4 ms·
Right, but then the problem is no longer linear and approximating is more difficult. I'm not exactly sure what you mean by "exponential weight" in a linear prog
by modalduality 9y ago
Right, but then the problem is no longer linear and approximating is more difficult. I'm not exactly sure what you mean by "exponential weight" in a linear program, do you have an example?
- Spivak 9y agoSo say you have a graph on distance, d, that you want to weight based on a preference, p, from 0-5 going from least preferred to most preferred. You might make your weight function as w(d, p) = d * (2 ^ p)
- modalduality 9y agoThat's fine, but `p` is a variable, not a constant, in the original post's case, right?
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- tgb 9y agoThat doesn't seem necessary. Picking a value of p just selects how much you prefer short hope to shorter total distance.
- Retric 9y agoI am not sure what your asking? Your just assigning constants for weights on each trip a>b = K1, a>c = K2. I am saying you may map K1 as √((x1-x2)^2+(y1-y2)^2) to find least distance, but TSP allows for arbitrary constants. So remove the √ and long trips will be strongly avoided. PS: As far as I know you can use any arbitrary set of constants then use a linear solver. Or am I forgetting about something?