This question is related to the fact one can't use Bellman-Ford to find max weight paths in directed graphs with cycles. The reason is that giving a new graph $\tilde{G}$ with negative weights (e.g. $\tilde{w}_{ij} = - w_{ij}$) will result in cycles with negative sum, thus no minimum cost walk exists in $\tilde{G}$.

General question

How well can we approximate the maximum weight path between $i$ and $j$ in $G$, using the shortest path algorithm on a graph $\tilde{G}$ with transformed weights $\tilde{w}_{ij}$ ? In my case, one has $w_{ij}\in\mathbb{N}\setminus\{0\}$. The transformed weights $\tilde{w}_{ij}$ can be obtained using any function, but I assume a decreasing function $f$ is well adapted so that $w < w' \iff f(w) > f(w')$.

My approach

I am currently trying to use $f(w) = \dfrac{1}{w}$, in this context we have two important paths given a fixed path length $L$.

Denote a path $p = (i_0i_1, i_1i_2, i_2i_3, \dots i_{L-1}i_L$) with weights $W = (w^{(1)}, w^{(2)}, w^{(3)}, \dots, w^{(L)})$,

denote the maximum weight path of length $L$ by $p^*$ with weights $W^*$ achieving $$ W^* = \arg\max_W\{\mathtt{Cost}(W)\} = \arg\max_W \sum_{k=1}^L w^{(k)}$$ and the path $p_*$ achieving minimum transformed cost with weights $W_*$ $$ W_* = \arg\min_W \sum_{k=1}^L \tilde{w}^{(k)} = \arg\min_W \sum_{k=1}^L f(w^{(k)}) = \arg\min_W \sum_{k=1}^L \dfrac{1}{w^{(k)}}$$

Specific question

In this specific context, do we have the approximation $\mathtt{Cost}(W^*) \approx \mathtt{Cost}(W_*)$ ?

Or is it "very wrong" to replace $W^*$ by $W_*$ ? What would be the distribution of the relative error defined as $$E_r = \dfrac{\lvert\mathtt{Cost}(W^*) - \mathtt{Cost}(W_*)\rvert}{\mathtt{Cost}(W^*)}$$

Statistical analysis

I have tried a quick statistical analysis as follows:

  • Letting $L\in\{3,4,5,6,7,8\}$
  • sampling $w \sim \mathrm{Uniform}(1,w_{\max})$
  • The value for $w_{\max}$ was also picked from the set $\{50, 100, 150, 200, 500\}$
  • for each combination of $L,w_{\max}$ I computed $10^5$ values of $W^*, W_*$ using $10^3$ candidate weights $W_i$

The results:

  • In $80.8\%$ of cases we have $\mathtt{Cost}(W^*) = \mathtt{Cost}(W_*)$ and even better $W^*=W_*$
  • When computing the relative error $E_r$ one finds that $82.6\%$ of error values are less than $0.001$, $89.7\%$ less than $0.01$ and $99.9\%$ less than $0.1$

The histogram for $E_r > 0$ is as follows

relative_error distribution

  • $\begingroup$ I can provide the python code for the test if anyone wishes to further the statistical study. $\endgroup$ – ArnoV Jan 16 at 13:16

You can't. That approach is a dead end. Finding the maximum (simple) path is NP-hard, by reduction from Hamiltonian cycle. See, e.g., How to prove NP-hardness of a longest-path problem?, How is the Longest Path Problem NP complete?, Longest cycle in a digraph, https://en.wikipedia.org/wiki/Longest_path_problem. It is also known to be hard to approximate, so you shouldn't expect any such simple approach to give a good approximation for all graphs.

  • $\begingroup$ Do you know any other method of approximation? I really need to give an approximate answer to this problem, any idea is welcome ! $\endgroup$ – ArnoV Jan 17 at 11:35
  • $\begingroup$ @user715586, "It is also known to be hard to approximate". If you have a practical problem, I suggest posting a new question with details of the problem, parameters, problem sizes, etc. $\endgroup$ – D.W. Jan 17 at 19:04
  • $\begingroup$ My problem involves finding the most popular path using the count/frequency of the edges. There shouldn't more than 1000 nodes and a few thousand edges. Cycles are very common. I posted another question which uses another approach $\endgroup$ – ArnoV Jan 17 at 19:07

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