Slides6 - Lecture 6 Critical Path Method Acyclic directed...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
1 Lecture 6 Critical Path Method Acyclic directed graphs So far we computed shortest paths in undirected graphs with non-negative edge costs. Let’s look at acyclic directed graphs. directed path not a directed path directed cycle not a directed cycle Acyclic directed graphs have no directed cycles.
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
2 Acyclic directed graphs Problem: Given an acyclic directed graph, find the length of shortest path from one node to all others. 1 4 3 2 7 6 5 8 4 12 2 5 13 1 11 8 1 6 7 9 3 Note: In acyclic directed graphs we can number the nodes 1,2,...,n such that implies i<j. Acyclic directed graphs i j Why? 1) there must be a node with no incoming arc (why?) 2) Label this with number 1, and delete all outgoing edges. Then, 1) applies again – number this node with the next higher number. Etc... 1 4 3 2 7 6 5 8 4 12 2 5 13 1 11 8 1 6 7 9 3 2 4 5 7
Image of page 2
3 1 2 3 4 5 6 7 8 4 12 2 5 13 1 11 8 1 6 7 9 3 Let’s find the shortest 1-8 path! Say, 1-3-6-8 is the shortest 1-8 path. Then, 1-3-6 also has to be the shortest 1-6 path (why?).
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern