CENTRE FOR EDUCATIONAL DEVELOPMENT
Developing practice around the
realigned Level 2 Mathematics
and Statistics standards
Workshop Four
Anne Lawrence, Alison Fagan , Cami Sawyer
Advisers in Secondary Numeracy & Mathematics
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Level 3 Consultation
In small groups pick 2-3 standards and
discuss what has changed
• Think about the pathways from last time
• Changes made to level 1 & 2
Resources
• Draft level 3 standards
• Draft Matrix
• Summary of what has changed
2
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Level 3 Consultation
Manawatu:
3.6 too much in it
3.8 evaluation – concern at higher literacy
needed
3.8 Using existing data sets – but this seems
to conflict with using each component of
PPDAC ie posing problem, collecting
data?
3.9 assume that making a prediction
3
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Level 3 Consultation
Hawkes Bay feedback:
4
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TKI Senior Secondary
Teaching and Learning Guides
AO M 8-7 (trigonometric, polynomial, and other non-linear equations)
What is new/changed?
* Manipulating logs will be new
AO M 8-11 (differentiation, integration, and anti-differentiation techniques)
What is new/changed?
* There is no integration at level 7
* This does not include related rates of change, integration of
relations, or volume of revolution.
What will happen to Simultaneous Equations and Linear
Programming?
5
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Achievement Objective M7- 5
In a range of meaningful contexts, students will be
engaged in thinking mathematically and statistically.
They will solve problems and model situations that
require them to:
Choose appropriate networks to find optimal
solutions.
Indicators
• Solves problems that can be modelled by networks
• Uses trial-and-improve methods to develop
algorithms for solving network problems
6
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Network Definitions
7
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Spot the Features
Identify the features or terminology from the last
activity that are shown in this tramping network of
huts linked by tracks.
Becks
Caps
Freddy
David
Ace
Gum
Eddy
Do the US 5249 Tasks
Happy
8
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Networks
AS2.5 –Use networks in solving problems
Look at the standard:
– What are the understandings required?
– What do you think should be the step up from
achieve to merit?
merit to excellence?
9
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TKI Senior Secondary
Teaching and Learning Guides
10
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TKI Senior Secondary
Teaching and Learning Guides
11
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Minimum Spanning Tree
Kruskal’s algorithm
1. Select the shortest edge
in a network
2. Select the next shortest
edge which does not
create a cycle
3. Repeat step 2 until all
vertices have been
connected
Prim’s algorithm
1. Select any vertex
2. Select the shortest edge
connected to that vertex
3. Select the shortest edge
connected to any vertex
already connected
4. Repeat step 3 until all
vertices have been
connected
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A cable company want to connect five villages
to their network which currently extends to the
market town of Avonford.
What is the minimum length of cable needed?
5
Brinleigh
Cornwell
3
4
6
8
8
Avonford
7
Donster
Fingley
5
4
2
Edan
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First model the situation as a network, then the
problem is to find the minimum connector for
the network
5
B
C
3
4
6
8
8
A
D
F
7
5
4
2
E
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Kruskal’s Algorithm
B
5
List the edges
in order of size:
C
3
4
6
8
8
A
D
F
7
5
4
2
E
ED
AB
AE
CD
BC
EF
CF
AF
BF
CF
2
3
4
4
5
5
6
7
8
8
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Kruskal’s Algorithm
B
5
Select the shortest
edge in the network
C
ED 2
3
4
6
8
8
A
D
F
7
5
4
2
E
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Kruskal’s Algorithm
B
5
Select the
next shortest
edge which
does not
create a cycle
C
3
4
6
8
8
A
D
F
7
ED 2
AB 3
5
4
2
E
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Kruskal’s Algorithm
B
5
Select the next
shortest edge
which does not
create a cycle
C
3
4
6
8
8
A
D
F
7
ED 2
AB 3
CD 4 (or AE 4)
5
4
2
E
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Kruskal’s Algorithm
B
5
Select the next
shortest edge
which does not
create a cycle
C
3
4
6
8
8
A
D
F
7
5
4
2
E
ED
AB
CD
AE
2
3
4
4
Kruskal’s Algorithm
B
5
Select the next
shortest edge
which does not
create a cycle
C
3
4
6
8
8
A
D
F
7
5
4
2
ED
AB
CD
AE
BC
EF
2
3
4
4
5 forms a cycle
5
E
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Kruskal’s Algorithm
B
5
All vertices have
been connected.
C
The solution is
3
4
6
8
8
A
D
F
7
5
4
ED
AB
CD
AE
EF
2
3
4
4
5
2
E
Total weight of
tree: 18
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Prim’s Algorithm
B
5
Select any
vertex
C
A
3
4
6
8
8
A
D
F
7
5
4
Select the
shortest edge
connected to
that vertex
2
AB 3
E
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Prim’s Algorithm
B
5
C
3
4
6
8
8
A
D
F
7
5
4
Select the
shortest
edge
connected to
any vertex
already
connected.
AE 4
2
E
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Prim’s Algorithm
B
5
C
3
4
6
8
8
A
D
F
7
5
4
Select the
shortest
edge
connected to
any vertex
already
connected.
ED 2
2
E
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Prim’s Algorithm
B
5
C
3
4
6
8
8
A
D
F
7
5
4
Select the
shortest
edge
connected to
any vertex
already
connected.
DC 4
2
E
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Prim’s Algorithm
B
5
C
3
4
6
8
8
A
D
F
7
5
4
CB 5 forms a
2
E
Select the
shortest
edge
connected to
any vertex
already
connected.
cycle
EF 5
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Prim’s Algorithm
B
5
All vertices
have been
connected.
C
3
6
8
The solution
is
4
8
A
D
F
7
5
4
2
ED
AB
CD
AE
EF
2
3
4
4
5
E
Total weight
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Prim’s and Kruskal’s
Algorithms
• Both algorithms will always give solutions with
the same length.
• They will usually select edges in a different order
– students need to show this in their working.
• Occasionally these algorithms will use different
edges – this may happen when you have to choose
between edges with the same length. In this case
there is more than one minimum connector for the
network.
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Dijkstra’s Algorithm
finds the shortest path from the start vertex to every other
vertex in the network. We will find the shortest path from A to G
4
B
4
F
1
2
7
4
D
A
7
3
3
2
G
C
5
E
2
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Order in
which
vertices are
labelled.
Dijkstra’s
Algorithm
1st
0
Working
label
4
B
4
1
2
7
F
4
D
A
Label vertex A
1st as it is the
first vertex
labelled
Permanent
label =
Distance
from A to
vertex
7
3
3
2
G
C
5
E
2
Dijkstra’s
Algorithm
We update each vertex adjacent to A
with a ‘working value’ for its distance
from A.
4
4
B
1st
0
4
F
1
2
7
7
4
D
A
7
3
3
2
G
C
5
E
2
3
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Dijkstra’s
Algorithm
4
1st
0
Order in
which
vertices are
labelled.
Look at ALL
the working
labels (no
ordinal yet).
Which is
smallest?
B
4
Permanent
label =
Distance
from A to
vertex
Working
label
4
1
2
7
7
F
4
D
A
7
3
3
2
G
Vertex C is closest
to A so we give it a
permanent label 3.
C is the 2nd vertex
to be permanently
labelled.
5
C
2nd
3
E
2
3
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Dijkstra’s
Algorithm
We update each vertex adjacent to C with a
‘working value’ for its total distance from A, by
adding its distance from C to C’s permanent
label of 3.
4
4
B
1st
4
0
6 < 7 so
replace the
t-label here
1
F
2
7 6
7
4
D
A
7
3
3
2
G
5
C
2nd
3
E
2
3
8
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Look at ALL the working labels (no ordinal yet).
Which is smallest?
Dijkstra’s
Algorithm
3rd
4
1st
0
4
The vertex with the smallest
temporary label is B, so make this
label permanent. B is the 3rd
vertex to be permanently labelled.
4
B
4
1
F
2
7 6
7
4
D
A
7
3
3
2
G
5
C
2nd
3
E
2
3
8
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Dijkstra’s
Algorithm
3rd
4
We update each vertex adjacent to B with a
‘working value’ for its total distance from A, by
adding its distance from B to B’s permanent
label of 4.
4
4
B
1st
0
4
5 < 6 so
replace the
t-label here
1
2
7 6 5
7
F
8
4
D
A
7
3
3
2
G
5
C
2nd
3
E
2
3
8
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Dijkstra’s
Algorithm
3rd
4
The vertex with the smallest
temporary label is D, so make
this label permanent. D is the
4th vertex to be permanently
labelled.
4
4
B
1st
0
4
4th 5
7 6 5
1
7
2
F
8
4
D
A
7
3
3
2
G
5
C
2nd
3
E
2
3
8
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Dijkstra’s
Algorithm
3rd
4
We update each vertex adjacent to D with a
‘working value’ for its total distance from A, by
adding its distance from D to D’s permanent
label of 5.
4
4
B
1st
0
4
F
4th 5
7 6 5
1
7
8 7
7 < 8 so
replace the
4 t-label here
2
D
A
7
3
3
5
C
2nd
3
2
E
7 < 8 so
replace the
t-label here
G
2
12
3
8 7
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Look at ALL the working labels (no ordinal yet).
Which is smallest?
Dijkstra’s
Algorithm
3rd
4
4
4
B
1st
0
4
F
4th 5
7 6 5
1
7
2
8 7
4
D
A
7
3
3
2
G
5
C
2nd
3
3
2
E
5th
8 7
7
12
The vertices with the
smallest temporary labels are E and
F, so choose one and make the label
permanent. E is chosen - the 5th
vertex to be permanently labelled.
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Dijkstra’s
Algorithm
3rd
4
We update each vertex adjacent to E with a
‘working value’ for its total distance from A, by
adding its distance from E to E’s permanent
label of 7.
4
4
B
1st
0
4
F
4th 5
7 6 5
1
7
2
8 7
4
D
A
7
3
3
2
G
5
C
2nd
3
3
2
E
5th
8 7
7
12 9
9 < 12 so
replace the
t-label here
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Look at ALL the working labels (no ordinal yet).
Which is smallest?
Dijkstra’s
Algorithm
3rd
4
The vertex with the smallest
temporary label is F, so make this
label permanent.F is the 6th vertex
to be permanently labelled.
4
4
B
1st
0
4
F
4th 5
7 6 5
1
7
2
6th 7
8 7
4
D
A
7
3
3
2
G
5
C
2nd
3
3
2
E
5th
8 7
7
12 9
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Dijkstra’s
Algorithm
3rd
4
We update each vertex adjacent to F with a
‘working value’ for its total distance from A, by
adding its distance from F to F’s permanent
label of 7.
4
4
B
1st
0
4
F
4th 5
7 6 5
1
7
2
6th 7
8 7
4
D
A
7
3
3
2
G
5
C
2nd
3
3
2
E
5th
8 7
7
12 9
11 > 9 so do
not replace
the t-label
here
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Dijkstra’s
Algorithm
Can you SEE the shortest path from A to G?
3rd
4
4
4
B
1st
0
4
F
4th 5
7 6 5
1
7
2
6th 7
8 7
4
D
A
7
3
3
2
G
5
C
2nd
3
3
2
E
5th
8 7
7
7th 9
12 9
G is the final vertex to
be permanently labelled.
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To find the shortest path from A to G, start
from G and work backwards, choosing arcs for
which the difference between the permanent
labels is equal to the arc length.
Dijkstra’s
Algorithm
3rd
4
4
4
B
1st
0
4
F
4th 5
7 6 5
1
7
2
6th 7
8 7
4
D
A
7
3
3
2
G
5
C
2nd
3
3
2
E
5th
8 7
7
7th 9
12 9
The shortest path is ABDEG, with length 9.
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Assessment Judgements
Using the assessment activity 2.5A
•
•
•
Complete the task
Examine the Assessment Schedule
Compare with your own solution
44
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Discussion
• Where will this learning fit in your curriculum
level 7 (NCEA level 2) courses?
• What prior knowledge will students need to
access this AO and standard?
• What are some of the new ideas in this
standard that you think are important?
• Where will it lead – careers and pathways?
45
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Where does this go?
http://www.newton.ac.uk/wmy2kposters/june/
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