Q1 The first known correct software solution to the critical-section problem for

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Q1 The first known correct software solution to the critical-section problem for n processes with a lower bound on waiting of n - 1 turns was presented by Eisenberg and McGuire. The processes share the following variables: enum pstate idle, want in, in cs; pstate flag[n]; int turn; All the elements of flag are initially idle. The initial value of turn is immaterial (betweenOand n-1). The structure of process Pi is shown below Prove that the algorithm satisfies all three requirements for the critical-section problem, namely: Mutual exclusion - Progress Bounded waiting while (true) ( flag[i] wt in; turn; while (j-i) { if (flag[j] !-idle) j-turn; else j (j + 1) % n; flag [i] = in cs; while ( (j )&&(turn-i flag [turn] -idle)) break; /critical section */ (turn + 1) % n; (j + 1) % n; while (flag [j]-idle) j turn flag[i]idle /*remainder section */ ) while (true)

Explanation / Answer

hi, Please go threw the code. If You have any query please comment.

#include <stdio.h>

#include <pthread.h>

#define n 10

pthread_mutex_t lock;

enum pstate{idle,want_in, in_cs};

enum pstate flag[n];

int turn =0;

void * thread(void * arg)

{

int i =0 , j=0;

do {

while(1){

flag[i] = want_in;

j = turn;

while(j != i){

if(flag[j] != idle) j = turn;

else j= (j+1)%n;

}

flag[i] = in_cs;

j=0;

while((j<n) && (j == i || flag[j] != in_cs)) j++;

if( (j >= n) && (turn == i || flag[turn] == idle)) break;

}

pthread_mutex_lock(&lock);

j = (turn + 1) %n;

while(flag[j] == idle) j = (j+1) % n;

turn =j;

flag[i] = idle;

pthread_mutex_unlock(&lock);

}while(1);

pthread_exit(NULL);

}

int main(void)

{

pthread_t t[n];

int i =0;

for(i=0; i <n ;i++)

{

pthread_create(&t[i], NULL, thread, NULL);

}

for(i=0; i <n ;i++)

{

pthread_join(t[i], NULL);

}

}

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