这被称为离散 p 色散 (maxmin) 问题。
在White (1991) 和Ravi et al. (1994) 中证明了最优性界限，为问题提供了因子 2 的近似值，后者证明了这种启发式方法是最好的（除非 P=NP）。
因子 2 近似
因子 2 的近似值如下：
Let V be the set of nodes/objects
Let i and j be two nodes at maximum distance
Let p be the number of objects to choose
p = set([i,j])
while size(P)<p:
Find a node v in V-P such that min_{v' in P} dist(v,v') is maximum
\That is: find the node with the greatest minimum distance to the set P
P = P.union(v)
Output P
你可以像这样在 Python 中实现它：
#!/usr/bin/env python3
import numpy as np
p = 50
N = 400
print("Building distance matrix...")
d = np.random.rand(N,N) #Random matrix
d = (d + d.T)/2 #Make the matrix symmetric
print("Finding initial edge...")
maxdist = 0
bestpair = ()
for i in range(N):
for j in range(i+1,N):
if d[i,j]>maxdist:
maxdist = d[i,j]
bestpair = (i,j)
P = set()
P.add(bestpair[0])
P.add(bestpair[1])
print("Finding optimal set...")
while len(P)<p:
print("P size = {0}".format(len(P)))
maxdist = 0
vbest = None
for v in range(N):
if v in P:
continue
for vprime in P:
if d[v,vprime]>maxdist:
maxdist = d[v,vprime]
vbest = v
P.add(vbest)
print(P)
精确解
您也可以将其建模为 MIP。对于 p=50、n=400，在 6000 秒后，最优差距仍为 568%。近似算法花费了 0.47 秒来获得 100%（或更少）的最优差距。一个朴素的 Gurobi Python 表示可能如下所示：
#!/usr/bin/env python
import numpy as np
import gurobipy as grb
p = 50
N = 400
print("Building distance matrix...")
d = np.random.rand(N,N) #Random matrix
d = (d + d.T)/2 #Make the matrix symmetric
m = grb.Model(name="MIP Model")
used = [m.addVar(vtype=grb.GRB.BINARY) for i in range(N)]
objective = grb.quicksum( d[i,j]*used[i]*used[j] for i in range(0,N) for j in range(i+1,N) )
m.addConstr(
lhs=grb.quicksum(used),
sense=grb.GRB.EQUAL,
rhs=p
)
# for maximization
m.ModelSense = grb.GRB.MAXIMIZE
m.setObjective(objective)
# m.Params.TimeLimit = 3*60
# solving with Glpk
ret = m.optimize()
缩放
显然，初始点的 O(N^2) 缩放是不好的。通过认识到这对必须位于数据集的凸包上，我们可以更有效地找到它们。这为我们提供了一种 O(N log N) 方法来找到该对。一旦我们找到它，我们就会像以前一样继续（使用 SciPy 进行加速）。
最好的缩放方法是使用 R*-tree 来有效地找到候选点 p 和集合 P 之间的最小距离。但这在 Python 中不能有效地完成，因为仍然涉及for 循环.
import numpy as np
from scipy.spatial import ConvexHull
from scipy.spatial.distance import cdist
p = 300
N = 16000000
# Find a convex hull in O(N log N)
points = np.random.rand(N, 3) # N random points in 3-D
# Returned 420 points in testing
hull = ConvexHull(points)
# Extract the points forming the hull
hullpoints = points[hull.vertices,:]
# Naive way of finding the best pair in O(H^2) time if H is number of points on
# hull
hdist = cdist(hullpoints, hullpoints, metric='euclidean')
# Get the farthest apart points
bestpair = np.unravel_index(hdist.argmax(), hdist.shape)
P = np.array([hullpoints[bestpair[0]],hullpoints[bestpair[1]]])
# Now we have a problem
print("Finding optimal set...")
while len(P)<p:
print("P size = {0}".format(len(P)))
distance_to_P = cdist(points, P)
minimum_to_each_of_P = np.min(distance_to_P, axis=1)
best_new_point_idx = np.argmax(minimum_to_each_of_P)
best_new_point = np.expand_dims(points[best_new_point_idx,:],0)
P = np.append(P,best_new_point,axis=0)
print(P)