Transportation problem solver in Python

transport.py

import numpy as np
from collections import Counter


def transport(supply, demand, costs):

    # Only solves balanced problem
    assert sum(supply) == sum(demand)

    s = np.copy(supply)
    d = np.copy(demand)
    C = np.copy(costs)

    n, m = C.shape

    # Finding initial solution
    X = np.zeros((n, m))
    indices = [(i, j) for i in range(n) for j in range(m)]
    xs = sorted(zip(indices, C.flatten()), key=lambda (a, b): b)

    # Iterating C elements in increasing order
    for (i, j), _ in xs:
        if d[j] == 0:
            continue
        else:
            # Reserving supplies in a greedy way
            remains = s[i] - d[j] if s[i] >= d[j] else 0
            grabbed = s[i] - remains
            X[i, j] = grabbed
            s[i] = remains
            d[j] -= grabbed

    # Finding optimal solution
    while True:
        u = np.array([np.nan]*n)
        v = np.array([np.nan]*m)
        S = np.zeros((n, m))

        _x, _y = np.where(X > 0)
        nonzero = zip(_x, _y)
        f = nonzero[0][0]
        u[f] = 0

        # Finding u, v potentials
        while any(np.isnan(u)) or any(np.isnan(v)):
            for i, j in nonzero:
                if np.isnan(u[i]) and not np.isnan(v[j]):
                    u[i] = C[i, j] - v[j]
                elif not np.isnan(u[i]) and np.isnan(v[j]):
                    v[j] = C[i, j] - u[i]
                else:
                    continue

        # Finding S-matrix
        for i in range(n):
            for j in range(m):
                S[i, j] = C[i, j] - u[i] - v[j]

        # Stop condition
        s = np.min(S)
        if s >= 0:
            break

        i, j = np.argwhere(S == s)[0]
        start = (i, j)

        # Finding cycle elements
        T = np.copy(X)
        T[start] = 1  
        while True:
            _xs, _ys = np.nonzero(T)
            xcount, ycount = Counter(_xs), Counter(_ys)

            for x, count in xcount.items():
                if count <= 1:
                    T[x,:] = 0
            for y, count in ycount.items():
                if count <= 1: 
                    T[:,y] = 0

            if all(x > 1 for x in xcount.values()) \
                    and all(y > 1 for y in ycount.values()):
                break
        
        # Finding cycle chain order
        dist = lambda (x1, y1), (x2, y2): abs(x1-x2) + abs(y1-y2)
        fringe = set(tuple(p) for p in np.argwhere(T > 0))

        size = len(fringe)

        path = [start]
        while len(path) < size:
            last = path[-1]
            if last in fringe:
                fringe.remove(last)
            next = min(fringe, key=lambda (x, y): dist(last, (x, y)))
            path.append(next)

        # Improving solution on cycle elements
        neg = path[1::2]
        pos = path[::2]
        q = min(X[zip(*neg)])
        X[zip(*neg)] -= q
        X[zip(*pos)] += q

    return X, np.sum(X*C)


if __name__ == '__main__':
    supply = np.array([200, 350, 300])
    demand = np.array([270, 130, 190, 150, 110])

    costs = np.array([[24., 50., 55., 27., 16.],
                      [50., 40., 23., 17., 21.],
                      [35., 59., 55., 27., 41.]])

    routes, z = transport(supply, demand, costs)
    assert z == 23540

    print routes

Here my parameters please try it
aCost = [[ 1, 8, 1, 5, 4]
,[ 5, 5, 3, 6, 7]
,[ 2, 9, 5, 9, 8]]

aDemand = [ 120, 130, 145, 125, 140]
aSupply = [ 240, 160, 260]

This won't work it there is a degenerate solution, because there will be less than m+n-1 non-zero numbers in the route and some cycle chain order won't be found and result in endless loops.

The generated cycle chain will be wrong for more complex chains because the dist function only considered distance between two node while ignoring that the two neighboring nodes should have exactly one common coordinate. My revised version is here.

bonjour,j'utilise actuellement votre code et ça me donne comme erreur ceci:

next = min(fringe, key=lambda x, y: dist(last, (x, y)))
TypeError: () missing 1 required positional argument: 'y'

Line while any(np.isnan(u)) or any(np.isnan(v)):
endless loop on my data