R IGraph計算定向網絡中的無向最短路徑嗎？

Question

關於IGraph最短路徑計算，我缺少一些東西。

假設我生成一個網絡（在stackoverflow中的某個地方找到）並執行簡單的計算：

library(igraph);
relations <- data.frame(from=c("Bob", "Cecil", "Cecil", "David", "David", "Esmeralda"), to=c("Alice", "Bob", "Alice", "Alice", "Bob", "Alice"));
g = simplify(graph_from_data_frame(d=relations, directed=T), remove.multiple = F, remove.loops = T);
#plotting the network in order to appreciate the directions of the edges
plot(g,edge.arrow.size=0.5);
#V(g)[5] is "Alice" which apparently should not be able to reach any node
print(all_shortest_paths(g,from=V(g)[5],to=V(g),mode="all")$res);

如您所見，找到的最短路徑是：

> print(all_shortest_paths(g,from=V(g)[5],to=V(g),mode="all")$res);
[[1]]
+ 2/5 vertices, named, from 823c15d:
[1] Alice Bob  

[[2]]
+ 2/5 vertices, named, from 823c15d:
[1] Alice Cecil

[[3]]
+ 2/5 vertices, named, from 823c15d:
[1] Alice David

[[4]]
+ 2/5 vertices, named, from 823c15d:
[1] Alice     Esmeralda

[[5]]
+ 1/5 vertex, named, from 823c15d:
[1] Alice

我期望的是，不應該返回最短的路徑，因為在有向圖中，愛麗絲沒有從其自身伸出的邊緣。 這是由於以下事實：當我計算最短路徑時，我正在使用以下選項：

mode="all"

而且這以某種方式甚至適用於有向圖？

當然，如果我更改圖形構造並設置：

directed=F

即

library(igraph);
relations <- data.frame(from=c("Bob", "Cecil", "Cecil", "David", "David", "Esmeralda"), to=c("Alice", "Bob", "Alice", "Alice", "Bob", "Alice"));

g = simplify(graph_from_data_frame(d=relations, directed=F), remove.multiple = F, remove.loops = T);

#plottin the network in order to appreciate the directions of the edges
plot(g,edge.arrow.size=0.5);

#V(g)[5] is "Alice" which apparently should not be able to reach any node
print(all_shortest_paths(g,from=V(g)[5],to=V(g),mode="all")$res);

返回相同的結果。

到底是怎么回事？ 我是否太累了，無法直截了當？

Answer 1

這就是mode="all"含義-使用所有邊緣而不管方向。

我將使用一個更簡單的圖形來輕松查看發生了什么。

rel2 <- data.frame(from=c("Bob", "Bob", "David"), 
        to=c("Alice", "Carol", "Carol"))
g = simplify(graph_from_data_frame(d=rel2, directed=T))
LO = layout_as_bipartite(g, types=c(F,F,T,T))
plot(g, layout=LO)

現在，用最短路徑聲明

print(all_shortest_paths(g,from=V(g)[3],to=V(g),mode="all")$res)
[[1]]
+ 2/4 vertices, named:
[1] Alice Bob  
[[2]]
+ 4/4 vertices, named:
[1] Alice Bob   Carol David
[[3]]
+ 1/4 vertex, named:
[1] Alice
[[4]]
+ 3/4 vertices, named:
[1] Alice Bob   Carol

即使邊沿相反的方向，我們也可以獲得將愛麗絲連接到另一個節點的所有路徑。

我認為您想要的是：

print(all_shortest_paths(g,from=V(g)[3],to=V(g),mode="out")$res)
[[1]]
+ 1/4 vertex, named:

這僅給出了從Alice到其自身的零長度路徑。

為了完整性，

print(all_shortest_paths(g,from=V(g)[3],to=V(g),mode="in")$res)
[[1]]
+ 2/4 vertices, named:
[1] Alice Bob  

[[2]]
+ 1/4 vertex, named:
[1] Alice

這僅使用進入的邊緣來遵循路徑，因此我們使用進入Alice的邊緣將路徑從“ Alice”到“ Bob”獲得，但是由於沒有進入Bob的邊緣，因此我們一無所獲。