CBMS-2023-09
题目来源:Problem 9 日期:2024-07-10 题目主题:CS-Algorithms-Graph Theory
具体题目
Let be a directed acyclic graph such that is the set of integers from to . We will consider the following sets of edges for .
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Consider a bijective function from to such that for any . For each of and , answer the number of all different bijective functions and the rationale.
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For each of and , answer the number of all different paths from to (or a recurrence for computing the number), and the rationale.
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For any directed acyclic graph , design an algorithm that calculates the number of paths from to in time for any , and explain the rationale.
 
正确解答
Part 1: Bijective Functions
Bijective Function Explanation
A bijective function is a function that is both injective (one-to-one) and surjective (onto). This means every element in maps to a unique element in , and every element in is mapped to by exactly one element in . For the given problem, we are interested in bijective functions that also respect the condition for any edge .
For
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Structure: The graph with edges has two types of edges: from vertex 1 to every other vertex , and from every vertex to vertex . Thus, must be the smallest and must be the largest.
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Ordering: The remaining values must be placed in increasing order between and .
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Number of bijective functions: Given that and are fixed as the smallest and largest respectively, the remaining vertices can be permuted freely.
 
For
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Structure: The graph with edges has edges from each vertex to and .
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Ordering: For to satisfy for all , must be in ascending order for all .
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Number of bijective functions: There is only one valid permutation: the natural ordering .
 
Part 2: Number of Paths from to
For
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Path Analysis: From to , the paths must pass through an intermediate vertex .
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Paths: Each vertex in provides a unique path .
 
For
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Path Analysis: From to , paths can be analyzed using dynamic programming:
Define as the number of paths from to .
 
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Base cases: (direct path from 1 to 1)
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For , there is also a direct path, so .
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Recurrence relation: Compute the number of paths to each vertex until .
 
Part 3: Algorithm for Number of Paths in a DAG
Algorithm
- Initialize an array of size to store the number of paths to each vertex.
 - Initialize for the starting vertex .
 - Traverse the vertices in topological order, and update the number of paths to each vertex based on the incoming edges.
- For each vertex , iterate over its incoming neighbors and update .
 
 - Once all vertices are processed, the number of paths to the destination vertex will be stored in .
 
Time Complexity
- The topological sort can be done in time.
 - Updating the number of paths for each vertex takes time.
 - Thus, the overall time complexity is .
 
知识点
难点解题思路
在 Part 1 中,需要理解图的结构及其对函数 的约束。在 Part 2 中,关键在于找出从起点到终点的所有可能路径数目,并用动态规划求解。在 Part 3 中,难点在于设计有效算法,通过拓扑排序和动态规划计算从起点到终点的路径数目。
解题技巧和信息
对于有向无环图 (DAG) 相关的问题,拓扑排序和动态规划是常用的解题技巧。在这个问题中,我们需要理解图的结构,找出有效的算法来计算从起点到终点的路径数目。
重点词汇
- directed acyclic graph (DAG) 有向无环图
 - topological order 拓扑排序
 - dynamic programming 动态规划
 - bijective function 双射函数
 - recurrence 递推关系
 
参考资料
- Cormen, T. H., Leiserson, C. E., Rivest, R. L., & Stein, C. (2009). Introduction to Algorithms (3rd ed.). MIT Press. Chap. 22 (Elementary Graph Algorithms), Chap. 24 (Single-Source Shortest Paths).