# IB DP Maths Topic 10.10 Deleted vertex algorithm for determining a lower bound. HL Paper 3

## Question

A graph G with vertices A, B, C, D, E has the following cost adjacency table.

$\begin{array}{*{20}{c|ccccc}} {}&{\text{A}}&{\text{B}}&{\text{C}}&{\text{D}}&{\text{E}} \\ \hline {\text{A}}& – &{12}&{10}&{17}&{19} \\ {\text{B}}&{12}& – &{13}&{20}&{11} \\ {\text{C}}&{10}&{13}& – &{16}&{14} \\ {\text{D}}&{17}&{20}&{16}& – &{15} \\ {\text{E}}&{19}&{11}&{14}&{15}& – \end{array}$

(a)     (i)     Use Kruskal’s algorithm to find and draw the minimum spanning tree for G.

(ii)     The graph H is formed from G by removing the vertex D and all the edges connected to D. Draw the minimum spanning tree for H and use it to find a lower bound for the travelling salesman problem for G.

(b)     Show that 80 is an upper bound for this travelling salesman problem.

## Markscheme

(a)     (i)     the edges are joined in the order

AC

BE

AB

ED     A2 A1

Note: Final A1 independent of the previous A2.

(ii) A1

the weight of this spanning tree is 33     A1

to find a lower bound for the travelling salesman problem, we add to that the two smallest weights of edges to D, i.e. 15 +16, giving 64     M1A1

[7 marks]

(b)     an upper bound is the weight of any Hamiltonian cycle, e.g. ABCDEA has weight 75 so 80 is certainly an upper bound     M1A1

[2 marks]

Total [9 marks]

## Examiners report

Part (a) was well done by many candidates although some candidates simply drew the minimum spanning tree in (i) without indicating the use of Kruskal’s Algorithm. It is important to stress to candidates that, as indicated in the rubric at the top of Page 2, answers must be supported by working and/or explanations. Part (b) caused problems for some candidates who obtained the unhelpful upper bound of 96 by doubling the weight of the minimum spanning tree. It is useful to note that the weight of any Hamiltonian cycle is an upper bound and in this case it was fairly easy to find such a cycle with weight less than or equal to 80.

## Question

Mathilde delivers books to five libraries, A, B, C, D and E. She starts her deliveries at library D and travels to each of the other libraries once, before returning to library D. Mathilde wishes to keep her travelling distance to a minimum.

The weighted graph $$H$$, representing the distances, measured in kilometres, between the five libraries, has the following table. Draw the weighted graph $$H$$.


a.

Starting at library D use the nearest-neighbour algorithm, to find an upper bound for Mathilde’s minimum travelling distance. Indicate clearly the order in which the edges are selected.


b.

By first removing library C, use the deleted vertex algorithm, to find a lower bound for Mathilde’s minimum travelling distance.


c.

## Markscheme complete graph on 5 vertices     A1

weights correctly marked on graph     A1

[2 marks]

a.

clear indication that the nearest-neighbour algorithm has been applied     M1

DA (or 16)     A1

AB (or 18) then BC (or 15)     A1

CE (or 17) then ED (or 19)     A1

$${\text{UB}} = 85$$     A1

[5 marks]

b.

an attempt to find the minimum spanning tree     (M1)

DA (16) then BE (17) then AB (18) (total 51)     A1

reconnect C with the two edges of least weight, namely CB (15) and CE (17)     M1

$${\text{LB}} = 83$$     A1

[4 marks]

c.

[N/A]

a.

[N/A]

b.

[N/A]

c.

## Question

Consider the following weighted graph G. State what feature of G ensures that G has an Eulerian trail.


a.i.

State what feature of G ensures that G does not have an Eulerian circuit.


a.ii.

Write down an Eulerian trail in G.


b.

State the Chinese postman problem.


c.i.

Starting and finishing at B, find a solution to the Chinese postman problem for G.


c.ii.

Calculate the total weight of the solution.


c.iii.

## Markscheme

G has an Eulerian trail because it has (exactly) two vertices (B and F) of odd degree      R1

[1 mark]

a.i.

G does not have an Eulerian circuit because not all vertices are of even degree      R1

[1 mark]

a.ii.

for example BAEBCEFCDF      A1A1

Note: Award A1 for start/finish at B/F, A1 for the middle vertices.

[2 marks]

b.

to determine the shortest route (walk) around a weighted graph      A1

using each edge (at least once, returning to the starting vertex)      A1

Note: Correct terminology must be seen. Do not accept trail, path, cycle or circuit.

[2 marks]

c.i.

we require the Eulerian trail in (b), (weight = 65)     (M1)

and the minimum walk FEB (15)     A1

for example BAEBCEFCDFEB    A1

Note: Accept EB added to the end or FE added to the start of their answer in (b) in particular for follow through.

[3 marks]

c.ii.

total weight is (65 + 15=)80      A1

[1 mark]

c.iii.

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

b.

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

c.iii.

## Question

The weights of the edges of a graph G with vertices A, B, C, D and E are given in the following table. Starting at A, use the nearest neighbour algorithm to find an upper bound for the travelling salesman problem for G .


a.

(i)     Use Kruskal’s algorithm to find and draw a minimum spanning tree for the subgraph obtained by removing the vertex A from G .

(ii)     Hence use the deleted vertex algorithm to find a lower bound for the travelling salesman problem for G .


b.

## Markscheme

using the nearest neighbour algorithm, starting with A,

$${\text{A}} \to {\text{E, E}} \to {\text{C}}$$     A1

$${\text{C}} \to {\text{D, D}} \to {\text{B}}$$     A1

$${\text{B}} \to {\text{A}}$$     A1

the upper bound is therefore 9 + 10 + 16 + 13 + 11 = 59     A1

[4 marks]

a.

(i)     the edges are added in the order CE     A1

BD     A1

BE     A1 A1

(ii)     the weight of the minimum spanning tree is 37     (A1)

we now reconnect A with the 2 edges of least weight     (M1)

i.e. AE and AB     A1

the lower bound is therefore 37 + 9 + 11 = 57     A1

[8 marks]

b.

[N/A]

a.

[N/A]

b.