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Posted 27 September 2012 - 04:03 AM

Exactly guys! You guys are all right on it... it would either require a massive expansion of nodes or probably a metaheuristic to solve! I'm just silly lol. Ever had that thought where you say ok I'm going to add a little twist to an existing problem to see if I can get better results only to find out that what you're trying to do is silly or impossible :-)? I'm surprised I couldn't find anything on google about this type of problem... "Graph" "shortest path" "multiple weights" "conditional weights" etc... turned up nothing!

Posted 27 September 2012 - 03:58 AM

Exactly guys! You guys are all right on it... it would either require a massive expansion of nodes or probably a metaheuristic to solve! I'm just silly lol. Ever had that thought where you say ok I'm going to add a little twist to an existing problem to see if I can get better results only to find out that what you're trying to do is silly or impossible :-)?

Posted 27 September 2012 - 03:58 AM

Exactly guys! You guys are all right on it... it would either require a massive expansion of nodes or probably a metaheuristic to solve! I'm just silly lol. Ever had that thought where you say ok I'm going to add a little twist to an existing problem to see if I can get better results only to find out that what you're trying to do is impossible :-)?

Posted 27 September 2012 - 03:58 AM

Exactly guys! You guys are all right on it... it would either require a massive expansion of nodes or probably a metaheuristic to solve! I'm just silly lol. Ever had that thought where you say ok I'm going to add a little twist to an existing problem to see if I can get better results only to find out that what you're trying to do is impossible :-)?

It's modeling terrain elevation as well as distance traveled. The same thing happens to mountain climbers: it's easier to climb two small mountains than one large one plus a tiny hill even though the distances traveled are the same. Though yeah... in this case elevation can be encoded per edge (as euclidean distance). Hmmm... going to have to think of a better way to explain this...

Posted 27 September 2012 - 03:54 AM

Exactly guys! You guys are all right on it... it would either require a massive expansion of nodes or probably a metaheuristic to solve! I'm just silly lol. Ever had that thought where you say ok I'm going to add a little twist to an existing problem to see if I can get better results only to find out that what you're trying to do is impossible :-)?

It's modeling terrain elevation as well as distance traveled. The same thing happens to mountain climbers: it's easier to climb two small mountains than one large one plus a tiny hill even though the distances traveled are the same. Though yeah... in this case elevation can be encoded per edge (as euclidean distance). Hmmm... going to have to think of a better way to explain this...

OK. Musical instrument, guitar. You play three notes. C -> B > E. Which of the two paths is better?

C 2nd string 1st fret -> B 2nd string 0 fret -> E 1st string 12th fret
or
C 4th string 10th fret -> B 2nd string 0 fret -> E 1st string 12th fret

It's obvious the last one is the better choice. The last weight is not dependent on the middle vertex, but the vertex before it.