This discussion is closed.
Bhaal85
Badges: 9
Rep:
?
#1
Report Thread starter 17 years ago
#1
Does anybody know the equation for working out how many different ways there are to connect nodes? Lost my notes for that, and need the answer. Thanks.
0
Expression
Badges: 14
Rep:
?
#2
Report 17 years ago
#2
(Original post by Bhaal85)
Does anybody know the equation for working out how many different ways there are to connect nodes? Lost my notes for that, and need the answer. Thanks.
I assume you mean the ways to pair up odd nodes (or vertices)...


Odd Verts | Ways of Pairing

2 | 1
4 | 3
6 | 15
8 | 105
10 | 945
12 | 10395
14 | 135135

There was no given equation, but we just learnt the numbers - for the sake of the exam, all you needed to know was that if you had 6 odd vertices you were going to end up having to find 15 results, and try and pair them to find the shortest connectors. Clearly you are going to have no more odd vertices than 6, as the computation becomes laborious, and unmanageable in an 80 minute exam.
0
Juwel
Badges: 18
Rep:
?
#3
Report 17 years ago
#3
Doesn't it have something to do with triangle numbers?
0
LH
Badges: 15
Rep:
?
#4
Report 17 years ago
#4
(Original post by Expression)
I assume you mean the ways to pair up odd nodes (or vertices)...


Odd Verts | Ways of Pairing

2 | 1
4 | 3
6 | 15
8 | 105
10 | 945
12 | 10395
14 | 135135

There was no given equation, but we just learnt the numbers - for the sake of the exam, all you needed to know was that if you had 6 odd vertices you were going to end up having to find 15 results, and try and pair them to find the shortest connectors. Clearly you are going to have no more odd vertices than 6, as the computation becomes laborious, and unmanageable in an 80 minute exam.
One question, what's a Chinese Postman got to do with all that?
0
Juwel
Badges: 18
Rep:
?
#5
Report 17 years ago
#5
(Original post by Lord Huntroyde)
One question, what's a Chinese Postman got to do with all that?
Read up on discrete maths and find out.
0
LH
Badges: 15
Rep:
?
#6
Report 17 years ago
#6
(Original post by ZJuwelH)
Read up on discrete maths and find out.
Will do.
0
Expression
Badges: 14
Rep:
?
#7
Report 17 years ago
#7
(Original post by Lord Huntroyde)
Will do.
Save you a job.


(Original post by AQA Discrete Mathematics One)

Chapter Six - Route Inspection

6.2 - The Chinese Postman Algorithm

It is useful to have a systematic proceedure to obtain a closed trail containing every edge of minimum length of weight. The following proceedure for finding the least-weight closed trail containing every edge was invented by the Chinese Mathematician Kuan Mei-Ko in 1962
The postman bit really is to do with the situation that a postman faces, he will invariably deliver to every street, and it would be useful to find a route by which time or length is the shortest.

When the network is set out for this, we are looking for the shortest route around the network, which takes in every edge (street).

Chinese Postman Algorithm

Step 1

Find all vertices of odd order

Step 2

For each pair of odd vertices, find connecting paths of minimum weight.

Step 3

Pair up all the odd vertices such that the sum of the weight of the connecting paths from step 2 is minimised.

Step 4

In the original graph, duplicate minimum weight paths found in Step 3.

Step 5

Find a trail containing every edge for the new (Eulerian) graph.
0
Bhaal85
Badges: 9
Rep:
?
#8
Report Thread starter 17 years ago
#8
(Original post by Expression)
I assume you mean the ways to pair up odd nodes (or vertices)...


Odd Verts | Ways of Pairing

2 | 1
4 | 3
6 | 15
8 | 105
10 | 945
12 | 10395
14 | 135135

There was no given equation, but we just learnt the numbers - for the sake of the exam, all you needed to know was that if you had 6 odd vertices you were going to end up having to find 15 results, and try and pair them to find the shortest connectors. Clearly you are going to have no more odd vertices than 6, as the computation becomes laborious, and unmanageable in an 80 minute exam.
Thanks, phew I thought there was some equation like all the other different algorithms. Are you certain of this???
0
LH
Badges: 15
Rep:
?
#9
Report 17 years ago
#9
(Original post by Expression)
Save you a job.




The postman bit really is to do with the situation that a postman faces, he will invariably deliver to every street, and it would be useful to find a route by which time or length is the shortest.

When the network is set out for this, we are looking for the shortest route around the network, which takes in every edge (street).

Chinese Postman Algorithm

Step 1

Find all vertices of odd order

Step 2

For each pair of odd vertices, find connecting paths of minimum weight.

Step 3

Pair up all the odd vertices such that the sum of the weight of the connecting paths from step 2 is minimised.

Step 4

In the original graph, duplicate minimum weight paths found in Step 3.

Step 5

Find a trail containing every edge for the new (Eulerian) graph.
And who says postmen (or postpeople) don't need Maths for their career?
0
Unregistered
Badges:
#10
Report 17 years ago
#10
i do
0
Expression
Badges: 14
Rep:
?
#11
Report 17 years ago
#11
(Original post by Bhaal85)
Thanks, phew I thought there was some equation like all the other different algorithms. Are you certain of this???
Pretty well certain.
0
Bhaal85
Badges: 9
Rep:
?
#12
Report Thread starter 17 years ago
#12
(Original post by Expression)
Pretty well certain.
......so........if there where 'n' odd nodes.................J/K
0
Camford
Badges: 12
Rep:
?
#13
Report 17 years ago
#13
I though Chinese Postman was to do with letter delivery. Like 5 letters in 3 postboxes or whatever.
0
Ditzy
Badges: 1
Rep:
?
#14
Report 17 years ago
#14
(Original post by Expression)
Pretty well certain.
Just looked the problem up in my Graph Theory text book, Expression was spot on! Just wanted to give you a bit of confirmation. x.
0
Unregistered
Badges:
#15
Report 17 years ago
#15
(Original post by ditzy blonde)
Just looked the problem up in my Graph Theory text book, Expression was spot on! Just wanted to give you a bit of confirmation. x.
You study Bsc maths? Surely u can confirm without looking in textboo?
0
Juwel
Badges: 18
Rep:
?
#16
Report 17 years ago
#16
(Original post by Bhaal85)
Thanks, phew I thought there was some equation like all the other different algorithms. Are you certain of this???
Could it be the sum of all the numbers from 1 to (2n-2)? I thought of this in my sleep and found that it worked for the first four, i.e. 2, 4, 6, and 8 vertices. I knew it had something to do with triangle numbers...
0
Ditzy
Badges: 1
Rep:
?
#17
Report 17 years ago
#17
(Original post by Unregistered)
You study Bsc maths? Surely u can confirm without looking in textboo?
Doesn't mean I can remember everything I've ever learnt tho, things kinda escape the second the exams over!
0
rae
Badges: 0
#18
Report 17 years ago
#18
this algorithm is straight forward - they cant ask you anything hard on it. learn the simplex algorithm and how to do a gantt cascade chart (!) and the algorithm that goes with it.
the exams are so damn hard!
0
X
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Do you feel comfortable going back to school on March 8th?

Yes (385)
32.11%
No (814)
67.89%

Watched Threads

View All