• Announcements

    • khawk

      Download the Game Design and Indie Game Marketing Freebook   07/19/17

      GameDev.net and CRC Press have teamed up to bring a free ebook of content curated from top titles published by CRC Press. The freebook, Practices of Game Design & Indie Game Marketing, includes chapters from The Art of Game Design: A Book of Lenses, A Practical Guide to Indie Game Marketing, and An Architectural Approach to Level Design. The GameDev.net FreeBook is relevant to game designers, developers, and those interested in learning more about the challenges in game development. We know game development can be a tough discipline and business, so we picked several chapters from CRC Press titles that we thought would be of interest to you, the GameDev.net audience, in your journey to design, develop, and market your next game. The free ebook is available through CRC Press by clicking here. The Curated Books The Art of Game Design: A Book of Lenses, Second Edition, by Jesse Schell Presents 100+ sets of questions, or different lenses, for viewing a game’s design, encompassing diverse fields such as psychology, architecture, music, film, software engineering, theme park design, mathematics, anthropology, and more. Written by one of the world's top game designers, this book describes the deepest and most fundamental principles of game design, demonstrating how tactics used in board, card, and athletic games also work in video games. It provides practical instruction on creating world-class games that will be played again and again. View it here. A Practical Guide to Indie Game Marketing, by Joel Dreskin Marketing is an essential but too frequently overlooked or minimized component of the release plan for indie games. A Practical Guide to Indie Game Marketing provides you with the tools needed to build visibility and sell your indie games. With special focus on those developers with small budgets and limited staff and resources, this book is packed with tangible recommendations and techniques that you can put to use immediately. As a seasoned professional of the indie game arena, author Joel Dreskin gives you insight into practical, real-world experiences of marketing numerous successful games and also provides stories of the failures. View it here. An Architectural Approach to Level Design This is one of the first books to integrate architectural and spatial design theory with the field of level design. The book presents architectural techniques and theories for level designers to use in their own work. It connects architecture and level design in different ways that address the practical elements of how designers construct space and the experiential elements of how and why humans interact with this space. Throughout the text, readers learn skills for spatial layout, evoking emotion through gamespaces, and creating better levels through architectural theory. View it here. Learn more and download the ebook by clicking here. Did you know? GameDev.net and CRC Press also recently teamed up to bring GDNet+ Members up to a 20% discount on all CRC Press books. Learn more about this and other benefits here.
Sign in to follow this  
Followers 0
Nicholas Kong

Finding strongly connected components in a graph

20 posts in this topic

[url=http://s33.photobucket.com/user/warnexus/media/stronglyConnected_zps8c6a6999.png.html]stronglyConnected_zps8c6a6999.png[/URL]

 

The answers listed strongly connected components in this graph. But the textbook said a strongly connected component is a strongly connected subgraph of the directed graph but not contained in larger strongly connected subgraphs.

 

I do not see how the subgraph of this graph is strongly connected because the textbook said strongly connected means there is a path from vertices a to b. But there is no such thing because the directed edges clearly shows that.

Edited by warnexus
0

Share this post


Link to post
Share on other sites

Strongly connected means if you can go from A->B you can also go from B->A? (EDIT: That doesn't mean you can go from A to B and vice versa in one hop. one way systems would suck then ;) )

 

The graphs don't have to be connected in that case. You can have 2 or more subgraphs which aren't connected. Disjointed graphs are fine.

 

EDIT2: I think it means you get rid of dead ends.

Edited by Paradigm Shifter
0

Share this post


Link to post
Share on other sites

Strongly connected means if you can go from A->B you can also go from B->A? (EDIT: That doesn't mean you can go from A to B and vice versa in one hop. one way systems would suck then ;) )

 

The graphs don't have to be connected in that case. You can have 2 or more subgraphs which aren't connected. Disjointed graphs are fine.

I'm a bit confused when the textbook says A to B are they saying A is the initial vertex and B is the terminal vertex? Or is A, a vertex labeled A and B, a vertex labeled B?

Edited by warnexus
0

Share this post


Link to post
Share on other sites

It means any 2 vertices (could even be the same vertex, unless implicitly stated this is disallowed). Don't be confused by "if you can get from A to B" thinking the meaning is "you can get from vertex "a" to vertex "b"".

 

Maths books hardly ever have numbers in them, they generalise.

 

In that graph you can go from a to b and vice versa though ;) - just follow the arrows between them.

 

EDIT: In the graph you posted, vertex "e" is a dead end so isn't strongly connected to the others, if I am correct in my guess at what strongly connected means. I'm not sure whether the strongly connected components are

 

{e} and G\{e}

 

or

 

{e} and G

 

(where G is the entire graph).

Edited by Paradigm Shifter
0

Share this post


Link to post
Share on other sites

It means any 2 vertices (could even be the same vertex, unless implicitly stated this is disallowed). Don't be confused by "if you can get from A to B" thinking the meaning is "you can get from vertex "a" to vertex "b"".

 

Maths books hardly ever have numbers in them, they generalise.

 

In that graph you can go from a to b and vice versa though ;) - just follow the arrows between them.

Yeah, it does go from vertex "a" to vertex "b" and vice-versa. which means it is strongly connected which would mean it is a strongly connected component. But the textbook did not list that as an answer.

0

Share this post


Link to post
Share on other sites

Ok well see my edit, the answer is either

 

{e} and G\{e}

 

or

 

{e} and G,

 

depending on the definition, which I'm too lazy too google (6th beer).

0

Share this post


Link to post
Share on other sites

Ok well see my edit, the answer is either

 

{e} and G\{e}

 

or

 

{e} and G,

 

depending on the definition, which I'm too lazy too google (6th beer).

the strongly connected components are {a, b, f }, {c, d, e}.

 

I am using Rosens' textbook.

0

Share this post


Link to post
Share on other sites

Mhmm ok, I can see now that once you go to vertex "c" you can never get back to {a, b, f}.

 

That's what it means then, in effect. Strongly connected means you can get from place to place but if you leave your neighbourhood and can never get back that is a new subgraph.

 

They are basically talking about dead ends and one way streets.

 

EDIT: I'm interested why {e} doesn't count on its own though? Once you get to {e} you are stuck...

 

EDIT2: I see you can get back to {c} from {e}. Ignore this drunkard ;) I guess we've both learned something today!

Edited by Paradigm Shifter
0

Share this post


Link to post
Share on other sites

Mhmm ok, I can see now that once you go to vertex "c" you can never get back to {a, b, f}.

 

That's what it means then, in effect. Strongly connected means you can get from place to place but if you leave your neighbourhood and can never get back that is a new subgraph.

 

They are basically talking about dead ends and one way streets.

 

EDIT: I'm interested why {e} doesn't count on its own though? Once you get to {e} you are stuck...

 

EDIT2: I see you can get back to {c} from {e}. Ignore this drunkard ;)

I am not sure what they mean by {a,b,f}

 

because once you go from vertex labeled a to vertex labeled b, after vertex b you cannot go to vertex f based off the graph.

0

Share this post


Link to post
Share on other sites

Sure you can, you go b->a->f. Like I said, you don't have to do it in one hop.

 

EDIT: Mind blowing time! You don't even need to do it in a finite number of hops, probably ;)

Edited by Paradigm Shifter
0

Share this post


Link to post
Share on other sites

Sure you can, you go b->a->f. Like I said, you don't have to do it in one hop.

 

EDIT: Mind blowing time! You don't even need to do it in a finite number of hops, probably ;)

but they did not say b->a->f. would b->a->f be {b,a,f}?

0

Share this post


Link to post
Share on other sites

{b, a, f}, b->a->f, whatever dude!

 

If you at b, you can get to a or f! No problem! If you at a, you can go to b or f! If you at f, you can go to a or b! Eventually!

 

Once you go to {c} (EDIT: or {d} or {e}) though you ain't showing your face in the {a, b, f} neighbourhood and better stick to {c, d, e}.

 

Street math 101

Edited by Paradigm Shifter
0

Share this post


Link to post
Share on other sites

EDIT2: I see you can get back to {c} from {e}. Ignore this drunkard ;) I guess we've both learned something today!
 

 

{b, a, f}, b->a->f, whatever dude!

 

If you at b, you can get to a or f! No problem! If you at a, you can go to b or f! If you at f, you can go to a or b! Eventually!

 

Once you go to {c} (EDIT: or {d} or {e}) though you ain't showing your face in the {a, b, f} neighbourhood and better stick to {c, d, e}.

 

Street math 101

Oh I see! I think I should think of streets and dead-ends like you stated before! It certainly does make it easier to understand now. Thanks Paradigm!

 

Haha! I'm glad this post refreshen your thoughts on strongly connected components!

Edited by warnexus
0

Share this post


Link to post
Share on other sites

OK. Now prove they divide the graph into disjoint subsets, or something ;)

 

Not really related to games, although opening doors for pathfinding seems somehow connected...

 

EDIT: Another problem with maths is they've heavily overloaded (in the C++/OOP way) words like "simple", "strongly", "fundamental", "connected" and "number".

Edited by Paradigm Shifter
0

Share this post


Link to post
Share on other sites

OK. Now prove they divide the graph into disjoint subsets, or something ;)

 

Not really related to games, although opening doors for pathfinding seems somehow connected...

 

EDIT: Another problem with maths is they've heavily overloaded (in the C++/OOP way) words like "simple", "strongly", "fundamental", "connected" and "number".

 

OK. Now prove they divide the graph into disjoint subsets, or something ;)

 

Not really related to games, although opening doors for pathfinding seems somehow connected...

 

EDIT: Another problem with maths is they've heavily overloaded (in the C++/OOP way) words like "simple", "strongly", "fundamental", "connected" and "number".

Yeah it is basically a lot of jargon to understand and really have a deep understanding of before the good stuff happens. I can understand that. CS is pretty about learning a new language to communicate new ideas.

 

Ah pathfinding that is interesting. I did remember pathfinding being in strategy games.Although there was a game review syaing how that game had horrible pathfinding. Does horrible pathfinding unnecessarily mean a "lack of understanding of graph theory"?

 

Well proving the graph can be divided to two disjoint sets can be done because each set has no common element from each other.

Edited by warnexus
0

Share this post


Link to post
Share on other sites

Lack of stack space/time is probably the most limiting factor?

 

Or simplifications (e.g. assuming spherical cows and whatnot).

0

Share this post


Link to post
Share on other sites

Lack of stack space/time is probably the most limiting factor?

 

Or simplifications (e.g. assuming spherical cows and whatnot).

 I never knew there was such type of limitations. I am still a CS student. It sure is interesting to bear in mind about those things.

0

Share this post


Link to post
Share on other sites

Well that's what tends to limit AI, you can't search an entire tree of 20 bazillion (rough estimate) nodes in real time, computers are not maths (where time is just another variable and an infinite set is fine).

 

Good convo though you'll go far!

 

Best go to bed now ;)

0

Share this post


Link to post
Share on other sites

Well that's what tends to limit AI, you can't search an entire tree of 20 bazillion (rough estimate) nodes in real time, computers are not maths (where time is just another variable and an infinite set is fine).

 

Good convo though you'll go far!

 

Best go to bed now ;)

Night. I still have to finish studying this last section for my midterm Midterm is tomorrow. This section was covered yesterday. Yeah awesome convo indeed. I will bear in mind about that for AI.

Edited by warnexus
0

Share this post


Link to post
Share on other sites


K. Now prove they divide the graph into disjoint subsets, or something ;)

 

I never heard of strongly connected components before this post, but it sounds to me that, given a directed graph, you build an undirected graph whose nodes are the nodes of the original graph and where there is an edge joining A and B if and only if there is a path from A to be and there is a path from B to A. Now you compute the connected components of that graph. So of course they will divide the graph into disjoint subsets... Did I miss anything?

0

Share this post


Link to post
Share on other sites

I dunno, it was a conjecture, I'd never heard of strongly connected components either! Sounds right to me though. I'm fairly certain the OPs book will go on to prove that in due course.

 

I think strongly connected is an equivalence relation and those partition a set into disjoint subsets anyway.

 

EDIT: Proof here http://www.ics.uci.edu/~eppstein/161/960220.html

Edited by Paradigm Shifter
0

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!


Register a new account

Sign in

Already have an account? Sign in here.


Sign In Now
Sign in to follow this  
Followers 0