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

这给出了以下情节：

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。所述meshgrid的两个相邻点之间所使用的度量我由下式给出(V[e]-V[cc])**2与cc当前小区和e相邻小区中的一个。邻居定义为完全连接：包括对角线的所有直接邻居。

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)

这给出了以下结果：

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精确长度start和之间的最短路径，并end为函数n提供n一个参数。

备注：如果我将使用的度量从 更改(V[e]-V[cc])**2为V[e]-V[cc]，这会增加负距离，我将获得看起来更好的图，因为它按预期通过了局部最小值：

Answer 1

由于我想获得对势中的盆地进行采样的合理路径，因此我编写了以下函数。为了完整起见，我记得dijkstra我编写的函数：

%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)

其中给出了以下情节：

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

然后，该extend_path函数背后的基本思想是扩展通过获取路径中使势能最小化的节点的邻居而获得的最短路径。集合保存在扩展过程中已经访问过的单元格的记录。

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长 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 时，我首先开始对更深的盆地进行采样n，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))