Grouping many points into closest pairs - Python + LP

Question

I have a list which contains a shop_id, latitude, and longitude. Assuming we have an even amount of points, I would like to match each shop to another shop such that our connections are unique and we minimise the overall distance.

I was curious if my approach is not too stupid and what a better one would be. This currently works in a reasonable amount of time (an hour) for 1000 points.

Imagine we have a list like this:

shops = [[1, '53.382072', '-2.7262165'],
[2, '52.478499', '-0.9222608'],
[3, '52.071258', '-1.2722802'],
...]

And that I use PuLP to create an .lp file using this workflow:

prob = LpProblem("Minimise distance",LpMinimize)
# Variables
x = []
xnames = []
for one in store_connect(shops):
    xnames.append(one)
    x.append(LpVariable(one,0,1,LpInteger)) 
print("Created %d variables" % len(x))
# Objective
prob += pulp.lpSum([i[1]*x[i[0]] for i in enumerate(distances(shops))])
print("Created objective")
# Constraints
counter = 0
for constraint_rw in all_constraints(shops):
    prob += pulp.lpSum(x[xnames.index(one_cn)] for one_cn in constraint_rw) == 1
    counter += 1
    if counter % 10 == 0:
        print("Added %d contraints out of %d" % (counter,len(shops)))   
# Save "LP" file for other solvers
prob.writeLP("for_cplex.lp")

Where I call a few generator functions (to help the RAM ...)

def store_connect(shops):
    """
    Cross stores and return connection e.g. "v1-2" by ID
    """
    for i in range(len(shops)-1):
        for j in range(i+1,len(shops)):
            yield 'v'+str(shops[i][0])+'-'+str(shops[j][0])

def distances(shops):
    """
    Return distance in miles for crosses
    """
    for i in range(len(shops)-1):
        for j in range(i+1,len(shops)):
            yield haversine([shops[i][1],shops[i][2]],
                            [shops[j][1],shops[j][2]])

def all_constraints(shops):
    """
    Return constraint row
    """
    for a in shops:
        cons = []
        for b in shops:
            if a[0]b[0]:
                cons.append("v"+str(b[0])+"-"+str(a[0]))
        yield cons

If I run this on 100 stores it's very quick and I can use the default PuLP solver. Otherwise I submit the LP file to CPLEX on NEOS servers. I created a small helper function which visualises the resulting matches:

Here is how my LP file looks for a trivial example of 10 stores (truncated):

Minimize
OBJ: 131.513 v1_10 + 97.686 v1_2 + 109.107 v1_3 + 36.603 v1_4 + 11.586 v1_5
 + 109.067 v1_6 + 113.862 v1_7 + 169.371 v1_8 + 220.098 v1_9 + 101.958 v2_10
 + 31.793 v2_3 + 61.792 v2_4 + 105.822 v2_5 + 73.055 v2_6 + 32.008 v2_7
 + 81.627 v2_8 + 122.493 v2_9 + 72.945 v3_10 + 72.983 v3_4 + 114.609 v3_5
 + 46.098 v3_6 + 5.555 v3_7 + 60.305 v3_8 ...
Subject To
_C1: v1_10 + v1_2 + v1_3 + v1_4 + v1_5 + v1_6 + v1_7 + v1_8 + v1_9 = 1
_C10: v1_10 + v2_10 + v3_10 + v4_10 + v5_10 + v6_10 + v7_10 + v8_10 + v9_10
 = 1
_C2: v1_2 + v2_10 + v2_3 + v2_4 + v2_5 + v2_6 + v2_7 + v2_8 + v2_9 = 1
_C3: v1_3 + v2_3 + v3_10 + v3_4 + v3_5 + v3_6 + v3_7 + v3_8 + v3_9 = 1
_C4: v1_4 + v2_4 + v3_4 + v4_10 + v4_5 + v4_6 + v4_7 + v4_8 + v4_9 = 1
_C5: v1_5 + v2_5 + v3_5 + v4_5 + v5_10 + v5_6 + v5_7 + v5_8 + v5_9 = 1
...
Binaries
v1_10
v...
End

The result:

Status: Optimal
Total minimised distance (miles):  190.575

Finally, here is a larger example on 1000 points:

Grouping many points into closest pairs - Python + LP

Answers (1)

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