Dining Philosophers Solution using Monitors

Рет қаралды 104,187

2 жыл бұрын

Operating System: Monitors
Topics discussed:
1. A Solution to the Dining Philosophers Problem using Monitors.
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Пікірлер: 32

@Lucy92735 2 ай бұрын

Thank you very much, you guys have been our constant support to pass my exams every semester!! Love you Neso Academy

@selviarora3093 Жыл бұрын

Thank you so much for the step by step explanation

@FarmanOfficial777 2 жыл бұрын

you are a lifesaver...in the world of operating system

@keshavchander2705 2 жыл бұрын

You are superb teacher 👍👍👍 You make me motivated to complete that very big book like galvin 😁😁 ty so much

@bhavaniraja5618 Жыл бұрын

Very clear explanation

@nehapol9862 Жыл бұрын

great explaination

@farahmohamed2104 6 ай бұрын

well done

@viswanadareddyindupuru5831 2 жыл бұрын

Nice Explanation Great Work

@deekshabhat5194 8 ай бұрын

thnk u

@vkatasonov 7 ай бұрын

Thx for a video. One question: will that solution let us make all of philosophers eat in the correct time? I mean if simultaneously odd and even ones can eat not to die?

@imranimmu4714 2 жыл бұрын

awesomme

@hamzafeghouli4297 2 жыл бұрын

you're awesome

@sukriti9702 2 жыл бұрын

plz complete data structure playlist

@surajsuryawanshi7835 2 жыл бұрын

agree

@sanathkumar6526 10 ай бұрын

Shouldn't putdown(i); come after test(i); in pickup function? Cus in this logic nothing is calling the putdown function

@anisksouri6905 6 ай бұрын

you will use those functions in your main code then you can call each functions however u need

@eyepatch8430 Ай бұрын

Bro this is pseudo code those are just functions

@iamright5818 2 жыл бұрын

When will be putdown function called?

@thinktank3613 Жыл бұрын

After eating

@iennguyenphuong7687 2 жыл бұрын

I have main function

@ScantaniouslyCombust 2 жыл бұрын

The code makes sense, but I don't understand how and where initialization-code is run or the monitor is accessed?

@fritz6600 2 жыл бұрын

Whenever a process (philosopher) wants to eat they would call the two functions (pickup and put down).

@physicslover6840 2 жыл бұрын

Initially assume all are hungry then it start clearly what code will do

@bronwyndrummer8326 Жыл бұрын

How does the putdown function work when testing the next left and right philosophers? The test function requires that the current philosopher is hungry but we never got to change the state of that current philosopher to hungry?

@bronwyndrummer8326 Жыл бұрын

We only got to change the current philosopher's state to hungry in the pickup function, so how can the current philosopher be hungry in the putdown function?

@SajidAli-ub6th Жыл бұрын

@@bronwyndrummer8326 Let's say we are calling putdown function for philosopher 1; we are setting its state back to thinking. Fine so far? Now since it is putting its chopsticks his left and right philosopher can eat now, we are calling test function on them. For ex, when we call test ((i+4)%5), the parameter of the test function is index of right philosopher, not the current one. Is your confusion because we are checking the state of i as hungry in the test function? if yes, please check the basic function concepts. When executing the test function, i is actually the right philosopher's index, not the current one.

@bronwyndrummer8326 Жыл бұрын

@@SajidAli-ub6th Yes I am a tiny bit confused because let's say Philosopher 1 initially wanted to eat. So his state is set to hungry when he picks up the chopsticks. When he is done he will put them down and the putdown function will test the philosopher to the right of him. The test function requires him (the philosopher that was on the right) to be hungry. But in the putdown function did not allow him to change his state to hungry as of yet. That only happens in the pickup function? So to my understanding that test function would evaluate to false for this philosopher because he never got to change his state to hungry yet?

@lindadeng3149 10 ай бұрын

@@bronwyndrummer8326 I have the same question. Also, I remember philosopher dinner only allows one philosopher to eat at a time, so if we test the left and the right both, will it result both of them can eat together?

@swarajya.55 2 жыл бұрын

First!

@surajsuryawanshi7835 2 жыл бұрын

6.33 test function{ if( i==1 ) state[(i+4)]!=eating) and state[(i+1)%5]!=eating else if(i==4) state[(i+4)%5]!=eating) and state[(i+1)]!=eating else state[(i+4)%5]!=eating) and state[(i+1)%5]!=eating }

@L3Moody Жыл бұрын

you have to remember that in an array, the first variable in the array is considered item 0 rather than item 1. meaning the array would have [0, 1, 2, 3, 4] therefor, the original solution is actually correct.

@jatinkumar6748 10 ай бұрын

This code is actually saying that only philosopher with i=1 can eat except other philosopher 0,2,3,4😂😂😂