Tokenownidownid 3 enter critical section

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(token[OwnID],OwnID)); //3. Enter Critical section critical_section(); //4. Leave Critical Section token[OwnID] = 0; } Why multiple waits? Because, have to wait for multiple processes
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7 Receive a token Receive a token Initially token[0] .. token[n-1] are set to zero Process i chooses token[i] as max (token[0],token[1],...,token[n-1]) + 1 ; Let n = 5; Let the order of execution be P0,P3,P4,P1,P2,P3,P4… P0 gets token[0] = max(0,0,0,0,0) + 1 = 0+1=1 P3 gets token[3] = max(1,0,0,0,0) + 1 = 1+1=2 P4 gets token[4] = max(1,0,0,2,0) + 1 = 2+1=3 P1 gets token[1] = max(1,0,0,2,3) + 1 = 3+1=4 P2 gets token[2] = max(1,4,0,2,3) + 1 = 4+1=5 P3 gets token[3] = max(1,4,5,2,3) + 1 = 5+1=6 P4 gets token[4] = max(1,4,5,6,3) + 1 = 6+1=7
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8 Wait for turn Wait for turn Pi waits until it has the lowest token of all the processes waiting to enter the critical section. Bakery Algorithm does not guarantee that two processes do not receive the same token In case of a tie, the process with the lowest ID is served first. for (OthersID = 0;OthersID < n ; OthersID++) while (token[OthersID]!=0 && (token[OthersID],OthersID)< (token[OwnID],OwnID)); (a,b) < (c,d) = TRUE if a < c or if both a = c and b < d token[OwnID] = 0 => Process is not trying to enter the critical section
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9 while (TRUE) { //1. Receive a token token[OwnID]= max(token[0],token[1],..,token[n-1])+1; //2. Wait for turn for (OthersID = 0;OthersID<n;OthersID++) while (token[OthersID]!=0 &&(token[OthersID],OthersID)< (token[OwnID],OwnID)); while (TRUE) { //1. Receive a token token[OwnID]= max(token[0],token[1],..,token[n-1])+1; //2. Wait for turn for (OthersID = 0;OthersID<n;OthersID++) while (token[OthersID]!=0 &&(token[OthersID],OthersID)< (token[OwnID],OwnID)); token[0] = 1 Timeout
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