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Suppose we have a 6 vertices graph. We also have 6 gates. Each gate is attributed a path.

For example,

Gate 'A' will have to go to 'B'- 'C' - 'D' and 'E'

Gate 'B' will have to go to 'D'

Gate 'C' will have to go to 'A' and 'B'

Gate 'D' will have to go to 'E' - 'A' and 'B'

Gate 'E' will have to go to 'C' - 'E' and 'A'

Gate 'B' will have to go to 'D'

How can I find the optimal placement of the gates on the graph, such that it makes the least distance overall, computing all the paths to be made?

For 6 gates, we could search all possible placements by brute force. However, in my current project I have 42 gates to dispatch, thus it sounds too complicated to brute force.

The following is the function used to calculate the distance between the two gates (stocked in a Pandas DataFrame)

import math
def calculateDistance(df, gate_a, gate_b):
    coord_a = [(x, df.columns[y]) for x, y in zip(*np.where(df.values == gate_a))][0] # GET COORDINATE OF GATE_A
    coord_b = [(x, df.columns[y]) for x, y in zip(*np.where(df.values == gate_b))][0] # GET COORDINATE OF GATE_B
    y1, x1 = coord_a[0], coord_a[1]
    y2, x2 = coord_b[0], coord_b[1]
    dist = math.sqrt((((x2 - x1) * scale_x) **2)  + ( ((y2 - y1) * scale_y) **2 ))
    return dist

calculateDistance(df, "A", "C")
John L.
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Achille G
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  • I've added the way I calculate the distance between two gates, finding the gate's coordinates within the dataframe. – Achille G Jul 25 '22 at 15:13
  • The least distance overall, computing all the paths to be made! Sorry if this is not clear – Achille G Jul 25 '22 at 15:21
  • No, in my current project I have 42 gates to dispatch, thus it sounds too complicated to brute force – Achille G Jul 25 '22 at 15:34
  • Travelling salesman problem is a very special case of your task. It is unlikely there is an algorithm in polynomial time of the number of gates. – John L. Jul 25 '22 at 15:57
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    Let me do my best attempt to rephrase the problem in CS terms: There's a metric space $X$, with $n$ points. Additionally, there is a collection of $n$ sequences of labels, where all labels are numbers in ${1, \ldots, n}$, and all sequences start with a different label. When the points are labeled in bijection to ${1, \ldots, n}$ then each sequence defines a path. The problem is to choose a bijective labeling of the nodes that minimizes the sum of the lengths of paths induced by the sequences? Is this a correct interpretation? If so your problem is clearly NP-hard as suggested by @JohnL. – Bernardo Subercaseaux Jul 25 '22 at 16:57

1 Answers1

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Based on the reformulation of your problem from Bernardo Subercaseaux, your problem is NP-hard (as John L explains), so you should not expect an algorithm that will be efficient in the worst case.

That said, 42 is a fairly small number, so there is some hope. I would suggest that you formulate this as an instance of integer linear programming (ILP). Add zero-or-one boolean variables $x_{v,i}$ for each vertex $v$ (i.e., point in the plane) and each gate $i$ (i.e., each index in $\{1,2,\dots,n\}$), where $x_{v,i}=1$ means that $v$ is associated with $i$. Add constraints to enforce that this is a bijective mapping, i.e., $\sum_v x_{v,i}=1$ for all $i$ and $\sum_i x_{v,i}=1$ for all $v$. Also, for each $v,w,i,j$ let $y_{v,i,w,j} = x_{v,i} \land x_{w,j}$, which can be enforced using linear inequalities as explained in Express boolean logic operations in zero-one integer linear programming (ILP). Finally, you now have an objective $\Phi$ defined as

$$\Phi = \sum_{i,j} \sum_v \sum_w d(v,w) y_{v,i,w,j}$$

where $d(v,w)$ is the distance between $v$ and $w$ (a constant), and you sum over all edges $(i,j)$ that appear in the input sequences, and you sum over all possible $v$ and $w$. This objective is a linear function, so you can use an ILP solver to find the solution that minimizes $\Phi$. You'll end up with an ILP instance with several million variables and tens of millions of inequalities, which might be small enough that an off-the-shelf ILP solver can solve it to find a decent solution.

D.W.
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