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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.
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#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.
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.
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.
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#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.
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.
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#5
(Original post by Lord Huntroyde)
One question, what's a Chinese Postman got to do with all that?
One question, what's a Chinese Postman got to do with all that?
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#7
(Original post by Lord Huntroyde)
Will do.
Will do.
(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
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
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.
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(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.
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.
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#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.
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.
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#11
(Original post by Bhaal85)
Thanks, phew I thought there was some equation like all the other different algorithms. Are you certain of this???
Thanks, phew I thought there was some equation like all the other different algorithms. Are you certain of this???
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(Original post by Expression)
Pretty well certain.
Pretty well certain.
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#13
I though Chinese Postman was to do with letter delivery. Like 5 letters in 3 postboxes or whatever.
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#14
(Original post by Expression)
Pretty well certain.
Pretty well certain.
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#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.
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.
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#16
(Original post by Bhaal85)
Thanks, phew I thought there was some equation like all the other different algorithms. Are you certain of this???
Thanks, phew I thought there was some equation like all the other different algorithms. Are you certain of this???
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#17
(Original post by Unregistered)
You study Bsc maths? Surely u can confirm without looking in textboo?
You study Bsc maths? Surely u can confirm without looking in textboo?
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#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!
the exams are so damn hard!
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