C++ 生成随机DAG
我正在解决一个关于有向无环图的问题 但是我在一些有向无环图上测试代码时遇到了困难。测试图应该是大的,并且(显然)是非循环的 我试着写了很多代码来生成无环有向图。但我每次都失败了C++ 生成随机DAG,c++,c,algorithm,graph,cycle,C++,C,Algorithm,Graph,Cycle,我正在解决一个关于有向无环图的问题 但是我在一些有向无环图上测试代码时遇到了困难。测试图应该是大的,并且(显然)是非循环的 我试着写了很多代码来生成无环有向图。但我每次都失败了 是否有一些现有方法可用于生成无环有向图?创建一个包含n节点以及每对节点n1和n2ifn1!=n2和n2%n1==0您可以生成一个随机有向图,然后对循环进行深度优先搜索。找到循环后,通过删除边来打断它 我认为这是最坏的情况。每个DFS取O(V),每个DFS至少删除一条边(因此最大E) 如果您通过均匀随机选择所有V^2可能的
是否有一些现有方法可用于生成无环有向图?创建一个包含
n
节点以及每对节点n1
和n2
ifn1!=n2
和n2%n1==0
您可以生成一个随机有向图,然后对循环进行深度优先搜索。找到循环后,通过删除边来打断它
我认为这是最坏的情况。每个DFS取O(V),每个DFS至少删除一条边(因此最大E)
如果您通过均匀随机选择所有V^2可能的边来生成有向图,并且您以随机顺序DFS并删除一条随机边-这将使您在所有可能的DAG上获得均匀分布(或至少接近它)。答案适用:如果您有图边的邻接矩阵表示,如果矩阵是下三角的,它必然是一个DAG
一种类似的方法是对节点进行任意排序,然后仅在x<y时考虑从节点X到Y的边。这个约束也应该通过构造来获得你的DAGness。如果使用结构来表示节点,那么内存比较将是对节点进行排序的一种任意方式
基本上,伪代码类似于:for(i = 0; i < N; i++) {
for (j = i+1; j < N; j++) {
maybePutAnEdgeBetween(i, j);
}
}
既然有
n*(n-1)/2
有序对(“N选择2”),我们可以选择它们之间是否有边缘。我编写了一个C程序来实现这一点。关键是对节点进行“排序”,并且只从排名较低的节点到排名较高的节点绘制边 我写的程序打印出来了 下面是代码本身,注释解释了它的含义:
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#define MIN_PER_RANK 1 /* Nodes/Rank: How 'fat' the DAG should be. */
#define MAX_PER_RANK 5
#define MIN_RANKS 3 /* Ranks: How 'tall' the DAG should be. */
#define MAX_RANKS 5
#define PERCENT 30 /* Chance of having an Edge. */
int main (void)
{
int i, j, k,nodes = 0;
srand (time (NULL));
int ranks = MIN_RANKS
+ (rand () % (MAX_RANKS - MIN_RANKS + 1));
printf ("digraph {\n");
for (i = 0; i < ranks; i++)
{
/* New nodes of 'higher' rank than all nodes generated till now. */
int new_nodes = MIN_PER_RANK
+ (rand () % (MAX_PER_RANK - MIN_PER_RANK + 1));
/* Edges from old nodes ('nodes') to new ones ('new_nodes'). */
for (j = 0; j < nodes; j++)
for (k = 0; k < new_nodes; k++)
if ( (rand () % 100) < PERCENT)
printf (" %d -> %d;\n", j, k + nodes); /* An Edge. */
nodes += new_nodes; /* Accumulate into old node set. */
}
printf ("}\n");
return 0;
}
#包括
#包括
#包括
#定义MIN_PER_秩1/*节点/秩:DAG的“胖”程度*/
#定义每个等级5的最大值
#定义最小等级3/*等级:DAG的“高度”应为多少*/
#定义最大等级5
#定义具有优势的概率为30/*的百分比*/
内部主(空)
{
int i,j,k,节点=0;
srand(时间(空));
整数秩=最小秩
+(rand()%(最大秩-最小秩+1));
printf(“有向图{\n”);
对于(i=0;i%d;\n”,j,k+节点);/*一条边*/
节点+=新的_节点;/*累积到旧节点集中*/
}
printf(“}\n”);
返回0;
}
下面是测试运行生成的图形:
所以,试着把所有这些合理的答案放在一起: (在下面,我使用V表示生成的图中的顶点数,E表示边数,我们假设E≤ V(V-1)/2.) 就个人而言,我认为最有用的答案是弗拉维乌斯(Flavius)的评论,他指向。这段代码非常简单,可以通过一条注释方便地描述,我复制了这条注释:
To generate a directed acyclic graph, we first
generate a random permutation dag[0],...,dag[v-1].
(v = number of vertices.)
This random permutation serves as a topological
sort of the graph. We then generate random edges of the
form (dag[i],dag[j]) with i < j.
要生成有向无环图,我们首先
生成一个随机排列dag[0],…,dag[v-1]。
(v=顶点数。)
这种随机排列可用作拓扑结构
有点像图表。然后,我们生成图像的随机边
带i
事实上,代码所做的是通过重复执行以下操作来生成请求数量的边:
// At the end of this snippet, a consists of a random sample of the
// integers in the half-open range [0, V(V-1)/2). (They still need to be
// converted to pairs of endpoints).
vector<int> a;
int N = V * (V - 1) / 2;
for (int i = 0; i < N; ++i) a.push_back(i);
for (int i = 0; i < E; ++i) {
int j = i + rand(N - i);
swap(a[i], a[j]);
a.resize(E);
//在这个代码段的末尾,一个由
//半开范围[0,V(V-1)/2]中的整数。(它们仍然需要
//转换为端点对)。
载体a;
int N=V*(V-1)/2;
对于(int i=0;i
这只需要O(E)时间,但需要O(N2)空间。事实上,可以通过一些技巧将其改进为O(E)空间,但SO代码段太小,无法包含结果,因此我将在O(E)空间和O(E log E)时间中提供一个更简单的代码段。我假设有一个类DAG,它至少包含:
class DAG {
// Construct an empty DAG with v vertices
explicit DAG(int v);
// Add the directed edge i->j, where 0 <= i, j < v
void add(int i, int j);
};
类DAG{
//使用v顶点构造空DAG
显式DAG(int-v);
//添加有向边i->j,其中0我最近尝试重新实现接受的答案,发现它是不确定的。如果不强制执行min_per_rank参数,可能会得到一个节点数为0的图
为了防止出现这种情况,我将for循环包装在一个函数中,然后检查以确保
class DAG {
// Construct an empty DAG with v vertices
explicit DAG(int v);
// Add the directed edge i->j, where 0 <= i, j < v
void add(int i, int j);
};
// Return a randomly-constructed DAG with V vertices and and E edges.
// It's required that 0 < E < V(V-1)/2.
template<typename PRNG>
DAG RandomDAG(int V, int E, PRNG& prng) {
using dist = std::uniform_int_distribution<int>;
// Make a random sample of size E
std::vector<int> sample;
sample.reserve(E);
int N = V * (V - 1) / 2;
dist d(0, N - E); // uniform_int_distribution is closed range
// Random vector of integers in [0, N-E]
for (int i = 0; i < E; ++i) sample.push_back(dist(prng));
// Sort them, and make them unique
std::sort(sample.begin(), sample.end());
for (int i = 1; i < E; ++i) sample[i] += i;
// Now it's a unique sorted list of integers in [0, N-E+E-1]
// Randomly shuffle the endpoints, so the topological sort
// is different, too.
std::vector<int> endpoints;
endpoints.reserve(V);
for (i = 0; i < V; ++i) endpoints.push_back(i);
std::shuffle(endpoints.begin(), endpoints.end(), prng);
// Finally, create the dag
DAG rv;
for (auto& v : sample) {
int tail = int(0.5 + sqrt((v + 1) * 2));
int head = v - tail * (tail - 1) / 2;
rv.add(head, tail);
}
return rv;
}
int pushed = 0
int addRank (void)
{
for (j = 0; j < nodes; j++)
for (k = 0; k < new_nodes; k++)
if ( (rand () % 100) < PERCENT)
printf (" %d -> %d;\n", j, k + nodes); /* An Edge. */
if (pushed < min_per_rank) return addRank()
else pushed = 0
return 0
}
import random
class Graph:
nodes = []
edges = []
removed_edges = []
def remove_edge(self, x, y):
e = (x,y)
try:
self.edges.remove(e)
# print("Removed edge %s" % str(e))
self.removed_edges.append(e)
except:
return
def Nodes(self):
return self.nodes
# Sample data
def __init__(self):
self.nodes = []
self.edges = []
def get_random_dag():
MIN_PER_RANK = 1 # Nodes/Rank: How 'fat' the DAG should be
MAX_PER_RANK = 2
MIN_RANKS = 6 # Ranks: How 'tall' the DAG should be
MAX_RANKS = 10
PERCENT = 0.3 # Chance of having an Edge
nodes = 0
ranks = random.randint(MIN_RANKS, MAX_RANKS)
adjacency = []
for i in range(ranks):
# New nodes of 'higher' rank than all nodes generated till now
new_nodes = random.randint(MIN_PER_RANK, MAX_PER_RANK)
# Edges from old nodes ('nodes') to new ones ('new_nodes')
for j in range(nodes):
for k in range(new_nodes):
if random.random() < PERCENT:
adjacency.append((j, k+nodes))
nodes += new_nodes
# Compute transitive graph
G = Graph()
# Append nodes
for i in range(nodes):
G.nodes.append(i)
# Append adjacencies
for i in range(len(adjacency)):
G.edges.append(adjacency[i])
N = G.Nodes()
for x in N:
for y in N:
for z in N:
if (x, y) != (y, z) and (x, y) != (x, z):
if (x, y) in G.edges and (y, z) in G.edges:
G.remove_edge(x, z)
# Print graph
for i in range(nodes):
print(i)
print()
for value in G.edges:
print(str(value[0]) + ' ' + str(value[1]))
get_random_dag()
def get_random_dag():
MIN_PER_RANK = 1 # Nodes/Rank: How 'fat' the DAG should be
MAX_PER_RANK = 3
MIN_RANKS = 15 # Ranks: How 'tall' the DAG should be
MAX_RANKS = 20
PERCENT = 0.3 # Chance of having an Edge
nodes = 0
node_counter = 0
ranks = random.randint(MIN_RANKS, MAX_RANKS)
adjacency = []
rank_list = []
for i in range(ranks):
# New nodes of 'higher' rank than all nodes generated till now
new_nodes = random.randint(MIN_PER_RANK, MAX_PER_RANK)
list = []
for j in range(new_nodes):
list.append(node_counter)
node_counter += 1
rank_list.append(list)
print(rank_list)
# Edges from old nodes ('nodes') to new ones ('new_nodes')
if i > 0:
for j in rank_list[i - 1]:
for k in range(new_nodes):
if random.random() < PERCENT:
adjacency.append((j, k+nodes))
nodes += new_nodes
for i in range(nodes):
print(i)
print()
for edge in adjacency:
print(str(edge[0]) + ' ' + str(edge[1]))
print()
print()