +
28 + 20 + 48 + 30 + 50 = 204
EXAMPLE #2: Start at B
A
12
E
19
24
B
17
29
21
19
27
C
9
D
Start at B:
B, A, E, C, D, B
B, D, C, E, A, B
EXAMPLE #3: Start at D
A
B
C
D
A
\$185 \$119 \$152
B \$185
\$121 \$150
C \$119 \$121
\$174
D \$152 \$150 \$174
E \$133 \$200 \$120 \$199
E
\$133
\$200
\$120
\$199
A – C = 119
E – C = 120
B – D = 150
A – D = 152
E – B = 200
741 = total cost
Start at D:
D, A, C, E, B, D or D, B, E, C, A, D
= 741
HOMEWORK: pp. 225 - 6 #43-45, 47
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