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Nuffield Free-Standing Mathematics Activity. Cable TV. Bride. 9.3. 4.6. 6.5. Ramsey. Kirk Michael. Ballaugh. 2.8. 9.3. 6.8. 15.4. 7.4. Peel. Laxey. St Johns. 7.7. 2.7. 8.2. 13.9. 9.3. Douglas. 10.2. 4.7. Port Erin. The Isle of Man. Castletown. Cable TV. - PowerPoint PPT Presentation

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© Nuffield Foundation 2011

Nuffield Free-Standing Mathematics Activity

Cable TV

Ramsey

Laxey

Bride

BallaughKirk Michael

Peel

St Johns

Douglas

Castletown

Port Erin The Isle of Man

A Cable TV company wants to connect all the towns shown on the map.

Cable TV

The company wants to use the minimum length of cable.

This is called a minimum connector problem.

6.5

9.3

10.2

2.8

4.6

9.3

7.7

15.4

8.2

9.313.9

4.7

2.7

6.8

7.4

This activity is about solving minimum connector problems

To solve the problem you need to find a spanning tree of minimum length.

A spanning tree is a tree that connects all the vertices together.

A minimum spanning tree is a spanning tree of minimum length.

Think aboutHow many edges will there be in a minimum spanning tree that connects n vertices?

Kruskal’s Algorithm

Step 1 List the edges in order of increasing length.

Step 2 Start with the shortest edge.

Step 3 From the remaining edges, select the shortest edge which does not form a cycle.

(If there are 2 shortest edges choose either.)

Step 4 Repeat Step 3 until all the vertices are connected.

Peel to St Johns 2.7

Edges in order of increasing length:

Ballaugh to Kirk Michael 2.8

Bride to Ramsey 4.6

Castletown to Port Erin 4.7

Ballaugh to Ramsey 6.5

Kirk Michael to Peel 6.8

Douglas to Laxey 7.7

Kirk Michael to St Johns 7.4

Douglas to St Johns 8.2

Ballaugh to BrideCastletown to St Johns 9.3Laxey to Ramsey

Castletown to Douglas 10.2

Peel to Port Erin 13.9

Douglas to Ramsey 15.4

Ramsey

Laxey

Bride

BallaughKirk Michael

Douglas

Castletown

Port Erin

The Isle of Man

4.7

Peel

St Johns2.7

2.8

4.66.5

6.8

7.7

8.2

9.3

Total length = 53.3 miles

Peel to St Johns 2.7

Ball to K Michael 2.8

Bride to Ramsey 4.6

Castletown to Port Erin 4.7

Ballaugh to Ramsey 6.5

Kirk Michael to Peel 6.8

Douglas to Laxey 7.7

Douglas to St Johns 8.2

Castletown to St Johns 9.3

Prim’s Algorithm

The method can be carried out using an adjacency matrix.

Step 1 Starting from any vertex, join it to the nearest adjacent vertex.

Step 2 Join the next nearest vertex to those already included, provided this does not form a cycle.

Step 3 Repeat Step 2 until all the vertices are included.

Writing the network as an adjacency network:

B1

B1

B2

C

D

K

L

P1

P2

R

S

B2 C D K L P1 P2 R S

-

-

-

-

-

-

-

-

-

-

9.3

9.3

2.8

2.8

6.5

6.5

4.6

4.6

10.2

10.2 7.7

7.7

15.4

15.4

8.2

8.2

6.8

6.8 13.9

13.9

2.7

2.7

4.7

4.7 -

-- -- --

-- - - -

-- - - - -

- -- - - - -

--- - -

- -- ---

-- --- --

- - - --

- -- --- -

--- - -

9.3

9.3

7.4

7.4

9.3

9.3

Starting from any vertex – say B1:

B1

B1

B2

C

D

K

L

P1

P2

R

S

B2 C D K L P1 P2 R S

-

-

-

-

-

-

-

-

-

-

9.3

9.3

2.8

2.8

6.5

6.5

4.6

4.6

10.2

10.2 7.7

7.7

15.4

15.4

8.2

8.2

6.8

6.8 13.9

13.9

2.7

2.7

-- - - - -

- -- - - - -

-- - - -

--- - 9.3 7.4 -

- - 7.4- ---

-9.3- --- --

9.34.7 -- - - --

-- -- --

- -- --4.7- -

--9.3- - -

1 2 34 5 67 89 10

Prim’s Algorithm results in the same minimum spanning tree, but the order the edges are added to the tree is different

Ballaugh to Kirk Michael 2.8Ballaugh to Ramsey 6.5Ramsey to Bride 4.6Kirk Michael to Peel 6.8Peel to St Johns 2.7St Johns to Douglas 8.2Douglas to Laxey 7.7St Johns to Castletown 9.3Castletown to Port Erin 4.7

Ramsey

Laxey

Bride

BallaughKirk Michael

Douglas

Castletown

Port Erin

The Isle of Man

4.7

Peel

St Johns2.7

2.8

4.66.5

6.8

7.7

8.2

9.3

Total length

= 2.8 + 6.5 + 4.6 + 6.8 + 2.7 + 8.2 + 7.7 + 9.3 + 4.7

= 53.3 miles

Think aboutCan you explain how the matrix method works?Which method do you prefer? Why?What else would need to be considered in the real situation?

© Nuffield Foundation 2011

Cable TVReflect on your work

What are the advantages or disadvantages of Kruskal’s and Prim’s algorithms?

What other sorts of things might need to be considered in a real situation?

How many edges will there be in the minimum spanning tree of a network with n vertices?

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