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Answer 1

A:

Each car should lock parts of the lane it's going to pass trough. If one of the parts is locked then car should wait until it will be released.

Nikita Prokopov 2009-11-02 08:05:38

ok - but how exactly will you implement it?Which object will you lock?You probably need to devide the junction to 4 parts, and to lock each time the part of the junction that you're going to go through. but can you suggest a good implementation for it?

2009-11-02 08:12:33

You should try to make a picture of it, just straight lines would do. Then you'll see four squares which will have to be locked/unlocked as the cars will go there.

Jonas 2009-11-02 08:25:16

I agree, but when getting down to the details, it gets complicated.lets say I have 4 objects - A,B,C,D . The let's say I have 4 cars, and mark them with the direction they're coming from: N (for north), S, E, W. now if car W wants to turn left, it should lock (before entering the junction) C, D, B. and free it only when it leaves the junction. When other cars come in, how should they wait? A queue is not good, because the availability of the road doesn't necessarily correspond to the arrival order.Can anyone suggect an implemetation, in c#, for example?

2009-11-02 08:38:09

After some car releasing part of the road, it notifies *all* the other cars so they check if they can move. First car which can move makes necessary locks and others wait again.Also, you can divide junction into more than 4 pieces. The more precise you are, the less traffic jams you will get.

Nikita Prokopov 2009-11-02 09:30:09

Answer 2

A:

What do you think about having 4 different queues for each part of the junction. each car that comes in enters the relevant queue (should enter to more than one queue?), and only after the car leaves all queue it can go through the junction.. Still not sure what is the best implementation for it though.

2009-11-02 09:06:42

Answer 3

A:

Create a circle buffer with two entries for each road (one for inbound, one for outbound) meeting at the intersection.

For each car that you have to route put it's name into the circle buffer for its source (inbound) and its destination (outbound). Then iterate through the circle buffer, if you get two instances of the same car together then that car may travel. After that pick at random from the other cars.

I have a feeling that's pretty unclear so consider an intersection with 4 roads that we'll call N, E, S and W. To that we'll have 3 cars, A coming from the East turning South, B from the South travelling North and C from the West travelling East.

Circle buffer can be built as such (i=inbound, o=outbound:

Ni No Ei Eo Si So Wi Wo
B  -  C  A  A  B  -  C

As we iterate through from left to right we realise that the two A's are adjacent so they can go but the B's and C's are not adjacent so these cars are block each other. Pick one at random for this light cycle and let the other go in the next light cycle. So either A and B can go or A and C can go.

Note1: testing adjacent ignores blanks so in the case of

Ni No Ei Eo Si So Wi Wo
D  E  -  -  E  D  -  -

which models a car travelling North and another travelling South both E's and D's are adjacent.

Note2: I've mapped this out for driving on the left because that's what I do. You'll have to mirror it for driving on the right.

Note3: You can't overwrite a position in the buffer, if two cars want the same destination they're automatically blocking and you should just leave the first one in there and consider the other one next time.

Daniel 2010-08-25 01:34:06

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