The ‘dining philosopher’ problem is a classic problem describing how various con
ID: 3753495 • Letter: T
Question
The ‘dining philosopher’ problem is a classic problem describing how various consumers who need multiple resources can be synchronized. In your class notes directory on absaroka at /home/605412/dining, there is one solution to the problem as described by Tanenbaum. Examine this source code to answer the following questions. Place your text responses in a word-processed document (.odt, .doc, .docx, .pdf):
a) How would you create an environment to run the five philosophers? (e.g. Could all five ‘philosophers’ run in a single threaded process? Five separate processes? Other ways?) Provide a program or programs (which will run on absaroka not Minix) to actually build an environment to demonstrate running the solution (note the ‘diners’ code has been implemented on absaroka in /home/605412/dining).
Explanation / Answer
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#include<stdio.h>
#define n 4
int compltedPhilo = 0,i;
struct fork{
int taken;
}ForkAvil[n];
struct philosp{
int left;
int right;
}Philostatus[n];
void goForDinner(int philID){ //same like threads concept here cases implemented
if(Philostatus[philID].left==10 && Philostatus[philID].right==10)
printf("Philosopher %d completed his dinner ",philID+1);
else if(Philostatus[philID].left==1 && Philostatus[philID].right==1){
printf("Philosopher %d completed his dinner ",philID+1);
Philostatus[philID].left = Philostatus[philID].right = 10; //remembering that he completed dinner by assigning value 10
int otherFork = philID-1;
if(otherFork== -1)
otherFork=(n-1);
ForkAvil[philID].taken = ForkAvil[otherFork].taken = 0; //releasing forks
printf("Philosopher %d released fork %d and fork %d ",philID+1,philID+1,otherFork+1);
compltedPhilo++;
}
else if(Philostatus[philID].left==1 && Philostatus[philID].right==0){ //left already taken, trying for right fork
if(philID==(n-1)){
if(ForkAvil[philID].taken==0){ //KEY POINT OF THIS PROBLEM, THAT LAST PHILOSOPHER TRYING IN reverse DIRECTION
ForkAvil[philID].taken = Philostatus[philID].right = 1;
printf("Fork %d taken by philosopher %d ",philID+1,philID+1);
}else{
printf("Philosopher %d is waiting for fork %d ",philID+1,philID+1);
}
}else{ //except last philosopher case
int dupphilID = philID;
philID-=1;
if(philID== -1)
philID=(n-1);
if(ForkAvil[philID].taken == 0){
ForkAvil[philID].taken = Philostatus[dupphilID].right = 1;
printf("Fork %d taken by Philosopher %d ",philID+1,dupphilID+1);
}else{
printf("Philosopher %d is waiting for Fork %d ",dupphilID+1,philID+1);
}
}
}
else if(Philostatus[philID].left==0){ //nothing taken yet
if(philID==(n-1)){
if(ForkAvil[philID-1].taken==0){ //KEY POINT OF THIS PROBLEM, THAT LAST PHILOSOPHER TRYING IN reverse DIRECTION
ForkAvil[philID-1].taken = Philostatus[philID].left = 1;
printf("Fork %d taken by philosopher %d ",philID,philID+1);
}else{
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