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City of London Academy 1

KRUSKAL AND PRIM

1.

The network in the diagram above shows the distances, in metres, between 10 wildlife

observation points. The observation points are to be linked by footpaths, to form a network along

the arcs indicated, using the least possible total length.

(a) Find a minimum spanning tree for the network in the diagram above, showing clearly the

order in which you selected the arcs for your tree, using

(i) Kruskal’s algorithm, (3)

(ii) Prim’s algorithm, starting from A. (3)

Given that footpaths are already in place along AB and FI and so should be included in the

spanning tree,

(b) explain which algorithm you would choose to complete the tree, and how it should be

adapted.

(You do not need to find the tree.) (2)

(Total 8 marks)


2.

The diagram above represents the distances, in metres, between eight vertices, A, B, C, D, E, F, G

and H, in a network.

(a) Use Kruskal’s algorithm to find a minimum spanning tree for the network.

You should list the arcs in the order in which you consider them. In each case, state whether

you are adding the arc to your minimum spanning tree. (3)

(b) Complete the matrix below, to represent the network.

A B C D E F G H

A – 31 30 – – – – –

B 31 – – – – – 38

C 30 – – 24 – –

D – – 18 – –

E – – 24 18 – 28 –

F – – – 28 – 21 –

G – – – – 21 –

H – 38 – – – –

(2)

(c) Starting at A, use Prim’s algorithm to determine a minimum spanning tree. You must

clearly state the order in which you considered the vertices and the order in which you

included the arcs. (3)

(d) State the weight of the minimum spanning tree. (1)

(Total 9 marks)


3. The table below shows the least costs, in pounds, of travelling between six cities, A, B, C, D, E

and F.

A B C D E F

A – 36 18 28 24 22

B 36 – 54 22 20 27

C 18 54 – 42 27 24

D 28 22 42 – 20 30

E 24 20 27 20 – 13

F 22 27 24 30 13 –

Vicky must visit each city at least once. She will start and finish at A and wishes to minimise the

total cost.

(a) Use Prim’s algorithm, starting at A, to find a minimum spanning tree for this network. (2)

(b) Use your answer to part (a) to help you calculate an initial upper bound for the length of

Vicky’s route. (1)

(c) Show that there are two nearest neighbour routes that start from A. You must make your

routes and their lengths clear. (3)

(d) State the best upper bound from your answers to (b) and (c). (1)

(e) Starting by deleting A, and all of its arcs, find a lower bound for the route length. (4)

(Total 11 marks)

4. Prim’s algorithm finds a minimum spanning tree for a connected graph.

(a) Explain the terms

(i) connected graph,

(ii) tree,

(iii) spanning tree. (3)

(b) Name an alternative algorithm for finding a minimum spanning tree. (1)


Cambridge London Norwich Oxford Portsmouth Salisbury York

Cambridge (C) – 60 62 81 132 139 156

London (L) 60 – 116 56 74 88 211

Norwich (N) 62 116 – 144 204 201 181

Oxford (O) 81 56 144 – 84 63 184

Portsmouth (P) 132 74 204 84 – 43 269

Salisbury (S) 139 88 201 63 43 – 248

York (Y) 156 211 181 184 269 248 –

Figure 1

Figure 1 shows the distances by road, in miles, between seven cities.

(c) (i) Use Prim’s algorithm, starting at London, to find the minimum spanning tree for

these cities. You must clearly state the order in which you selected the edges of your

tree, and the weight of the final tree.

(ii) Draw your tree using the vertices given in Figure 2 below.

Figure 2 (5)

(Total 9 marks)


5.

A B C D E F

A – 135 180 70 95 225

B 135 – 215 125 205 240

C 180 215 – 150 165 155

D 70 125 150 – 100 195

E 95 205 165 100 – 215

F 225 240 155 195 215 –

The table shows the lengths, in km, of potential rail routes between six towns, A, B, C, D, E and F.

(a) Use Prim’s algorithm, starting from A, to find a minimum spanning tree for this table. You

must list the arcs that form your tree in the order that they are selected. (3)

(b) Draw your tree using the vertices given in the diagram below.

(1)

(c) State the total weight of your tree. (1)

(Total 5 marks)


6.

A B C D E F

A – 24 – – 23 22

B 24 – 18 19 17 20

C – 18 – 11 14 –

D – 19 11 – 13 –

E 23 17 14 13 – 21

F 22 20 – – 21 –

The table shows the distances, in metres, between six vertices, A, B, C, D, E and F, in a network.

(a) Draw the weighted network using the vertices given in Diagram 1 below.

Diagram 1 (3)

(b) Use Kruskal’s algorithm to find a minimum spanning tree. You should list the edges in the

order that you consider them and state whether you are adding them to your minimum

spanning tree. (3)

(c) Draw your tree on Diagram 2 below and find its total weight.


Diagram 2 (2)

(Total 8 marks)

7.

(a) State two differences between Kruskal’s algorithm and Prim’s algorithm for finding a

minimum spanning tree. (2)

(b) Listing the arcs in the order that you consider them, find a minimum spanning tree for the

network in the diagram above, using

(i) Prim’s algorithm,

(ii) Kruskal’s algorithm. (6)

(Total 8 marks)

8. (a)

18 20 11 7 17 15 14 21 23 16 9

The list of numbers shown above is to be sorted into ascending order. Apply quick sort to

obtain the sorted list. You must make your pivots clear. (5)


The diagram above represents a network of paths in a park. The number on each arc represents the

length of the path in metres.

(b) Using your answer to part (a) and Kruskal’s algorithm, find a minimum spanning tree for

the network in the diagram above. You should list the arcs in the order in which you

consider them and state whether you are adding it to your minimum spanning tree. (4)

(c) Find the total weight of the minimum spanning tree. (1)

(Total 10 marks)

9.

M A B C D E

M – 215 170 290 210 305

A 215 – 275 100 217 214

B 170 275 – 267 230 200

C 290 100 267 – 180 220

D 210 217 230 180 – 245

E 305 214 200 220 245 –

The table shows the cost, in pounds, of linking five automatic alarm sensors, A, B, C, D and E, and

the main reception, M.

(a) Use Prim’s algorithm, starting from M, to find a minimum spanning tree for this table of

costs. You must list the arcs that form your tree in the order that they are selected. (3)

(b) Draw your tree using the vertices given in the diagram below.

21

G

9

I

16

H

23

F

14

11

117

C20A

18

B

17

D

1115

E


(1)

(c) Find the total weight of your tree. (1)

(d) Explain why it is not necessary to check for cycles when using Prim’s algorithm. (2)

(Total 7 marks)

10.

A B C D E F G

A — 48 117 92 — — —

B 48 — — — — 63 55

C 117 — — 28 — — 85

D 92 — 28 — 58 132 —

E — — — 58 — 124 —

F — 63 — 132 124 — —

G — 55 85 — — — —

The table shows the lengths, in metres, of the paths between seven vertices A, B, C, D, E, F and G

in a network N.

(a) Use Prim's algorithm, starting at A, to solve the minimum connector problem for this table

of distances. You must clearly state the order in which you selected the edges of your tree,

and the weight of your final tree. Draw your tree using the vertices given in the diagram

below.

M A

E

D C

B


(5)

(b) Draw N using the vertices given in the diagram below.

(3)

(c) Solve the Route Inspection problem for N. You must make your method and working clear.

State a shortest route and find its length.

(The weight of N is 802)

Shortest route:

.........................................................................................................................

Length:

.................................................................................................................................... (7)

(Total 15 marks)

A

BC

D

E

F

G

A

BC

D

E

F

G


11.

The network in the diagram above shows the distances, in metres, between 10 wildlife

observation points. The observation points are to be linked by footpaths, to form a network along

the arcs indicated, using the least possible total length.

(a) Find a minimum spanning tree for the network in the diagram above, showing clearly the

order in which you selected the arcs for your tree, using

(i) Kruskal’s algorithm, (3)

(ii) Prim’s algorithm, starting from A. (3)

Given that footpaths are already in place along AB and FI and so should be included in the

spanning tree,

(b) explain which algorithm you would choose to complete the tree, and how it should be

adapted. (You do not need to find the tree.) (2)

(Total 8 marks)

A

DG

C F I

JHB

15

17 21

32

30

1920

24

38

21

25

1227

31

3945

E


12.

A B C D E F

A – 7 3 – 8 11

B 7 – 4 2 – 7

C 3 4 – 5 9 –

D – 2 5 – 6 3

E 8 – 9 6 – –

F 11 7 – 3 – –

The matrix represents a network of roads between six villages A, B, C, D, E and F. The value in

each cell represents the distance, in km, along these roads.

(a) Show this information on the diagram below.

(2)

(b) Use Kruskal’s algorithm to determine the minimum spanning tree. State the order in which

you include the arcs and the length of the minimum spanning tree. Draw the minimum

spanning tree.

(5)

A B

C

DE

F

A B

C

DE

F


(c) Starting at D, use Prim’s algorithm on the matrix given below to find the minimum

spanning tree. State the order in which you include the arcs.

A B C D E F

A – 7 3 – 8 11

B 7 – 4 2 – 7

C 3 4 – 5 9 –

D – 2 5 – 6 3

E 8 – 9 6 – –

F 11 7 – 3 – –

(3)

(Total 10 marks)

13. (a) Define the terms

(i) tree,

(ii) spanning tree,

(iii) minimum spanning tree. (3)

(b) State one difference between Kruskal’s algorithm and Prim’s algorithm, to find a minimum

spanning tree. (1)

Figure 1

B

A

C

F

EN

H5

10

116

10

15

13

14

14

9 127

8


(c) Use Kruskal’s algorithm to find the minimum spanning tree for the network shown in Fig.

1. State the order in which you included the arcs. Draw the minimum spanning tree in

Diagram 1 below and state its length.

Diagram 1

Order arcs included……………………………………………………………………………….

Length of minimum spanning tree………………………………………………………………... (4)

Figure 2

Figure 2 models a car park. Currently there are two pay-stations, one at E and one at N. These two

A

B

C

F

E

H

N

B

A

C

F

EN

H50

100

110

60

100

150

130

140

140

90 12070

80


are linked by a cable as shown. New pay-stations are to be installed at B, H, A, F and C. The

number on each arc represents the distance between the pay-stations in metres. All of the

pay-stations need to be connected by cables, either directly or indirectly. The current cable

between E and N must be included in the final network. The minimum amount of new cable is to

be used.

(d) Using your answer to part (c), or otherwise, determine the minimum amount of new cable

needed. Use Diagram 2 to show where these cables should be installed. State the minimum

amount of new cable needed.

Diagram 2

(3)

(Total 11 marks)

14. (a) Describe the differences between Prim’s algorithm and Kruskal’s algorithm for finding a

minimum connector of a network. (3)

(b) Listing the arcs in the order that you select them, find a minimum connector for the

network in the diagram above, using

(i) Prim’s algorithm; (ii) Kruskal’s algorithm. (4)

(Total 7 marks)

A

B

C

F

E

H

N

A B

C

D

E

F

G

15

17

18 16

24

32

25

35 14

20

21


15.

The diagram above shows a number of satellite towns A, B, C, D, E and F surrounding a city K.

The number on each edge give the length of the road in km.

(a) Use Dijkstra’s algorithm to find the shortest route from A to E in the network.

Show your working in the boxes provided on the answer sheet. (8)

It is planned to link all the sites A, B, C, D, E and F and K by telephone lines laid alongside the

roads.

(b) Use Kruskal’s algorithm to find a minimum spanning tree for the network and hence obtain

the minimum total length of cable required. Draw your tree. (6)

(Total 14 marks)

A

B C

D

E

F

K

2010

36

30 10

1215

815

28

30

14


Sheet for use in answering this question

(a)

Length of Shortest Path:

...…………………………………………………………………………………….

Determination of shortest route:

...…………………………………………………………………………………….

...…………………………………………………………………………………….

...…………………………………………………………………………………….

...…………………………………………………………………………………….

...…………………………………………………………………………………….

(b) Minimum spanning tree

Total weight of minimum spanning tree: ……………………………………………

2010

36

30 10

1215

815

28

30

14

A

B C

D

E

F

K

Vertex

Temporary Labels

PermanentLabels

KEY:


MARK SCHEME

1. (i) FH, AD, DE, CE, (not DC), (not AC), CF, HI, (not FI), IJ M1A1A1 3

(ii) AD, DE, EC, CF, FH, HI, IJ stop M1A1A1 3

(b) Start off the tree with AB and FI, then apply Kruskal M1 A1 2 [8]

2. (a) DE GF DC EG (not EF not CF) AC (not AB) GH M1 A1 A1 3

Note

1M1: Kruskal’s algorithm – first 4 arcs selected chosen correctly.

1A1: All seven non-rejected arcs chosen correctly.

2A1: All rejections correct and in correct order and at correct time.

(b)

A B C D E F G H

A - 31 30 - - - - -

B 31 - - 24 - - - 38

C 30 - - 22 24 29 - - B2, 1, 0 2

D - 24 22 - 18 - - 34

E - - 24 18 - 28 26 -

F - - 29 - 28 - 21 -

G - - - - 26 21 - 33

H - 38 - 34 - - 33 -

Note

1B1: condone two (double) errors

2B1: cao

(c) AC CD DE BD GE GF GH M1 A1 A1 3

Note

1M1: Prim’s algorithm – first four arcs chosen correctly, in order,

or first five nodes chosen correctly, in order.{A,C,D,E,B….}

1A1: First six arcs chosen correctly or all 8 nodes chosen correctly,

in order. {A,C,D,E,B,G,F,H}

2A1: All correct and arcs chosen in correct order.

(d) Weight: 174 B1 1

Note

,

EG

BC

,

EG

BC

not CE

BD


1B1: cao

Starting at Minimum arcs required

for M1

Nodes order

A AC CD DE DB ACDEB(GFH) 15234(768)

B BD DE DC BDEC(GFAH) (7)1423(658)

C CD DE DB CDEB(GFAH) (7)4123(658)

D DE DC DB DECB(GFAH) (7)4312(658)

E ED DC DB EDCB(GFAH) (7)4321(658)

F FG GE ED DC DB FGEDCB(AH) (7)654312(8)

G GF GE ED DC DB GFEDCB(AH) (7)654321(8)

H HG GF GE HGFE(DCBA) (8765)4321

[9]

3. (a)

M1 A1 2

Notes

1M1: Spanning tree found. Allow 1×2×43 across top of table or 93

1A1: CAO must see tree or list of arcs

(b) Minimum Spanning tree length 93, so upper bound is £186 B1ft 1

Note

1B1ft: 186 their ft93 × 2

(c) A C F E B D A M1

18 24 13 20 22 28 Length 125 A1

A C F E D B A

18 24 13 20 22 36 Length 133 A1 3

Notes

1M1: One Nearest Neighbour each vertex visited at least

once (condone lack of return to start)

1A1: One correct route and length CAO – must return to start.

2A1: Second correct route and length CAO – must return to start.

(d) Best upper bound is £125 B1ft 1

Note

1B1ft: ft but only on three different values.


(e) Delete A

M1 A1

RMST weight = 77

Lower bound = 77 + 18 + 22 = £117 M1 A1 4

Notes

1M1: Finding correct RMST (maybe implicit) 77 sufficient,

or correct numbers. 4 arcs.

1A1: CAO tree or 77.

2M1: Adding 2 least arcs to A, 18 and 22 or 40 only

2A1: CAO 117 [11]

4. (a) (i) All pairs of vertices connected by a path,

but not describing complete graph. B1

(ii) No cycles B1

(iii) All nodes connected (accept definition of

minimum spanning tree) B1 3

(b) Kruskal’s (algorithm) B1 1

(c) (i) L-O 56 Using Prim. first 2 correct M1

L-C 60

C-N 62

O-S 63 Next 2 A1

S-P 43

C-Y 156 Finish A1

Total length 440 (miles) Total A1 =B1

(ii) Tree correct


B1 5

Note

Accept weights as indicating arcs.

Misreads – award M1 A0 A0 for these:

• Vertices, not edges given L O C N S P Y

• Numbers across top, edges either incorrect

or not given: 3 1 4 2 6 5 7.

Also accept these, misreading And not starting

at L – again M1A0A0

Started at Minimum arcs nodes Numbers

C CL,LO,CN,….. CLONSPY 1243657

N NC,CL,LO,OS,SP,CY NCLOSPY 2314657

O OL,LC,CN,OS,…. OLCNSPY 3241657

P PS,SO,OL,LC,CN.CY PSOLCNY 5463127

S SP.SO,… SPOLCNY 5463217

Y YC,CL,LO,CN,.. YCLONSP 2354761

[9]

5. (a) AD, AE, DB; DC, CF M1 A1; A1 3

Note

1M1: Using Prim – first 2 arcs probably but

condone starting from another vertex.

1A1: first three arcs correct

2A1: all correct.

(b)

B1 1

Note


1B1: CAO

(c) Weight 595 (km) B1 1

Note

1B1: CAO condone lack of km.

Apply the misread rule, if not listing arcs or not starting at A.

So for M1 (only)

Accept numbers across the top (condoning absence of 6)

Accept full vertex listing

Accept full arc listing starting from vertex other than A

[AD AE DB DC CF] {1 4 5 2 3 6} ADEBCF

BD AD AE CD CF {3 1 5 2 4 6} BDAECF

CD AD AE BD CF {3 5 1 2 4 6} CDAEBF

DA AE DB CD CF {2 4 5 1 3 6} DAEBCF

EA AD DB DC CF {2 4 5 3 1 6} EADBCF

FC CD AD AE BD {4 6 2 3 5 1} FCDAEB [5]

6. (a)

M1

A1

A1 3

Note

1M1: More than 10 arcs

1A1: all arcs correct

2A1: all values correct

(b) CD, DE, reject CE, BE, reject BC, reject BD, BF, reject EF, AF M1 A1

11 13 14 17 18 19 20 21 22 A1 3

Note

1M1: First three arcs correctly chosen

1A1: All used acrs selected correctly

2A1: All rejected arcs selected in correct order


(c)

B1

Weight of tree 83 (m) B1 2

Note

1B1: CAO for arcs – numbers not needed. NO ft.

2B1: CAO 83, condone units [8]

7. (a) e.g.

• Prims starts with any vertex, Kruskal starts with the shortest arc.

• It is not necessary to check for cycles when using Prim.

• Prims adds nodes to the growth tree, Kruskal adds arcs.

• The tree ‘grows’ in a connected fashion when using Prim.

• Prim can be used when data in a matrix form.

Other correct statements also get credit. B2,1,0 2

1B1: Generous one coprrect difference. If bod give B1

2B1: Generous two distinct, correct differences.

(b) (i) e.g. AC, CF, FD, DE, DG, AB. M1, A1, A1 3

(ii) CF, DE, DF, not CD, not EF, DG, not FG, not EG, AC, not AD, AB.

[18, 19, 20, not 21, not 21, 22, not 23, not 24, 25, not 26, 27]

M1, A1, A1 3

1M1: Prim’s algorithm – first three arcs chosen correctly, in order,

or first four nodes chosen correctly, in order.

1A1: First five arcs chosen correctly; all 7 nodes chosen correctly,

in order.

2A1: All correct and arcs chosen in correct order.

2M1: Kruskal’s algorithm – first 4 arcs selected chosen correctly.

1A1: All six non-rejected arcs chosen correctly.

2A1: All regions correct and in correct order and ta correct time.


[8]

8. (a) e.g.

M1

A1

A1ft

A1ft

A1 5

(b) CF M1

GI

{BC or BF(*) – accept one, reject one

{CD (*) A

EF (*)

DF (*)

HI (*) A1

BE (*)

AB (*)

AC (*)

EG (*)

Tree complete A1 4

(*) Needs letters

(c) 107 m B1 1 [10]

9. (a) MB, BE, MD, DC, CA M1A1A1 3

(b)

B1ft 1

(c) 170 + 200 + 210 + 180 + 100 = 860 B1 1

B

A

C F

DG

E

2719

22

20

18

25

18

11

11

7

7

20

7

7

11

9

9

11

14

9

9

11

11

11

7

9

14

14

17

15

15

15

18

18

16

16

16

14

20

20

17

17

21

17

17

18

18

18

18

23

21

16

20

20

20

16

23

21

21

9

16

23

23

23

E

D C

A

B

M (170)

(230)

(200) (110)

(180)


(d) (A cycle is formed when an arc is used that connects two vertices

already connected to each other in the tree) B2,1,0 2

Prim’s algorithm always selects arcs that bring a vertex not in

the tree into the tree, so cycles can’t happen [7]

10. (a) AB, BG, BF | GC, CD, DE {1 2 5 6 7 4 3} M1 A1 A1 3

weight 337 m B1

B1ft 2

(b)

M1 A1 A1 3

(c) AB + CF = 48 + 160 = 208 M1 A1

AC + BF = 117 + 63 = 180 A1

AF + B C = 111 + 140 = 251 A1 4

e.g. A1

length 802 + 180 = 982 m M1 A1ft 3 [15]

11. (a) (i) FH,AD,DE,CE, (not DC), , (not AC), CF, HI,

(not FI), IJ stop M1 A1 A1 3

(ii) AD, DE, EC, , CF, FH, HI, IJ stop M1 A1 A1 3

(b) Start off the tree with AB and FI, then apply Kruskal B2, 1, 0 2 [8]

A

B C

D

EFG

A

BC

D

EFG

48

5563

117

85

92

132

124

58

28

A B F B G C A C D E F D A

EG

BC

EG

BC


12. (a) B1

B1 2

(b) M1

BD, , BC, Not CD, DE A1, A1

B1

Length = 18 km B1 5

(c) DB, DF, BC, CA, DE [5,2,4,1,6,3,] M1 A1 A1 3 [10]

13. (a) (i) A connected graph with no cycles, loops or multiple edges B1

(ii) A tree that includes all vertices B1

(iii) A spanning tree of minimum total length B1 3

(b) E.g.

In Kruskal the shortest arc is added (unless it completes a

cycle), in Prim the nearest unattached vertex is added

There is no need to check for cycles when using Prim,

but there is when using Kruskal B1 1

In Prim the tree always “grows” in a connected fashion

Kruskal starts with the shortest edge, Prim with any vertex

(c) BH, NF, HN, HA, BE, NC; length = 48 M1 A1; B1

B1 4

(d)

A B

C

DE

F

11

7

74

3

3

2

9

8

5

6

A B

C

DE

F

F D

CA

5

7

8

613

A

B

C

E

F

H

9N


B1

New cable – 390m M1 A1 3 [11]

14. (a) e.g.: In Prim the tree always ‘grows’ in a connected fashion; B3,2,1,0 3

In Kruskal the shortest arc is added (unless it completes a cycle),

in Prim the nearest unattached vertex is added;

There is no need to check for cycles when using Prim;

Prim can be easily used when network given in matrix form

Prim stats with a vertex, Kruskal with an edge

(b) (i) Either AC, AB, BD, BE, EF, EG (if starts at A or C)

or BD, BA, AC, BE, EF, EG (if starts at B or D)

or EF, EG, BE, BD, BA, AC (if starts at E or F)

or GE, EF, BE, BD, BA, AC (if starts at G) M1 A1

(ii) EF, AC, BD, BA, EG, BE M1 A1 4 [7]

15. (a)

Length of Shortest Path: 57 km B1

Determination of shortest route:

Label E – Label D = 57 – 43 = 14 (DE)

Label D – Label C = 43 – 35 = 8 (CD) M1 A1

Label C – Label B = 35 – 20 = 15 (BC)

Label B – Label A = 20 – 0 = 20 (AB)

Hence shortest route is A B C D E A1 8

50

70

80

60130

2010

36

3010

1215

815

28

30

14

A

B C

D

E

F

K

Vertex

Temporary Labels

PermanentLabel

0

0

20

20

30

30

30

30

35

35

36

45

43

43

60 58 57

57


(b) Minimum spanning tree

CD (8)

BK (10)

KF (10)

CK (12)

DE (14)

Not KD (15) Cycle

Not BC (15) Cycle

AB (20)

Total weight of minimum spanning tree: 74 km (tree) B1

(weight) B1 6 [14]

EXAMINERS' REPORTS

1. No Report available for this question.

2. Around 45% of the candidates gained full marks on this question. Candidates were directed to list the arcs in the order in which they

included them in the tree, but many candidates did not do so.

In part (a) a number of candidates only stated the arcs they were including in their tree and did not state the arcs that they rejected, as

they rejected them. Some candidates only referred to the length of the arc rather than by its end vertices, this makes it difficult for the

examiners to determine which arc is being considered.

Part (b) was usually completed correctly.

In part (c) many candidates showed their working on the table but then did not list the arcs in the correct order, often adding BD too late.

Many candidates completed part (d) correctly.

3. This proved a good first question with over a third of the candidates scoring full marks and over 50% getting at least 10 marks.

In part (a) many candidates did not list the arcs or draw the tree, BD was often included. Almost all were able to correctly find an initial

upper bound based on their tree.

In (c) most candidates were able to find the two nearest neighbour routes but a surprisingly large number did not return to A, others

found the NN route from A to B and A to D and then, alarmingly, doubled it.

Those who completed (c) correctly usually completed (d) correctly.

Many completed part (e) correctly but BD was, once again, often included in the residual minimum spanning tree.

4. The definitions in part (a) challenged many, particularly that of a ‘connected graph’ which challenged even the better candidates,

with some defining a complete graph and others being unclear about the nature of the connection between nodes. There were the usual

confusions of technical terms, ‘nodes connected by vertices’ etc. Part (b) was often correct, although often misspelled, but Prim and

Dijkstra were also popular incorrect answers. Most candidates were able to make progress with part (c), but many lost marks because

they did not list the edges in order of selection, or did not start at L as demanded in the question. The correct tree and its weight were

often seen.

5. This was a good starter and was well answered. Some candidates headed the columns correctly but then did not list the arcs correctly;

they should be encouraged to list each arc as it is selected. As expected the commonest error was to look for the smallest entry in the

latest column rather than in all numbered columns.

6. Almost all the candidates gained full marks in part (a), with only a few ambiguous numbers on the diagonal arcs. There were also

many fully correct solutions seen to part (b), although a few did not make the rejected arcs clear. Good presentation can be a great time

saver here, many gave an ordered list of all arcs with just ticks and crosses by the rejections, others wrote long sentences about their

decision to reject, due to cycles forming. Not all candidates stated the weight of the tree.

A

B C

D

K

E

F

20

10

10

128

14


7. Part (a) caused problems for some candidates. Candidates were asked for differences between the two algorithms and many simply

made statements such as Prim uses nodes without going on to say the Kruskal uses edges. A popular answer was to mention ordering

the arcs into ascending order for Kruskal; if this used as a difference candidates must indicate why this is important when using the

algorithm. There was the usual muddled use of basic technical terminology. Some candidates muddled the two algorithms. Part (b) as

usually well-answered although some just drew their MST and did not list the arcs, in order, as instructed. Some wasted time in Prim by

converting to the matrix form. Some did not make their rejections clear when applying Kruskal’s algorithm.

8. (a) Many excellent answers were seen here and the vast majority used the pivot selection seen on the mark scheme. There were

however a number of flawed solution seen also. The most common errors were; random or inconsistent pivot choices; misordering of

21, 23, 16 on the first pass, misordering of 11 and 9 on the third pass and not selecting 20 as a pivot; only selecting one pivot per line.

Some candidates used colour to indicate their pivots which should be discouraged. A very few bubble sorts were seen. In part (b) almost

all the candidates were able to select the first two arcs, CF and GI but the three arcs of length 11 challenged many. There was often a

lack of clarity concerning the order in which edges were rejected. Those who listed all the arcs, in order, and then ticked or crossed them

gained the marks most efficiently. Some candidates wasted a great deal of time (and space) drawing numerous diagrams, each

successive one having an extra arc.

9. There were many fully correct answers to (a) seen, but also many who applied the nearest neighbour algorithm instead. Many seemed

to confuse arcs with vertices and did not list their arcs as required. Most candidates drew a tree in part (b) and almost all who got the

correct tree also correctly found its weight. Many candidates were able to describe the mechanics of the algorithm in part (d), with the

commonly seen reference to ‘crossing out rows’, but only the best were able to think about HOW the algorithm worked and secure both

marks.

10. This proved an accessible question, but careless slips resulted in few gaining full credit. Not all candidates used Prim’s algorithm

correctly and of those that did, some did not clearly state the order in which the edge were selected – as directed in the question. A

disappointing number of candidates used the nearest neighbour algorithm instead of Prim’s algorithm. Many omitted to state the weight

of the tree. The vast majority were able to draw the network in part (b). The majority did three pairings but many did not find all the

shortest routes between the pairs. Some discarded any route involving more than one edge and some did not state the totals of their

pairings. A few did not state a route and some found a semi-Eulerian solution.

11. There was a varied response to this question, with some very good and some quite poor responses seen. Part (a) was generally better

attempted than part (b). In part (a) the candidates were asked to make the order in which they selected the arcs clear, many did not do

this. Many candidates wasted time by drawing a succession of diagrams showing the addition of one arc at a time. Many candidates lost

marks in Kruskal’s algorithm by not showing the rejection of the arcs that created loops. In some attempts at Prim’s algorithm, many

did not list the arcs and others referred to rejecting arcs to prevent cycles from forming. Many wasted time in drawing a matrix for Prim.

Part (b) was often poorly done with many candidates opting for Prim despite the two arcs not being connected. Of those who correctly

selected Kruskal, only a few were able to give a coherent explanation.

12. Almost all of the candidates were able to draw the arcs correctly but it was not always clear which numbers went with which arcs,

which lead to mistakes later. Most candidates were able to perform Kruskal’s algorithm correctly although a number omitted to state

that arc CD was rejected. Most went on to correctly draw the minimum spanning tree, but units were often omitted when stating its

weight. Prim’s algorithm was less well done. Some did not show that they had used the matrix, some started at A and many listed

vertices instead of arcs.

13. The majority of candidates found part (a) tricky. There were some very poor definitions, poor use of technical terms and muddled

thinking seen. Candidates often did not write in complete sentences, making it very difficult to understand the point they were trying to

make. Part (b) was usually better done although candidates confused arcs, vertices, nodes and edges and made some worrying remarks

about cycles. Part (c) was usually completed correctly. Part (d) caused problems for some candidates, with some incorrect diagrams and

incorrect answers of 54m, 540m and 39m often seen.

14. Very few candidates attempted a comparison in part (a). Most just described the two methods. A significant number confused the

two algorithms. There was some poor use of technical terms seen. Part (b) was very well done with most candidates gaining full marks,

but a few did not list the arcs as requested. A surprising number of candidates seem to think that Prim’s algorithm can only be applied

to a matrix and wasted a lot of time in creating the matrix. Others wasted time (and often went onto at least two extra sides of paper) by

drawing a ‘cartoon sequence’ showing the state of the tree after each arc addition, rather than simply listing the arcs in order.

15. No Report available for this question.

kruskal and prim - · pdf filekruskal and prim 1. the network in the diagram above shows the...

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