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D1 - Chinese Postman
I have a couple of questions about 'Chinese Postman' questions:
- What's the general formula for working out how many ways the odd vertices can be paired together, eg: 4 odd vertices can be paired in 6 ways
- When the question asks for 'an optimal Chinese postman route' what do you have to do? do you still try and find the shortest route, despite there being 6 odd vertices, thus meaning it will take a long time to work out
thanks
- What's the general formula for working out how many ways the odd vertices can be paired together, eg: 4 odd vertices can be paired in 6 ways
- When the question asks for 'an optimal Chinese postman route' what do you have to do? do you still try and find the shortest route, despite there being 6 odd vertices, thus meaning it will take a long time to work out
thanks
doesn't anyone go on the maths forum anymore? it used to be the one place i was guaranteed help
well do you happen to know the answer cos im stuck.
kinda off topic but i can now see why so few people do decision maths - i absolutely hate it!! im so not a logical person
kinda off topic but i can now see why so few people do decision maths - i absolutely hate it!! im so not a logical person
it's tedious and not even that easy. the whole following the rules thing is easy, but some of the random questions are hard (im not in a very eloquent mood right now lol)
2 odd vertices can only be paired once.
4 odd vertices can only be paired six ways.
6 odd vertices can only be paired fifteen ways.
See the pattern? It's every other triangle number.
As a formula, I think it's n odd vertices can be paired (n-1)+(n-2)+...+2+1 ways, or n(n-1)/2 ways. n is of course necessarily even.
4 odd vertices can only be paired six ways.
6 odd vertices can only be paired fifteen ways.
See the pattern? It's every other triangle number.
As a formula, I think it's n odd vertices can be paired (n-1)+(n-2)+...+2+1 ways, or n(n-1)/2 ways. n is of course necessarily even.
Just a small reminder that Chinese Postman is a another fancy name for the not so fancy ROUTE INSPECTION
You have the equation slightly incorrect, just try it (I initially made the same mistake).
Several students ask for it and I haven't found it anywhere, so had to derive it, unsimplified its the product of the odd values less than n (if there are n odd nodes):
(n-1)(n-3)(n-5)x...x3 x1
As an expression this is:
n! / [2^(n/2) *(n/2)!]
For 12 this would be:
12!/[2^6*6!]
479001600 / [64*720]
479001600 / 46080
10395
As required
Several students ask for it and I haven't found it anywhere, so had to derive it, unsimplified its the product of the odd values less than n (if there are n odd nodes):
(n-1)(n-3)(n-5)x...x3 x1
As an expression this is:
n! / [2^(n/2) *(n/2)!]
For 12 this would be:
12!/[2^6*6!]
479001600 / [64*720]
479001600 / 46080
10395
As required
Original post by djmcguigan
You have the equation slightly incorrect, just try it (I initially made the same mistake).
Several students ask for it and I haven't found it anywhere, so had to derive it, unsimplified its the product of the odd values less than n (if there are n odd nodes):
(n-1)(n-3)(n-5)x...x3 x1
As an expression this is:
n! / [2^(n/2) *(n/2)!]
For 12 this would be:
12!/[2^6*6!]
479001600 / [64*720]
479001600 / 46080
10395
As required
Several students ask for it and I haven't found it anywhere, so had to derive it, unsimplified its the product of the odd values less than n (if there are n odd nodes):
(n-1)(n-3)(n-5)x...x3 x1
As an expression this is:
n! / [2^(n/2) *(n/2)!]
For 12 this would be:
12!/[2^6*6!]
479001600 / [64*720]
479001600 / 46080
10395
As required
Since, you're new, it's understandable you bumping a 6 year old thread
Btw, welcome to TSR
Original post by boromir9111
Since, you're new, it's understandable you bumping a 6 year old thread
Btw, welcome to TSR
Btw, welcome to TSR
Threads this old are spooky.
Original post by thats_my_poison
'- What's the general formula for working out how many ways the odd vertices can be paired together, eg: 4 odd vertices can be paired in 6 ways'.
The solution for this problem is: 1 x 3 x 5 x .... (next odd number until n-1);
e.g. 4 odd vertices = 1 x 3 = 3 pairs
6 odd vertices = 1 x 3 x 5 = 15 pairs
- When the question asks for 'an optimal Chinese postman route' what do you have to do? do you still try and find the shortest route, despite there being 6 odd vertices, thus meaning it will take a long time to work out
The optimal Chinese Postman Route = Total + Repeat Route.
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