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}?

{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

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!

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".

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.

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

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

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.

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 ;)

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.

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?