igraph R 中有向树图中从根到叶的所有路径

Question

给定一棵树：

library(igraph)

# setup graph
g= graph.formula(A -+ B,
                 A -+ C,
                 B -+ C,
                 B -+ D,
                 B -+ E
)
plot(g, layout = layout.reingold.tilford(g, root="A"))

顶点"A"是树的根，而顶点"C", "D", "E"被认为是终端叶子。

问题：

任务是找到根和叶之间的所有路径。我使用以下代码失败，因为它只提供最短路径：

# find root and leaves
leaves= which(degree(g, v = V(g), mode = "out")==0, useNames = T)
root= which(degree(g, v = V(g), mode = "in")==0, useNames = T)

# find all paths
paths= lapply(root, function(x) get.all.shortest.paths(g, from = x, to = leaves, mode = "out")$res)
named_paths= lapply(unlist(paths, recursive=FALSE), function(x) V(g)[x])
named_paths

输出：

$A1
Vertex sequence:
[1] "A" "C"

$A2
Vertex sequence:
[1] "A" "B" "D"

$A3
Vertex sequence:
[1] "A" "B" "E"

问题：

如何找到包括顶点序列的所有路径："A" "B" "C"？

我的理解是，缺少的序列 "A" "B" "C" 不是由 get.all.shortest.paths() 提供的，作为从 "A" 到 "C" 通过顶点序列的路径："A" "C"（在列表元素 $A1 中找到）更短。所以 igraph 正常工作。 尽管如此，我正在寻找一种代码解决方案，以 R list.

的形式获取从根到所有叶子的所有路径

评论：

我知道对于大树，覆盖所有组合的算法可能会变得昂贵，但我的实际应用相对较小。

Answer 1

基于 Gabor 的评论：

all_simple_paths(g, from = root, to = leaves)

产量：

[[1]]
+ 3/5 vertices, named:
[1] A B C

[[2]]
+ 3/5 vertices, named:
[1] A B D

[[3]]
+ 3/5 vertices, named:
[1] A B E

[[4]]
+ 2/5 vertices, named:
[1] A C

igraph R 中有向树图中从根到叶的所有路径

All paths in directed tree graph from root to leaves in igraph R

r

igraph