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Home/ Questions/Q 3216988
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Editorial Team
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Editorial Team
Asked: May 17, 20262026-05-17T15:21:32+00:00 2026-05-17T15:21:32+00:00

If you’re unfamiliar with the problem, it’s something like this . I didn’t come

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If you’re unfamiliar with the problem, it’s something like this.

I didn’t come to ask for an answer, I have actually finished all of my coding. I have just found that my solution doesn’t solve it the best way possible, because my solution only allows one car at a time on the bridge. I was hoping I could get some tips about how to go about using sem_wait and sem_post to solving this problem. I hope to allow traffic flowing the same direction to flow together and not one at a time.

My solution currently looks something like:

(default sem_t north and south = 1 for unlocked for 1 car)

IF northcar then sem_wait(south), sem_wait(north). Cross the bridge, and then sem_post(north), sem_post(south). This is obviously wrong because it’s locking the bridge from all cars other than the one on it. I want to enable traffic to flow together. Any ideas?

I am using randomly generated traffic which adds a bit of complexity to it.

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  1. Editorial Team
    Editorial Team
    2026-05-17T15:21:33+00:00Added an answer on May 17, 2026 at 3:21 pm

    In real life you would solve this problem with a traffic light which is either red-north/green-south, green-north/red-south, or red-north/red-south, and sensors in the approach and exit lanes at both ends of the bridge.

    Suppose the light starts out red-north/green-south. Cars can flow from south to north without stopping. When a car approaches from the north, it stops at the red light and triggers the north approach sensor. That makes the light go red-south, but it’s still red-north as well. When all the cars currently on the bridge have left (triggering the north exit sensor on their way out) the light can change to green-north. That state persists until another car comes from the south and triggers the south approach sensor.

    Think about how you’d translate that into code. (You’ll need to use the counting property of semaphores.)

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