python - 计算网格上两点之间恰好有 `n` 个节点的最短路径

Question

我在网格上定义了以下 3D 表面:

%pylab inline
def muller_potential(x, y, use_numpy=False):
    """Muller potential
    Parameters
    ----------
    x : {float, np.ndarray, or theano symbolic variable}
    X coordinate. If you supply an array, x and y need to be the same shape,
    and the potential will be calculated at each (x,y pair)
    y : {float, np.ndarray, or theano symbolic variable}
    Y coordinate. If you supply an array, x and y need to be the same shape,
    and the potential will be calculated at each (x,y pair)

    Returns
    -------
    potential : {float, np.ndarray, or theano symbolic variable}
    Potential energy. Will be the same shape as the inputs, x and y.
    Reference
    ---------
    Code adapted from https://cims.nyu.edu/~eve2/ztsMueller.m
    """
    aa = [-1, -1, -6.5, 0.7]
    bb = [0, 0, 11, 0.6]
    cc = [-10, -10, -6.5, 0.7]
    AA = [-200, -100, -170, 15]
    XX = [1, 0, -0.5, -1]
    YY = [0, 0.5, 1.5, 1]
    # use symbolic algebra if you supply symbolic quantities
    exp = np.exp
    value = 0
    for j in range(0, 4):
        if use_numpy:
            value += AA[j] * numpy.exp(aa[j] * (x - XX[j])**2 + bb[j] * (x - XX[j]) * (y - YY[j]) + cc[j] * (y - YY[j])**2)
        else: # use sympy
            value += AA[j] * sympy.exp(aa[j] * (x - XX[j])**2 + bb[j] * (x - XX[j]) * (y - YY[j]) + cc[j] * (y - YY[j])**2)
    return value

给出了以下情节:

minx=-1.5
maxx=1.2
miny=-0.2
maxy=2
ax=None
grid_width = max(maxx-minx, maxy-miny) / 50.0
xx, yy = np.mgrid[minx : maxx : grid_width, miny : maxy : grid_width]
V = muller_potential(xx, yy, use_numpy=True)
V = ma.masked_array(V, V>200)
contourf(V, 40)
colorbar();

我编写了以下代码来定义该网格上两点之间的最短路径。我在 meshgrid 的两个相邻点之间使用的度量由 (V[e]-V[cc])**2 给出，其中 cc 当前单元格和 e 相邻小区之一。邻居定义为全连通性:包括对角线在内的所有直接邻居。

def dijkstra(V):
    mask = V.mask
    visit_mask = mask.copy() # mask visited cells
    m = numpy.ones_like(V) * numpy.inf
    connectivity = [(i,j) for i in [-1, 0, 1] for j in [-1, 0, 1] if (not (i == j == 0))]
    cc = unravel_index(V.argmin(), m.shape) # current_cell
    m[cc] = 0
    P = {}  # dictionary of predecessors 
    #while (~visit_mask).sum() > 0:
    for _ in range(V.size):
        #print cc
        neighbors = [tuple(e) for e in asarray(cc) - connectivity 
                     if e[0] > 0 and e[1] > 0 and e[0] < V.shape[0] and e[1] < V.shape[1]]
        neighbors = [ e for e in neighbors if not visit_mask[e] ]
        tentative_distance = [(V[e]-V[cc])**2 for e in neighbors]
        for i,e in enumerate(neighbors):
            d = tentative_distance[i] + m[cc]
            if d < m[e]:
                m[e] = d
                P[e] = cc
        visit_mask[cc] = True
        m_mask = ma.masked_array(m, visit_mask)
        cc = unravel_index(m_mask.argmin(), m.shape)
    return m, P

def shortestPath(start, end, P):
    Path = []
    step = end
    while 1:
        Path.append(step)
        if step == start: break
        step = P[step]
    Path.reverse()
    return asarray(Path)

D, P = dijkstra(V)
path = shortestPath(unravel_index(V.argmin(), V.shape), (40,4), P)

结果如下:

contourf(V, 40)
plot(path[:,1], path[:,0], 'r.-')

路径长度为112:

print path.shape[0]
112

我想知道是否有可能计算出 start 和 end 之间精确长度 n 的最短路径，其中 n 给函数的参数。

备注:如果我将使用的指标从 (V[e]-V[cc])**2 更改为 V[e]-V[cc] ，这增加了负距离我得到的图看起来更好，因为它按预期通过局部最小值:

Answer 1

由于我想获得一个合理的路径，在潜力中采样盆地，我写了下面的函数。为了完整起见，我记得我写的 dijkstra 函数:

%pylab
def dijkstra(V, start):
    mask = V.mask
    visit_mask = mask.copy() # mask visited cells
    m = numpy.ones_like(V) * numpy.inf
    connectivity = [(i,j) for i in [-1, 0, 1] for j in [-1, 0, 1] if (not (i == j == 0))]
    cc = start # current_cell
    m[cc] = 0
    P = {}  # dictionary of predecessors 
    #while (~visit_mask).sum() > 0:
    for _ in range(V.size):
        #print cc
        neighbors = [tuple(e) for e in asarray(cc) - connectivity 
                     if e[0] > 0 and e[1] > 0 and e[0] < V.shape[0] and e[1] < V.shape[1]]
        neighbors = [ e for e in neighbors if not visit_mask[e] ]
        t.ntative_distance = asarray([V[e]-V[cc] for e in neighbors])
        for i,e in enumerate(neighbors):
            d = tentative_distance[i] + m[cc]
            if d < m[e]:
                m[e] = d
                P[e] = cc
        visit_mask[cc] = True
        m_mask = ma.masked_array(m, visit_mask)
        cc = unravel_index(m_mask.argmin(), m.shape)
    return m, P

start, end = unravel_index(V.argmin(), V.shape), (40,4)
D, P = dijkstra(V, start)

def shortestPath(start, end, P):
    Path = []
    step = end
    while 1:
        Path.append(step)
        if step == start: break
        step = P[step]
    Path.reverse()
    return asarray(Path)

path = shortestPath(start, end, P)

给出了以下情节:

contourf(V, 40)
plot(path[:,1], path[:,0], 'r.-')
colorbar()

然后，extend_path 函数背后的基本思想是扩展通过获取路径中节点的邻居使潜力最小化而获得的最短路径。一组记录在扩展过程中已经访问过的单元格。

def get_neighbors(cc, V, visited_nodes):
    connectivity = [(i,j) for i in [-1, 0, 1] for j in [-1, 0, 1] if (not (i == j == 0))]
    neighbors = [tuple(e) for e in asarray(cc) - connectivity 
                 if e[0] > 0 and e[1] > 0 and e[0] < V.shape[0] and e[1] < V.shape[1]]
    neighbors = [ e for e in neighbors if e not in visited_nodes ]
    return neighbors

def extend_path(V, path, n):
    """
    Extend the given path with n steps
    """
    path = [tuple(e) for e in path]
    visited_nodes = set()
    for _ in range(n):
        visited_nodes.update(path)
        dist_min = numpy.inf
        for i_cc, cc in enumerate(path[:-1]):
            neighbors = get_neighbors(cc, V, visited_nodes)
            next_step = path[i_cc+1]
            next_neighbors = get_neighbors(next_step, V, visited_nodes)
            join_neighbors = list(set(neighbors) & set(next_neighbors))
            if len(join_neighbors) > 0:
                tentative_distance = [ V[e] for e in join_neighbors ]
                argmin_dist = argmin(tentative_distance)
                if tentative_distance[argmin_dist] < dist_min:
                    dist_min, new_step, new_step_index  = tentative_distance[argmin_dist], join_neighbors[argmin_dist], i_cc+1
        path.insert(new_step_index, new_step)
    return path

下面是我将最短路径延伸250步得到的结果:

path_ext = extend_path(V, path, 250)
print len(path), len(path_ext)
path_ext = numpy.asarray(path_ext)
contourf(V, 40)
plot(path[:,1], path[:,0], 'w.-')
plot(path_ext[:,1], path_ext[:,0], 'r.-')
colorbar()

正如预期的那样，当我增加 n 时，我首先开始对较深的盆地进行采样，如下所示:

rcParams['figure.figsize'] = 14,8
for i_plot, n in enumerate(range(0,250,42)):
    path_ext = numpy.asarray(extend_path(V, path, n))
    subplot('23%d'%(i_plot+1))
    contourf(V, 40)
    plot(path_ext[:,1], path_ext[:,0], 'r.-')
    title('%d path steps'%len(path_ext))