在R中使用igraph生成k-正则图时，每个k-正则图都是唯一的和非随机的吗？

我想使用

R

中的

igraph

生成一个k-正则图。我将使用一个示例函数，首先获得游戏，然后转换为邻接矩阵：

game <- sample_k_regular(no.of.nodes=3000, k=30)
game_adj <- as.matrix(as_adj(game))

并计算出差值：

sum(game2_adj-game_adj)

我们得到的正好是0

---

一旦我们指定了节点的数量和每个节点的度（邻居的数量），它会是唯一的吗？（这意味着在函数调用顶部添加

set.seed（）

不会产生任何影响？

确实存在随机性。让我们看一些更容易理解的东西：

game1 <- sample_k_regular(no.of.nodes = 5, k = 2)
game2 <- sample_k_regular(no.of.nodes = 5, k = 2)
game_adj1 <- as.matrix(as_adj(game1))
game_adj2 <- as.matrix(as_adj(game2))

game_adj1
#      [,1] [,2] [,3] [,4] [,5]
# [1,]    0    0    0    1    1
# [2,]    0    0    1    1    0
# [3,]    0    1    0    0    1
# [4,]    1    1    0    0    0
# [5,]    1    0    1    0    0
game_adj2
#      [,1] [,2] [,3] [,4] [,5]
# [1,]    0    0    1    0    1
# [2,]    0    0    1    1    0
# [3,]    1    1    0    0    0
# [4,]    0    1    0    0    1
# [5,]    1    0    0    1    0
game_adj1 - game_adj2
#      [,1] [,2] [,3] [,4] [,5]
# [1,]    0    0   -1    1    0
# [2,]    0    0    0    0    0
# [3,]   -1    0    0    0    1
# [4,]    1    0    0    0   -1
# [5,]    0    0    1   -1    0

---

但这是意料之中的，因为

sum（game_adj1）

和

sum（game_adj2）

是

节点数*k

。

@user321627，它能回答你的问题吗？

game1 <- sample_k_regular(no.of.nodes = 5, k = 2)
game2 <- sample_k_regular(no.of.nodes = 5, k = 2)
game_adj1 <- as.matrix(as_adj(game1))
game_adj2 <- as.matrix(as_adj(game2))

game_adj1
#      [,1] [,2] [,3] [,4] [,5]
# [1,]    0    0    0    1    1
# [2,]    0    0    1    1    0
# [3,]    0    1    0    0    1
# [4,]    1    1    0    0    0
# [5,]    1    0    1    0    0
game_adj2
#      [,1] [,2] [,3] [,4] [,5]
# [1,]    0    0    1    0    1
# [2,]    0    0    1    1    0
# [3,]    1    1    0    0    0
# [4,]    0    1    0    0    1
# [5,]    1    0    0    1    0
game_adj1 - game_adj2
#      [,1] [,2] [,3] [,4] [,5]
# [1,]    0    0   -1    1    0
# [2,]    0    0    0    0    0
# [3,]   -1    0    0    0    1
# [4,]    1    0    0    0   -1
# [5,]    0    0    1   -1    0

sum(game_adj1 - game_adj2)
# [1] 0