设图G内有20个节点和50条边。G图中有两个点vi和vj，它们的度分别为4和5。假设完全随机生成图G，vi 到 vj的边预期有多少条？

Consider a graph G, with 20nodes and 50edges. If there are two nodes vi and vj in the graph, degree of vi is 4, and degree for vj is 5. What is the expected number of edges from vi to vj?

Question

设图G内有20个节点和50条边。G图中有两个点vi和vj，它们的度分别为4和5。 假设完全随机生成图G，vi 到 vj的边预期有多少条？

Consider a graph G, with 20nodes and 50edges. If there are two nodes vi and vj in the graph, degree of vi is 4, and degree for vj is 5. What is the expected number of edges from vi to vj?

BlackTom AI · Accepted Answer

In the configuration model for a uniformly random graph with fixed degrees, the expected number of edges between two vertices equals (degree(vi) × degree(vj)) divided by the total number of stubs (2m). This gives the key formula used below.

1) Relevant knowledge points and concepts
- Each edge contributes two stubs, so total stubs = 2m. Here m = 50, so total stubs = 100.
- Under a random matching of stubs (configuration model), each stub of vi has probability proportional to degree(vj) of being paired to a stub of vj.
- The expected number of vi–vj edges is the sum over vi’s stubs of the probability each connects to vj’s stubs, giving E = d_i · d_j / (2m) (common approximation).

2) Step-by-step reasoning
- deg(vi) = d_i = 4, deg(vj) = d_j = 5, number of edges m = 50.
- Total stubs = 2m = 100.
- Expected number of edges between vi and vj ≈ (d_i × d_j) / (2m) = (4 × 5) / 100.

3) Intermediate calculations
- 4 × 5 = 20
- 20 / 100 = 0.20
- 0.20 expressed as a fraction = 1/5

(Note: a slightly more exact calculation using conditional probabilities without the 2m approximation gives 4 × 5 / (2m − 1) = 20/99 ≈ 0.20202, but the multiple-choice options use the simpler 1/5 = 0.20.)

4) Why the correct answer is correct
The formula captures that each of vi’s four stubs has about a 5/100 chance of attaching to one of vj’s five stubs; summing those four independent expectations yields 4×(5/100)=20/100=1/5.

5) Why the other options are incorrect
- 4/5 (0.8) is far too large; it would imply nearly one edge per stub, inconsistent with the sparse stub pool.
- 2/20 = 1/10 (0.1) and 1/20 (0.05) are too small given the product 4×5 = 20 and 100 total stubs; they do not reflect the correct ratio 20/100.

6) Conclusion
Using the standard random-stub argument gives expected edges = (4×5)/100 = 1/5. This matches the provided correct answer: 1/5.

Similar Questions

在一个图中，给定一个节点 v 的邻居中的连通对(pair)的数量与 v 的邻居中的所有对(pair)的数量之比率是 The ratio between the number of connected pairs of v's neighbors and the total number pairs of v's neighbors is called

在无向图 G(V, E) 中，|G| = n 和 |E| = m，度的总和是 In an undirected graph G(V, E), where |G| = n and |E| = m, the summation of degrees is

How many vertices does the Shiba model have?

Select the set that corresponds to the relation given in the arrow diagram below:

Given the adjacency matrix \( M \) of a directed graph, the number of paths of length 3 from vertex i to vertex j is found in matrix Answer Question 5[input]. Use proper notation (e.g., M^3).

A tree with 10 vertices has how many edges?

在社交媒体网络中，连通任意两个人的最短路径介于： In a social media network, the shortest path that connect any two people in the network is between

More Practical Tools for Students Powered by AI Study Helper

Homework AI Solver

Stylized AI Paper Writer

Plagiarism Checker Assistant

Citation AI Academic Writing Tool

In-Class Translation Assistant

AI Note Generator

AI Quiz Answers

Past Exam Questions from University Test Bank

Smart Practice Assistant

Adaptive Practice

Making Your Study Simpler