拉普拉斯矩阵，解决电阻网格问题答案

【问题标题】：Laplacian matrix, solving a resistor mesh problem拉普拉斯矩阵，解决电阻网格问题
【发布时间】：2021-05-06 17:39:55
【问题描述】：

我正在尝试使用拉普拉斯矩阵求解网络上的流量。我开始在这里测试这个问题：https://rosettacode.org/wiki/Resistor_mesh#Python

当所有权重都为 1 时，它完美地得出了解决方案，R = 1.6089whatever。我希望能够解决不等于 1 的电阻！然后能够获得每个电阻器上的电流，所以我尝试为电导（权重）生成一组随机值并检查它是否正常工作，看看来自注入电流的节点的电流总和是否相等注入应有的电流。不幸的是，事实并非如此。我检查了拉普拉斯算子，它看起来与我的预期相符，但除此之外，我完全迷失了，有人能说明我是否在这里找到了正确的拉普拉斯算子吗？或者如果我遗漏了一些非常明显的东西，例如一些不正确的索引？

代码如下（请原谅我生成网格的方式非常业余，对此也有任何提示和技巧！）：

library(igraph)
library(SparseM)

# setup edgelist + grid to look at
sample_grid = function(N_x, N_y, xlim = c(-1,1), ylim = c(-1,1)) {

  
  # bounding box 
  min_x <- xlim[1]
  max_x <- xlim[2]
  min_y <- ylim[1]
  max_y <- ylim[2]

  
  x_locs <- seq(0, N_x)
  y_locs <- seq(0, N_y)
  
  N_grid <- N_x * N_y
  
  x <- rep(0, length(N_grid))
  y <- rep(0, length(N_grid))
  
  for(i in 1:N_x){
    for(j in 1:N_y) {
      x[N_y * (i-1) + j] <- x_locs[i]
      y[N_y * (i-1) + j] <- y_locs[j]
    }
  }
  
  locations <- cbind(x,y)
  
  return(locations)
}

N_x <- 10
N_y <- 10

# node locations
Vert <- sample_grid(N_x,N_y)

# edges...

# horizontally...
fromH <- c()
for(i in 1:(N_x-1)){
  fromH <- c(fromH, seq(0,(N_y*N_x-N_x), length.out = (N_y)) + i)
}
toH <- fromH + 1

# vertically...
fromV <- 1:(N_x*N_y-N_x)
toV <- fromV + N_x

# --------------------------------------------------------------------------------------------
# crux

Edges <- data.frame(from = c(fromH, fromV), to = c(toH, toV)) # , weights = rep(1,(2*N_x*N_y - (N_x+N_y))))     

# change the weights up
set.seed(1)
weight <- rlnorm(n = nrow(Edges), meanlog = 0, sdlog = 1)
Edges$weight <- weight

the_graph <- graph_from_data_frame(Edges, directed = FALSE)
lo <- layout.norm(as.matrix(Vert))
plot(the_graph, layout = lo, directed = FALSE, edge.arrow.size=0)


# solving
L <- laplacian_matrix(the_graph, weights = weight)

# boundary conditions on 68, 12: 
# draw 1 amp @ 12
# inject 1 amp @ 68

q <- matrix(rep(0, nrow(Vert)), ncol = 1)
q[68,] <- +1
q[12,] <- -1

# solve
p <- solve(L,q)


R <- p[68,] - p[12,]


# investigating why the weights aren't working! (wrong first attempt)
# neighbours <- c(78,58,67,69)  # neighbours of node 68
# neighbours_weights <- weight[neighbours]
# neighbours_potential_diffs <- p[68,] - p[neighbours,]
# neighbours_currents <- neighbours_potential_diffs * neighbours_weights

neighbours <- which(Edges$from == 68 | Edges$to == 68)
potential_diffs <- p[Edges$from,] - p[Edges$to,]
currents <- potential_diffs * Edges$weight
what <- cbind(Edges,p[Edges$from,], p[Edges$to,], currents)
what[neighbours,]



# exact solution for effective resistance between 68 and 12 with 10x10 and all 1ohm
exact <- 455859137025721/283319837425200

问题在于计算的电流与预期的总和不等于一：

> neighbours_currents
[1] 0.05035087 0.09044874 0.03309549 0.21782845

【问题讨论】：

标签： r igraph physics laplacian

【解决方案1】：

答案：laplacian_matrix()函数将节点按降序重新排序，这意味着所有节点都被打乱了。

解决方法是指定节点标签的顺序

the_graph <- graph_from_data_frame(Edges, directed = FALSE)

成为

Verts <- data.frame(label = 1:(N_x*N_y))
the_graph <- graph_from_data_frame(Edges, directed = FALSE, vertices = Verts)

给出总和为一的电流：

> what[neighbours,]
    from to    weight p[Edges$from, ] p[Edges$to, ]    currents
67    67 68 0.1644813       0.2582772     0.8279230 -0.09369607
77    68 69 0.6419198       0.8279230     0.4943076  0.21415437
148   58 68 1.0175478       0.3902397     0.8279230 -0.44536367
158   68 78 0.5372635       0.8279230     0.3685843  0.24678589

【讨论】：